Discrete Time Signal processing Solution Manual

474 Pages·1996·7.105 MB·English

by  Alan V. Oppenheim, Ronald W. Schafer

1
Solutions - Chapter 2
Discrete-Time Signals and Systems
3
=
2.1. (a) T(:z:[n]) g[nl*l
1*11
• Stable: Let $ M then IT[*] I :;; lg[n]IM. So, it is stable if lg[n]l is bounded.
• Causal: y,[n] = g[n]z [n] and 112[n] = g[n]:z:2[n], so if :z:,[n] = :z:2[n] for all n < no, then
= 1
y,[n] 112[n] for all n <no, and the system is causal.
• Linear:
T(a:z: [n] + b:z:2[n]) = g[n](= [n] + bz2[n]
1 1
= og[n]:z: [n] + bg[n]:z:2[n]
1
=
oT(:z: [n]) + bT(z2[n])
1
So tbis is linear.
• Not time-invariant:
=
T(:z:[n - no]) g[n]:z:[n - no]
cF y[n - no] = g[n - no]z[n - no]
which is not TI.
=
• Memoryless: y[n] T(z[n]) depends only on the n" value of z, so it is memoryless.
=I::._
(b) T(:z:[n]) :z:[k]
L::=,.,
• Not Stable: l:z:[n]l:;; M-+ IT(:z:[n])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:
•
L:
T(o:z: [n] + bz [n]) = =•lkJ + b:z:.[kJ
1 2
t=no
L• Ln
= a :z:,[n] + b z2[n]
A:=no .t=no
= aT(z,[n]) + bT(z2[n])
•
The system is linear.
• Not Tl:
•
L
T(z[n-no]) = ...... :z:[k-no]
•-no
= .L... z[k]
·-
...
= L
"' y[n-no] ....... :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
L
T(az1[n] + bz2[n]) = azt[k] + bz2[k]
t=n-no
n+no n+no
L L =
= a :tt[k] +b :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:
T(az,[n] + bz•[n]) = az,[n- no]+ bx.[n-no]
=
aT(:t,[n]) + bT(x.[n])
This is linear.
• 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:
T(az [n] + b:t2[n]) = eu•I•J+i>z•l•l
1
= eAZt(n)eb,[n]
# aT(:t [n]) + bT(z.[n])
1
This is not linear.
= =
• TI: T(:t[n -no]) e•l•-nol 11[n - no], so this is TI.
n••
• Memory]ess: 11[n] depends on the 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:
=
T(e:t1[n] + d:t2[n]) acz1[n] + t>d:t•[n] + b
# eT(:t [n]) + d7'(:t2(n])
1
This is not linear.
5
• 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:
T(az1(n] + b:r2(n]) = c:z:,[-n] + b:r2:z:[-n]
= aT(z (n]) + bT(z2[n])
1
This is linear.
• Not Tl:
=
T(z(n - no]) z( -n-no]
# y(n-no] = z[-n +no]
This is not Tl.
• Not memoryless: For n # 0, it depends on a value of z other than the n'h value, so it is not
memoryl ess.
(h) T(z(n]) = z[n] + u(n + 1]
• Stable: IT(z(n])l $ M + 3 for n;::: -1 and IT(:z:(n])l $ M for n < -I, so it is stable.
• Causal: Since it doesn't use future values of z(n], it is causal.
• Not linear:
=
T(az,(n] + b:r2(n]) a:r,[n] + b:r2(n] + 3u(n + 1]
# aT(:z:1(n]) + bT(r2(n])
This is not linear.
• Not Tl:
T(:r(n-no] = :r(n -no]+ 3u(n + 1]
=
y(n-no]
= r(n-no]+ 3u(n-no+ 1]
This is not TI.
• Memoryless: y(n] depends on then"' value of :r only, so this is memoryless.
2.2. For an LTI system, the output is obtained from the convolution of the input with the impulse response
of the system:
= L00
y(n] h(k]z(n - k]
b-oo
(a) Since h[k] # 0, for (No $ n $ N1),
LN,
y(n] = h[k]z(n - k]
t=No
The input, r(n] # 0, for (N2 $ n $ N3), so
r[n- k] # 0, for N2 $ (n- k) ::; N3
6
Note that the minimum value of (n - l:) is N2. Thus, the lower bound on n, which occurs for
k =No is
Using a similar argument,
=
N• N +N,.
1
Therefore, the output is nonzero for
(No +N2):,; n:,; (N, + N,).
(b) H :r[n] # 0, for some n n:,; (no+ N- 1), and h[n] # 0, for some n, :,; n:,; (n, + M -1), the
0 :,;
results of part (a) imply that the output is nonzero for:
(n + n,) :,; n :,; (n + n + M + N-2)
0 0 1
So the output sequence is M + N-1 samples long. This is an important quality of the convolution
for finite length sequences as we shall see in Chapter 8.
2.3. We desire the step response to a system whose impulse response is
=
h[n] 4-nu[-n], for 0 < 4 < 1.
The convolution sum:
L00
y[n] = h[k]:r[n - k]
.k=-~
The step response results when the input is the unit step:
= = { 1 forn2:0
:r[n] u[n] •
0, for n < 0
Substitution into the convolution sum yields
L00
y[n] = 4-•u[-k]u[n-k]
b-oo
For n :,; 0:
00
y[n] = I: 4-·
1=-CX)
I00:
= 4•
b-n
4-n
=
1-4
Forn>O:
0
y[n] = .I. : 4-·
•=-em
= r:4•
"="
1
=
1-4
7
2.4. The difference equation:
y(n]- 43 11!n- 1] + S1 y[n-2] = 2l:(n- 1]
To solve, we take the Fourier transform of both sides.
The system function is given by:
=
The impulse response (for :r[n] 6[n]) is the inverse Fourier transform of H(eiw).
H(eiw)- -8 + 8
- 1 + le-Jw 1- le-Jw
• 2
Thus,
= 1 1 y
h[n] -8( )"u[n] + 8( )"u[n]. :::: [>1)
4 2
2.5. (a) The homogeneous difference equation:
=
y[n]- 5y[n-1] + 6y[n- 2] 0
Taking the Z-transform,
1-sz-l +6z-• = 0
(1- 2z-1)(1-3z-1) = 0.
The homogeneous solution is of the form
ll•[n] = A (2)" + A,(3)".
1
(b) We take the z-transform of both sides:
Y(z)[1-sz-1 + 6z-2] = 2z-1 X(z)
Thus, the system function is
Y(z)
H(z) =
X(z)
=
1 Sz-1 + 6z-2
= -2 2
+::---;:--,.
1 - 2z 1 1 - 3z-l '
where the region of convergence is outside the outermost pole, ~ the system is causal. Hence
the ROC is jzj > 3. Taking the uiverse z-transform, the impulse response is
h(n) = -2(2)"u(n] + 2(3)"u(n].
8
=
(c) Let z[n] u[n] (unit step), then
1
X(z) = 1 z 1
and
Y(z) = X(z) · H(z)
=
(1-z 1)(1-2z ')(I-3z-1f
Partial fraction expansion yields
1 4 3
Y(z) = 1 ' -:--7=+ ,.
-z l-2z 1 l-3z
The inverse transform yields:
y[n] = u[n] - 4(2)"u[n] + 3(3)"u[n].
2.6. (ar The difference equation:
y[n] - 21 11[n- 1] = z[n] + 2z[n- 1] + z[n- 2]
Taking the Fourier transform of both sides,
=
Y(eiw)[l- !e-jw] X(.,.iw)[l + 2e-)w +.-;:a...].
2
Hence, the frequency response is
1 + 2e-;... +.-;:a...
=
1-!e-iw
(b) A system with frequency response:
H(eiw)
cross multiplying,
Y(eiw)[l + ~·-jw + ~e-i2wJ = X(eiw)[l - ~e-iw + e-i3w],
and the inverse transform gives
1 3 1
y[n] + v[n- 1] + 11[n-2] = z[n] - z[n- 1] + z[n - 3].
4 2
2
2.7. z[n] is periodic with period N if z[n] = :z:[n + NJ for some integer N.
(a) :z:[n] is periodic with period 12:
.;< fn) = .;< f )(n+N) =. ;< fn+. .k )
~ 2?rk =iN, for integers k,N
= =
Making k 1 and N 12 shows that z[n] has period 12.
9
(b) :r[n] is periodic with period 8:
= =
eJ'( 'i'nJ eJ'( 'i' )(n+N) eJ'( 'i'n+2d)
= 3
27Tk = : N, for integers k, N
=
N = ~k,for integers k,N
The smallest k for which both k and N are integers are is 3, resulting in the period N being 8.
=
(c) :r[n] [sin(..n/5)]/(7Tn) is not periodic because the denominator term is linear inn.
(d) We will show that :t[n] is not periodic. Suppose that :t[n] is periodic for some period N:
= =
,i!?,n) ,i!?,)(n+N) ,i!?,n+2••J
= =
2..-k ~N,for integers k,N
=
N = 2v'2k, for some.integers k, N
There is no integer k for which N is an integer. Hence :r[n] is not periodic.
2.8. We take the Fourier transform of both h[n] and :t[n], and then use the fact that convolution in the time
domain is the same as multiplication in the frequency domain.
5
H(e'w) =
1 +le-i""'
2
Y(e"w) = H(e"w)X(e"w)
= 5 1
1 +le2- i~>~ 1- fe jw
3 2
= 1 + ~e jw + 1 - ie J·w
1 1
y[n] = 2(3)nu[n] + 3(-2)nu[n]
2.9. (a) First the frequency response:
Y(eJ'w) _ ~.-iwy(eJ'w) + .!, -2iwy(e"w) = .!..-2jw X(e'w)
0
6 6 3
Y(eiw}
H(.,JW) =
X(eiw)
= le-2jw
1 - !•e -3i w + l•e 2jw
Now we take the ;,;...,rse Fourier transform to find the impulse response:
H(eJ'W) =
h(n]
10
For the step response s[n]:
L00
s[n] = h[k]u[n - A:]
•
L:
= h[kJ
1:=-oo
1-(1/3)•+• 1 - (1/2)n+'
= -2 1-1/3 u[n] + 2 1-1/2 u[n]
= (1 + (~)"-2(~)")u[n]
(b) The homogeneous solution ll>[n] solves the difference equation when :[n] = 0. It is in the form
· ll>[n] = 2: A(c )", where the c's solve the quadrMic equation
5 1
c"--c+-=0
6 6
So for c = 1/2 and c = 1/3, the general form for the homogeneous solution is:
=A,(~)"+ A,(~)"
ll>[n]
(c) The total solution is the sum of the homogeneous and particular solutions, with the particular
solution being the impulse response found in part (a):
y[n] = !l>[n] + y [n]
9
= A,(~)"+ A2(~ )" + -2(~ )"u[n] + 2(~ )"u[n]
Now we use the constraint y[O] = y[1] = 1 to solve for A and A,:
1
y[O] = A + A, - 2 + 2 = 1
1
y[1] = A,/2 + A,f3-2/3 + 1 = 1
A +A = 1
1 2
A,f2+ A,/3 = 2/3
With A, = 2 and A,= -1 solving the simultaneous equations, we find that the impulse response
is
lf[n] = 2(1- )"- (-1 )" + -2(1- )"u[n] + 2(1- )"u[n]
2 3 3 2
2.10. (a)
=
y[n] h[n]• z[n]
= L00
a•u[-k- 1]u[n-A:]
lo=-oo
{ t ..
n::; -1
_,
lo=-oo
=
L
....... a•, n> -1
a•
{
n$ -1
1-1/a'
=
1/a
n > -1
1-1/a'
11
(b) First, let us define u[n] = 2nu(-n- 1]. Tben, from part (a), we know that
= 2'*1 nn ;< -_1
w(n] = u[n] ov(n] , '
{ 1 1
Now,
y(n] = u(n - 4] • u(n]
= w{n-4]
{ 2n-3 n$3
= 0
1, n>3
(c) Given the same definitions for u(n] and w(n] from part(b), we use the fact that h(n] = 2n-1u(-(n-
1) - 1] = u(n-1] to reduce our work:
y[n] = r(n] • h(n]
= r[n]• u(n- 1]
= w{n -1]
= { 2n, n$0
1, n>O
(d) Again, we use u(n] and w[n] to help us.
y[n] = r[n] • h[n]
= (u[n]-u[n- 10]) • u[n]
= w[n]-w[n - 10]
= (2n+lu[-(n + 1)] + u[n])- (2n-9u[-(n- 9)] + u[n- 10])
2(n+l) - 2(n-1). n $ -2
= 1-2(n-l), -1 $ n $8
{
0, n~9
2.11. First we re-write :r[n] as a sum of complex e:rponentials:
Since comple¥ exponentials are eigenfunctions of LTI systems,
H(.J•I<).J"'i'- H(e-i•l')e-iwn/4
=
y[n] 2j
Evaluating the frequency response at w = ±7: /4:
=
H(eif)
We get:
2../2£-i•l•ei•n/4 _ 2,.T2.J•I<e-i•n/4
y(n] =
2j
=
2J2sin(7:n/4-•/4).