Ch 4: The Continuous-Time Fourier Transform

Ch 4: The Continuous-Time Fourier Transform Fourier Transform of x(t) Inverse Fourier Transform jt X ( j) x ( t ) e dt jt x ( t ) X ( j) e d 2 Ghulam Muhammad, King Saud University

Continuous-time aperiodic signals Example 4. at Consider the signal x ( t ) e u ( t ), a Find its Fourier transform. jt X ( j) x ( t ) e dt ( aj ) t at jt ( a j) t e = e u ( t ) e dt e dt ( a j) t a j 2 2 a j a X ( j) 2 2 2 2 a j a 2 2 a 2 2 2 2 a a a X ( j) tan / tan 2 2 2 2 a a a Ghulam Muhammad, King Saud University 2

Example 4.2 x t e a at ( ), for x(t) Time-domain t frequency-domain The Fourier transform of the signal is: jt a t jt X ( j) x( t) e dt e e dt = ( a j) t ( a j) t e e = ( a j ) ( a j ) 2 2 a j a j = 2 2 a j a j a = a at jt at jt e e dt e e dt 2a 2/a X(jw) /a -a a w Ghulam Muhammad, King Saud University 3

Example 4.4 x() t, t T, t T x(t) -T T t T jt jt e jt X ( j). e dt e e j j T jt 2 sin( jt j T ) 2sin( T ) e e j j T T jt 2T X(jw) -/T /T Ghulam Muhammad, King Saud University 4

Example 4.5 X ( j), W, W Complete by yourself. W jt sin( Wt) x ( t ). e d 2 t W Draw the signals. Ghulam Muhammad, King Saud University 5

Problem 4. Use the Fourier transform analysis equation to calculate the Fourier transforms of (a) e u( t ) (b) e 2( t) 2 t (a) jt 2( t) jt X ( j) x( t) e dt e u( t ) e dt 2 j ( ) j 2 j = ( ) ( ) e u e d e e u e d (2 j ) j (2 j ) j e = e e d e j e = 2 j (2 j) = t- (b) jt 2 t jt X ( j) x( t) e dt e e dt = j 2 j j 2 j = e e. e d e e. e d = e j e 2 j ( ) j 2 j e e d e e e d j 4e = 4 2 (2 j ) (2 j ) e (2 j ) (2 j ) Ghulam Muhammad, King Saud University 6

Problem 4.2 Use the Fourier transform analysis equation to calculate the Fourier transform of (a) ( t ) ( t ) jt X ( j) x ( t ) e dt jt [ ( t ) ( t )] e dt jt jt ( t ) e dt ( t )] e dt = e e e e j ( ) j () j j j j e e =2 2cos 2 X ( j) 2 cos( ) Some useful Fourier transform: x ( t ) ( t ) X ( j) x ( t ) ( t ) X ( j) e x ( t ) ( t ) X ( j) e j j Ghulam Muhammad, King Saud University 7

2 d u t u t dt (b) ( 2 ) ( 2) 2 Problem 4.2 d jt X ( j) u ( 2 t ) u ( t 2) e dt dt jt jt ( 2 t ) e dt ( t 2) e dt I I jt I ( 2 t ) e dt, Let 2 t, d dt j ( 2) 2 j j 2 j ( ) e ( d ) e ( ) e d e I ( t 2) e jt dt e X ( j) e e 2j 2j 2j 2j e e 2 j 2 j 2j sin(2 ) X ( j) 2 sin(2 ) 2 j Ghulam Muhammad, King Saud University 8

Fourier Transform for Periodic Signals Example 4.6 X ( j) a 2 ( k ) k jk x () t a e k k k t The Fourier series coefficients of the above periodic signal: a k sin( k T ) k sin( k T ) sin( k T ) X ( j) 2 ( k ) 2 ( k ) k k k k Ghulam Muhammad, King Saud University 9

Example 4.6 - continued Using T = 4T k= k=2 Ghulam Muhammad, King Saud University

x ( t ) sin( t ) Example 4.7 x ( t ) sin( t ) e e 2j 2j Fourier series coefficients: j t j t a, a, ak for k 2j 2j Find Fourier transform: X(jw). X ( j) a 2 ( k ) k k 2 2 ( ) ( ) 2j 2j ( ) ( ) j j -/j X(j) /j - Do for x(t) = cos( t) Ghulam Muhammad, King Saud University

x ( t ) ( t kt ) k Example 4.8 Impulse train with a period of T. The Fourier series coefficients for this signal are: T /2 T T T T jk t jk t a x ( t ) e dt ( t ) e dt e k T Therefore, the Fourier transform is: 2 X ( j) 2 ak ( k ) ( k ) T k 2 2 ( k ) T T k k T /2 Ghulam Muhammad, King Saud University 2

Example 4.8 contd. X(j) = 2/T.. -6/T -4/T -2/T 2/T = 4/T = 2 6/T = 3 Ghulam Muhammad, King Saud University 3

Problem 4.3 Determine the Fourier transform of each of the following periodic signals. (a) sin 2t 4 (b) cos6t 8 Solution: (a) 2 x ( t ) sin 2 t ; comparing with x ( t ) sin t, and T 4 4 2 ; T j ( t /4) j ( t /4) j/4 j/4 e e e e x () t e e 2 j 2 j 2 j j t j t a a - all other k a k =, for Ghulam Muhammad, King Saud University 4

The Fourier transform of the periodic signal Problem 4.3-contd. X j a k e e j/4 j/4 ( ) k 2 ( ) ( 2 ) ( 2 ) k j j (b) e e x ( t ) cos6t e 8 2 j () t j /8 j 6 t j /8 j 6t. e e e e e 2 2 j (6 t /8) j (6 t /8) a a a - All other a k = X ( j) a 2 ( k ) k k j/8 j/8 2 ( ) e ( 6 ) e ( 6 ) 6 Ghulam Muhammad, King Saud University 5

Problem 4.4 Use the Fourier synthesis equations to determine the Fourier transform of (a) X ( ) 2 ( ) ( 4 ) ( 4 ) j 2, 2 (b) X 2( j) 2, 2, Solution: (a) jt x ( t ) 2 ( ) ( 4 ) ( 4 ) e d 2 jt jt jt ( ) e d ( 4 ) e d ( 4 ) e d 2 2 e e e 2 2 cos(4 t ) j 4t j 4t Ghulam Muhammad, King Saud University 6

Problem 4.4 contd. Solution: (b) 2 jt jt x 2( t ) 2e d 2e d 2 2 2 jt jt e e j 2t j 2t ( e ) ( e ) jt jt jt 2 jt jt 2 2 2 jt jt jt jt 2 jt jt e e e e e e jt 2 j jt jt jt 4 e e 4 sin 2 ( t ) Ghulam Muhammad, King Saud University 7

4.3 Properties of Cont. Time Fourier Trans. Linearity If x ( t ) X ( j) y ( t ) Y ( j) Then ax ( t ) by ( t ) ax ( j) by ( j) Time shifting x t t e X j jt ( ) ( ) If a signal is time-shifted, the magnitude of the Fourier transform does not change; only there is a phase-shift in the Fourier transform. Proof: jt x ( t ) X ( j) e d 2 j ( t t ) x ( t t ) X ( j) e d jt jt x ( t t ) e X ( j) e d jt ( x ( t t )) e X ( j) Ghulam Muhammad, King Saud University 8 2 2

.5 x(t) Example 4.9 By shifting 2.5.5 x(t) 2 3 4 t -.5 -.5.5.5 t x (t) +.5 x 2 (t) -.5 -.5.5.5 t -.5 -.5.5.5 t x(t) = ½ x (t - 2.5) + x 2 (t - 2.5) Ghulam Muhammad, King Saud University 9

Example 4.9 contd. From Example 4.4 and the signals of the previous slide, we get: 2sin( / 2) 2sin(3 / 2) X ( j) and X 2( j) By linearity and time shifting properties: X ( j) { x ( t 2.5)} { x 2( t 2.5)} 2 2.5 j 2.5 j e { x ( t )} e { x 2( t )} 2 2.5 j 2sin( / 2) 2.5 j 2sin(3 / 2) e e 2 2.5 j sin( / 2) 2sin(3 / 2) e Ghulam Muhammad, King Saud University 2

Conjugation and Conjugate Symmetry - If x ( t ) X ( j) then x *( t ) X *( j) If x(t) is real, x(t) = x*(t), hence X *( j) X ( j) Im X(j) = Re{X(j) + j Im{X(j)} [Re{X(j) + j Im{X(j)}]* = Re{X(-j)} + j Im{X(-j)} Re X*(j) = Re{X(j) - j Im{X(j)} Ghulam Muhammad, King Saud University 2

Conjugation and Conjugate Symmetry -2 For real x(t): Re{ X ( j)} Re{ X ( j)} Im{ X ( j)} Im{ X ( j)} Even function of Odd function of In polar form: X ( j) X ( j) X ( j) X ( j) Even function of Odd function of For real and even x(t): x ( t ) X ( j) Even{ x ( t )} Re{ X ( j)} Odd{ x ( t )} j Im{ X ( j)} Ghulam Muhammad, King Saud University 22

Example 4. Evaluate Fourier transform of x(t) = e -a t for a > Example 4.: at x ( t ) e u ( t ), a e at u () t a j at at a t at at e u ( t ) e u ( t ) at x ( t ) e e u ( t ) e u ( t ) 2 2Even{ e u ( t )} 2 t > t < e at u () t is real; from symmetric property, at at 2Even{ e u ( t )} 2 Re{ e u ( t )} a j X ( j) 2 Re 2 Re 2 2 a j a 2a 2 2 a x t e a at ( ), for x(t) t Ghulam Muhammad, King Saud University 23

Differentiation and Integration jt x ( t ) X ( j) e d 2 Differentiating both sides w.r.t. time dx ( t ) d jt d jt = X ( j) e d X ( j) e d dt 2 dt 2 dt jt jt X ( j) je d jx ( j) e d 2 2 dx () t dt t x ( t ) X ( j) jx ( j) Similarly, x ( ) d X ( j) X () ( ) j Ghulam Muhammad, King Saud University 24

Example 4. u(t) Determine the Fourier transform of the unit step function. x ( t ) u ( t ) X ( j)? t du dt Now, ( t ); For unit impulse, g ( t ) ( t ) G ( j) t G( j) x ( t ) ( ) d X ( j) G () ( ) j X ( j) ( ) j (t) G(j) t Also, we observe that du ( t ) j ( ) j ( ) dt j Ghulam Muhammad, King Saud University 25

Time and Frequency Scaling Proof: jt { x ( at )} x ( at ) e dt Let, = at, and we get two cases: If x ( t ) X ( j) j Then x ( at ) X ( ) a a Where a is real and nonzero i) When a > : at d adt dt d a At t, ; at t, j / a { ( )} ( ) / x at x e a d j ( / a) j x ( ) e d X a a a Ghulam Muhammad, King Saud University 26

Time and Frequency Scaling contd. ii) When a > : at d adt dt d a At t, ; at t, j / a { ( )} ( ) / x at x e a d j ( / a) j x ( ) e d X a a a Therefore, j X, for a a a j { x ( at )} X j a a X, for a a a In particular, x ( t ) X ( j) Reversing a signal in time reverses its Fourier transform also. Ghulam Muhammad, King Saud University 27

Duality Ghulam Muhammad, King Saud University 28

Example 4.3 What is the Fourier transform, G(j) of g(t) = 2 / ( + t 2 )? From example 4.2: at x ( t ) e, a X ( j) a 2a 2 2 For a =, t 2 x ( t ) e X ( j) 2 jt x ( t ) X ( j) e d 2 t 2 jt 2 2e e d g(t) Interchanging t and, and then substituting t by t, e e dt e dt e dt 2 jt 2 jt 2 jt 2 ( ) 2 2 2 t ( t ) t G ( j) 2 e G(j) Ghulam Muhammad, King Saud University 29

Convolution For an LTI system: y ( t ) h( t )* x ( t ) Y ( j) X ( j) H ( j) A convolution in time domain implies a multiplication in Fourier domain. Example 4.5 An impulse response of an LTI system: h( t ) ( t t ) The frequency response of the system: j t j t H ( j ) ( t t ) e dt e The Fourier transform of the output: This is consistent with the time shifting property. jt jt Y ( j) y ( t ) e dt x ( t t ) e dt jt j t jt e x e d e X j ( ) ( ) Y j H j X j e X j Ghulam Muhammad, King Saud University 3 jt ( ) ( ) ( ) ( ) y ( t ) x ( t t ) Slide 8

x(t) LTI System y(t) dx () t y() t dt From differential property, This implies that H ( j) j Examples 4.6, 4.7 Differentiator Y ( j) jx ( j) Frequency response of a differentiator. t y ( t ) x ( ) d Integrator The impulse response of the system is a unit step, u(t). jt h( t ) u ( t ) H ( j) u ( t ) e dt ( ) X ( j) j X ( j) ( ) X ( j) j = X ( j) ( ) X () j Ghulam Muhammad, King Saud University 3 From example 4. H( j) ( ) j Y ( j) H ( j) X ( j)

Example 4.9 bt at Find the response, y(t) of an LTI system, if x ( t ) e u ( t ), h( t ) e u ( t ); a, b at jt at jt ( a j ) t H ( j) e u ( t ) e dt e e dt e dt a j bt jt X ( j) e u ( t ) e dt b j Y ( j) H ( j) X ( j) a j b j A B a j b j a j b j A ( b j) B ( a j) Ab Ba j( A B ) Ab Ba ; A B A B b a Y ( j) b a a j b j Ghulam Muhammad, King Saud University 32

Table 4. We know, at bt e u ( t ) and e u ( t ) a j b j This implies, at bt If b = a, Y Example 4.9 contd. y ( t ) e e u ( t ), b a b a ( j) a j b j a j 2 du dv v u d u dx dx d ( a j)() ( )( j ) 2 2 dx v v d a j a j d j Y 2 d a j a j ( j) d tx ( t ) j X ( j) d ( at d t e u ( t )) j d a j at y ( t ) te u ( t ), a b Ghulam Muhammad, King Saud University 33

Some Important Cont.-Time Relationship Unit Impulse: (t) Input: x(t) Continuous Time LTI System Unit Impulse Response: h(t) Output: y(t) Time Domain Unit Impulse: Input: X(j) Continuous Time LTI System Frequency Response: H(j) Output: Y(j) Frequency Domain Ghulam Muhammad, King Saud University 34

TABLE 4. PROPERTIES OF FOURIER TRANSFORM Property Aperiodic Signal Fourier Transform x t y t X(jω) Y(jω) Linearity ax t + by(t) ax(jω) + by(jω) Time Shifting x t t e jωt X(jω) Frequency Shifting e jω t x t X(j ω ω ) Conjugation x t X ( jω) Time Reversal x t X( jω) Time and Frequency Scaling x at Convolution x t y(t) X(jω)Y(jω) a X jω a Multiplication Differentiation Integration න x t y(t) t dx t dt x t dt + 2π න X jθ Y j ω θ jωx(jω) X jω + πx δ(ω) jω Differentiation in Frequency tx t j d dω X(jω) Conjugate Symmetry for Real Signals x t real ቊ x e t = Ev x(t), x(t) real Re X(jω) x Ghulam Muhammad, King Saud University 35 o t = Od x(t), x(t) real j Im X(jω) dθ X(jω) = X ( jω) Re X(jω) = Re X( jω) Im X(jω) = Im X( jω) X(jω) = X( jω) X jω = X( jω) Real and Even Signals x t real and even X(jω) real and even Real and Odd Signals x t real and odd X(jω) purely imaginary and odd Even-Odd Decomposition of Real Signals

TABLE 4.2 BASIC FOURIER TRANSFORM PAIRS Selected Signal Fourier Transform x t = ቊ,, δ t u t t < T t > T 2 sin ωt ω + π δ(ω) jω δ t t e jωt e at u t, te at u t, Re a > Re a > a + jω a + jω 2 t n n! e at u t, Re a > a + jω n Ghulam Muhammad, King Saud University 36