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

# 7 big s haded b lock i n f ig 820b is g iven b y hz

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Unformatted text preview: response for t he e xponential i nput f[kJ = r k is given by C 2=-¥ Therefore r y",[k] (9.70) Q[EJy[kJ = P [EJf[kJ 2Cl = - 12} = - 6k - ¥ f (9.71) "'Ii where P[rJ H[rJ = Q[rJ The total response is + y",[k] y[k] =Yn[k] = Bl(2)k ¥ + B2(3)k - 6k - k ?O (9.67) To determine arbitrary constants B l and B2 we set k = 0 and 1 and substitute the initial conditions y[O] = 4, y [l] = 13 to obtain 4 ¥} = B l + B2 - 13 = 2 Bl + 3B2 - ¥ Bl (9.72) T he p roof follows from t he f act t hat if t he i nput f[kJ = r k, t hen from Table 9.2 ( Pair 4), y¢[kJ = c rk. T herefore = f [k + iJ = r k+i = r irk = y¢[k + IJ = c rk+j = c rjr k E i f[kJ E jy¢[kJ a nd a nd P [EJf[kJ = P[rJr k Q[EJy[kJ = cQ[rJr k so t hat E q. (9.70) reduces t o = 28 = B2 = -::3 Therefore (9.68) and y[k] = 28(2)k - ¥(3)k - 6k - ¥ '-....--' Yn[kJ o y~[kJ C omputer E xample C 9.6 Solve Example 9.10 using MATLAB. ) (c96.m) Y =[4 13]j Y =Y'j k=O:10j k =k'j f =[5 8 ];f=f'j f or m =1:length(k)-2 y =5*Y(m+1)-6*Y(m)+f(m+1)-5*f(m)j Y =[Yj yJ; F =3*(m+1)+5jf=[fjF]j e nd s tem(k,Y) o (9.69) '-v-' • which yields c = P[rJlQ[rJ = H[rJ. T his r esult is valid only i f r is n ot a c haracteristic root of t he s ystem. I f r is a characteristic r oot, t he forced response is c kr k where c is determined by s ubstituting y¢[kJ i n t he s ystem e quation a nd e quating coefficients of similar t erms o n t he two sides. Observe t hat t he e xponential r k i ncludes a wide variety o f s ignals such as a c onstant C , a sinusoid cos (13k + 0), a nd a n e xponentially growing o r d ecaying sinusoid hlk cos (13k + 0). 1. A Constant Input f[kJ = C T his is a special case of exponential (9.71), we have y¢[kJ = C C rk ~~~l w ith r 1. T herefore, from Eq. (9.73) = C H [lJ 2. A Sinusoidal Input T he i nput e j (3k is an exponential rk w ith r = e j (3. Hence a 6 02 9 Time-Domain Analysis of Discrete-Time Systems Similarly, for t he i nput e -if3k 9.6 S ystem S tability 603 Here H yq,[k] = H [e-if3]e-if3k Consequently, if the input ! [k] = cos 13k = ! (eif3k yq,[k] = + e -if3k ), For t he i nput cos (2k + ~ lurk], P[r] r + 0.32 [r] = Q[r] = r 2 _ r + 0.16 t he forced response is ! {H[ejf3]eif3k + H[e-if3]e-if3k} S ince t he two terms on t he r ight-hand side are conjugates i2 H [ i2] _ e + 0.32 e - (ei2)2 _ ei2 + 0.16 If ( -0.416 + jO.909) + 0.32 ( -0.654 - jO.757) - (-0.416 + jO.909) + 0.16 = 0.548ei3.294 T herefore IH[~2]1 yq,[k] = Re { IH[e if3 J/ ei (f3k+LH[e = IH [e if3 l1 cos (13k j a nd i'J)} + L H [e if3 J) (9.74a) y.p[k] = 0.548 cos (2k = 0.548 cos (2k U sing a similar argument, we can show t hat for t he i nput I [k] = cos (13k = 0.548 + ~ + 3.294)u[k] + 4.34)u[k] • + 8) Assessment o f t he Classical Method (9.74b) • E xample 9 .11 For a s ystem specified by t he e quation 9 .6 System Stability (E2 - 3E + 2)y[k] = ( E + 2 )f[k] find t he forced response for t he i nput f [k] = (3)ku[k]. In t his case H r _ P [r] _ r +2 [ ] - Q[r] - r2 - 3r + 2 a nd t he forced response t o i nput (3)k u [k] is H[3](3)k; t hat is, y.p[k] T he r emarks in C hapter 2 concerning t he classical m ethod for sol...
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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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