Discrete -Time Signals And Systems Continuous-Time Signals & Systems

Discrete - Time Signals and Systems Continuous-Time signals & systems Fourier Transform Yogananda Isukapalli

Fourier Transform Motivation: We need to define the frequency spectrum for a more general class of continuous time signals. Fourier Transform is at the heart of modern communication systems We have defined the spectrum for a limited class of signals such as sinusoids and periodic signals. These kind of spectrum are part of Fourier series representation of signals

Fourier Transform : Definition Forward continuous - time Fourier Transform X ( jw ) ò x(t )e - jwt dt (1) - Inverse continuous - time Fourier Transform 1 x(t ) 2p ò X ( jw )e jwt dt - Time domain x(t ) Frequency domain F ¾ X ( jw ) (2)

Example Determine the Fourier Transform of the one - sided exponential signal ; x(t ) e -7t u (t ) Forward continuous - time Fourier Transform X ( jw ) ò x(t )e - jwt dt - ò e -7t u (t )e - jwt dt - ò e -7t e - jwt dt (notice the change in limits) 0

Example contd òe - ( 7 jw )t e dt - ( 7 jw ) 0 ( - 7 jw ) - 7 jw ) 0 e ( -e ( e ) 0 - ( 7 jw ) 1 (7 jw ) Time domain -7 t - ( 7 jw )t F ¾ ( - 7 jw ) !e ( 0 ) Frequency domain 1 (7 jw )

Fig.18.1 Fourier Series & Fourier Transform Interpretation of Fourier Transform from Fourier Series xT0 (t ) periodic signal, with period T0 xT0 (t ) - 5T0 2 - 2T0 - 3T0 2 - T0 - T0 2 0 T0 2 T0 Time ‘t’ Note that the shape of the signal is drawn arbitrarily

Fourier Series & Fourier Transform contd The periodic signal can be approximated by summing up harmonically related periodic exponential signals Fourier Analysis equation; ak 1 T0 T0 2 ò -T0 2 xT0 (t )e - jwk t dt , w k k w0 k 2p T0 And the Fourier synthesis equation, xT0 (t ) å k - ak e jwk t ak . - 3w0 - w0 0 w0 Spectrum 3w0 . Frequency,w kw0

Fourier Series & Fourier Transform contd Now, assume that the period of the signal x(t) is infinite x(t) Fig.18.3 - T0 2 0 T0 2 Time ‘t’ lim xT0 (t ) x(t ) T0 Note that the above equation also implies that all non - periodicsignals can be considered periodic with a period of ' '

Fourier Series & Fourier Transform contd Consider the Fourier Analysis equation 1 ak T0 T0 2 ò -T0 2 xT0 (t )e - jwk t dt Now define; X ( jwk ) X ( jkw0 ) T0 2 ò -T0 2 \ X ( jwk ) T0 ak , X ( jwk ) ak T0 xT0 (t )e - jwk t dt T0 ak

Fourier Series & Fourier Transform contd Then from the Fourier synthesis equation, xT0 (t ) å k - ak e jwk t æ X ( jwk ) ö jwk t åç e T0 ø k - è pay attention to the term, e jwk t e 1 \ xT0 (t ) T0 å k - X ( jwk )e jwk t æ 2p jk çç è T0 ö t ø ; lim e T0 æ 2p jk çç è T0 ö t ø 1

Fourier Series & Fourier Transform contd 1 xT0 (t ) T0 å k - X ( jwk )e jwk t Define; 2p (k 1) 2p k 2p Dwk wk 1 - wk T0 T0 T0 2p T0 Dwk substitute the result in the sysnthesis equation, Dwk xT0 (t ) 2p 1 X ( jwk ) å 2p k - å k - X ( jwk )e jwk t Dwk

Fourier Series & Fourier Transform contd lim T0 Þ Dwk 0 T0 æ 2p ö ç! Dwk T0 ø è 2p k wk k Dwk T0 lim T0 Þ wk w T0 The above equation implies that the discrete - variable ' wk ' can now be replaced with a continuous variable ' w ' Going back to the original equation connecting the periodic and non - periodic signals; lim xT0 (t ) x (t ) T0

Fourier Series & Fourier Transform contd 1 jwk t xT0 (t ) X ( j w ) e Dwk å k 2p k - lim xT0 (t ) x(t ) Þ lim xT0 (t ) x(t ) Dwk 0 T0 1 lim Dwk 0 2p å k - 1 x(t ) 2p X ( jwk )e jwk t Dwk ò X ( jw )e jwt d w - Because; ò x(u )du - lim Du 0 å X (k Du )Du k -

Fourier Series & Fourier Transform contd Fourier Transform integrals X ( jw ) ò x(t )e - jwt d w : Analysis (1) - 1 x(t ) 2p ò X ( jw )e jwt d w : Synthesis (2) - equations (1) and (2) exist if and only if , lim x(t ) 0; t that is ò x(t ) dt - Sufficient condition for the existance of X ( jw )

Example The periodicity is increased to show the effect of higher period. Notice that for fig.c, no repetition is visible in the plot. Also the spectrum plot for the three cases shows an interesting phenomenon.

Example contd Fourier Analysis equation; 1 ak T0 T0 2 ò -T0 2 T 2 ak T0 ò xT0 (t )e - jwk t dt e - jwk t dt -T 2 ( notice the change in limits for the present case ) - jwk t e ak T0 - jwk T 2 -T 2 é e - jwk T 2 - e jwk T ê - jwk ë sin(wk T 2) wk 2 2 ù ú û

Example contd The frequecncies, kw0 wk , get closer and closer as T0 and eventually become dense in the interval - w . The quatities 'ak T0 ' approach a continuous envelope function sin(wk T 2) X ( jw ) lim kw0 w wk 2 sin(w T 2) w 2 17 Fig.18.5

Fourier Transform Pair 1 Fig.18.6 x(t ) e - at u (t ) Forward continuous - time Fourier Transform - - X ( jw ) ò x(t )e - jwt dt ò e - at u (t )e - jwt dt ò e - at e - jwt dt (notice the change in limits) 0

Fourier Transform Pair 1 contd . òe - ( a jw )t dt 0 ( e - ( a jw ) - a jw ) - a jw ) 0 e ( -e ( 0 ( 1 (a u (t ) ) - ( a jw ) Time domain - at e - ( a jw )t - a jw ) !e ( 0 jw ) ) Frequency domain F ¾ 1 (a jw ) , Fourier transform is unique

Fourier Transform Pair 1 contd . X ( jw) 1 (a X ( jw) (a jw ) 1 2 w 2 ) ; -1 æ 1ö ÐX ( jw) - tan ç èw ø æ a - jw ö 1 a - jw X ( jw) ç 2 ( a jw ) è a - jw ø a w 2 ( Âe { X ( jw)} a 2 a w 2 ) ; Ám { X ( jw)} -w a2 w 2

Fourier Transform Pair 1 contd . Âe { X ( jw)} Ám { X ( jw)} a 2 a w 2 -w 2 a w 2

Fourier Transform Pair 2 x (t ) [ u (t T 2) - u (t - T 2) ] Forward continuous - time Fourier Transform X ( jw ) ò x (t ) e - jw t dt - ò [ u (t T - 2) - u (t - T 2) ] e - jwt dt Fig.18.8

Fourier Transform Pair 2 contd . T 2 ò e - ( jw t ) dt -T 2 ( e - ( jw t ) T 2 ( - jw ) -T - jw T 2 ) jw T 2 ) e ( - e( ) 2 - ( jw ) -2 j sin(w T 2) - jw sin(w T 2) w 2 , 'sinc' function; Time domain [u (t T 2) - u (t - T 2) ] sin(pq ) Frequency domain F ¾ sin(w T 2) w 2 pq

Fourier Transform Pair 2 contd . X ( jw ) Fig.18.9 sin(w T 2) w 2 Note : X ( j 0) lim w 0 sin(w T 2) w 2 , using L ' Hospitals rule; X ( j 0) lim w 0 T 2 cos(w T 2) 12 T

Fourier Transform Pair 2 -- Synthesis X ( jw ) 1 -wb wb w Fig.18.10 The frequency domain information is given above, Find the corresponding time - domain signal X ( jw ) [u (w wb ) - u (w - wb ) ] 1 x(t ) 2p ò - X ( jw )e jwt d w : Synthesis equation

Fourier Transform Pair 2 – Synthesis contd x (t ) 1 2p jw t u ( w w ) u ( w w ) e dw [ ] ò b b - wb 1 2p e ò e jw t d w ( notice the change in limits ) -wb ( jwt ) wb ( 2p jt ) -w ( b 2 j sin(wb t ) ( j 2p t ) e( jwb t ) - jw t -e ( b ) ( j 2p t ) sin(wb t ) (p t ) )

Fourier Transform Pair 2 – Synthesis contd Fig.18.11 Frequency domain [u (w wb ) - u (w - wb ) ] Time domain F ¾ sin(wb t ) (p t ) Note : finite - time signals are infinite in frequeny domain finite - frequency signals are infinite in time domain

Fourier Transform Pair 3 Fig.18.12 Fig.18.13 x (t ) Ad (t ) X ( jw ) ò x (t ) e - jw t dt - Ae - jwt F Ad (t ) ¾ ò - t 0 A A Ad (t )e - jw t dt

Fourier Transform Pair 3 -- Synthesis X ( jw ) 2pd (w ) x(t ) (2p ) 1 w Fig.18.14 1 x(t ) 2p 1 2p e jwt Fig.18.15 ò X ( jw )e jwt d w : Synthesis equation ò 2pd (w )e jwt d w - - w 0 1 t

Fourier Transform Pair 4 Let X ( jw ) 2pd (w - w0 ) 1 jwt x(t ) X ( j w ) e d w : Synthesis equation ò 2p - 1 jwt 2 pd ( w w ) e dw ò 0 2p - e jwt w w0 e jw0t Frequency domain Time domain 2pd (w - w0 ) F ¾ e jw0t 2pd (w w0 ) F ¾ e - jw0t

Fourier Transform Pair 5 Find the Fourier Transform for x(t ) sin(w0 t ) æ e jw0t - e - jw0t x(t ) sin(w0 t ) ç ç 2j è ö ø F x(t ) ¾ X ( jw ), use the identities from previous example e jw0t e - jw0t F ¾ 2pd (w - w0 ) F ¾ 2pd (w w0 ) Time domain ax1 (t ) bx2 (t ) sin(w0 t ) Frequency domain F ¾ F ¾ aX 1 ( jw ) bX 2 ( jw ) - jpd (w - w0 ) jpd (w w0 )

Fourier Transform Pair 5 contd .

Fourier Transform Pair 6 Find the Fourier Transform for x(t ) cos(w0t ) æ e jw0t e - jw0t x(t ) cos(w0 t ) ç ç 2 è ö ø F x(t ) ¾ X ( jw ), use the identities from previous example e jw0t e - jw0t F ¾ 2pd (w - w0 ) F ¾ 2pd (w w0 ) Time domain ax1 (t ) bx2 (t ) cos(w0 t ) Frequency domain F ¾ F ¾ aX 1 ( jw ) bX 2 ( jw ) pd (w - w0 ) pd (w w0 )

Periodic Signals x(t ) å ak e jwk t k - Where : x(t ) x(t T0 ) k 2p wk kw0 T0 1 ak T0 x(t ) F ¾ T0 2 - jkw0t x ( t ) e dt , ò -T0 2 X ( jw ) ì jkw0t ü F { x(t )} F í å ak e ý î k - þ

Periodic Signals contd e jw0t \e F ¾ jkw0t 2pd (w - w0 ) F ¾ 2pd (w - kw0 ) X ( jw ) å 2p ak d (w - kw0 ) k - Time domain å ak e k - jwk t Frequency domain F ¾ å 2p ak d (w - kw0 ) k -

Example: Square Wave Fig.18.17 1 ak T0 1 T0 T 2 ò x(t )e - jkw0t dt , -T 2 - jkw0t æe çç è - jkw0 T 2 ö ø -T 2 æ e - jkw0 T 2 - e jkw0 T ç ç - jkw0T0 è 2 ö ø

Example: Square wave contd Assume a 50% duty cycle, which means, 2p T T0 2 ; \ w0T p and w0T0 T0 2p T0 æ 2 sin( kw0 T 2) ö æ sin(p k 2) ö ç ç k w T k p è ø 0 0 è ø 1 a0 T0 T 2 ò dt -T 2 T 1 T0 2 ì sin(p k 2) k¹0 ï ï kp ak í ï1 k 0 ï î2

Example: Square wave contd Fig.18.18 jwk t a e å k F ¾ k - å 2p ak d (w - kw0 ) k - æ 2 sin(p k 2) ö X ( jw ) pd (w ) å ç d (w - kw0 ) k ø k - è

Example: Impulse Train Fig.18.19 p(t ) å d (t - nTs ) n - Ts 2 1 T2 1 - jkw0t - jkws t ak x ( t ) e dt d ( t ) e dt ò ò T0 -T 2 Ts -Ts 2

Example: Impulse Train contd 1 Ts Ts 2 ò d (t )e -Ts 2 - jkws t 1 dt Ts 2p P ( jw ) å 2p ak d (w - kws ) å d (w - kws ) k - k - Ts Fig.18.20

Table 1: Fourier Transform Pairs Time domain : x (t ) e - at e u ( -t ), bt ( a 0) u (t ), ( b 0) Frequency domain : X ( jw ) 1 (a jw ) 1 (b - jw ) u (t T 2) - u (t - T 2) sin(w T 2) w 2 sin(wb t ) (p t ) u (w wb ) - u (w - wb ) d (t ) 1 d (t - td ) e - jwtd

Table 2: Fourier Transform Pairs 1 Frequency domain : X ( jw ) 1 pd (w ) jw 2pd (w ) e jw0t 2pd (w - w0 ) A cos(w0 t f ) cos(w0 t ) sin(w0 t ) p Ae jf d (w - w0 ) p Ae - jf d (w w0 ) pd (w - w0 ) pd (w w0 ) - jpd (w - w0 ) jpd (w w0 ) Time domain : x (t ) u (t ) å ak e k - jkw0t å d (t - nT ) n - å 2p ak d (w - kw0 ) k - 2p d (w - kw ) å k - T

Reference James H. McClellan, Ronald W. Schafer and Mark A. Yoder, “ 11.1- 11.4 “Signal Processing First”, Prentice Hall, 2003

a more general class of continuous time signals. Fourier Transform is at the heart of modern communication systems We have defined the spectrum for a limited class of signals such as sinusoids and periodic signals. These kind of spectrum are part of Fourier series representation of signals ( ) ( )