Exam (elaborations) TEST BANK FOR Discrete Time Signal processing 3rd Edition By Alan V. Oppenheim, Ronald W. Schafer (Solution Manual)-Converted

2.1. (a) T(:z:[n]) = g[nl*l • • Stable: Let 1*11 $ M then IT[*] I :;; lg[n]IM. So, it is stable if lg[n]l is bounded. • Causal: y,[n] = g[n]z1[n] and 112[n] = g[n]:z:2[n], so if :z:,[n] = :z:2[n] for all n no, then y,[n] = 112[n] for all n no, and the system is causal. • Linear: So tbis is linear. • Not time-invariant: which is not TI. T(a:z:1[n] + b:z:2[n]) = g[n](=1 [n] + bz2[n] = og[n]:z:1[n] + bg[n]:z:2[n] = oT(:z:1[n]) + bT(z2[n]) T(:z:[n - no]) = g[n]:z:[n - no] cF y[n - no] = g[n - no]z[n - no] • Memoryless: y[n] = T(z[n]) depends only on the n" value of z, so it is memoryless. (b) T(:z:[n]) =I::._ :z:[k] • Not Stable: l:z:[n]l:;; M-+ IT(:z:[n])l :;; L::=,., l:z:[k]l :;; In- noiM. A$. n-+ oo, T-+ oo, so not stable. • Not Causal: T(:z:[n]) depends on tbe future values of :z:[n] when n no, so tbis is not causal. • Linear: • T(o:z: 1 [n] + bz2 [n]) = L: =•lkJ + b:z:.[kJ t=no • n = a L :z:,[n] + b L z2[n] A:=no .t=no = aT(z,[n]) + bT(z2[n]) The system is linear. • Not Tl: • T(z[n- no]) = L :z:[k-no] ...... •-no = .L... z[k] ·-... "' y[n- no] = L :z:[l:] ....... The system is not TI. • Not Memoryless: Values of y[n] depend on past values for n no, so tbis is not memoryless. (c) T(:z:[n]) L:::~ ... :z:[k] • Stable: IT(z[n])l :;; I:::.~ ... lz[k]l $ L::.!..~ ... :z:[k]M $ l2no +liM for lz[n]l $ M, so it is stable. • Not Causal: T(:z:[n]) depends on future values of :z:[n], so it is not causal. 4 • Linear: n+no T(az1 [n] + bz2[n]) = L azt[k] + bz2[k] t=n-no n+no n+no = a L :tt[k] +b L :t2[k] = aT(z,[n]) + bT(:t2[n]) This is linear. • TI: a+no T(z[n- no] = L: z[k- nol .t=n-ne n = L: :t[k] t=n-no = 11!n- no] This is TI. • Not memoryless: The values of 11[n] depend on 2no other values of :t, not memoryless. (d) T(:t[n]) = z[n- no] • Stable: IT(z[n])l = [z[n- no]!~ M if [z[n] ~ M, so stable. • Causality: If no ~ 0, this is causal, otherwise it is not causal. • Linear: This is linear. T(az,[n] + bz•[n]) = az,[n- no]+ bx.[n- no] = aT(:t,[n]) + bT(x.[n]) • TI: T(:t[n- nd] = :t[n-no- nd] = 11[n- n•l· This is TI. • Not memoryless: Unless no = 0, this is not memoryless. (e) T(:t[n]) = e•l•l • Stable: jz[n]l ~ M, ]T(x[n])l = [e•l•lj ~ el•l•ll ~eM, this is stable. • Causal: It doesn't use future values of z[n], so it causal. • Not linear: This is not linear. T(az1[n] + b:t2[n]) = eu•I•J+iz•l•l = eAZt(n)eb,[n] # aT(:t1[n]) + bT(z.[n]) • TI: T(:t[n -no]) = e•l•-nol = 11[n - no], so this is TI. • Memory]ess: 11[n] depends on the n•• value of :t only, so it is memoryless. (f) T(:t[n]) = az[n] + b • Stable: IT(z[n])l = [az(n] + bl ~ t[MI + [b(, wbicb is stable for finite a aDd b. • Causal: This doesn't use future values of z[n], so it is causal. • Not linear: This is not linear. T(e:t1[n] + d:t 2[n]) = acz1[n] + td:t•[n] + b # eT(:t1 [n]) + d7'(:t2(n]) • Tl: T(:z:(n -noll = c:z:(n - no] + b = y(n -no]. It is Tl. • Memoryless: y(n] depends on the n'h value of :z:(n] only, so it is memoryless. (g) T(:z:(n]) = :z:(-n] • Stable: IT(:z:(n])l :l:z:(-n]l $ M, so it is stable. • Not causal: For n 0, it depends on the future value of :z:(n], so it is not causal. • Linear: This is linear. • Not Tl: This is not Tl. T(az1(n] +