TEST BANK FOR Discrete Time Signal processing 3rd Edition By Alan V. Oppenheim, Ronald W. Schafer

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+i>z•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] + t>d: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] +