17 Interpolation
Solutions to
Recommended Problems
S17.1
It is more convenient to solve this problem in the time domain than in the frequency
domain. Since x,(t) = x(t)p(t) and p(t) is an impulse train, x,(t) is a sampled ver­
sion of x(t), as shown in Figure S17.1-1.
x,(t)
t
xP\
4T
0
-4T
Figure S17.1-1
x(t)
=
x(t)
bt
-
nT)
n= -oo
=
Z
n=
x(nT) 5(t - nT)
-oO
Since y(t) = x,(t) * h(t) and x,(t) is impulsive, the convolution carried out in the
time domain is as shown in Figure S17.1-2.
y(t)
\t
-4T
0
5T
Figure S17.1-2
Here we have that x(t) is sampled by a rectangular pulse train as opposed to an
impulse train.
S17-1
Signals and Systems
S17-2
Since h(t) * (1/T)h(t), shown in Figure S17.1-3, is wider than the sampling
period T, the resultant w(t) is not a triangularly sampled version of x(t). w(t) con­
sists of the superposition of waveforms shown in Figure S17.1-4.
1- ­
t
\
0
T
2T
Figure S17.1-3
w(t)
--4T
i
6T
Figure S17.1-4
We note that this superposition is actually a linear interpolation between the sam­
ples of x(t). For example, Figure S17.1-5 convolved with Figure S17.1-6 equals
Figure S17.1-7.
x,(t)
2
1
0
Ji1
1
2
Figure S17.1-5
t
Interpolation / Solutions
S17-3
1
1
t
t
1
0
2
3
Figure S17.1-7
Figure S17.1-6
Now adding the shifted and scaled triangles yields Figure S17.1-8, which we see is
the linear interpolation between samples of x,(t).
0
3
Figure S17.1-8
Now since (1/T)h(t) * h(t) is Figure S17.1-9, we expect that w(t) is the linear inter­
polation of x,(t) shifted right by T, shown in Figure S17.1-10.
w (t)
/000
2T
Figure S17.1-9
-4T
1'0
Figure S17.1-10
6T
Signals and Systems
S17-4
S17.2
y(t) in all cases is the superposition of two signals.
(a)
y 1 (t)
Figure S17.2-1
(b)
Y
2
(t)
0
-
3
2
-
-
-
-_ --
Figure S17.2-2
(c)
y
3
(t)
sinin(t
T (t -1)
1)
sin
2
Figure S17.2-3
n (t - 2)
(t - 2)
Interpolation / Solutions
S17-5
S17.3
(a) |
P(W)
4
XP(wj)
A'-Z
2Ir
\
WCm
TT
T
Figure
~+
WO
S17.3-1
We have
22r
27 -
COM> Win,
T
CO,>
2
wm,
where w, is the sampling frequency. To ensure no aliasing we require that
T < 7r/om.
(b) To recover x(t) from x,(t), we must interpolate. We have previously shown that
by lowpass filtering, the spectrum is recovered, assuming that sampling has
been performed at a sufficiently high rate. The interpolation may be done in
many different ways, however, depending on the cutoff frequency we choose
for the lowpass filter. For example, any of the filters HI(o), H 2(w), and H3(w) in
Figures S17.3-3 to S17.3-5 may be used to interpolate x,(t), whose Fourier
transform is shown in Figure S17.3-2, to yield x(t).
Signals and Systems
S17-6
Xp(o)
2T
2T
T
T7-
Figure S17.3-2
HI(o)
T
2T
2T
Figure S17.3-3
H2(w)
T
7T
7
T
2T
ITT7
2T
T
Figure S17.3-4
H3(w)
T
2T
2T
Figure S17.3-5
T
Interpolation / Solutions
S17-7
(c) If
x,(t)
=
x(nT) b(t -
(
n=
nT)
-00
and H(w) is an ideal lowpass filter whose Fourier transform is shown in Figure
S17.3-6, then
TWC sine c
==ct
7r
h(t) = T sin
irt
and
x(t) = xp(t) * h(t) =
Two
--
irn=-o
x(nT) sine (t - sr nT)
H(w)
T
-Coc
'oc
Figure S17.3-6
S17.4
xXt
X
X t
g (t)
No
H(cj)
- y (t)
p(t)
Figure S17.4
We want to choose H(w) so that the cascade of the two filters is an ideal lowpass
filter. In this example,
G(x) = 2 sin
A/ e ~jwA2
so that
H(w) =
2 sin wA/2
H(w) = 0
e jA/2
otherwise
IcoI
< r/A,
Signals and Systems
S17-8
S17.5
X,(Go)
1
T
-41r X 104
0
rX 104
47X 104
X(E2)
1
T
0
-27r
27r
7T
Y(2)
1
T
-27r
2r
4 0
Yc()
0
X104
Figure S17.5
Solutions to
Optional Problems
S17.6
(a) We want ho[n] such that x,[n] * ho[n] = x 0[n]. We sketch ho[n] in Figure
S17.6-1.
ho [n]
0
N-I
Figure S17.6-1
Interpolation / Solutions
S17-9
(b) xO[n] = x,[n] * ho[n]
If there is no aliasing, we can recover x[n] from xO[n] by proper filtering. Since
Ho(Q) = 1
--
1
e
jQN .
e -"
jQ(N- 1)/2
-u
sin N/2
sin 0/2
we require
N/HO(Q),
|A|
H(Q) =
0,
N(ejU(N
101 -<Iir
-- 1)/2
sinW/2'
~N
0,
N< |
(c)' hl[n] = (1/N) (ho[n] * ho[-n]), so hi[n] is a triangular discrete-time pulse, as
shown in Figure S17.6-2.
h, [n]
ii I
..?TII
-N it..
N
0
Figure S17.6-2
(d) We require that
N
H(Q)
forI|I <­
~N'
=
0,
-7 <
for
|Q
I
<
ir
From part (c),
H ()
N
= N
[
)1
|HO(U)|2
2
=H(
(sin 9/2
N'
sin N/2)'2
2
o
N
< I0I < r
S17.7
(a) Taking Fourier transforms of both sides of the LCCDE yields
jY(wO) + Ye(W)
=
1,
Ye(o)
=
1
.,
1 + Jw'
Signals and Systems
S17-10
so yc(t) = e-'u(t). By examining the sampler followed by conversion to an
impulse train, we note that
y[n] = yc(nT) = e -" Tu[n]
(b) Since y[n] = e -nTu[n], we can take the Fourier transform to yield
Y(Q)
e -"T e -jfn
=
n=O
1
-
e
(T+jQ)
In order for w[n] = 6[n], we require W(R) = 1 for all Q. Thus,
Y(Q)H(Q) = W(Q) = 1,
H(Q) = 1 - e -(T+jQ)
Now since H(Q) = 2=_oo h[n]e -", we see by inspection that
h[O]
h[1]
= 1,
=-e
h[n]= 0,
n # 0, 1
S17.8
(a) Since X,(w) = X(w)P(w), we conclude that X,(w) is as shown in Figure S17.8-1.
X,(w)
K
HwH
s
Figure S17.8-1
X,(w)
=
X(w)
=
Z
6(co - no,)
X(nwo) 6(w - nwo)
(b) From the convolution theorem,
y(t) = x(t)*p(t)
and
2 rn
p(t) =
-
-
Ws
Interpolation / Solutions
S17-11
The fact that
6
{pxt)}=
-
nw)
71= -OD
can be easily verified. Therefore,
1
02,7rn\
y(t) = x(t) *- T 5
(s n =-OD
Kt- n,
(c) We see from the sketch in Figure S17.8-2 that for no time-domain aliasing,
T
27r
-W,
-
so
T,
W_s
­
T
y (t)
I
WS
27
0
T
'//
a
WS
Figure S17.8-2
(d) x(t) may be recovered from y(t), assuming that no time-domain aliasing
has occurred, by low-time filtering y(t) from t = - T to T and applying a gain
of oS.
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Resource: Signals and Systems
Professor Alan V. Oppenheim
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