19 Discrete-Time Sampling
Solutions to
Recommended Problems
S19.1
x[n] is given by
x[n]
(-1)" = e'"
=
Hence, the Fourier transform of x[n] is
b(O -
57
X(Q) =
r -
2-xk)
k= -c
Now p[n] can be written as
p[n]
=
1 + (-1)"
2
Hence, its Fourier transform is given by
P
=
0
-
21rk) +
6(9 -
r
-
2x7rk)
2k= -oo
It is clear that x,[n] = p[n]. Hence
X,(Q) = P(Q)
S19.2
(a) x,[n] is x[n] "stretched" by interspersing with zeros, as indicated in Figure
S19.2-1.
x,[n ]
x [n]
n
0
1.
0
0
Figure S19.2-1
X,()=
(
n
xjnle
-1""
=-00
x,[2nle -ji2n +
=
x,[2n
+ l]e -j"l**+
nt= -o
=
(
x[nje -j(21)n
+ 0
= = -X
=X(29)
S19-1
Signals and Systems
S19-2
(b)
xd[n]
= x[2n],
x[n]e -jfln
Xd(Q) =
=
x[2n]e-jn
(
L
=
=
__
(x[n] + (-1)"x[n])e -jln/2
x[n]e
X
(2n=
-j(fl/2)n
__
00 _xne
: x[n]e
+
j[(12/2)-In
+
(
X
(c) X8,(Q)
+
(
-
,r)
=X(2Q)
XS(92)
-7r
8
IT
-9
S
Figr
IT
Figure S19.2-2
Xd(Q)
=
+
X
x
X
- ,r)
'X(Q/2) is indicated in Figure S19.2-3. Therefore,
S19.2-4.
Xd(Q)
is as shown in Figure
1 X(2)
2
2
1
2
7rI
IT
7T
Figure S19.2-3
Xd(E2 )
-27r
-I
-
0
T
Figure S19.2-4
27r
Discrete-Time Sampling / Solutions
S19-3
S19.3
(a) For N = 1, p[n] = 1. Hence
P(Q) = 2(r
b(Q -
2rk),
k= -- o
as shown in Figure S19.3-1.
P(Q)
214
2-n
0
-27
Figure S19.3-1
For N = 2,
p[n] =
6[n -
2k]
k=­
Hence
P(Q)
( O - 7rk),
=
k= -o
shown in Figure S19.3-2.
iT
0
-IT
Figure S19.3-2
ForN =L
p[n]
=
(
b[n - LkI
k = -w
Hence
-2rk
P(Q) =
shown in Figure S19.3-3.
k=
Signals and Systems
S19-4
(b) X,(Q), the spectrum of x,[n], is proportional to the periodic convolution of P(Q)
and X(Q). Consequently, with P(Q) as indicated in Figure S19.3-3 and X(Q) as
indicated in Figure S19.3-4, X,(Q) is shown in Figure S19.3-5. In order that x[n]
be reconstructible from x,[n] using an ideal lowpass filter, aliasing must be
avoided, which requires that
9M <N
N
UM <
or
-M,
N
X(92)
-21 2T
M
-M
Figure S19.3-4
P(92)
FiM u
M
2.3-5
N
Figure S19.3-5
(i)
Qm = 37r/10. Therefore, to avoid aliasing,
> 37
or
N10
<
10
Since N must be an integer, we require that N s 3. For N = 3, the cutoff
frequency of the lowpass filter must be greater than 3r/10 and less than
3
(ii)
10
3)10
5
Q, = 37r/ . To avoid aliasing,
> 3w
N
5
or
N<
3
Since N must be a positive integer, this requires that N = 1, i.e., x[n]
cannot be sampled.
Discrete-Time Sampling / Solutions
S19-5
S19.4
(a) The sampling period Ti is 3 ms for the system in Figure P19.4-2 to be equivalent
to the one in Figure P19.4-1.
(b) X(Q) is sketched in Figure S19.4-1.
IX(92)I
-2T2 2T
Figure S19.4-1
From the result of part (a), Y(Q) is as shown in Figure S19.4-2.
|Y(G)|
l-
|
-27T
e
'
I
­
31T 37T
2
2
2n
Figure S19.4-2
S19.5
(a) Consider xd,[n] and Xdfn], and let
Xd3[n] = Xd,[n] + aXd in]
Then
xP3 [n]
=
XdXnl/N]
± aXdn/N],
n = 0, +N,
otherwise
0,
But
x,,[n] =
otherwise
0,
And
ax,2[n]
=I
d[/
,
otherwise
Signals and Systems
S19-6
Hence,
x,[n] + axP2[n] =
Xd[f/N] + aXd2 [n/N],
n = 0, ±N,...,
0,
otherwise
and
xP3 [n] = xp,[n] + ax 2 [n]
So system A is linear.
(b) Take xd,[n] as shown in Figure S19.5-1, with N
-1
0
1
=
4.
2
Figure S19.5-1
Then x,,[n] is as shown in Figure S19.5-2.
0
4
8
Figure S19.5-2
Take Xd2[n]
=
Xd,[n
+ 1]. Then x,,[n] is as shown in Figure S19.5-3.
-4
0
4
Figure S19.5-3
Hence, system A is not time-invariant.
(c) x,[n] =
xd[n/N],
n = 0,
0,
otherwise
N,... ,
Hence
X,()=
as shown in Figure S19.5-4.
xd[n]e -jl(Nn) = Xd(NQ),
Discrete-Time Sampling / Solutions
S19-7
X (E2)
7T IT
3
3
Figure S19.5-4
(d) X(Q) is as shown in Figure S19.5-5 for exact bandlimited interpolation.
X(Q)
3
F-
Figure S19.5-5
S19.6
(a) x,(t) is sketched in Figure S19.6-1, and Y,(w) is sketched in Figure S19.6-2.
--2To
- .....x~(t)
-TO To
2 To
CO
2coo
Figure S19.6-1
Y,(w)
-H­
-2wo
-- 0
Figure S19.6-2
Signals and Systems
S19-8
(b) X,(w) is sketched in Figure S19.6-3, and y,(t) is sketched in Figure S19.6-4.
X,(o)
1
To
2
17
7T
TO
2
7T
2TO
7r
To
Figure S19.6-3
(c) Yes, y,(t) is periodic and this is reflected in Y,(w), which contains impulses.
Solutions to
Optional Problems
S19.7
(a)
n =0,
x,[n] =x[n,
0,
2, +4,.
n =±1,
. .,
35,...
This is sketched in Figure S19.7-1.
Ip
[nn
0
Figure S19.7-1
Similarly, xd[n] = x[2n], as shown in Figure S19.7-2.
Discrete-Time Sampling / Solutions
S19-9
0
Figure S19.7-2
(b) X,(Q) is obtained as follows:
x,[n] =
-
XCQ)
x[n] + 2(- 1)"x[n]
iX(Q)
+ jX(Q -
ir)
and
1
Xa(Q)
X-
=
x-
1 ),
which are shown in Figures S19.7-3 and S19.7-4. See Problem P19.2(b).
Xp(E2)
4
4
4
4
Figure S19.7-3
Xd(n)= Xp ()
2
2
Figure S19.7-4
S19.8
(a) We know that the Fourier transform of p[n] is given by
27rk)
k= -0
Signals and Systems
S19-10
Aliasing will be just avoided when the sampled spectra will look as shown in
Figure S19.8-1.
Hence, we require that
2>r
67r
N> 11'
or
N<
2= N - 3
6
Consequently, aliasing is avoided if 1 S N
shown in Figure S19.8-2.
3. X,(Q) for N = 1, 2, and 3 are
N ='1A
121
XP(2)
N1
=1227
N=2
1
2
1
-27r
--2
_42
31
37r_
I7
F
I
-
27r
i
_2\ 3n
1
3l
2n
3In
2r
3
2
N=3
N
-21T
3
47r
3
-27r
3
-
3
7r
11
0
11
Figure S19.8-2
4_
3
21T
Discrete-Time Sampling / Solutions
S19-11
(b) An appropriate H(Q) is shown in Figure S19.8-3.
H(2)
N
3nr
37r
Figure S19.8-3
S19.9
x[3n] + x[3n + 1] + x[3n + 2]
3
y]
(a)
y[-4]=
0
y[-3] = 0
y[-2]
=x[-4]
y[-1]
=
[0]
=
=1
x[-3] + x[-2] + x[-1]
2
3
x[0] + x[1] + x[2]
_
3
x[3] + x[4] + x[5]
1
y[1] =
y[2] = 0
y[3] = 0
Hence, y[n] can be sketched as in Figure S19.9.
4
3
-4
-3
-2
-1
0
1
2
3
Figure S19.9
(b) If
z[n] = [x[nj + x[n + 11 + x[n + 2]],
for all n
Signals and Systems
S19-12
and
y[n] = z[3n],
we have expressed the processing as a combination of filtering and decimati
S19.10
If h[O] = 1 and h[n] = 0 for n = kN, k # 0, it is easy to see that the samples
xo[n] that came from x[n] will be unaffected. Hence,
y[kN] = x[k],
for all k
MIT OpenCourseWare
http://ocw.mit.edu
Resource: Signals and Systems
Professor Alan V. Oppenheim
The following may not correspond to a particular course on MIT OpenCourseWare, but has been
provided by the author as an individual learning resource.
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.