MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Electrical Engineering and Computer Science
6.003: Signals and Systems — Spring 2007
Tutorial 9 Solutions
April 16, 2007
Problem 9.1 (O&W Problem 8.36)
(a) Let us assume that we are working with real signals, so each of the transforms depicted in this problem
is symmetric about the ω = 0 axis. We are given Y (jω) and H1 (jω):
Y (jω)
2ωM-
C
C
C
C
C
C
2ωM-
1
CC
−ωc
C
C
C
C
C
C
CC
ω
ωc
H1 (jω)
2ωM-
C
2ωT
2ωM-
K
C
C
C
C
C
CC
−ωc
C
2ωT
C
C
C
C
C
CC
ωc
Let’s the call output of the coarse tunable filter s(t) so that S(jω) = Y (jω)H1 (jω):
1
ω
S(jω)
2ωM-
C
C
CC
C
C
C
2ωM-
K
C
C
C
C
2ωT C -C
CC
C
C
C
CC
C
C
C
C
C
C
C
2ωT C -C
CC
C
−ωc
ωc
ω
As we’ve seen in previous problems, multiplication by cos((ωc + ωf )t) shifts S(jω) by ±(ωc + ωf ) and
scales it down by a factor of 2 to obtain Z(jω) (only ω > 0 shown here):
Z(jω)
2ωM-
K
2
C
C
CC
C
C
C
2ωMC
C
C
C
C
CC
C
C
C
C
C
C
C
2ωT C -C
CC
C
C
2ωT C -C
CC
C
ωf
2ωc + ωf
ω
(b) We know that X1 (jω) is:
X1 (jω)
2
C
−ωM
C
C
C
C
C
CC
ωM
ω
The height of the triangle is 2 because after modulating x1 (t) by cos ωf t, the spectrum scaled down by
a factor of 2 to 1. So, R(jω) is:
2
R(jω)
2ωM-
C
C
C
C
C
C
1
CC
2ωM-
−ωf
C
C
C
C
C
C
CC
ωf
ω
ωT must be such that the “non-Y1 ” portion of Y (jω) does not enter the non-zero region of the spectrum
above. Looking at Z(jω), we see that the “tail” cannot interfere with the spectrum we want, which
requires that:
−ωf + ωT
=⇒ ωT
<
<
ωf − ωM
2ωf − ωM
(c) We need the cutoffs α and β to be the same as the boundaries of the information we want and a
compensating gain to bring the height of the triangle to 1, as required by the diagram of R(jω). So,
α
= ωf − ωM ,
= ωf + ωM ,
2
G =
.
K
β
Problem 9.2 (O&W Problem 8.37)
First, we rewrite z(t) using only the first four terms of the power series,
1
1
z(t) = 1 + y(t) + y(t)2 + y(t)3
2
6
1
1
= 1 + (x(t) + cos ωc t) + (x(t) + cos ωc t)2 + (x(t) + cos ωc t)3
2
6
5 5
1
1
1
9
2
3
= + x(t) + x(t) + x(t) +
+ x(t) + x(t)2 cos ωc t
4 4
2
6
8
2
1
1
cos 3ωc t
+ (1 + x(t)) cos 2ωc t +
4
24
(a) Using the expression for z(t) above we can sketch the spectrum of z(t). Note that the bandwidth at
baseband is 6ω1 , at ωc is 4ω1 , and at 2ωc is 2ω1 . At 3ωc there is only a tone.
3
Z
(
j
ω
)
ω
3
ω
1
ω
c
2
ω
1
ω
c
ω
c
+
2
ω
1
2
ω
c
3
ω
c
0
2
ω
c
ω
1
2
ω
c
+
ω
1
(b) In order to get an amplitude modulated version of x(t) we must filter out everything except for the
components centered around 2ωc . This means that
ωc + 2ω1 <α < 2ωc − ω1
2ωc + ω1 <β < 3ωc .
Problem 9.3 (O&W Problem 8.48)
(a) Since p[n] is periodic, we can use the DT FourierPSeries analysis equation to find the general form for ak
2πk
and then use the relationship that P (ejω ) = 2π +∞
k=−∞ ak δ(ω − N ):
ak
=
=
1
N
X
2π
p[n]e−jk N n
n=<N >
M
1 X −jk 2π n
N
e
N n=0
2π
ak
=
=
=
=
1 1 − e−jk N (M+1)
2π
N
1 − e−jk N
1 e
N
M +1
−jk 2π
N ( 2 )
(e
since:
al =
l=0
M +1
jk 2π
N ( 2 )
2π 1
2π 1
e−jk N ( 2 ) (ejk N ( 2 )
"
B
X
−e
−
M +1
−jk 2π
N ( 2 )
1 − aB+1
1−a
)
2π 1
e−jk N ( 2 ) )
sin( kπ
N (M +
2j sin( kπ
N )
1 −jk 2π ( M ) 2j
N
2
e
N
#
"
(M + 1))
sin( kπ
π
−j kM
N
N
e
N sin( kπ
N )
1))
#
Note: You can also use Tables 5.1 and 5.2 on pp391 - 392 to find this result. Therefore:
P (ejω )
= 2π
+∞
X
ak δ(ω −
k=−∞
ak
=
(
e−j
kM π
N
h
2πk
)
N
sin( kπ
N (M+1))
N sin( kπ
N )
M+1
N
For N = 4, M = 1, P (ejω ) will look like:
4
where
i
, k 6= N i for i ∈ Z
, k = N i for i ∈ Z
P (ejω )
6
π
6
j π4
−j π4
.71πe
.71πe
6
6
π
π
.71πej 4
.71πe−j 4
6
6
-
−π
−3π
2
(b) Let ωM =
π
2N .
0
−π
2
P (e ) =
2π
=
(
+∞
X
ak
=
π
N
<
2π
N
kπ
e−j N
2πk
)
where
N
h
i
sin( 2πk
N )
, k 6= N i for i ∈ Z
kπ
N sin(
)
ak δ(ω −
k=−∞
π
2N
ω
3π
2
Since M = 1, then:
jω
Since 2 ·
π
π
2
N
2
N
, k = N i for i ∈ Z
(no aliasing) and Y (ejω ) =
Y (ejω ) =
+∞
X
1
jω
2π X(e )
ak X(ej(ω−
∗ P (ejω ), then:
2πk
N )
)
k=−∞
For N = 4, Y (ejω ) will look like:
Y (ejω )
6
2
1
π
.35ej 4
π
.35e−j 4
-
−π
−π
2
π
8
−π
8
π
2
π
(c) For this problem, we are given:
X(ejω ) =
nonzero , |ω| < ωM
0
, ωM < |ω| ≤ π
where X(ejω ) repeats every 2π. In order to make sure that there is no aliasing:
2ωM < ωs =
Therefore, ωM <
π
N
2π
N
where N ∈ Z+ . Notice that this calculation does not depend on M.
5
ω
(d) To get back the original signal, we need to low-pass filter y[n]:
N
M+1
y[n]
-
-
−π −π
N
x[n]
π π
N
Problem 9.4
Here we divide each system into sub components as shown in system 1 and determine the output at each
sub system (see next page).
6
• System 1
1
a(t)
b(t)
.
-
J
J
6
c(t)
-
-
ω
?
A s(t)
A
6
A
A
A
−ωm
- J
J
6
p(t) = cos(ωm t)
q(t)
ωm
1
-
-
ω d(t)
−ωm
ωm
p(t) = cos(ωm t)
From the diagram above, s(t) = a(t) + b(t) cos(ωm t)
S(jw) is shown below
Clearly, filtering this signal with H(jω) would not
give us something proportional to a(t), thus c(t)
is NOT proportional to a(t).
S(jω)
& %
−2ωm
2ωm
ω
Modulating s(t) by a cosine gives us q(t). Q(jω) is shown below
Q(jω)
−3ωm
Again, filtering this signal with H(jω) would not
give us something proportional to b(t). Thus, d(t)
is NOT proportional to b(t).
% & 6
3ωm
ω
ωm
7
• System 2
Using the same technique in system 1, we divide the system into small parts again. This time from
figure H and I, we can conclude that only d(t) = K2 b(t) where K2 = 2.
a(t)
J
6
G(jω)
? C s(t)
6
C
C
C
r(t) = cos(2ωm t)
b(t)
-
?
J
G(jω)
j
f(t)
j
-
ω
-j
-
c’(t)
J
6
1
ω
−ωm ωm
p(t) = sin(ωm t)
?
- J
ω
1
d’(t)
-j
ω
−ωm ωm
The real and imaginary parts of S(jω) are given below:
Re{S(jω)}
T
T
T
T
−2ωm
T
T
T
T
Im{S(jω)}
2ωm
ω
ω
2ωm
−2ωm
Passing s(t) through G(jω) gives f(t):
Re{F (jω)}
Im{F (jω)}
ω
−2ωm
2ωm
T
T
T
−2ωm
8
c(t)
T
2ωm
T
T
T T
ω
d(t)
A
J
a(t)
D ? CC
6
C
C
C
C 6
r(t) = cos(2ωm t)
?
J
B b(t)
E F J
6
G(jω)
G(jω)
j
ω
-j
j
-
ω
-j
H -
1
ωm
−ωm
ω
p(t) = sin(ωm t)
?
- J
G I - d(t)
1
ωm
−ωm
c(t)
ω
Recall that P (jω) = −jπδ(ω − 2ωm ) + jπδ(ω + 2ωm )
C ′ (jω) is shown below:
Im{C ′ (jω)}
Re{C ′ (jω)}
−4ωm
@
@
@
A
A
A
AA
−4ωm
4ωm
ω
@
2ωm @
−2ωm
@
4ωm
−2ωm
2ωm
ω
Filtering C ′ (jω) with H(jω) would then give something proportional to a(t). Thus
c(t) IS proportional to a(t).
D ′ (jω) is shown below:
Re{D′ (jω)}
Im{D′ (jω)}
−4ωm
4ωm
−2ωm
2ωm
ω
A
A
A
−4ωm
AA
4ωm
−2ωm
Filtering D ′ (jω) with H(jω) would then give something proportional to b(t).
Thus d(t) IS proportional to b(t)
9
2ωm
AA
A
A A ω