Signal Processing and Linear Systems-B.P.Lathi copy

# Therefore to enhance the frequency response a t this

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 3 I f t he two systems are t o b e equivalent, y (kT) in Eq. (12.29c) must be equal t o y[k] i n Eq. (12.30). T herefore y (t) (a) h[k] = lim T ha(kT) T _O (12.31) This is t he t ime-domain criterion for equivalence of t he two systems. according t o t his criterion, h[k], t he u nit impulse response of H[z] in Fig. 12.8a, must be equal t o T t imes t he s amples of halt), t he u nit impulse response o f t he s ystem in Fig. 12.8b, assuming t hat T .... O. T his is known as t he i mpulse i nvariance c riterion of filter design. t I{. (s) (b) 1 2.4-2 F ig. 1 2.8 A nalog f ilter r ealization w ith a d igital f ilter. 12.8b. Therefore y (kT), t he samples of t he o utput in Fig. 12.8b, are identical t o y[k], t he o utput of H[z] in Fig. 12.8a. T he o utput y(t) of t he s ystem in Fig. 12.8b i st T he Frequency-Domain Equivalence Criterion I n Sec. 2.4-3 [Eq. (2.47)J, we proved t hat for a n analog system with transfer function Ha(s), t he s ystem response y(t) t o t he everlasting exponential i nput f (t) = e st is also a n e verlasting exponential (12.32) y(t) = i :f(r)ha(t-r)dr Similarly, in Eq. (9.57a), we showed t hat for a discrete-time system with transfer function H[z], t he s ystem response y[kJ t o a n everlasting exponential i nput f[k] = zk is also a n everlasting exponential H[zJz k : An integral is a sum in t he limit. Therefore, t he above equation can be expressed as y[k] = H[z]zk 00 y(t) = lim . 6.7_0 " f(ml::.r)ha(t - ml::.T)l::.r L -t (12.29a) m =-oo (12.33) I f t he s ystems in Figs. 12.8a a nd 12.8b are equivalent, t hen t he response of b oth s ystems t o a n e verlasting exponential i nput f (t) = e st s hould be t he same. A continuous-time signal f (t) = e st s ampled every T seconds results in a discretetime signal F or o ur purpose i t is convenient t o use t he n otation T for l::.r in t he above equation. S uch a change of notation yields f[kJ = e skT = zk w ith z = e sT 00 y(t) = lim T T -O " f (mT)h,,(t - mT) ~ (12.29b) m =-oo T his discrete-time exponential zk is applied a t t he i nput of H[zJ i n Fig. 12.8a, whose response is y[kJ = H[zJzklz=e.T T he response a t t he k th sampling i nstant is y (kT) o btained by setting t = k T in t he above equation: y (kT) = lim T T -O " ~ f (mT)h,,[(k - m)TJ (12.29c) = H [esT]eskT (12.34) Now, for t he s ystem in Fig. 12.8b, y (kT), t he k th sample o f t he o utput y(t) in Eq. (12.32), is m =-oo I n Fig. 12.8a, t he i nput t o H[zJ is f (kT) = f[kJ. I f h[kJ is t he u nit impulse response o f H[zJ, t hen y[k], t he o utput of H[z], is given b y (12.35) I f t he two systems are t o b e equivalent, a necessary condition is t hat y[kJ in Eq. (12.34) must be equal t o y (kT) in Eq. (12.35). This condition means t hat 00 y[kJ = L f[m]h[k - mJ (12.30) (12.36) m =-oo t For t he s ake o f generality, we a re a ssuming a noncausal system. T he a rgument a nd t he results a re a lso valid for causal systems. This is t he frequency-domain criterion for equivalence of t he two sys...
View Full Document

## This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.

Ask a homework question - tutors are online