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.

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 5 2 sin S 0= 2 /T =k 0=k2 TTakk Fourier transform As T , discrete harmonic amplitudes a continuum E( ). 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.

3 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. 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.

4 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. 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.

5 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. 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.

6 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.

7 22x2(t)1tb 0X2(j ) Which is true? 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.

8 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. 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.

9 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 !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.

10 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. 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.