The Z-Trasform Sampled Data The geeralied fuctio (t) (also kow as the impulse fuctio) is useful i the defiitio ad aalysis of sampled-data sigals. Figure below shows a simplified graph of a impulse. (t-t ) For a impulse, it ca be show that Figure : Simplified graph of a impulse fuctio f(t) δ(t t) dt f(t) t This property is called the siftig property ad may be used to defie a sampled sigal f * (t) as show i Figure below. The sampled sigal is basically f(t) modulated by the pulse trai p(t) give by p(t) = δ(t T) () t () f(t) f * (t) t t p(t) t Figure : Ideal impulse samplig Therefore f * (t) = f(t) p(t) = f(t) δ(t T) () So that the sampled sigal is a amplitude modulated trai of pulses. If each pulse is replaced with the umber f(t), it is the called discrete-time sigal. Fially, if f(t) is defied oly over t, the summatio i Eq. () is take over [,].
The -Domai The Laplace ad -trasforms are closely related techiques. To demostrate this, we will start with the Laplace trasform ad show how it ca be chaged ito the -trasform. The sigle-sided Laplace trasform of the time domai f(t) is defied as: F(s) f(t) e st dt (4) where F(s) is the s-domai represetatio of the sigal f(t). The above equatio aalyes the time domai sigal i terms of sie ad cosie waves that have a expoetially chagig amplitudes. This ca be uderstood by usig the substitutio s = +j so that the Laplace trasform becomes: (5) t jt F(, ) {f(t) e } e dt Figure : A iterpretatio of F(,) for a uit step u(t) Now, take the Laplace trasform of the sampled-data sigal f * (t) give by Eq. () F * (s) * st f (t) e dt f(t) δ(t T) e Ts f(t) e f() + f(t) e -st + f(t) e -st + f(t) e -st + (6) This equatio is our startig poit to defie the -trasform of a sampled sigal. It also relates i a direct way the s-trasform ad the -trasform as will be soo illustrated. Defie the forward delay operator = e st ad rewrite Eq. (6) as follows: F() f(t) f() + f(t) - + f(t) - + f(t) - + (7) where F() is called the -trasform of the discrete sigal f(t) or Z{f(t)}. st dt
Example: Fid the -trasform for the uit step u(t)= for t ad is ero for egative time. SOLUTION: Substitute for u(t) i Eq. (7) to get U() u(t) + + + + = = give that > Example: Fid the -trasform for the uit ramp fuctio r(t)=t for t. SOLUTION: Substitute for r(t) i Eq. (7) to get R() ( T) = T ( + + + 4 + ) Also, ote that R() R() = T + + T + = T Hece, R() = ( ) T ( ) = Properties of the -trasform: () Liearity: Give f(t) ad g(t) with correspodig F() ad G(), the for arbitrary costats α, βiɍ or () Traslatio: For m > Z{α f(t)+ β g(t)}= αf()+ βg() (8) Z{f(T+mT)}= m F() m f m- f f m- (9) PROOF: ad Z{f(TmT)}= m F() () Z{f(T+mT)}= f(t mt) = m ( m) f(t mt) = m m ( m) f m f = m F() m f m f f m- m Proof for the secod case is similar except that f(t) is assumed ero for egative idices. () Complex Differetiatio: Z {T f(t)}= T df()/d () PROOF: Note that df()/d = f f f 4
Example: Fid the -trasform for the ramp fuctio r(t) = t for t usig the complex differetiatio property. SOLUTION: Usig complex differetiatio property R() = T du()/d = T d( )/d = T ( ) T ( ) = (4) Frequecy Scalig: Z{ f(t)}=f( ) () PROOF: Usig traslatio property Z{ f(t)}= f ( ) = F( ) (5) Iitial Value Theorem: f() = lim F() () (6) Fial Value Theorem: f() = lim ()F() (4) PROOF: Usig traslatio property Z {f + f }= lim N N (f - f ) = F() f() F() Now, let = ad observe that lim () F() f() = lim {(f f )+ (f f )+ +(f N f N- )} = f() f(). N Note that this fial value exists oly if (-)F() has its poles withi uit circle. Example: Calculate the iitial ad fial values for the followig complex fuctios ( ) (a) F() = ( )( ) SOLUTION: (b) F() = ( 5 6) (a) Calculate iitial value Ad the fial value (b) Iitial value f() = f() = f() = lim F() = lim ()F() = lim F() = lim O ( ) = lim ( ) ( ) lim O ( ) = While the fial value ( ) f() = lim ()F() = lim ( 5 6 ) = But this value is ot true sice has roots outside the uit circle. I fact f() =. = 4
The followig table summaries the Laplace ad -trasforms for a few of the stadard fuctios that may be useful. Table: Stadard Laplace ad -trasforms Time Fuctio f(t) t > Laplace Trasform F(s) -trasform F() Kroecker delta (t) Step fuctio u(t) Power fuctio a t s Ramp t s Parabolic fuctio t s k! t k k e at s s a te at lim( ) a T ( ) ( ) T ( ) k k [ k e e at at Te at ( s a) ( e ) at e at a ( e ) at s( s a) ( )( e ) t(e at )/a Damped sie e at si(t) T at a T ( e ) at s ( s a) ( ) a ( )( e ) ( s a) e e at at sit cost e ] at Damped Cosie e at cos(t) s a ( s a) at e cost at e cost e at 5
The Iverse -trasform There are several methods to obtai the iverse of the -trasforms. We shall discuss two simple methods; log divisio ad partial fractio expasio. Both are illustrated though the followig examples. Example: Fid the iverse -trasform for. F() = (.5)( ) SOLUTION: (a) Usig log divisio +.7 +.85 +.775 4 +.5.5 +..5.5 Or, simply F() = (b) Usig partial fractios.7+.5.7.5.5.85 +.5.85.45.45.775 +.45 f(t) = +.7 +.85 +.775 4 + F() ca be expaded as F() = F() =..8 so that (.5) ( )..8 (.5) ( ) ad from -trasform table, we get f + =.(.5) +.8 for ad f = lim. (.5)( ) = as cofirmed i (a). (c) Computatio usig Matlab The followig Matlab code may be used to cofirm the above results: >> delta=[ eros(,5)]; >> um=[.]; >> de=cov([.5],[ -]); >> f=filter(um,de,delta) f =.7.85.775.85.797.8.7984.88.7996.8.7999.8.8.8 Note that the sequece f is see as the impulse respose of a filter with F() as its trasfer fuctio. 6
Example: Fid the iverse -trasform for F() = SOLUTION: (a) Usig log divisio F() = ( 5 6) f(t) = +5 + 9 + 65 4 + Also, F()/ ca be expaded as F()/ = Hece F() = ( ) ( ) So that f((t) = () (), for. ( = 5 6) ( ) ( ) (b) Usig Matlab: >> delta=[ eros(,5)]; >> um=[ ]; >> de=[ -5 6]); >> f=filter(um,de,delta) f = 5 9 65 (c) Usig Symbolic Toolbox: >> syms f F >> F=/(^-5*+6); >> f=itras(f) f = -^+^ Example: Fid the value of the ifiite series f = k k (.5) SOLUTION: First, we may ote that f is simply F() evaluated at = (provided that F() has o poles outside the uit circle). Let us evaluate F() as the complex d differetiatio of i.e. F() = ( ) =.5 d.5 (.5) Hece, the sum f = F() = 8. Example: Usig -trasform table ad partial fractio expasio, fid the iverse - trasform for ( ) F() = ( )( ) SOLUTION: First, let us expad F() usig partial fractios i the form ( ) a b c F()/ = = = ( )( ) ( ) ( ) ( ) ( ) k 7
So that F() = ( ) (.5 ) Comparig the secod term with table etry for cos(t) (a=), we fid cos(t)= /, so that T = / ad f((t) = ( cos(/)) u(t) The followig matlab code was used to verify the values of f((t): >> delta=[ eros(,9)]; >> um=[ ]; >> de=[ - -]); >> f=filter(um,de,delta); >> stem(f),grid >> ylabel('\bf f(t)'),xlabel('\bf T') 5 4.5 4.5 f(t).5.5.5 4 6 8 4 6 8 T Figure 4: Plot of the periodic sequece f(t). Useful Hits: The followig formulas may be useful ad are used quite ofte N r N (i) r r < (Fiite geometric series) r (ii) r r < (Ifiite geometric series) r (iii) (a+b) = a k b k where k = ()(k+)/k! (Biomial expasio) k k (iv) (a+b) = a a b + a b a 4 b + Special case of (iii). (v) Z{ -k }= k k ( ) 8
Exercises : () Fid the -trasform for the followig fuctios: (a) f(t) = u(tt)k, t (b) f(t) = u(t)(.5) (c) f(t) = e T cos( T) u(t) ( Hit: use the fact that cos(t) is the real part of e jt ) (d) f(t) = u(t)(4) + () Fid the iitial value f() ad fial value f() for the followig sigle sided -trasforms: (a) F() = (b) F() = (c) F() = (d) F() = ( ) (5 ) ( ) () Fid the iverse -trasform for: (a) F() = e (b) F() = ( T 4 4 4)( ) (c) F() = (.5) (.5) (4) Fid the -trasform for the Fiboacci sequece give by the followig recursive equatio: x k+ = x k+ + x k give x = x =. i.e. the discrete sequece x k = {,,,, 5, 8,,, 4, }. Also fid the golde ratio defied by lim x k+ /x k k 9