University of Maryland at College ParkDepartment of Electrical and Computer EngineeringENEE 624 Advanced Digital Signal ProcessingProblem Set 2 SolutionsSpring 2003Monday, February 17, 2003Problem 2.1All three systems are linear, as each system is a cascade of linear systems.System A: Let v[n], w[n], and y[n] denote the outputs of the upsample-by-L system, the LTI systemwith transfer function H(z), and the overall system output, respectively, when the inputis a fixed but arbitrary signal x[n]. Then, when the system input is x[n − k] (for somefixed but arbitrary integer k), the output of the upsample-by-L system is v[n − kL], theoutput of the LTI system with transfer function H(z) is w[n − kL], and the output of thedownsample-by-L system is y[n − k]. Hence system A is (linear and) time-invariant.x[n]-↑ L-v[n]H(z)-w[n]↓ L-y[n]x[n−k]-↑ L-v[n−kL]H(z)-w[n−kL]↓ L-y[n−k]Figure 2.1-1 System AWe haveW (z) = H(z) V (z) = H(z) X(zL)andY (z) =1LL−1Xk=0W³z1/LWkL´=1LL−1Xk=0H³z1/LWkL´Xµ³z1/LWkL´L¶=1LL−1Xk=0H³z1/LWkL´X³z WkLL´= X (z)"1LL−1Xk=0H³z1/LWkL´#so the transfer function of System A isHA(z) =1LL−1Xk=0H³z1/LWkL´= H(z)¯¯¯↓L.Problem Set 2 Solutions 2Consequently the impulse response of System A is given by hA[n] = h[Ln], and its fre-quency response viaHA¡ejω¢=1LL−1Xk=0H³ejω−2kπL´.System B: Clearly, if L = 1 or H(z) = 0, System B is LTI. So, assume L > 1 and H(z) 6= 0. Letv[n], w[n], and y[n] denote the outputs of the downsample-by-L system, the LTI systemwith transfer function H(z), and the overall system output, respectively, when the inputis a fixed but arbitrary signal x[n]. We have (see Figure 2.1-2)W (z) = H(z) V (z) = H(z)"1LL−1Xk=0X³z1/LWkL´#Finally,Y (z) = W¡zL¢= H¡zL¢"1LL−1Xk=0X³z WkL´#=1LH(zL) X(z)+H¡zL¢"1LL−1Xk=1X³z WkL´#Hence, System B cannot LTI for L > 2 (unless H(z) = 0 for all z) since the responseof the system at any z = z0depends on the input X(z) at values of z different from z0(e.g., z = z0WL). Since System B is linear, it cannot be time-invariant (unless H(z) = 0,or L = 1). However, System B is periodically time-varying. Specifically, given as inputx[n − k L] (for some fixed but arbitrary integer k), the output of the downsample-by-Lsystem is v[n − k], the corresponding output of the system with transfer function H(z) isw[n − k], and the corresponding overall system output is y[n − kL].x[n]-↓ L-v[n]H(z)-w[n]↑ L-y[n]x[n−kL]-↓ L-v[n−k]H(z)-w[n−k]↑ L-y[n−kL]Figure 2.1-2 System BSystem C: Let v[n], w[n], s[n], t[n] and y[n] denote the outputs of the upsample-by-2L system, theLTI system with transfer function H(z), the downsample-by-L system, the LTI systemwith transfer function G(z), and the overall system output, respectively, when the input isa fixed but arbitrary signal x[n]. Then, if the input is x[n−k] (for some fixed but arbitraryinteger k): the output of the upsample-by-2L system is v[n − 2kL]; the output of the LTIsystem with transfer function H(z) is w[n − 2kL]; the output of the downsample-by-Lsystem s[n−2k]; the output of the LTI system with transfer function G(z) is t[n−2k], andthe overall system output is y[n − k]. Hence the System C is (linear and) time invariant.Problem Set 2 Solutions 3x[n]-↑ 2L-v[n]H(z)-w[n]↓ L-s[n]G(z)-t[n]↓ 2-y[n]x[n−k]-↑ 2L-v[n−2kL]H(z)-w[n−2kL]↓ L-s[n−2k]G(z)-t[n−2k]↓ 2-y[n−k]Figure 2.1-3 System CWe haveV (z) = X¡z2L¢W (z) = V (z) H(z) = X¡z2L¢H(z)andS(z) =1LL−1Xk=0W³z1/LWkL´=1LL−1Xk=0H³z1/LWkL´Xµ³z1/LWkL´2L¶=1LL−1Xk=0H³z1/LWkL´X³z2W2kLL´= X¡z2¢"1LL−1Xk=0H³z1/LWkL´#.In addition,T (z) = S(z) G(z) = X¡z2¢"1LL−1Xk=0G(z) H³z1/LWkL´#,andY (z) =12T³z1/2´+12T³−z1/2´= X(z)"12LG³z1/2´L−1Xk=0H³z1/(2L)WkL´+12LG³−z1/2´L−1Xk=0H³W2Lz1/(2L)WkL´#.Thus, the system function of System C isHC(z) =12L"G³z1/2´L−1Xk=0H³z1/(2L)WkL´+ G³−z1/2´L−1Xk=0H³W2Lz1/(2L)WkL´#and its frequency response isHC¡ejω¢=12L"G³ejω /2´L−1Xk=0H³ejω−4kπ2L´+ G³−ejω /2´L−1Xk=0H³ejω−2π(2k+1)2L´#Problem Set 2 Solutions 4Problem 2.2We havehk[n] = h0[n] cos(2πkn/L)Z←→ Hk(z) =12H0³z Wk´+12H0³z W−k´where W = e−j2π /L. ThenY0(z) = X¡zL¢Hk(z) = X¡zL¢·12H0³z Wk´+12H0³z W−k´¸.x[n]-↑ L-v[n]H0(z)-s[n]h×6-y1[n]cos(2πnk/L)Figure 2.2-1 Right-hand side systemFor the second system we have the following relationshipsS(z) = X¡zL¢H0(z)Y1(z) =12S³z Wk´+12S³z W−k´=12Xµ³z Wk´L¶H0³z Wk´+12Xµ³z W−k´L¶H0³z W−k´.Since WkL= W−kL= 1, we getY1(z) = X¡zL¢·12H0³z Wk´+12H0³z W−k´¸= Y0(z) .Next we turn to the example, with L = 3 and k = 1.π−πV(e )jω10−π/3 π/3π−πS(e )jω1ω0−π/3 π/3 π−π1/20−π/3 π/3 2π/3−2π/3Y (e ) =jω1Y (e )jω0π−πX(e )jω1ω0Figure 2.2-2Problem Set 2 Solutions 5Problem 2.3We haveY0(z) = X(z)·12H0³z Wk´+12H0³z W−k´¸Y1(z) =12X³z Wk´H0³z Wk´+12X³z W−k´H0³z W−k´.These are, in general, different. In fact, the system with response y0[n] is LTI, while the onewith response y1[n] is not. For instance, let k = 1, L = 2. Let the input to both systems bex[n] = ejωonwith ωo= 0. We havey0[n] = C ejωon= C ,whereC =12H0¡ejωoW¢+12H0¡ejωoW−1¢=12H0¡ejπ¢+12H0¡e−jπ¢= H0(−1) ,whiley1[n] = C1ej(ωo−π) n+ C2ej(ωo+π) n= (C1+ C2) (−1)n.Clearly, y0[n] is not equal to y1[n].Problem Set 2 Solutions 6Problem 2.4x[n]-↓ M-y[n] = x[nM]Figure 2.4-1 M-fold decimation systemLet y[n] denote the response of the M -fold decimator system to an arbitrary N -periodic signalx[n]. Then, the response to x[n − M N] is y[n − N]. But since x[n] is N-periodic, we havex[n] = x[n − MN ]. Hence y[n − N ] = y[n], i.e., y[n] is also periodic with period N. It onlyremains to determine as much as we can about the fundamental period of y[n]. We distinguishbetween the following two cases:Case 1. M & N are relatively prime (i.e., gcd(M, N) = 1):We already know y[n] is N-periodic, so we just need to check whether it is also periodicwith some other period N0< N. We will assume that y[n] is N0-periodic with N0< Nand arrive to a contradiction. Since y[n] is also N-periodic, N must be an integer mul-tiple of N0, that is N = N0p for some integer p ≥ 2. For any integer no, let Ay(no) ,{y[no], y[no+ 1], · · · , y[no+ N − 1]}, and Ax(no) , {x[no], x[no+ 1], · · · , x[no+ N − 1]}.Since M and N are relatively prime, the sets Ay(0) and Ax(0) are identical