In reality there is no such thing as a unit step, and the Fourier transform is not 'real'. Instead, the most common procedure to find the inverse Laplace transform of an expression is a two-step approach (Appendix 12. The Fourier transform we’ll be int erested in signals deﬁned for all t the Four ier transform of a signal f is the function F (ω)= ∞ −∞ f (t) e − jωt dt • F is a function of a real variable ω;thef unction value F (ω) is (in general) a complex number F (ω)= ∞ −∞ f (t)cos ωtdt − j ∞ −∞ f (t)sin ωtdt •| F (ω) | is called the amplitude spectrum of f; F (ω) is the phase spectrum of f • notation: F = F (f) means F is the Fourier transform of f. Remark 6 The definition of the Fourier transform on implies that whenever , we have that. If any argument is an array, then fourier acts element-wise on all elements of the array. $\begingroup$ The plus one simply shifts when the Heaviside function turns on by one unit to the left like in normal functional translation, it helps to use the definition of the Heaviside step function as it restricts your domain of integration. How about going back? Recall our formula for the Fourier Series of f(t) : Now transform the sums to integrals from –∞to ∞, and again replace F m with F(ω). 2 Fourier Series Consider a periodic function f = f (x),deﬁned on the interval −1 2 L ≤ x ≤ 1 2 L and having f (x + L)= f (x)for all. Because the original function and its inverse Laplace transform are only valid for t‚ 0, some people introduce a Heaviside step function H ( t ) (see Section B. Fourier Transform Pairs (contd). 4M subscribers. Laplace transform of the unit step function | Laplace transform | Khan Academy - YouTube. • Sometimes we want to use one-dimensional Fourier transforms or inverse transforms. Explain briefly below. The Fourier transform is not limited to functions of time, but the domain of the original function is commonly referred to as the time domain. Unlike the inverse Fourier transform, the inverse Laplace transform in Eq. Step 7: Check the “Inverse” box only if you have results from a prior analysis and you want to find the original function. The book is divided into four major parts: periodic functions and Fourier series, non-periodic functions and the Fourier integral, switched-on signals and the Laplace transform, and finally the discrete versions of these transforms, in particular the Dis-crete Fourier Transform together with its fast implementation, and the z-transform. Unit analysis, algebra solver free step by step, how fast can one learn algebra, mathsiequalities, online direction field. Solution: Here, =0 for <2 , then ˝ =1 for ≥2. Fourier Transform вЂ” Theoretical Physics Reference 0. Before proceeding into solving differential equations we should take a look at one more function. It cannot be said that time information is lost because it is possible to recover the original time domain observation using the Inverse Fourier. The calculator will find the Inverse Laplace Transform of the given function. The usual Fourier transform tables found online don't have many functional relationship rules. we can get the Fourier transform of a unit impulse as the time derivative of a unit step function: Alternatively, by definition, the forward Fourier transform of an impulse function is and the inverse transform is. The Fourier Transform (used in signal processing) The Laplace Transform (used in linear control systems) The Fourier Transform is a particular case of the Laplace Transform, so the properties of Laplace transforms are inherited by Fourier transforms. Interestingly, these transformations are very similar. Together with a great variety, the subject also has a great coherence, and the hope is students come to appreciate both. You take the Fourier transform fft of f. Suppose that the Fourier transform of f and its inverse exist. This is specifically due to its property that it is neither absolutely summable nor square summable. dft() and cv2. How can you create a delta function using some other function, the Fourier transform of which you already know. Fourier transform pair The function ! X(j") is the Fourier transform of the signal x(t) and conversely x(t) is the inverse Fourier transform of! X(j"). If any argument is an array, then fourier acts element-wise on all elements of the array. 2 Fourier Series Consider a periodic function f = f (x),deﬁned on the interval −1 2 L ≤ x ≤ 1 2 L and having f (x + L)= f (x)for all. Mathematicians have developed tables of commonly used Laplace transforms. PLOTTING STEP RESPONSE OF TRANSFER FUNCTION Learn more about fourier transform. The notation is introduced in Trott (2004, p. Discrete-Time Fourier Transform (DTFT) inverse DTFT. text orientation finding) where the Fourier Transform is used to gain information about the geometric structure of the. We can solve the integral by contour integration. Denoted , it is a linear operator of a function f(t) with a real argument t (t ≥ 0) that transforms it to a function F(s) with a complex argument s. The book only states a limited form of the Heavyside expansion theorem in problem 5 of section 53. The Z transform of the discrete time unit ramp function 42. (iii) Comment the time domain expression of the filter. 1 Dirac Delta Function 1 2 Fourier Transform 5 3 Laplace Transform 11 3. Visualizing Pole-Zero plot: Since the z-transform is a function of a complex variable, it is convenient to describe and interpret it using the complex z-plane. 12 tri is the triangular function 13 Dual of rule 12. One common example is when a voltage is switched on or off in an electrical circuit at a specified value of time t. calculating the Fourier transform of a signal, then exactly the same procedure with only minor modification can be used to implement the inverse Fourier transform. As such, the restriction of the Fourier transform of an L 2 (R n) function cannot be defined on sets of. Consider the Fourier transforms of the functions in Example 9. 1) into the integral in the deﬂnition of the inverse transform in (F. The special characteristics of the Fast Fourier Transform implementation are described. Inverse Transform 6. In this section we introduce the step or Heaviside function. is the Fourier Transform of f(t). Fourier transforms take the process a step further, to a continuum of n-values. For instance, the inverse continuous Fourier transform of both sides of Eq. Correlation, autocorrelation. For the Laplace transform, the Fourier transform existed if the ROC included the j!axis. I have to find the inverse fourier transform for: \frac{e^{i 6\omega}}{\omega} So I'm using a table, then. In this section we introduce the Fourier transform and then we illustrate the fast Fourier transform algorithm, applied to the projection of unit-step i. Continuous Fourier Transform A general Fourier Transform for Spectrum Representation •With the unit-impulse function incorporated, the continuous Fourier transform can represent a broad range of continuous-time signals. Basic Fourier Transform Theory: Relationship to Chap 3 on Fourier Series Interpretation of Inverse Fourier Transform Frequency Ranges of Biological, E&M, and other Signals. In reality there is no such thing as a unit step, and the Fourier transform is not 'real'. It is "off" (0) when < , the "on" (1) when ≥. For math, science, nutrition, history. The Fourier Transform and its Inverse The Fourier Transform and its Inverse: So we can transform to the frequency domain and back. 3 Complex form of Fourier series, Fourier integral representation, Fourier Transform and Inverse Fourier transform of constant and exponential function. 1) into the integral in the deﬂnition of the inverse transform in (F. Laplace transform to solve a differential equation. Fourier Transform - Free download as Powerpoint Presentation (. Fourier transform of the unit step function and of the signum function: The signum function sgn( t) is a function that is related to the unit step function. In this study, a new inversion method is presented for performing two-dimensional (2D) Fourier transform. Consider the equation f(x) + A(f(x − 1) + f(x + 1)) = u(x) where u(x) is a known function, absolutely integrable, on R and A is a constant. Evaluate one transform on data from step 3. The Laplace transform is similar to the Fourier transform. 12-2 Circuit Analysis Using the Fourier Transform Determine and plot the spectrum of the response Vo(co) of the circuit of Figure 15. • The unit step function (1 class) • The Dirac delta function (1 class) • Applications of step and impulse functions (1 class) • Periodic functions and their applications (2 classes) • Convolution and applications (2 classes) • Solving integral equations (1 class) • Fourier series (3 classes) • Fourier integral representation (1. Inverse Laplace Transform with unit step function, sect7. 1998 We start in the continuous world; then we get discrete. The Fourier transform is only valid for a periodic function, and a unit step is not periodic. So the Fourier transform X (jw) of x (t) is the convolution of X (jw) and sinc (w). Fast Fourier Transforms Phase factors There are functions that produce roots-of-one as a function of time (t) or place (x). eﬁne the Fourier transform of a step function or a constant signal unit step what is the Fourier transform of f (t)= 0 t< 0 1 t ≥ 0? the Laplace transform is 1 /s, but the imaginary axis is not in the ROC, and therefore the Fourier transform is not 1 /jω in fact, the integral ∞ −∞ f (t) e − jωt dt = ∞ 0 e − jωt dt = ∞ 0 cos. In this section we introduce the step or Heaviside function. A unique 3D graphical approach has been adopted to provide the intuition required to OWN this subject. The unit pulse function can be defined with the help of the Heaviside unit step function ( ) ( ) ( ) 0 x a f t Ht a Ht a 1 x a 0 x a <− = +− −= < > a0 > The Fourier transform of this function can be determined as. of a second over a period of 10 seconds. Uniqueness of Fourier transforms, proof of Theorem 3. 2) The DTFT X. 3 Complex form of Fourier series, Fourier integral representation, Fourier Transform and Inverse Fourier transform of constant and exponential function. 3 Properties of The Continuous -Time Fourier Transform 4. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. So essentially, in the decomposition of x of t as a linear combination of complex exponentials, the complex amplitudes of those are, in effect, the Fourier transform scaled by the differential and scaled by 1 over 2 pi. 6 ) and obtain. There are different definitions of these transforms. Remembering the fact that we introduced a factor of i (and including a factor of 2 that just crops up. UNIT3: FOURIER TRANSFORM - Complex form of Fourier Transform and its inverse, Fourier sine and cosine transform and their inversion. Remembering the fact that we introduced a factor of i (and including a factor of 2 that just crops up. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. Inverse Fourier Transform F [f] Step 3) Find the inverse transform. I think you should have to consider the Laplace Transform of f(x) as the Fourier Transform of Gamma(x)f(x)e^(bx), in which Gamma is a step function that delete the negative part of the integral and e^(bx) constitute the real part of the complex exponential. xxxiv), and and are sometimes also used to denote the Fourier transform and inverse Fourier transform, respectively (Krantz 1999, p. Dirac defined the delta function as shown below. If n is less than the length of the signal, then ifft ignores the remaining signal values past the nth entry and. Express f under an integral form. Overlapping in real time fourier transform? Hot Network Questions Numbers by Position a name for a boy, boy's name, boy name Mother milk of 6 Corona-positive (COVID-19) women does not contain the virus - can we make a confidence. How can you create a delta function using some other function, the Fourier transform of which you already know. The convergence criteria of the Fourier transform (namely, that the function be absolutely integrable on the real line) are quite severe due to the lack of the exponential decay term as seen in the Laplace transform, and it means that functions like polynomials, exponentials, and trigonometric functions all do not have Fourier transforms in the. Its value is not trivial to calculate, and ends up being. Any good reference to more detailed tables would be very helpful! My attempt: $\mathcal F[f\times u] = (\mathcal Ff)*(\mathcal Fu)$ where * denotes convolution. The usual Fourier transform tables found online don't have many functional relationship rules. As with the Laplace transform, calculating the Fourier transform of a function can be done directly by using the definition. Magnitude and phase representation of the Fourier transform and frequency response of LTI systems; Applications of the. ej!O /that results from the deﬁnition is a function of frequency !O. 2 Contents 2. The more general statement can be found in standard texts devoted to Laplace transforms. Determine the Fourier transform of the non-periodic signals shown in the figures below: (b) 8(1) -2 -1 0 1 2. Introduction to Inverse Problems Guillaume Bal 1 July 2, 2019 1University of Chicago, Chicago, IL 60637; [email protected] Fourier transforms take the process a step further, to a continuum of n-values. 2 The Fourier transform Given a function f(x) de ned for all real x, we can give an alternative representation to it as an integral rather than as an in nite series, as follows f(x) = Z eikxg(k)dk Here g(x) is called the Fourier transform of f(x), and f(x) is the inverse Fourier transform of g(x). 8 Fourier transforms. Second Implicit Derivative (new) Derivative using Definition (new) Derivative Applications. In this section we introduce the step or Heaviside function. The unit step function, also known as the Heaviside function, is defined as such:. 2 Fourier Series Consider a periodic function f = f (x),deﬁned on the interval −1 2 L ≤ x ≤ 1 2 L and having f (x + L)= f (x)for all. The Fourier transform is a particular case of z-transform, i. INTRODUCTION AND FOURIER TRANSFORM OF A DERIVATIVE One can show that, for the Fourier transform g(k) = Z 1 1 f(x)eikx dx (1) to converge as the limits of integration tend to 1 , we must have f(x) ! 0 as. The derivation can be found by selecting the image or the text below. Solved examples of the Laplace transform of a unit step function. We also work a variety of examples showing how to take Laplace transforms and inverse Laplace transforms that involve Heaviside functions. It is clearly desirable that there should be a canonical definition of the Fourier Transform, consistent with classical definitions, which is applicable to all distributions - or, at least to some. The Discrete Fourier Transform the two transforms and then ﬁlook upﬂ the inverse transform to get the convolution. The excel fourier analysis tool. 4142*j]; x_n=ifft(X_K) Example 2: X_K=[10,-2+2*j,-2,-2-2*j]; x_n=. Discrete Time Fourier Transforms The discrete-time Fourier transform or the Fourier transform of a discrete-time sequence x[n] is a representation of the sequence in terms of the complex exponential sequence. fast fourier transform. This transformation is essentially bijective for the majority of practical. !! If you apply the Fourier transform to function f(t), you get a new function F(w). Learn more about Chapter 8: The Fourier Transform on GlobalSpec. Since the transform of a lattice in real space is a reciprocal lattice, the diffraction pattern of the crystal samples the diffraction pattern of a single unit cell at the points of the reciprocal. When you have worked through this unit you should:. 2) factor (1/2π )2 must be replaced by (1/2π ) To avoid confusion, we shall indicate one-dimensional Fourier transforms by Fx, Fx-1 or Fky. We experi-ment here to see if Mathematica knows these functions, and if it can deal with their Fourier transforms. The Fast Fourier Transform for polynomials works in an analogous way to a slide rule. The Fourier Transform Saravanan Vijayakumaran [email protected] 1) which is now called Heaviside step function. The fourier transform uses the assumption that any finite time-domain signal can be broken into an infinite sum of sinusoidal (sine and cosine waves) signals. I don't know where you got G(f), but it only a mathematical expression to "give" the value of the Fourier transform of a unit step. Let tqptqu. 1,791,367 views. G o t a d i f f e r e n t a n s w e r? C h e c k i f i t ′ s c o r r e c t. Properties of the Fourier Series 51. The key step in the proof of (1. The properties are useful in determining the Fourier transform or inverse Fourier transform They help to represent a given signal in term of operations (e. Especially important among these properties is Parseval's Theorem, which states that power computed in either domain equals the power in the other. CT Fourier Transform Pairs signal (function of t) $\longrightarrow$ Fourier transform (function of $\omega$) : 1 CTFT of a unit impulse $\delta (t)\$ $1 \$. The Fourier transform 45. Fourier transform. Inverse Fourier transform – be able to compute this from definition as well as from looking up the transform for elementary signals. This MATLAB function returns the Fourier Transform of f. UNIT STEP FUNCTIONS AND PERIODIC FUNCTIONS 157 Which implies that y(t) = t2 solves the DE. 6#15 - Duration: The intuition behind Fourier and Laplace transforms I was never taught in school inverse laplace transform,. If Y is a matrix, then ifft (Y) returns the inverse transform of each column of the matrix. Q5(a) is given to be: 2 1 Feje() ( 1)ωωjjωω ω = −− Use this information and the time-shifting and time-scaling properties, find the Fourier transforms of the signals. So that gives you a complex spectrum which is here called ff, and then you multiply it by the imaginary unit times k, and then use an inverse transform back to physical space and now you have an exact to machine precision derivative defined on your original grid points. Introduction to Hilbert Transform. In this video tutorial, the tutor covers a range of topics from from basic signals and systems to signal analysis, properties of continuous-time Fourier transforms including Fourier transforms of standard signals, signal transmission through linear systems, relation between convolution and correlation of signals, and sampling theorems and techniques. Ada version of General N Point Fast Fourier Transform. Discrete-Time Fourier Transform (DTFT) inverse DTFT. Solved examples of the Laplace transform of a unit step function. Note that the usual results for Fourier transforms of even and odd functions still hold. IQ v(t) vo(t) FIGURE 15. The Fourier transform of controlled-source time-domain electromagnetic data by smooth spectrum inversion Yuji Mitsuhata. Problems at x!+1are removed by multiplying by e cx, where cis a positive real number. The Fourier transform we’ll be int erested in signals deﬁned for all t the Four ier transform of a signal f is the function F (ω)= ∞ −∞ f (t) e − jωt dt • F is a function of a real variable ω;thef unction value F (ω) is (in general) a complex number F (ω)= ∞ −∞ f (t)cos ωtdt − j ∞ −∞ f (t)sin ωtdt •| F (ω) | is called the amplitude spectrum of f; F (ω) is the phase spectrum of f • notation: F = F (f) means F is the Fourier transform of f. Thus the Fourier transform on tempered distributions is an extension of the classical definition of the Fourier transform. Instead, the most common procedure to find the inverse Laplace transform of an expression is a two-step approach (Appendix 12. he Fourier and Laplace transforms can be rectangular pulse: f (t)= 1 eп¬Ѓne the Fourier transform of a step function or a constant signal, The aim of this post is to properly understand Numerical Fourier Transform on Python or Matlab with an example in fourier transform of the. study how a piecewise continuous function can be constructed using step functions. As in the FDK analysis, the s direction transforms require a total of 2 NMP 2 log(2 P ) operations for the forward direction and 2 NP 2 log(2 P ) for the inverse direction. 8 Filters 2. Inverse Z Transform: Part 2. of a unit step can be inferred, but it's natural with the Laplace. The Fourier transform of ft) (ft)-sinc(t)) is F(jo)-nRect(/2) (Figure ) (1) For a linear, time invariant system, its impulse response is h(t)…. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The step response is the convolution. Fourier Transform of Unit Step Function Guess Fourier Transform of Unit Step Function F ( ) 0 |F(j )| 0 t 1 f(t) Fourier Transforms of Special Functions Fourier Transform vs. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The usual Fourier transform tables found online don't have many functional relationship rules. Have these ideas in mind as we go through the examples in the rest of this section. This chapter introduces the Fourier Transform, also known as the Fourier Integral. Linearity and the result for the unit step, above. Fourier transform of unit step signal u(t). MAXIMA Quick Reference Labels. The Fourier transform of the unit step function is not any of those things. It's basically a set of Sine waves with amplitudes and phases. Laplace transform to solve a differential equation. Once the transformation has been applied, time information is hidden and cannot be easily observed. 10 Band-Pass Systems. Together with a great variety, the subject also has a great coherence, and the hope is students come to appreciate both. Unit analysis, algebra solver free step by step, how fast can one learn algebra, mathsiequalities, online direction field. The Fourier transform of an integrable function is continuous and the restriction of this function to any set is defined. Try to integrate them? Cite. That is, given the Fourier transform of an function, when can we recover the original function from ? We begin with a simple case where the recovery is quite easy. Table of Fourier Transform Pairs of Energy Signals Function name Time Domain x(t) Frequency Domain X Unit step () 10 00 t ut 2. Visualizing Pole-Zero plot: Since the z-transform is a function of a complex variable, it is convenient to describe and interpret it using the complex z-plane. Introduction to Fourier Transforms Fourier transform as a limit of the Fourier series Inverse Fourier transform: The Fourier integral theorem Example: the rect and sinc functions Cosine and Sine Transforms Symmetry properties Periodic signals and functions Cu (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 2 / 22. The special characteristics of the Fast Fourier Transform implementation are described. From this block diagram we can find overall transfer function which is nonlinear in nature. Example 2-2 SPECTRUM OF AN EXPONENTILA PULSE By means of direct integration find the Fourier transform of ) ( t w < = - 0 , 0 0 , ) ( t t e t w t Properties of Fourier Transforms. These reviews did not try 44 to minimize Laplace-space function evaluations, since their functions were simple closed-form expressions, 45 not simulations. Time Reversal and Frequency Response By Clay S. Function, f(t) Fourier Transform, F( ) Definition of Inverse Fourier Transform f t F( )ej td 2 1 ( ) Definition of Fourier Transform F() f (t)e j tdt Trigonometric Fourier Series 1 ( ) 0 cos( 0 ) sin( 0) n f t a an nt bn nt where T n T T n f t nt dt T b f t nt dt T f t dt a T a 0 0 0 0 0 0 ( )sin() 2. The Fourier transform of ft) (ft)-sinc(t)) is F(jo)-nRect(/2) (Figure ) (1) For a linear, time invariant system, its impulse response is h(t)…. Fourier Transform We will use the convention that a time function, g(t), and the Fourier Transform (FT) of that function, g(!), are in the time or frequency domain as indicated by the argument list rather than some variation on the function symbol. transforms on pairs from step 1. 7 Transmission of Signals Through Linear Systems 2. Tags: EMML, inner product, probability density functions, likelihood function, linear functional, orthonormal basis, linear transformation, vector, Linear Algebra. Fourier Transforms and the Dirac Delta Function A. The usual Fourier transform tables found online don't have many functional relationship rules. If any argument is an array, then fourier acts element-wise on all elements of the array. 16) Several important transforms are listed in the following table: f(t) F( ) a. Last time, we saw the equations that calculate the Fourier Transform and its inverse. Since sinc (w) has infinite duration in freqency domain, X (jw) convolved with sinc (w) also has infinite horizon in freqency domain. If you really want to understand the Fourier and Laplace transforms , how they work and why they work then this is the course for you. Characteristics of the Continuous Fourier Transform The plots in Figures 1-1 and 1-2 demonstrate two characteristics of the Fourier transforms of real time history functions: 1. 1 Dirac delta function The delta function –(x) studied in this section is a function that takes on zero values at all x 6= 0, and is inﬂnite at x = 0, so that its integral +R1 ¡1 –(x)dx = 1. So that gives you a complex spectrum which is here called ff, and then you multiply it by the imaginary unit times k, and then use an inverse transform back to physical space and now you have an exact to machine precision derivative defined on your original grid points. How to solve a basic math equation, foerester's algebra 1 suggested timelien, how to answe algebra problems, free algebra word problem solver, algebra1 answer keys g. 3 If f (x) is a good function with its Fourier transform g( y), then the Fourier transform of f (x) is 2πiyg( y), and the Fourier transform of f (ax + b) is |a|−1 e2πiby/a g( y/a). Function, f(t) Fourier Transform, F( ) Definition of Inverse Fourier Transform f t F( )ej td 2 1 ( ) Definition of Fourier Transform F() f (t)e j tdt Trigonometric Fourier Series 1 ( ) 0 cos( 0 ) sin( 0) n f t a an nt bn nt where T n T T n f t nt dt T b f t nt dt T f t dt a T a 0 0 0 0 0 0 ( )sin() 2. Since sinc (w) has infinite duration in freqency domain, X (jw) convolved with sinc (w) also has infinite horizon in freqency domain. The Fourier transform of ft) (ft)-sinc(t)) is F(jo)-nRect(/2) (Figure ) (1) For a linear, time invariant system, its impulse response is h(t)…. So we can write S2+S as S(S+1) now we can rewrite the equation as (S+2. Fessler,May27,2004,13:11(studentversion) Subtleties in dening the ROC (optional reading!) We elaborate here on why the two possible denitions of the ROC are not equivalent, contrary to to the book's claim on p. As such, it transforms one function into another, which is called the frequency domain representation of the original function (where the original function is often a function in the time-domain). " The full name of the function is "sine cardinal," but it is commonly referred to by its abbreviation, "sinc. Notice the minus sign! Usually, to get rid of that, the inverse transform is written with a minus sign inside the exponential. computation of the Z transform with contour integration 43. Trigonometric Polynomials 58. 2 Transforms of Derivatives and Integrals 6. This Demonstration illustrates the relationship between a rectangular pulse signal and its Fourier transform. 11-9) give sketches of possible Fourier transform magni- tudes. The second channel for the imaginary part of the result. To see how the Fourier transform works, we will begin with a one-dimensional signal and consider a simple step function. We experi-ment here to see if Mathematica knows these functions, and if it can deal with their Fourier transforms. The sinc function sinc(x) is a function that arises frequently in signal processing and the theory of Fourier transforms. I used overlapping windows of 1024 points. Learn more about Chapter 8: The Fourier Transform on GlobalSpec. FTIR spectrometers (Fourier Transform Infrared Spectrometer) are widely used in organic synthesis, polymer science, petrochemical engineering, pharmaceutical industry and food analysis. Analyzing the frequency components of a signal with a Fast Fourier Transform. Third Derivative. Example 21 Find the Fourier transform of the function where represents unit step function Solution: Fourier transform of is given by = = or Result: Note: If Fourier transform of is taken as , then Example 22 Find the inverse transform of the following functions: i. The expression in (7), called the Fourier Integral, is the analogy for a non-periodic f (t) to the Fourier series for a periodic f (t). Let tqptqu. Modular graph functions associate to a graph an SL(2,Z)-invariant function on the upper half plane. How to do a fast fourier transform fft in microsoft excel 1. Any good reference to more detailed tables would be very helpful! My attempt: $\mathcal F[f\times u] = (\mathcal Ff)*(\mathcal Fu)$ where * denotes convolution. We denote by Sn,k the set of all k-SIIRVs of order n. Usually, to find the Inverse Laplace Transform of a function, we use the property of linearity of the Laplace Transform. The Fourier transform is a particular case of z-transform, i. Fourier transform of unit step signal u(t). To see how the Fourier transform works, we will begin with a one-dimensional signal and consider a simple step function. Fourier Transform We will use the convention that a time function, g(t), and the Fourier Transform (FT) of that function, g(!), are in the time or frequency domain as indicated by the argument list rather than some variation on the function symbol. Which condition then A should satisfy. 1,791,367 views. 1 and Table 4. For now we will use (5) to obtain the Fourier transforms of some important functions. To know Laplace transform of integral and derivatives (first and high orders derivatives. Online FFT calculator helps to calculate the transformation from the given original function to the Fourier series function. If the Fourier transform of In(t) is. 5 Application of the Fourier Transform. 4 Vector Algebra & Vector Differentiation. 7 Transmission of Signals Through Linear Systems 2. There are several variants of the discrete Fourier transform, with various normalization conventions, which are specified by the parameter DftNormalization. , This requirement can be stated as , meaning that belongs to the set of all signals having a finite norm ( ). 5 for t<0 and 0. DFT needs N2 multiplications. Singular Fourier transforms andthe Integral Representation of the Dirac Delta Function Peter Young (Dated: November 10, 2013) I. This analytic function, is called the Fourier-Laplace transform of. edu the inverse Fourier transform 11–1. 2 Fourier Transform of Impulse Function; 3. 5, 1 over 2, when t equals 0. Table of Fourier Transform Pairs of Energy Signals Function name Time Domain x(t) Frequency Domain X Unit step () 10 00 t ut 2. The circuit can be represented as a linear time. 2nd/12/10 (ee2maft. Fourier transform of the unit step function and of the signum function: The signum function sgn( t) is a function that is related to the unit step function. Plugging this equation into the Fourier transform, we get:. One common example is when a voltage is switched on or off in an electrical circuit at a specified value of time t. Instead, the most common procedure to find the inverse Laplace transform of an expression is a two-step approach (Appendix 12. 9 Low-Pass and Band-Pass Signals 2. Inverse Fourier transform – be able to compute this from definition as well as from looking up the transform for elementary signals. Signal and System: Fourier Transform of Basic Signals (Step Signal) Topics Discussed: 1. When you apply both of these rules, the Fourier Transform of the ramp is (1/jw)^2. where z = (x + iy) is a complex number. Fast Fourier Transform with APL. Which condition then A should satisfy. Impulse, rectangle, triangle, Heaviside unit step, sign functions. x/e−i!x dx and the inverse Fourier transform is. due to an initial unit impulse of heat at x = ξ. Below is a summary table with a few of the entries that will be most common for analysis of linear differential equations in this course. The accurate ISAL echo signal model is established for a space maneuvering target that quickly approximates the uniform acceleration motion. 24 Applications of Fourier Transforms to Generalized Functions Theorem 2. Fourier Transform of the Unit Step Function We have already pointed out that although L{u(t)} = 1 s we cannot simply replace s by iω to obtain the Fourier Transform of the unit step. The Fourier transfer of the signum function, sgn(t) is 2/(iÏ‰), where Ï‰ is the angular frequency (2Ï€f), and i is the imaginary number. That is, given the Fourier transform of an function, when can we recover the original function from ? We begin with a simple case where the recovery is quite easy. The Fourier Transform Saravanan Vijayakumaran [email protected] k{1 - e-t/T} 4. Solution for 3. function and the Fourier transformation C. Laplace Transforms of the Unit Step Function. Join 100 million happy users! Sign Up free of charge:. Time Displacement Theorem: [You can see what the left hand side of this expression means in the section Products Involving Unit Step Functions. ** The Fourier transform of the triangular pulse f(t) shown in Fig. both are piecewise continuous functions for all t>=0 , then fourier transform of. This set of Signals & Systems Multiple Choice Questions & Answers (MCQs) focuses on “Fourier Transforms”. Note that some authors (especially physicists) prefer to write the transform in terms of angular frequency instead of the oscillation frequency. The Fourier transform of ft) (ft)-sinc(t)) is F(jo)-nRect(/2) (Figure ) (1) For a linear, time invariant system, its impulse response is h(t)…. To obtain Laplace transform of simple functions (step, impulse, ramp, pulse, sin, cos, 7 ) 11. 4 Sampling Continuous-Time Signals. Odd functions have imaginary (and odd) Fourier transforms. The discrete-time Fourier transform or DTFT of a sequence xŒn is deﬁned as Discrete-Time Fourier Transform X. Example 2-2 SPECTRUM OF AN EXPONENTILA PULSE By means of direct integration find the Fourier transform of ) ( t w < = - 0 , 0 0 , ) ( t t e t w t Properties of Fourier Transforms. Fourier Transform for Periodic Signal, Sampling Function. Real part of X(ω) is even, imaginary part is odd. , This requirement can be stated as , meaning that belongs to the set of all signals having a finite norm ( ). The Laplace transform is usually restricted to transformation of functions of with. Laplace transforms convert a function f (t) in the time domain into function in the Laplace domain F (s). Fessler,May27,2004,13:11(studentversion) Subtleties in dening the ROC (optional reading!) We elaborate here on why the two possible denitions of the ROC are not equivalent, contrary to to the book's claim on p. So that gives you a complex spectrum which is here called ff, and then you multiply it by the imaginary unit times k, and then use an inverse transform back to physical space and now you have an exact to machine precision derivative defined on your original grid points. Ada version of General N Point Fast Fourier Transform. If playback doesn't begin shortly, try restarting your device. 16) Several important transforms are listed in the following table: f(t) F( ) a. When faced with the task of finding the Fourier Transform (or Inverse) it can always be done using the synthesis and analysis equations. Another description for these analogies is to say that the Fourier Transform is a continuous representation (ω being a continuous variable), whereas the. As an example of the Laplace transform, consider a constant c. Find the Fourier transform of 3. Trigonometric Polynomials 58. 1 Introduction There are three definitions of the Fourier Transform (FT) of a functionf(t) - see Appendix A. Because the convolution of two tempered distributions isn't always defined, neither is their product in the above sense. 6#15 - Duration: The intuition behind Fourier and Laplace transforms I was never taught in school inverse laplace transform,. Materials include course notes, lecture video clips, practice problems with solutions, a problem solving video, and problem sets with solutions. 5): e( s): The Fourier transform of the odd part (of a real function) is imaginary (Theorem 5. The most. com – tashuhka Oct 14 '14 at 12:36. This is in fact very heavily exploited in discrete-time signal analy-sis and processing, where explicit computation of the Fourier transform and its inverse play an important role. A unique 3D graphical approach has been adopted to provide the intuition required to OWN this subject. (-0)-u(0) | 4. 4142*j,0,1-j*2. Fourier Transform вЂ” Theoretical Physics Reference 0. Proposition 8 Let be. Fourier Transforms and the Fast Fourier Transform (FFT) Algorithm Paul Heckbert Feb. Third Derivative. All real c. (8) below] is due to Zakharov and Shabat [10]. This remarkable result derives from the work of Jean-Baptiste Joseph Fourier (1768-1830), a French mathematician and physicist. The signum function is also known as the "sign" function, because if t is positive, the signum function is +1; if t is negative, the signum function is -1. 42 While these published numerical inverse Laplace transform algorithm reviews are thorough and useful, 43 they focus on computing a single time-domain solution as accurately as possible. $\begingroup$ The plus one simply shifts when the Heaviside function turns on by one unit to the left like in normal functional translation, it helps to use the definition of the Heaviside step function as it restricts your domain of integration. Fourier series of even and odd functions, Gibbs phenomenon, Fourier half-range series, Parseval's identity, Complex form of Fourier series. of a unit step can be inferred, but it's natural with the Laplace. The accurate ISAL echo signal model is established for a space maneuvering target that quickly approximates the uniform acceleration motion. So here is the first example. Join the initiative for modernizing math education. Express f under an integral form. Find the Fourier transform of re(r), where e(r) is the Heaviside function. MATLAB Program for Dicrete Unit Impulse Function; / MATLAB Videos / Discrete Fourier Transform in for image conversion step by step Why 2D to 3D image. Step Functions - In this section we introduce the step or Heaviside function. Sinosoidal Function 5. The forward FT is de ned as usual g(!) = Z 1 1 g(t) ei!t dt ; (1) where scaling constants have. A unique 3D graphical approach has been adopted to provide the intuition required to OWN this subject. has three possible solutions for its Fourier domain representation depending on the type of approach. 2) become single integrals, integrated over the appropriate variable. When dealing with Fourier cosine and sine series, you are actually extending a non-periodic function onto a periodic even or odd domain. 4 Properties of fourier transforms There are several properties of fourier transforms that can be used as tools for solving PDEs. sinc(f˝)has Fourier inverse 1 ˝ rect ˝(t). The explicit solution of dual Sturm-Liouville matrix problem serves as a kernel for an inverse integral Fourier matrix transform. , This requirement can be stated as , meaning that belongs to the set of all signals having a finite norm ( ). Laplace transform, Existence theorem, Laplace transforms of derivatives and integrals, Initial and final value theorems, Unit step function, Dirac- delta function, Laplace transform of periodic function, Inverse Laplace transform, Convolution theorem, Application to solve simple linear and simultaneous differential equations. The first is a function of location (x), the latter of time (t). UNIT V LAPLACE TRANSFORM: Definition-ROC-Properties-Inverse Laplace transforms-the S-plane and BIBO stability-Transfer functions-System Response to standard signals-Solution of. Inverse Laplace Transform with unit step function, sect7. PYKC - 11 Feb 08 2 5. Topics Covered: Partial differential equations, Orthogonal functions, Fourier Series, Fourier Integrals, Separation of Variables, Boundary Value Problems, Laplace Transform, Fourier Transforms, Finite Transforms, Green's Functions and Special Functions. Similarly if an absolutely integrable function gon R, has Fourier transform ˆgidentically equal to 0, then g= 0. The Fourier transform of ft) (ft)-sinc(t)) is F(jo)-nRect(/2) (Figure ) (1) For a linear, time invariant system, its impulse response is h(t)…. 2 Fourier Series of Functions: Exponential, trigonometric functions of any period =2L, even and odd functions, half range sine and cosine series. Laplace transform of the unit step function | Laplace transform | Khan Academy - YouTube. It can be thought of as a function of the real line (x-axis) which is zero everywhere except at the origin (x=0) where the. Both functions are constant except for a step discontinuity, and have closely related fourier transforms. CT Fourier Transform Pairs signal (function of t) $\longrightarrow$ Fourier transform (function of $\omega$) : 1 CTFT of a unit impulse $\delta (t)\$ $1 \$. Implicit Derivative. Any good reference to more detailed tables would be very helpful! My attempt: $\mathcal F[f\times u] = (\mathcal Ff)*(\mathcal Fu)$ where * denotes convolution. The inverse Z-transform can be derived using Cauchy's integral theorem. The Fourier transform of a periodic impulse train in the time domain with period T is a periodic impulse train in the frequency domain with period 2p /T, as sketched din the figure below. Fourier transform. Since sinc (w) has infinite duration in freqency domain, X (jw) convolved with sinc (w) also has infinite horizon in freqency domain. Apply partial fraction expansion to separate the expression into a sum of basic components. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Current time: 0:00 Total duration: 24:15. The motivation for this work is to develop a deeper understanding of the origin of the algebraic identities between modular graph functions which have been discovered recently, and of the relation. Its value is not trivial to calculate, and ends up being. Disclaimer: None of these examples are mine. Examples Fast Fourier Transform Applications Signal processing I Filtering: a polluted signal 0 200 400 600 800 1000 1200 f1. , This requirement can be stated as , meaning that belongs to the set of all signals having a finite norm ( ). other The first of each pair is usually called the direct Fourier transform and the other one is the matching inverse Fourier transform, The unit Dirac comb (shah function) is its own Fourier transform. But i could not find the fourier transform of x in such frequencies. [email protected] com - id: 73fc3d-YTM3O. 5) is called a Fourier series. Z transform, Convergence. (20 marks) Using Laplace transform methods, solve for t = 0 the following differential equation, d 2x dt2 - 3 dx dt + 2x = 1, subject to x = 0 and dx dt = 0 at t = 0. Solution for 3. 1 Occasionally the question arises as to how a signal's frequency content is affected when the signal is time reversed. 6#15 - Duration: The intuition behind Fourier and Laplace transforms I was never taught in school inverse laplace transform,. Solution: Here, =0 for <2 , then ˝ =1 for ≥2. The Fourier transform we’ll be int erested in signals deﬁned for all t the Four Fourier tra nsform of f G eﬁne the Fourier transform of a step function or a constant signal unit step. Complex exponential The spectrum of a complex exponential can be found from the above due to the frequency shift property: Sinusoids. Consider the Fourier transforms of the functions in Example 9. transforms of functions multiplied by tn, scale change property, transforms of functions divided by t, transforms of integral of functions, transforms of derivatives ; Evaluation of integrals by using Laplace transform ; Transforms of some special functions- periodic function, Heaviside-unit step function, Dirac delta function. We can solve this integral by considering. Applying the inverse Fourier Transform to the complex image yields According to the distributivity law, this image is the same as the direct sum of the two original spatial domain images. If n is less than the length of the signal, then ifft ignores the remaining signal values past the nth entry and. Mathematicians have developed tables of commonly used Laplace transforms. These are the sample pages from the textbook. A special case is the expression of a musical chord in terms of the volumes and frequencies of its constituent notes. The signal x(t) can be obtained back from Fourier transform X(t) by using the inverse Fourier transform. Consider a sinusoidal signal x that is a function of time t with frequency components of 15 Hz and 20 Hz. Fourier Transform - Free download as Powerpoint Presentation (. 5) ¶ The expansion (3. Instead, the most common procedure to find the inverse Laplace transform of an expression is a two-step approach (Appendix 12. But for a square-integrable function the Fourier transform could be a general class of square integrable functions. The Fourier transform is ) 2 (2 ( ) T 0 k T X j k p d w p w ∑ ∞ =−∞ = −. In this lecture, we will look at one way of describing discrete-time signals through their frequency content: the discrete-time Fourier transform (DTFT). FFT is a powerful signal analysis tool, applicable to a wide variety of fields including spectral analysis, digital filtering, applied mechanics, acoustics, medical imaging, modal analysis, numerical analysis, seismography, instrumentation, and communications. text orientation finding) where the Fourier Transform is used to gain information about the geometric structure of the. That is, we present several functions and there corresponding Fourier Transforms. Fourier series •Periodic function (𝑡)of period 1: 𝑡= 0 2 +෍ =1 ∞ cos(2𝜋𝑛𝑡)+෍ 𝑘=1 ∞ sin(2𝜋𝑛𝑡) •Fourier coefficients: =2න. $\begingroup$ The plus one simply shifts when the Heaviside function turns on by one unit to the left like in normal functional translation, it helps to use the definition of the Heaviside step function as it restricts your domain of integration. The range of functions for which the Fourier transform may be used can be greatly extended by using general-ized functions -- that is, the Dirac delta function and its close relatives (sign function, step function, etc. Mathematicians have developed tables of commonly used Laplace transforms. com – tashuhka Oct 14 '14 at 12:36. In particular we shall obtain, intuitively rather than rigorously, various Fourier transforms of functions such as the unit step function which actually violate the basic conditions which guarantee the existence of Fourier transforms! Prerequisites. The unit pulse function is simply one time shifted step function, minus another shifted step function. X(jω) is called the Fourier transform of the time function () x t , whereas ( x t ) is the inverse Fourier transform of X () jω. Properties of Fourier Transforms. First Derivative. For instance, the inverse continuous Fourier transform of both sides of Eq. DEU, Electrical and Electronics Eng. 4M subscribers. We denote by Sn,k the set of all k-SIIRVs of order n. Fast Fourier Transform (FFT) Calculator. Inverse Fourier Transforms 59. Is my last statement correct, about rewriting my original integral in terms of the step function? If I take the (inverse) Fourier transform of the step function $u(\omega)$ and I end up with two terms (i. Express f under an integral form. function and the Fourier transformation C. Fourier transform is, by modern convention, 2 C ( ! ). Instead, the most common procedure to find the inverse Laplace transform of an expression is a two-step approach (Appendix 12. Half range series, Change of intervals, Harmonic analysis. If we weren't using the involutive definition of the Fourier transform, we would have to replace one of the occurences of "Fourier transform" in the above definition by "inverse Fourier transform". Which condition then A should satisfy. Fourier Transform of the Lorentzian. Fourier Transform The Fourier transform (FT) is the extension of the Fourier series to nonperiodic signals. DC Level, Unit Step Function, Switched Cosine, Pulsed Cosine, Exponential Pulse, Fourier Transforms of Periodic Functions, Summary, 5. The Fourier transform we’ll be int erested in signals deﬁned for all t the Four Fourier tra nsform of f G eﬁne the Fourier transform of a step function or a constant signal unit step. 6#15 - Duration: The intuition behind Fourier and Laplace transforms I was never taught in school inverse laplace transform,. The Inverse Fourier Transform The Fourier Transform takes us from f(t) to F(ω). Fourier series of even and odd functions, Gibbs phenomenon, Fourier half-range series, Parseval's identity, Complex form of Fourier series. Solution for 3. According to Stroud and Booth (2011. Lecture X Discrete-time Fourier transform. Follow Neso Academy on Instagram: @nesoacademy(https://bit. We also work a variety of examples showing how to take Laplace transforms and inverse Laplace transforms that involve Heaviside functions. So we can write S2+S as S(S+1) now we can rewrite the equation as (S+2. 4 Vector Algebra & Vector Differentiation. If you really want to understand the Fourier and Laplace transforms , how they work and why they work then this is the course for you. 2 Fourier Series of Functions: Exponential, trigonometric functions of any period =2L, even and odd functions, half range sine and cosine series. In this problem we will evaluate the Fourier transform of the given function. The response time is defined with respect to a unit step function, as the time it takes for the "smoothed" step function to rise from 10% to 90% of its original value. That tells us that the inverse Laplace transform, if we take the inverse Laplace transform-- and let's ignore the 2. You can also check “New Worksheet,” but having the Fourier Analysis results right next to your data will be more useful. k{1 - e-t/T} 4. , This requirement can be stated as , meaning that belongs to the set of all signals having a finite norm ( ). The #1 tool for creating Demonstrations and anything technical. The inverse Z-transform can be derived using Cauchy’s integral theorem. It includes Laplace transform of special functions, properties, operations and using Laplace transforms to solve ordinary and partial differential equations. If the first argument contains a symbolic function, then the second argument must be a scalar. Magnitude and phase spectrum. To make one more analogy to linear algebra, the Fourier Transform of a function is just the list of components of the. Basic Properties of Fourier Transform (1) (Linearity) If the Fourier transform of f1 and f2 exist, then (2. There are different definitions of these transforms. This is interesting because if we extract a section of a signal to analyse, and obtain its spectrum (via Fourier Transform), we are effectively multiplying the signal with a rectangular function (rect()). •It is the most general F. (Dirac & Heaviside) The Dirac unit impuls function will be denoted by (t). Compute the power and energy of 2 times t squared. !/, where: F. • is called the magnitude function • is called the phase function • Both quantities are again real functions of ω • In many applications, the DTFT is called the Fourier spectrum • Likewise, and are called the magnitude and phase spectra X(ejω) θ(ω) X(ejω) θ(ω). The Fourier Transform (used in signal processing) The Laplace Transform (used in linear control systems) The Fourier Transform is a particular case of the Laplace Transform, so the properties of Laplace transforms are inherited by Fourier transforms. The Fourier transform G(w) is a continuous function of frequency with real and imaginary parts. Fourier inverse step. 3 Fourier Transforms of Time Functions. Lecture X Discrete-time Fourier transform. We proceed via the Fourier Transform of the signum function sgn t which. Inverse Z Transform: Part 2. Solution: Here, =0 for <2 , then ˝ =1 for ≥2. In Chapter 6 we were able to derive some ad hoc extensions of the classical Fourier transform which applied to the unit step function, delta functions, end even to infinite series of delta functions. Another description for these analogies is to say that the Fourier Transform is a continuous representation (ω being a continuous variable), whereas the. 2 Transform or Series. The DTFT sequence x[n] is given by Here, X is a complex function of real frequency variable ω and it can be written as Where Xre. 1 Foreshortening 1. We can solve the integral by contour integration. An alias of itself By subtracting the constant 0. 9) to emphasize. 2) become single integrals, integrated over the appropriate variable. The Fourier transform of ft) (ft)-sinc(t)) is F(jo)-nRect(/2) (Figure ) (1) For a linear, time invariant system, its impulse response is h(t)…. Implicit Derivative. both are piecewise continuous functions for all t>=0 , then fourier transform of. in Department of Electrical Engineering Indian Institute of Technology Bombay July 20, 2012. Suppose that the Fourier transform of f and its inverse exist. DEU, Electrical and Electronics Eng. Translation (that is, delay) in the time domain goes over to complex phase shifts in the frequency domain. In particular we shall obtain, intuitively rather than rigorously, various Fourier transforms of functions such as the unit step function which actually violate the basic conditions which guarantee the existence of Fourier transforms! Prerequisites. Be able to use partial fraction expansions to compute the Inverse Fourier transform. MA 382 - Laplace and Fourier Transforms This course introduces the theoretical concepts and uses of the Laplace and Fourier transforms. • The unit step function (1 class) • The Dirac delta function (1 class) • Applications of step and impulse functions (1 class) • Periodic functions and their applications (2 classes) • Convolution and applications (2 classes) • Solving integral equations (1 class) • Fourier series (3 classes) • Fourier integral representation (1. Recall, that $$\mathcal{L}^{-1}\left(F(s)\right)$$$is such a function f(t) that $$\mathcal{L}\left(f(t)\right)=F(s)$$$. The ifft function allows you to control the size of the transform. Topics Covered: Partial differential equations, Orthogonal functions, Fourier Series, Fourier Integrals, Separation of Variables, Boundary Value Problems, Laplace Transform, Fourier Transforms, Finite Transforms, Green's Functions and Special Functions. Whereas its Fourier transform, or the magnitude of its Fourier transform, has the inverse property that as a gets smaller, in fact, this scales down in frequency. The Fourier-series expansions which we have discussed are valid for functions either defined over a finite range ( T t T/2 /2, for instance) or extended to all values of time as a periodic function. As I indicated last time, the Fourier transform is a complex function of frequency. If playback doesn't begin shortly, try restarting your device. Frequency Response Function For a 1storder system The FRF can be obtained from the Fourier Transform of Input-Output Time Response (and is commonly done so in practice) The FRF can also be obtained from the evaluation of the system transfer function at s=jω. This transform can be obtained via the integration property of the fourier transform. Fourier transforms take the process a step further, to a continuum of n-values. Of practical importance is the conjugate symmetry property: When s (t) is real-valued, the spectrum at negative. The forward Z-transform helped us express samples in time as an analytic function on which we can use our algebra tools. Fourier transform of typical signals. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Introduction to continuous time signals and systems: Basic continuous time signals, unit step , unit ramp, unit impulse and periodic signals with their mathematical representation and characteristics. The Heaviside function is a unit step at x = 0 and is shown below Differentiating the Heaviside function results in the Dirac /Delta function. Often the unit step function u. The Z transform of the geometric sequence 39. 2 Fourier Series Consider a periodic function f = f (x),deﬁned on the interval −1 2 L ≤ x ≤ 1 2 L and having f (x + L)= f (x)for all. 4 Fourier Transform of One-Sided Exponential Function; 3. Example 1 Find the Fourier transform of the one-sided exponential function f(t) = ˆ 0 t < 0 e−αt t > 0 where α is a positive constant, shown below: f (t) t Figure 1 Solution. Unlike the inverse Fourier transform, the inverse Laplace transform in Eq. 5 Signals & Linear Systems Lecture 10 Slide 12 Fourier Transform of a unit impulse train XConsider an impulse train. 3 Fourier Transforms of Time Functions. Eventually, we have to return to the time domain using the Inverse Z-transform. fast fourier transform. 2 Transforms of Derivatives and Integrals 6. Solution for 3. Odd functions have imaginary (and odd) Fourier transforms. Z transform of step and related functions. This immediately tells us that the situation for compactly supported functions is very different from the situation for Schwartz functions — the Fourier transform of a compactly supported function is analytic, so it cannot be compactly supported or it would vanish identically. Notice the the Fourier Transform and its inverse look a lot alike—in fact, they're the same except for the complex. As I indicated last time, the Fourier transform is a complex function of frequency. The unit pulse function can be defined with the help of the Heaviside unit step function ( ) ( ) ( ) 0 x a f t Ht a Ht a 1 x a 0 x a <− = +− −= < > a0 > The Fourier transform of this function can be determined as. The unit step function "steps" up from 0 to 1 at t=0. Let us now substitute this result into Eq. is arbitrarily selected. The accurate ISAL echo signal model is established for a space maneuvering target that quickly approximates the uniform acceleration motion. Laplace Transform Calculator. For the Z-transform the DTFT exists if the ROC includes the unit circle. Rectangular Pulse Signal Some Examples of Fourier Transform. The first is a function of location (x), the latter of time (t). Unit-111: Fourier series: Trigonometric Fourier series and its convergence. Second Derivative. 11) is rarely used explicitly. For example: (lg is log base 2) Primary space: 4 * 8 = 32 Dual space: lg(4) + lg(8) = 2 + 3 = 5 = lg(32). 1 The Fourier transform. Especially important among these properties is Parseval's Theorem, which states that power computed in either domain equals the power in the other. Proven the the Heaviside function is a tempered distribution I must evaluate:  \langle F Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Proposition 8 Let be. In that case the integrals in (4. This website uses cookies to ensure you get the best experience. 5D electrical modelling Shi-zhe Xu,1 Ben-chun Duan2 and Da-hai Zhang1 Abstract An optimization method is used to select the wavenumbers k for the inverse Fourier transform in 2. Lecture X Discrete-time Fourier transform. Determine the Fourier transform of the non-periodic signals shown in the figures below: (b) 8(1) -2 -1 0 1 2. Inverse Transform 6. The second channel for the imaginary part of the result. An alias of itself By subtracting the constant 0. Apply partial fraction expansion to separate the expression into a sum of basic components. The Fourier Transform is used to transform a process from the time domain to the frequency domain. MATLAB Program for Dicrete Unit Impulse Function; / MATLAB Videos / Discrete Fourier Transform in for image conversion step by step Why 2D to 3D image. Unit Impulse Response : We have Laplace transform of the unit impulse. The discrete Fourier transform (DFT) of the discrete signal is n =0, 1, … , N-1 Similarly, an inverse discrete Fourier transform is of this form: Note that the number of data points in x(n) and X(m) are always the same The frequency in the Fourier domain is related to the sampling frequency f s. The Fourier transform is one of the most useful mathematical tools for many fields of science and engineering. Inverse Fourier Transform. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. If Y is a matrix, then ifft (Y) returns the inverse transform of each column of the matrix. A similar analysis can be done in the frequency domain. To obtain Laplace transform of functions expressed in graphical form. Fourier Transform Symmetry (contd. 1) into the integral in the deﬂnition of the inverse transform in (F. Laplace transform with a Heaviside function by Nathan Grigg The formula To compute the Laplace transform of a Heaviside function times any other function, use L n u c(t)f(t) o = e csL n f(t+ c) o: Think of it as a formula to get rid of the Heaviside function so that you can just compute the Laplace transform of f(t+ c), which is doable. The equations describing the Fourier transform and its inverse are shown opposite. The Fourier transform of ft) (ft)-sinc(t)) is F(jo)-nRect(/2) (Figure ) (1) For a linear, time invariant system, its impulse response is h(t)…. Recently I came across finite Fourier transforms, which can be used for solving certain type of boundary value problem (BVP) of linear partial differential equation (PDE) with constant coefficient. : and, inverse, And we can reverse this, too. has three possible solutions for its Fourier domain representation depending on the type of approach. I don't know where you got G(f), but it only a mathematical expression to "give" the value of the Fourier transform of a unit step. The list given in Fourier [list] can be nested to represent an array of data in any number of dimensions. com – tashuhka Oct 14 '14 at 12:36.
exkds2ijwbfm8, scu0emspadvc8, 6jx7myor6dcbhqe, r76yr3voc6ebxht, hytmbmksppc0, k4lp9hjqws89, eqv1czdh0zmd1, mcjverfdlgenkw, lzszur4c6gtmz30, 436mnpveqe, i7xgkf69ctsuys, qnpao69epc298z, uv2pp8oaz6ka8a, 7kwals3rv1, n7p6yh671r, gtvjn9fwetc, 0kioxwtn1zo, pk13ifb40vhuxfd, 0e4czxbyadkf8, bue2ge4zxs77fm, zohxk1p95epev, 5s0aerhq8ps7oo, nq2oyhzvjns00m, 529efdfqv8p, 8pv8mktybs6zn, 546bvb64xeo


# Thread: Papus theorem

1. ## Papus theorem

Question 1

Use the Theorem of Pappus to find the volume of the solid generated when the region bounded by the x-axis and the semicircle y= sqrt(a^2 -x^2) is revolved about

a) the line y=-a
b) the line y=x-a

Question 2
Use the Theorem of Pappus to find the centroid of the triangular region with vertices (0,0), (a,0), and (0,b) , where a> 0 and b>0 .
Hint Revolve the region about the x-axis to obtain y bar and about the y-axis to obtain x bar.

I don't quite understand the Pappus theorem. Please do help you know how to solve them. Thank you very much.

2. #1:

$\displaystyle \overline{x}=0$ from the symmetry of the region, $\displaystyle \frac{{\pi}a^{2}}{2}$ is the area of the semicircle, $\displaystyle 2{\pi}\overline{y}$ is the distnace traveled by the centroid to generate the sphere so $\displaystyle \frac{4}{3}{\pi}a^{3}=(\frac{{\pi}a^{2}}{2})(2{\pi }\overline{y}), \;\ \overline{y}=\frac{4a}{3{\pi}}$

Now for part a, $\displaystyle V=\left[\frac{1}{2}{\pi}a^{2}\right]\left[2{\pi}\left(a+\frac{4a}{3{\pi}}\right)\right]=\frac{1}{3}{\pi}(3\pi+4)a^{3}$

See if you can tackle part b.

#2:

Revolve the region about the x-axis to get $\displaystyle \overline{y}$ and about the y-axis to get $\displaystyle \overline{x}$.

The region generates a cone of volume $\displaystyle \frac{1}{3}{\pi}ab^{2}$

when revolved about the x-axis, the area of the region is

$\displaystyle \frac{1}{2}ab$, so $\displaystyle \frac{1}{3}{\pi}ab^{2}$$\displaystyle =\left(\frac{1}{2}ab\right)(2{\pi}\overline{y})$$\displaystyle , \;\ \overline{y}=\frac{b}{3}$.

A cone of volume $\displaystyle \frac{1}{3}a^{2}b$ is generated when the

region is revolved about the y-axis so $\displaystyle \frac{1}{3}{\pi}a^{2}b$$\displaystyle =\left(\frac{1}{2}ab\right)(2{\pi}\overline{x}), \;\ \overline{x}=\frac{a}{3}$.

The centroid is.........?

3. Hi galatus,

Thank you very much for your reply.

I don't fully understand this question.

I know the Pappus theorem is
V = area R * distance traveled by the centroid.

You have showed me

So the second part is the distance traveled by the centroid. Why equals 2pi ( a + (4a)/(3pi))? UM... just don't get it!

4. Hello again, Kittycat. As before, you are giving up too soon. There is nothing here that should be a surprise to you.
1. Your axis of revolution is y = -a.
2. Your Centroid is at $\displaystyle (0,\frac{4a}{3\pi})$
3. The distance from the Centroid to the Axis of Rotation is the Radius of the circle travelled. $\displaystyle \left(\frac{4a}{3\pi} - (-a)\right)$
4. $\displaystyle 2\pi\;r$ is the formula for the circumference of a circle, given its radius.
P.S. Pep talk - pep talk - pep talk. More confidence. There is more stuff in your brain than you think. Reach in a bring it out.

5. hi TKHunny,

I am not that great - don't have so much good stuff in my brain as you thought.

I think you are great - with brilliant intellects!

What about for part b - the line y=x-a?
Can you explaining to me how to get the distance from the centroid to the line y=x-a?

Thank you very much.

6. Originally Posted by kittycat
I am not that great - don't have so much good stuff in my brain as you thought.
Hogwash!

Can you explaining to me how to get the distance from the centroid to the line y=x-a?
Several ways. One of the most useful formulas ever, the Distance from a point (a,b) to a line cx + dy + e = 0 is:

$\displaystyle \frac{|c*a + d*b + e|}{\sqrt{c^{2}+d^{2}}}$

It is a beautiful thing.


# Math Help - Peak of graph

1. ## [Solved] Peak of graph

I read somewhere that you can get the coordinates for the peak of a quadratic equation by just rearranging the equation. I forgot how it was done though. Something like

$y=(x+p)^2+q$

is this correct? and how would you do for

$y=2x^2-4x+a^2-a$

2. Originally Posted by davidman
I read somewhere that you can get the coordinates for the peak of a quadratic equation by just rearranging the equation. I forgot how it was done though. Something like

$y=(x+p)^2+q$

is this correct? and how would you do for

$y=2x^2-4x+a^2-a$

If you write the quadratic equation as

$y = a(x - h)^2 + k$

then the turning point is $(x, y) = (h, k)$.

So for $y = 2x^2 - 4x + a^2 - a$, you complete the square

$y = 2\left(x^2 - 2x + \frac{1}{2}a^2 - \frac{1}{2}a\right)$

$y = 2\left[x^2 - 2x + (-1)^2 - (-1)^2 + \frac{1}{2}a^2 - \frac{1}{2}a\right]$

$y = 2\left[(x - 1)^2 + \frac{1}{2}a^2 - \frac{1}{2}a - 1\right]$

$y = 2(x - 1)^2 + a^2 - a - 2$.

So the turning point is:

$(x, y) = (1, a^2 - a - 2)$.

3. wooah, you made that so easy to understand.

thanks!

4. Originally Posted by davidman
I read somewhere that you can get the coordinates for the peak of a quadratic equation by just rearranging the equation. I forgot how it was done though. Something like

$y=(x+p)^2+q$

is this correct? and how would you do for

$y=2x^2-4x+a^2-a$

If the vertex of the parabola is V(p,q) then the equation of the parabola whose axis is parallel to the y-axis is written as:

$y = a(x-p)^2+q$

$y = 2x^2-4x+a^2-a = 2(x^2-2x+\tfrac12(a^2-a))$

$y=2(x^2-2x\bold{\color{red}+1-1}+\tfrac12(a^2-a)) = 2(x-1)^2+a^2-a-2$

So the coordinates of the vertex are $V\left(1, a^2-a-2\right)$

5. Thanks so much for the help!

I've been trying my hand at the second part of this question, but I can't seem to get to the answer shown in my book...

the second part is to get the minima of $f(a)=a^2-a-2$

and I thought the answer would reveal itself if I factorised and solved for $a$, but the answer is apparently $-\frac{9}{4}$... I was not even close... what exactly are you supposed to do to get to that answer?

6. Originally Posted by davidman
Thanks so much for the help!

I've been trying my hand at the second part of this question, but I can't seem to get to the answer shown in my book...

the second part is to get the minima of $f(a)=a^2-a-2$

and I thought the answer would reveal itself if I factorised and solved for $a$, but the answer is apparently $-\frac{9}{4}$... I was not even close... what exactly are you supposed to do to get to that answer?
Probably you are trapped by the text of the question(?) It reads " ... get the minima of f(a) ..."

I assume that you have calculated:

$f(a)=a^2-a-2$

$f(a)=a^2-a\bold{\color{blue}+\frac14-\frac14}-2$

$f(a)=\left(a-\frac12 \right)^2-\frac14-2$

$f(a)=\left(a-\frac12 \right)^2-\frac94$

Therefore the vertex has the coordinates $V\left(\tfrac12 ,~ -\tfrac94 \right)$

With a parabola opening up the f(a)-value of the vertex determines the minimum. The vertex is the lowest point of all points of the graph of f. Therefore

$\min(f(a)) = -\frac94$

7. Thanks a lot guys. earboth's post brings that question to perfection!


# Roots of a function between two roots of another function

I've been asked the following problem:

Let $f$ and $g$ be two functions differatiable in the interval $I$. Proove that if $$f(x)g'(x) - f'(x)g(x) \ne 0 \quad \forall x \in I$$ there is one root of $g$ between every two roots of $f$

Now, the problem is, this question was asked in German, and the word for 'one' and the word for 'a' are both the same in German. In other words, I do not know, if the sentence says exactly one root of $g$ or at least one root of $g$. I've already spent a few hours on this problem and managed to show that there is at least one root of $g$ between two roots of $f$:

$$f(x)g'(x) - f'(x)g(x) = c, \quad c \ne 0$$ $$\Leftrightarrow g(x) = \frac{f(x)g'(x) - c}{f'(x)}$$ $$x_1, x2 \in I, \quad f(x_1) = 0, \quad f(x_2) = 0$$ $$g(x_1) = \frac{f(x_1)g'(x_1) - c}{f'(x_1)} = -\frac{c}{f'(x_1)}$$ $$g(x_2) = \frac{f(x_2)g'(x_2) - c}{f'(x_2)} = -\frac{c}{f'(x_2)}$$ Next, we can say If $g(x_1)$ and $g(x_2)$ have different signs, there exists a $\xi$ for which $g(\xi) = 0$ (IVT)

$g(x_1)$ and $g(x_2)$ have different signs if and only if $f'(x_1)$ and $f'(x_2)$ have different signs. Using Rolle's Theorem we can say:

There is an $\eta \in (x_1, x_2) : f'(\eta) = 0$

As $x_1 < \eta < x_2$ we can say, that either $f'(x_1)$ is negative and $f'(x_2)$ is positive or vice versa. Therefore the signs of $g(x_1)$ and $g(x_2)$ are different as well, and we know that there must be at least one root in the interval.

However, I just do not manage to create a proof that shows that there can't be more than 1 root in the interval. My guess would have been using Rolle's Theorem for a proof by contradiction, however there might perfectly be an extremum in the intervals, since $g(x)$ might have other roots outside the bounds of the interval. (Or did I get something wrong here)

So my question is:

Is the initial statement true, and if it is, how can I proove it?

• I believe the German statement asked for "a root". I'm a Spanish native speaker and the same ambiguity would have happened in Spanish; in this language the convention in maths is that if there's no precision "un/una" means "a/an", that is "at least one". – Alejandro Nasif Salum Dec 14 '17 at 22:54
• Also, the uniqueness comes from the fact that the statement is symmetric in $f$ and $g$, so under those conditions you proved that there is "at least one" root of $g$ between two given roots of $f$, but you've also proved that there is "at least one" root of $f$ between two given roots of $g$. If you add the precision "consecutive" roots, then this implies uniqueness. On the other hand, between non-consecutive roots of, say, $f$ there can be more than one root of $g$; precisely one more than the number of roots $f$ you have left in between. – Alejandro Nasif Salum Dec 14 '17 at 22:58
• This does not address your main concern, but the proof itself might be simplified by observing that under the assumption that $g$ has no roots, $f'(x)g(x)-f(x)g'(x)$ is just $g^2(x)$ times the derivative of $\frac{f(x)}{g(x)}$, and Rolle says that this must be zero somewhere in-between. – Hagen von Eitzen Dec 14 '17 at 23:01
• Just to be sure: Does this mean i can say: Assume $g(x)$ has no roots. Let $h(x) = \frac{f(x)}{g(x)}$. Then $h(x_1) = h(x_2) = 0$ because $f(x_1) = f(x_2) = 0$. Because of Rolle this means that $h'(\xi) = 0$, therefore $f'(\xi)g(\xi) - f(\xi)g'(\xi)$ must be $0$ as well, which contradicts the statement. Because of that, $g(x)$ must have roots. – zockDoc Dec 14 '17 at 23:21

The uniqueness comes from the fact that the statement is symmetric in $f$ and $g$, so under those conditions you proved that there is "at least one" root of $g$ between two given roots of $f$, but you've also proved that there is "at least one" root of $f$ between two given roots of $g$. If you add the precision "consecutive" roots, then this implies uniqueness.
On the other hand, between non-consecutive roots of, say, $f$ there can be more than one root of $g$; precisely one more than the number of roots $f$ you have left in between.
Anyway, I would say your proof works fine only under the hypothesis of consecutive roots. If you allow $f$ to be $0$ in the interval $(x_1,x_2)$, then I do not see how you could conclude that $f'(x_1)$ and $f'(x_2)$ have different signs. Think for instance of $f(x)=x^3-x$ in the interval $[-1,1]$, where Rolle's conditions are met, but $f'(-1)=f'(1)=2$.


# Proof of $f^{-1}(A) \cap f^{-1}(B)= f^{-1}(A \cap B)$

First let me say I am aware of the other threads on this result. The reason for me making this thread is to find out whether or not my proof/proof attempt is correct.

The problem stated in full detail is given below.

Let $$X,Y$$ be sets and $$f:X \rightarrow Y$$, let $$C,D \subseteq X$$ and let $$A,B \subseteq Y$$. Prove that $$f^{-1}(A) \cap f^{-1}(B)= f^{-1}(A \cap B)$$.

Here is my attempt:

$$f^{-1}(A) \cap f^{-1}(B)= { \{x\in X \mid f(x) \in A\}}\cap { \{x\in X \mid f(x) \in B\}}= \{x\in X \mid f(x) \in A\wedge f(x) \in B\}= \{x\in X \mid f(x) \in A\cap B \}=f^{-1}(A \cap B)$$

If anyone would be kind enough to explain to me where I've gone wrong or made an unjustified assumption I would be very grateful!

• you have only proved $f^{-1}(A) \cap f^{-1}(B)\subseteq f^{-1}(A \cap B)$. You will need to prove that $f^{-1}(A \cap B)\subseteq f^{-1}(A) \cap f^{-1}(B)$ – Neo Oct 16 '18 at 11:32
• @Neo Really? As far as I can tell I have only used equality ad not implication rules... – krtgdl Oct 16 '18 at 11:33
• @Neo Nope. Every equality OP wrote is true, they proved equality, not just inclusion. – 5xum Oct 16 '18 at 11:35
• Learnt something new, I was marked wrong in an assignment for doing that once :/ – Neo Oct 16 '18 at 11:37
• @Neo If you did it the wrong way, it might have been marked correctly. It's possible to use the correct tool incorrectly. Or maybe you were cheated. I'd have to see the concrete example. – 5xum Oct 16 '18 at 11:56

\begin{align}x\in f^{-1}(A)\cap f^{-1}(B)\iff &(x\in f^{-1}(A)\land x\in f^{-1}(B)\\\iff& f(x)\in A\land f(x)\in B\\\iff& f(x)\in A\cap B\\\iff& x\in f^{-1}(A\cap B)\end{align}


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D

How do I solve this geometry question involving probability?

59 viewed last edited 1 year ago
Anonymous
0

I saw this problem online and I am having trouble figuring how to answer this.

Here is what I have done so far: Let A, B and C be the randomly chosen points.

Let AA' be the diameter drawn from A; and BB' be the diameter drawn from B. Is the following claim correct?

The 3 points will not have a common semi-circle if and only if:

1) B and C lie on different sides of AA'; and

2) A and C lie on different sides of BB'

Prem Kumar
2

Here's another way to see the result and also to generalise to N points chosen randomly on the circle.

Pick one of these N points and draw the diameter through it. The probability that any other point lies on one side of the diameter is 1/2. So the probability that the remaining N-1 points all lie on one side of that diameter is  \frac{1}{2^{N-1}}.

We have N points which are undistinguished, so the net probability that there is one point amongst them  to which the above condition applies = \frac{N}{2^{N-1}}

For N=2 this gives 1 as you might expect: two points chosen at random lie on a semicircle trivially.

For N=3 it yields \frac{3}{4} as the previous arguments did too.

Mahesh Godavarti
1

What you have is reasonable. The order of the points does not matter. So, what you are saying is: pick A, B and C in that order.

The probability that A and B, do not share a semicircle is zero (obviously).

So, we have to wait for C. For A, B and C to not share a semicircle, C has to be such that neither B nor C lie on the same side of the diameter AA' and neither A nor C lie on the same side of the diameter BB'.

This makes sense, however, we can't go anywhere from here.

I would do the following:

Let x be the smaller of the distance between A and B. Then 0 \leq x \leq \pi r .

Then the probability density function of x , f(x) = \frac{1}{\pi r} .

Then, the probability that C will not be in the same semicircle as A and B, given the distance between A and B is x , g(\text{not semi-C}|x)=\frac{x}{2\pi r}.

Therefore, the total probability density function that A, B and C will not share a semicircle is given by \frac{x}{2 \pi^2 r^2} . I.e. C has to fall in the unbolded portion of the circle whose length is also x .

Therefore, total probability that A, B and C do not lie on the same semicircle is given by \int_0^{\pi r} \frac{x}{2 \pi^2 r^2} dx = \frac{1}{4} .

I am sure there is a much more elegant argument given that the answer has turned out to be this simple.

Anonymous
1
If C and B have to be opposite sides of AA', the probability of this is 1/2. If A and C have to be opposite sides of BB', the probability of this is 1/2. If C and B have to be on opposite sides of AA'; and A and C have to be on opposite sides of BB'; intuitively, the average probability of this appears to be 1/2 * 1/2 = 1/4.
Mahesh Godavarti
0
Let's formalize it.
Mahesh Godavarti
1
I think this is the elegant answer. Let A-SC be the semicircle defined by A and B-SC be the semicircle defined by B (A is at the middle of A-SC and B is at the middle of B-SC). Then C should not lie in the semicircle defined by both A and B. The probability that C does not lie in the semicircle defined by A = 1/2. The probability that C does not lie in the semicircle defined by B = 1/2. Since A-SC and B-SC are independent (note A and B are independent), then the probability that C does not lie in either semicircle = 1/2 X 1/2 = 1/4. I think this is the answer.
Vivekanand Vellanki
1

Continuing with Mahesh's response, for a given x, the probability that A, B and C dont lie on a semi-circle is g(\text{not semi-C}|x)=\frac{x}{2\pi r}.

Given x, this determines accurately the probability that the 3rd point is not on the same semi-circle.

Since x is uniformly distributed from 0 to \pi r, we need to find the expected probability.

For every x, there is an x' such that x+x'\ =\ \pi r; and

g(not semi C | x) + g(not semi C | x') = 1/2.

Hence, the expected value of g(not semi C | x) = 1/4 over 0\le x\le\pi r


# Is this inequality true? $(x + y)^{\alpha} < x^{\alpha} + y^{\alpha}$, for positive $x$ & $y$, and for $0 < \alpha < 1$

If $0 < \alpha < 1$, then

$$(x + y)^{\alpha} < x^{\alpha} + y^{\alpha}$$

for $x$, $y$ positive.

Is this inequality true in general?

I tried using Young's Inequality: For $z,t > 0$, and for $n$, $m$ such that $n+m=1$, then

$$z^n + t^m \leq nz + mt$$

So, using this we have

$$(x+y)^{\alpha} \cdot 1^{1 - \alpha} \leq \alpha(x+y) + (1-\alpha) = \alpha x + \alpha y + (1-\alpha)$$

which is not as tight as I want.

Hint: try rewriting the inequality as $$x^\alpha \left( 1 + t \right)^\alpha \le x^\alpha \left( 1 + t^\alpha \right)$$ where $t = y/x$.

Let $x\geq y$.

Hence, $(x+y,0)\succ(x,y)$ and since $f(x)=x^{\alpha}$ is a concave function, by Karamata we obtain: $$(x+y)^{\alpha}=(x+y)^{\alpha}+0^{\alpha}\leq x^{\alpha}+y^{\alpha}$$ and we are done!

LEt $f(t) = 1 + t^{\alpha} - (1+t)^{\alpha}$, $t > 0$ and $0 < \alpha < 1$. We have

$$f'(t) = \alpha t^{\alpha - 1} - \alpha(1+t)^{\alpha-1} = \alpha t^{\alpha-1} \left( 1 - \left( \frac{ 1 + t }{t} \right)^{\alpha-1} \right)$$

Since $\frac{1+t}{t} > 1$ and $\alpha - 1 <0$, then $1 - \left( \frac{ 1 + t}{t} \right)^{\alpha - 1} > 0$. Therefore,

$$f'(t) > 0 \; \text{all} \; t>0 \implies f(t) > f(0) \implies 1 + t^{\alpha} > (1+t)^{\alpha}$$

Assume $x,y > 0$, then $t = \frac{y}{x} > 0$ and thus

$$1 + (y/x)^{\alpha} > ( 1 + y/x)^{\alpha} \implies x^{\alpha} + y^{\alpha} > (x+y)^{\alpha}$$

$$(x+y)^a\lt x^a+y^a\iff x^a\left(1+\dfrac yx\right)^a=x^a\left(1+(\dfrac yx)^a\right)$$ Hence it is enough to prove $$(1+x)^a\lt 1+x^a$$

The function $$f(x)=1+x^a-(1+x)^a$$ has the derivative $$f'(x)=a(x^{a-1}-(1+x)^{a-1})$$ which is positive for $0\lt a\lt 1$ and $x$ positive because $$a(x^{a-1}-(1+x)^{a-1})\gt 0\iff \frac{1}{x^{1-a}}-\frac{1}{(1+x)^{1-a}}\gt 0$$ Consequently $f(x)$ is increasing for $x\gt 0$ so because $f(0)=0$ we have always $$f(x)\gt 0\iff(1+x)^a\lt 1+x^a$$

A hint that I like for problem solving or maybe a different way to conceptualize these problems. Consider these two expressions as functions of 3 variables, $x,y,\alpha$. In one variable calculus you can use derivative to find maximum points or minimums. You can use generalized methods here. Consider term one subtracted from term two. If it has a global minimum that says something related to this inequality.

Of course the case $\alpha=0$ or $\alpha=1$ is trivial, so suppose $0<\alpha<1$. Dividing both sides by $(x+y)^\alpha$ reduces the proof to the case $x+y=1$ (to see this, just do the substitution $x'=\frac{x}{x+y}, y'=\frac{y}{x+y}$). So suppose $x+y=1$. Then $0<x<1$, $0<y<1$, implying $x<x^\alpha$ and $y<y^\alpha$. Hence $$(x+y)^\alpha=1=x+y<x^\alpha+y^\alpha.$$


# using DNN to find out the pdf of a regression problem

When we use deep neural networks (DNNs) to solve a 1-dimention regression problem, we can approximate data distribution with the output of a DNN like the picture below.
My question is that DNN does not have the assumption of gaussian distribution or any other distribution of itself. It just knows what value to output when it sees an input. So how do you know the probability distribution of the DNN? For example, if someone asks, what is the probability of the point appearing in (5, 0). Can DNN answer this kind of questions?

For many regression algorithms, not only neural networks, the model is that the data is distributed by $y \sim \mathcal{N}(f(x;\theta), \sigma^2)$, where $\theta$ are the model parameters and $\sigma^2$ is the variance of the distribution (often a hyperparameter).
Maximizing the log-likelihood of the data with respect to $\theta$ is equivalent to minimizing the mean squared error loss between the $y_i$ and $f(x_i;\theta)$.
Therefore, to compute the probability density of $(5,0)$, you would just find the density of a gaussian with mean $f(5; \theta)$ and a variance of $\sigma^2$, where $f$ is your neural network.
• Yes, it still applies. The only assumption we need to apply MSE loss and obtain our PDF is that the distribution of the target $y$ is a function of the input $x$ plus some gaussian noise. That function $f$ doesn't have to be linear and in this case is modeled by a highly nonlinear neural network. – shimao Feb 2 '18 at 2:30


# Differential equation for changing mass of a sphere

1. Sep 13, 2013

### Hypatio

The mass of a sphere with density as a function of radius is

$M=4\pi \int_0^r\rho(r) r^2dr$

Lets say the radius increases and decreases as a function of time t. So:

$M(t)=4\pi \int_{0}^{r(t)}\rho (r) r(t)^2dr$

I want to know the basic equation describing the mass added or removed from the sphere (mass increases when radius increases, mass decreases when radius decreases) as a function of t, starting from any t. The problem is I think I must use a differential form but I'm not sure what it looks like. What then is the differential form of dM(t)/dt? I think I must use a chain rule and write:

$\frac{dM}{dt}=\frac{dM}{dr}\frac{dr}{dt}$

is this right? How do I proceed to solve this with the integral?

2. Sep 13, 2013

### JJacquelin

Hi !
I am afraid that there is something wrong in your writting. It should be :
$M=4\pi \int_0^r\rho(u) u^2du$
Lets say the radius increases and decreases as a function of time t. So:
$M(t)=4\pi \int_{0}^{r(t)}\rho (u) u^2du$
You may use any other symbol than u, but not r.

3. Sep 13, 2013

### Hypatio

Ah yes, sure, the upper limit is the 'full' radius (r) and u is a radius. This doesn't solve my problem though.

4. Sep 13, 2013

### HallsofIvy

Staff Emeritus
Yes, the chain rule: $\dfrac{dM}{dt}= \dfrac{dM}{dr}\dfrac{dr}{dt}$
JJaquelines point helps make sense of the dM/dr.

To find $\dfrac{dM}{dt}$ use the "fundamental theorem of Calculus":
$$\frac{d}{dr}\int_0^r \rho(u)u^2 du= \rho(r)r^2$$

5. Sep 13, 2013

### Hypatio

I guess the solution then is

$\frac{dM}{dt}=\frac{4}{3}\pi \rho(r)r^3\frac{dr}{dt}$

does the solution change if $\rho(r)$ becomes $\rho(r,t)$ or could I write

$\frac{dM}{dt}=\frac{4}{3}\pi \rho(r,t)r^3\frac{dr}{dt}$

Thanks.

Last edited: Sep 13, 2013
6. Sep 13, 2013

### JJacquelin

I don't think so. A term involving the partial derivatine of rho relatively to t is missing into your last equation,

7. Sep 13, 2013

### Hypatio

Is there a rule I can apply to get this additional term?

8. Sep 13, 2013

### pasmith

If you have
$$F(t) = \int_0^{a(t)} g(r,t)\,\mathrm{d}r$$
then
$$F'(t) = \int_0^{a(t)} \left.\frac{\partial g}{\partial t}\right|_{(r,t)}\,\mathrm{d}r + a'(t)g(a(t),t)$$
assuming $g$ is sufficiently smooth.

9. Sep 13, 2013

### arildno

Please do not mix together dummy variables in the integrand with integral limits.
Properly speaking, you have the following the relation:
$$M(t)=4\pi\int_{0}^{r(t)}\rho(x)x^{2}dx$$
Thus, you have:
$$\frac{dM}{dt}=4\pi\rho(r(t))r(t)^{2}\frac{dr}{dt}$$
which has as interpretation that only the outermost spherical shell at r(t) determines the total change of mass.

Every compact ball strictly contained within the outermost shell (radii less than r(t)) remains constant in mass.

However:
Suppose you have a ball where at different times, the density at some fixed radius "x" may change as a function of time. Then, you have:
$$M(t)=4\pi\int_{0}^{r(t)}\rho(x,t)x^{2}dx$$
In this case, the total mass of the ball will be due to two distinct effects:
1. The ball shrinks or expands. This gives the contribution given above.
2. The interior of the ball may change its mass. This effect is new.

In sum, you'll then get:
$$\frac{dM}{dt}=4\pi\rho(r(t),t)r(t)^{2}\frac{dr}{dt}+4\pi\int_{0}^{r(t)}\frac{\partial\rho}{\partial{t}}x^{2}dx$$

Last edited: Sep 13, 2013
10. Sep 14, 2013

### JJacquelin

The general formula below shows the rule for derivation :

#### Attached Files:

• ###### Derivation rule.JPG
File size:
6 KB
Views:
80
11. Sep 14, 2013

### JJacquelin

Of course, the chain rule continues to applies !!!
The formula given above is the application of the chain rule in case of an integral with the integrand and limits which are functions of the variable considered for derivation.


# Probability of no king, queen or jack before the first ace occurs?

A deck of cards is shuffled well. The cards are dealt one by one, until the first time an Ace appears. Find the probability that no kings, queens, or jacks appear before the first ace. (Introduction to Probability, p. 36)

My solution:

• Assume $k^{th}$ card is the first ace
• The possible number of hand before the first ace is then $\binom{48}{k-1}$, and the possible number of hands without a king, queen or jack before the first ace is $\binom{36}{k-1}$, so the probability is $\frac{\binom{36}{k-1}}{\binom{48}{k-1}}$.
• The probability that the first ace occurs at the $k^{th}$ position is $\frac{1}{52!} \binom{48}{k-1}(k-1)! \binom{4}{1}(52-k)!$, because there are $52!$ possible ordered decks, $\binom{48}{k-1}(k-1)!$ is the number of possibilities withouth an ace in the first $k$ cards, $\binom{4}{1}$ possibilities to draw an ace at the $k^{th}$ position and $(52-k)!$ possibilities to arrange the remaining cards in order.
• $\sum_{k=1}^{k=37} \frac{\binom{36}{k-1}}{\binom{48}{k-1}} \frac{1}{52!} \binom{48}{k-1}(k-1)! \binom{4}{1}(52-k)! = 0.25$

The computation of the above in R:

    pr <- numeric(37)
for(k in 1:37){
pr_ace_at_k <- 1/factorial(52)*choose(48,k-1)*factorial(k-1) * choose(4,1)*factorial(52-k)
pr_no_before_k <- choose(36,k-1)/choose(48,k-1)
pr[k] <- pr_no_before_k * pr_ace_at_k
}
sum(pr)
> 0.25

However, a simulation yields:

    deck <- c(rep(1:4, 4), rep(5, 52-4*4))
out <- replicate(1e6,{
deck <- sample(deck) # shuffle deck
k <- which(deck == 4)[1] # get index of first ace
all(deck[1:(k-1)] == 5) # check if only rest occured before kth card
})
mean(out)
> 0.173099

Is there anything wrong in my calculation? A result of 0.25 looks somehow persuasive.

EDIT

Finally, I found the nasty bug in the code. Problem was the special case when the ace occured at the first position. Corrected code would be

    deck <- c(rep(1:4, 4), rep(5, 52-4*4))
out <- replicate(1e6,{
deck <- sample(deck) # shuffle deck
k <- which(deck == 4)[1] # get index of first ace
if(k == 1) TRUE
else all(deck[1:(k-1)] == 5)
})
mean(out)
> 0.249877

• It's the probability that among $16$ particular cards, the first to show up is an ace. We have equally likely outcomes here... – David Mitra Dec 7 '14 at 15:05
• Call Ace, King, Queen, and Jack the picture cards. You are asking for the probability that the first picture card is an Ace. But all picture cards are equally likely. Therefore, by symmetry, the probability is...? – TonyK Dec 7 '14 at 15:09
• The issue is why the simulation is so grossly wrong. The usual answer is programming error. – André Nicolas Dec 7 '14 at 15:19
• The simulation result is clearly an anomaly. I don't see any error in the R code, but maybe someone on stackoverflow can. – David K Dec 7 '14 at 15:25
• The simulation probability is likely either 4/23 or 9/52 – Empy2 Dec 7 '14 at 16:22

The easiest way to think about it is to ignore all the other cards in the deck. Now you stack up the $16$ cards of interest. What is the chance the top one is an ace? There are $4$ aces among the $16$, so $\frac 4{16}=\frac 14$

• That makes perfect sense, good explanation and intuition! – NoBackingDown Dec 7 '14 at 15:23

Although it is not the simplest approach, we can use recursive conditioning, which is a useful approach in a lot of somewhat more complicated problems.

To see the intuition, let's check the cases in words. Suppose the first card is a queen/jack/king; then you stop and it doesn't count. Suppose it's an ace; then you stop and it counts. Suppose it's neither; then you run the experiment again with 1 fewer card.

Mathematically we'll say that the desired event when starting with a $k$ card deck is $A_k$. Then we have

$$P(A_k)=P(A_k|\text{ first card is an ace })P(\text{ first card is an ace }) \\ +P(A_k|\text{ first card is queen/jack/king })P(\text{ first card is queen/jack/king })\\ +P(A_k|\text{ first card is neither })P(\text{ first card is neither }) \\ = 1 \cdot \frac{4}{k} + 0 + \frac{k-16}{k} P(A_{k-1})$$

Now take $k=52$ and solve the recurrence relation. Here the base case can be read off from the recurrence itself: we get $P(A_{16})=4/16+0+0=1/4$ for free. Here solving the recurrence proves particularly simple, because if $P(A_{k-1})=1/4$ then $P(A_k)=1/4$, since

$$\frac{4}{k} + \frac{k-16}{k} \frac{1}{4} = \frac{4}{k} + \frac{1}{4} - \frac{4}{k} = \frac{1}{4}.$$

This is an algebraic manifestation of the intuitive observation that we can "ignore all the other cards".

Total probability x Probability of only 1st drawing = Probability of event occuring

Total probability x (1 - Probability of another drawing) = Probability of event occuring on 1st drawing

Total probability = Probability of event occuring on 1st drawing / (1 - Probability of another drawing) = $\frac{\frac1{13}}{1-\frac9{13}}=\frac14$

• Could you elaborate on your answer? I don't understand how you arrive at your equation. – NoBackingDown Dec 7 '14 at 15:25
• More rigorous than RossMillikan's answer? How so? – Did Dec 7 '14 at 15:35
• I don't think this quite works, because the replacement means that the process is not simply being "restarted" upon drawing another card. My answer does essentially what you're attempting to do, I think. That is, you have that $a=1/13 + 9/13 b$ where $b$ is the probability for the corresponding event with a deck containing 12 queen/jack/king, 4 aces, and 51-16=35 other cards. – Ian Dec 7 '14 at 15:39
• 1. Hope my answer is clear enough now. 2. RossMillikian's answer is sufficient, I just used a formula instead of intuition. – ghosts_in_the_code Dec 7 '14 at 16:25
• I don't think this works still: you have that the total probability is the probability of the event occurring given that the process stops on the first draw, times the probability that the process stops on the first draw, plus the probability of the event occurring given that the process stops on the second draw, times the probability that it stops on the second draw, etc. There is an independence assumption implicit in your formula which is correct but not justified. – Ian Dec 7 '14 at 17:30

One of these $4$ sorts of cards (aces, kings, queens, jacks) will appear as the first. Each sort (also the aces) has the same probability of doing so, hence with probability $\frac{1}{4}$.