Lecture 16: Fourier transform - MIT OpenCourseWare

1 : Signals and Systems Fourier transform November 3, 2011 1 Last Time: Fourier Series Representing periodic signals as sums of sinusoids. new representations for systems as filters. Today: generalize for aperiodic signals. 2 Fourier transform An aperiodic signal can be thought of as periodic with infinite period. Let x(t) represent an aperiodic signal. x(t)t S S 0 Periodic extension : xT (t) = x(t + kT ) k= xT(t)t S STThen x(t) = lim xT (t). T 32 sin S 0= 2 /T =k 0=k2 TTakk Fourier transform Represent xT (t) by its Fourier series. xT(t)t S STak = 1 T T/2 T/2 xT (t)e j 2 Tktdt = 1 T S S e j 2 Tktdt = sin 2 kS T k = 2 T sin S 4 2 sin S 0= 2 /T =k 0=k2 TTakk Fourier transform Doubling period doubles # of harmonics in given frequency interval. xT(t)t S STak = 1 T T/2 T/2 xT (t)e j 2 Tktdt = 1 T S S e j 2 Tktdt = sin 2 kS T k = 2 T sin S 5 2 sin S 0= 2 /T =k 0=k2 TTakk Fourier transform As T , discrete harmonic amplitudes a continuum E( ).

2 XT(t)t S STak = 1 T T/2 T/2 xT (t)e j 2 Tktdt = 1 T S S e j 2 Tktdt = sin 2 kS T k = 2 T sin S lim T T ak = lim T T /2 T /2 x(t)e j tdt = 2 sin S = E( ) 6 Fourier transform As T , synthesis sum integral. xT(t)t S ST2 sin S 0= 2 /T =k 0=k2 TTakk lim T T ak = lim T T /2 T /2 x(t)e j tdt = 2 sin S = E( ) 1 00 0 1 2 x(t) = E( ) e jTkt = 2 E( )e j t E( )e j td 2 T k= k= ak 7 Fourier transform Replacing E( ) by X(j ) yields the Fourier transform relations. E( ) = X(j ) Fourier transform j tdtX(j )= x(t)e ( analysis equation) 1 j td x(t)= X(j )e( synthesis equation) 2 Form is similar to that of Fourier series provides alternate view of signal. 8 Relation between Fourier and laplace Transforms If the laplace transform of a signal exists and if the ROC includes the j axis, then the Fourier transform is equal to the laplace transform evaluated on the j axis.

3 laplace transform : X(s) = x(t)e stdt Fourier transform : X(j ) = x(t)e j tdt = X(s)|s=j 9 Relation between Fourier and laplace Transforms Fourier transform inherits properties of laplace transform . Property x(t) X(s) X(j ) Linearity ax1(t) + bx2(t) aX1(s) + bX2(s) aX1(j ) + bX2(j ) st0 X(s) j t0 X(j )Time shift x(t t0) ee 1 s 1 j Time scale x(at) XX|a| a|a| adx(t)Differentiation sX(s) j X(j )dt d 1 d Multiply by t tx(t) X(s) X(j )ds j d Convolution x1(t) x2(t) X1(s) X2(s) X1(j ) X2(j ) 10 Relation between Fourier and laplace Transforms There are also important differences. tx(t)tCompare Fourier and laplace transforms of x(t) = e u(t). laplace transform t (s+1)tdt =1 X(s) = e u(t)e stdt = e ; Re(s) > 1 0 1 + s a complex-valued function of complex domain. Fourier transform t j tdt = (j +1)tdt =1 X(j ) = e u(t)ee 0 1 + j a complex-valued function of real domain.

4 11 laplace transform The laplace transform maps a function of time t to a complex-valued function of complex-valued domain s. x(t)t-101-101010 Real(s)Imaginary(s)Magnitude|X(s)|= 11 +s 12 Fourier transform The Fourier transform maps a function of time t to a complex-valued function of real-valued domain . x(t)t0 1 X(j) = 11 +j Frequency plots provide intuition that is difficult to otherwise obtain. 13 Check Yourself Find the Fourier transform of the following square pulse. 11x1(t)1t1. X1(j ) = 1 e e 2. X1(j ) = 1 sin 3. X1(j ) = 2 e e 4. X1(j ) = 2 sin 5. none of the above 14 Fourier transform Compare the laplace and Fourier transforms of a square pulse. 11x1(t)1tLaplace transform : X1(s) = 1 e ste stdt = s 1 1 = 1 s e s [function of s = + j ] e 1 s Fourier transform 11 j t e j tdt = 2 sin X1(j ) = [function of ] = e j 1 1 15 Check Yourself Find the Fourier transform of the following square pulse.

5 4 11x1(t)1t1. X1(j ) = 1 e e 2. X1(j ) = 1 sin 3. X1(j ) = 2 e e 4. X1(j ) = 2 sin 5. none of the above 16()() laplace transform laplace transform : complex-valued function of complex domain. 11x1(t)1t-505-5050102030|X(s)|= 1s(es e s) 17 Fourier transform The Fourier transform is a function of real domain: frequency . Time representation: 11x1(t)1tFrequency representation: 2 X1(j ) =2 sin 18 Check Yourself Signal x2(t) and its Fourier transform X2(j ) are shown below. 22x2(t)1tb 0X2(j ) Which is true? 1. b = 2 and 0 = /2 2. b = 2 and 0 = 2 3. b = 4 and 0 = /2 4. b = 4 and 0 = 2 5. none of the above 19 Check Yourself Find the Fourier transform . 22 j t e2 sin 2 4 sin 2 j tdt = ==X2(j ) = 2 e j 2 2 4 /2 20 Check Yourself Signal x2(t) and its Fourier transform X2(j ) are shown below. 22x2(t)1tb 0X2(j ) Which is true?

6 3 1. b = 2 and 0 = /2 2. b = 2 and 0 = 2 3. b = 4 and 0 = /2 4. b = 4 and 0 = 2 5. none of the above 21 Fourier Transforms Stretching time compresses frequency. 11x1(t)1t2 X1(j ) =2 sin 22x2(t)1t4 /2X2(j ) =4 sin 2 2 22 Check Yourself Stretching time compresses frequency. Find a general scaling rule. Let x2(t) = x1(at). If time is stretched in going from x1 to x2, is a > 1 or a < 1? 23 Check Yourself Stretching time compresses frequency. Find a general scaling rule. Let x2(t) = x1(at). If time is stretched in going from x1 to x2, is a > 1 or a < 1? x2(2) = x1(1) x2(t) = x1(at) Therefore a = 1/2, or more generally, a < 1. 24 Check Yourself Stretching time compresses frequency. Find a general scaling rule. Let x2(t) = x1(at). If time is stretched in going from x1 to x2, is a > 1 or a < 1? a < 1 25 Fourier Transforms Find a general scaling rule.

7 Let x2(t) = x1(at). j tdt = j tdtX2(j ) = x2(t)ex1(at)e Let = at (a > 0). j /a 11 j X2(j ) = x1( )e d = X1 aaa If a < 0 the sign of d would change along with the limits of integra tion. In general, 1 j x1(at) X1 . |a| a If time is stretched (a < 1) then frequency is compressed and ampli tude increases (preserving area). 26 ( )( ) Moments The value of X(j ) at = 0 is the integral of x(t) over time t. j tdt =X(j )| =0 = x(t)ex(t)ej0tdt = x(t) dt 11x1(t)1tarea= 22 X1(j ) =2 sin 27 Moments The value of x(0) is the integral of X(j ) divided by 2 . 1 1 j td =x(0) = X(j ) eX(j ) d 2 2 11x1(t)1t++ ++ 2 X1(j ) =2 sin area2 = 128 Moments The value of x(0) is the integral of X(j ) divided by 2 . 1 1 j td =x(0) = X(j ) eX(j ) d 2 2 11x1(t)1t++ ++ 2 X1(j ) =2 sin area2 = 12 equal areas !

8 29 Stretching to the Limit Stretching time compresses frequency and increases amplitude (preserving area). 11x1(t)1t2 X1(j ) =2 sin 221t4 1t2 New way to think about an impulse! 30 Fourier transform One of the most useful features of the Fourier transform (and Fourier series) is the simple inverse Fourier transform . j tdtX(j )= x(t)e ( Fourier transform ) 1 j td x(t)= X(j )e( inverse Fourier transform ) 2 31 Inverse Fourier transform sin 0t = t 0 Find the impulse reponse of an ideal low pass filter. 0 0H(j )1 h(t) = 1 2 H(j )ej td = 1 2 0 0 ej td = 1 2 ej t jt 0 0/ 0h(t)tThis result is not so easily obtained without inverse relation. 32 Fourier transform The Fourier transform and its inverse have very similar forms. j tdtX(j )= x(t)e ( Fourier transform ) 1 j td x(t)= X(j )e( inverse Fourier transform ) 2 Convert one to the other by t t scale by 2 33 Duality The Fourier transform and its inverse have very similar forms.

9 J tdtX(j ) = x(t)e 1 j td x(t) = X(j )e2 Two differences: minus sign: flips time axis (or equivalently, frequency axis) divide by 2 (or multiply in the other direction) x1(t) =f(t) X1(j ) =g( )x2(t) =g(t) X2(j ) = 2 f( ) tt ;flip; 2 34 Duality Using duality to find new transform pairs. x1(t) =f(t) X1(j ) =g( )x2(t) =g(t) X2(j ) = 2 f( ) tt ;flip; 2 f(t) = (t)t1g( ) = 1 g(t) = 1t2 f( ) = 2 ( ) 2 The function g(t) = 1 does not have a laplace transform ! 35 More Impulses Fourier transform of delayed impulse: (t T ) e j T . x(t) = (t T)tT1X(j ) = (t T)e j tdt=e j T X(j ) = 1 1 X(j ) = T T36 Eternal Sinusoids Using duality to find the Fourier transform of an eternal sinusoid. (t T)e jtTe j 0t e j T2 ( +T)2 ( + 0)T 0: tt ;flip; 2 k= 0 ake2 CTFS j kt x(t) = x(t + T ) = {ak}T 2 k 2 CTFT j kt x(t) = x(t + T ) = 2 ak 0 k= k= 0 akeT T 37 x(t) = k= xp(t kT)t0T akk X(j ) = k= 2 ak ( k2 T) 02 TRelation between Fourier transform and Fourier Series Each term in the Fourier series is replaced by an impulse.

10 38 Summary Fourier transform generalizes ideas from Fourier series to aperiodic signals. Fourier transform is strikingly similar to laplace transform similar properties (linearity, differentiation, ..) but has a simple inverse (great for computation!) Next time applications (demos) of Fourier transforms 39 MIT Signals and SystemsFall 2011 For information about citing these materials or our Terms of Use, visit.