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

# I lolla w hen t he i nput i s a u nit s tep f unction

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Unformatted text preview: F[zJ a nd H[zJ is 0.8 &lt; Izl &lt; 2. Therefore Y [z J = _Z2(Z + 0 .4) z--........,..0.,.::5:.,..(zo=._-&quot;':&quot;&quot;'=0.:':&quot;8)::.(-(z-_--::&quot;7 .) 2) 0.8 &lt; Izl &lt; 2 7&quot;( Expanding Y[zJ into modified partial fractions yields I nverse T ransform by Expansion o f F[z] in P ower Series o f z k F [z] = f [-I]z + f [-2]z2 + f [-3]z3 + . .. W e c an find t he i nverse z -transform o f F [z] b y d ividing t he n umerator p olynomial w ith t he d enominator p olynomial, b oth i n a scending p owers o f z , t o o btain a p olynomial i n a scending p owers o f z. T hus, t o find t he i nverse t ransform o f z /(z - 0.5) ( when t he r egion o f c onvergence is Izl &lt; 0.5), we d ivide z w ith - 0.5 + z t o o btain - 2z - 4z2 - 8z 3 - .... H ence, f [-I] = - 2, f [-2] = - 4, f [-3] = - 8 a nd so on. • B ecause t he b ilateral z -transform c an h andle n oncausal s ignals, we c an a nalyze n oncausallinear s ystems u sing t his t ransform. T he z ero-state r esponse y[k] is given by y[k] = Z -l { F[z]H[z]} p rovided t hat F [z]H[z] e xists. T he r egion of convergence o f F [z]H[z] is t he r egion w here b oth F [z] a nd H [z] e xist, a f act w hich m eans t hat t he r egion is c ommon t o t he c onvergence o f b oth F [z] a nd H [z]. f[kJ = (0.8)ku[kJ + (0.6)k u [-(k + I)J ' ---v--' I t [k] &quot;---....----' z Ft[zJ = z - 0.8 Observe t hat a common region of convergence for F I [zJ a nd F2 [zJ does not exist. Therefore F[zJ does not exist. I n such a case we t ake advantage of t he s uperposition principle and find yt[kJ a nd Y2[kJ, t he system responses t o /I[kJ a nd h[kJ, separately. T he desired response y[kJ is t he s um of yt[kJ a nd Y2[kJ. Now H[zJ = z _z 0.5 Izl &gt; 0.5 Yt[zJ = Ft[zJH[zJ = (z _ 0.5)(z - 0.8) + 2 (2)ku[-(k + I)J - z(z + 0.4) (z - 0.8)(z - 2) 0.8 0.5 &lt; Izl &lt; 0.6 Expanding Yt[zJ a nd Y2[ZJ i nto modified partial fractions yields YI[ZJ = &lt; Izi &lt; 2 _Z2(Z + 0 .4) Y[zJ = F[zJH[zJ = (z _ 0.5)(z _ 0.8)(z - 2) Izl &gt; 0.8 Y2[ZJ = F2[ZJH[zJ = (z _ 0.5)(z - 0.6) z':0.5 T he z-transform of this signal is found from Example 11.12 ( part c) as Therefore Izl &gt; 0 .8 Izl &lt; 0.6 find the zero-state response t o i nput F[zJ = f2 [k] T he z-transforms of t he causal and anticausal components /I[kJ a nd h[kJ o f t he o utput are E xample 1 1.13 For a causal system specified by t he transfer function f[kJ = (0.8)ku[kJ • E xample 1 1.14 For t he s ystem in Example 11.13 find t he zero-state response t o i nput Analysis o f LTID S ystems Using t he B ilateral Z-Transform H[zJ = &lt; Izl &lt; 2 y[kJ = [_(0.5)k + ~(0.8)kl u[kJ + ~(2)kU[-(k + I)J F or a n a nticausal s equence, which e xists o nly for k :S - 1, t his e quation b ecomes • 0.8 T he poles a t 0.5 and 0.8 a re enclosed within t he ring of convergence a nd therefore correspond t o t he causal p art, a nd t he pole a t 2 is outside t he ring of convergence and corresponds t o t he anticausal p art of Y[zJ. Therefore F [z] = L f[k]z-k 1 1.7-1 8( z - z 0.8 ) - :3 ( z...
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## This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.

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