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

# 56e 081yk 923a f ind t he t otal r esponse i f t he

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Unformatted text preview: this method, we specify the system and the input by a, b, and f vectors. The response is p lotted for 20 points. This program can be used to find the discrete-time system response to any input. Determine t he unit impulse response h[kJ for a system in Example 9.4 specified by the equation y[k [ (-0.2)k +4(0.8)k] u[kJ • . Example 9 .5 y[kJ - 0.6y[k - IJ - 0.16y[k - 2J = Therefore, according to Eq. ( 9.34) + 6y[k) = + 4y[k) = S /[k + 1) - 19J[k) 2 /[k + 2) - 2/[k + 1) Answers: (9.38) To determine C i a nd C2, we need to find two values of h[kJ iteratively. This step is already taken in Example 9.4, where we determined t hat h[OJ = 5 and h[IJ = 3. Now, setting k = 0 and 1 in Eq. ( 9.38) and using the fact t hat h[OJ = 5 and h[IJ = 3 , we o btain 5 3 = Cl + C2 } = - 0.2Cl + 0.8C2 Cl =1 =- C2 = 4 :j:We showed that h[k) consists of characteristic modes only for k > O. Hence, the characteristic mode terms in h[k) s tart at k = 1. To reflect this behavior, they should be expressed in the form "'Iju[k - 1). But because u[k - 1) = u[k) - ark), A nju[k - 1) = A nju[k) - Aj 8[kJ, and h[k) can be expressed in terms of exponentials "'Iju[k) (which start at k = 0), plus an impulse at k = O. 9 .4 System Response t o External Input: The Zero-State Response T he z ero-state r esponse y[k] is t he s ystem r esponse t o a n i nput f [k] w hen t he s ystem is in zero s tate. I n t his s ection we s hall a ssume t hat s ystems a re i n zero s tate u nless m entioned o therwise, so t hat t he z ero-state r esponse will b e t he t otal r esponse o f t he s ystem. H ere we follow t he p rocedure p arallel t o t hat u sed i n t he c ontinuous-time c ase by expressing a n a rbitrary i nput f [k] a s a s um o f i mpulse 586 9 T ime-Domain Analysis of Discrete-Time S ystems 9.4 S ystem r esponse t o E xternal I nput: T he Z ero-State Response 587 I f we k new t he s ystem r esponse t o i mpulse 6[k]' t he s ystem r esponse t o a ny a rbitrary i nput could b e o btained b y s umming t he s ystem response t o v arious impulse components. L et h[k] b e t he s ystem r esponse t o i mpulse i nput 6[k]. We shall use t he n otation f [kl J[k] (a) -2 - I 4 k- t o i ndicate t he i nput a nd t he c orresponding response of t he s ystem. T hus, if 6[k] f[-2]o[k+2] = h[k] t hen b ecause o f t ime i nvariance 6[k - m] (b) -2 = y[k] = h[k - m] k- a nd b ecause of linearity f[-I]o[k+l] J[m]6[k - m] a nd a gain because o f l inearity (c) -I k- rO'' '1 00 L J[m]h[k - m] v y[k] T he l eft-hand side is J[k] [see Eq. (9.40)], a nd t he r ight-hand s ide is t he s ystem r esponse y[k] t o i nput J[k]. T herefore 00 L J[mJh[k - m] (9.41) m =-oo (e) T he s ummation o n t he r ight-hand side is known as t he c onvolution s um o f J[k] a nd h[k], a nd is represented symbolically by J[k] * h[k] 00 J[k] * h[k] = L J[m]h[k - m] (9.42) m =-oo ( t) k- J[k] in terms of impulse components. components. F igure 9.2 shows how a signal J[k] i n Fig. 9.2a c an b e e xpressed as a s um o f i mpulse components such as those depicted in Figs....
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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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