?
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Electrical Engineering and Computer Science
6.003,Signals and Systems—Fall 2003
Problem Set 7 Solution
Exercise for home study
O&W 7.28
(a) Using the Fourier series coe?cients of x(t),we can write,
k∞
1
| |
x(t)= e
jkw
0
t
,
2
k=?∞
where
2π
w
0
= =20π rad/sec,
0.1sec
The lowpass?lter H(jw) has a cuto? frequency w
c
= 205π rad/sec,Thus,x
c
(t)is x(t)
where all terms with frequency above w
c
are removed by the lowpass?lter,The terms
which are kept have |kw
0
|≤205π rad/sec =?|k|≤10.25,so the output,x
c
(t),is
k10
1
| |
x
c
(t)= e
jkw
0
t
2
k=?10
To obtain x[n],we sample x
c
(t)every T =5 × 10
3
seconds with an impulse train,
The sampling frequency is
2
T
π
= 400π =2 × maximum frequency in x
c
(t),Therefore,
we can write,
x[n]= x
c
(nT )
k10
1
| |
jkw
0
(nT )
= e
2
k=?10
k10
1
| |
= e
jkw
D
n
,where w
D
= w
0
T =0.1π rad,(1)
2
k=?10
Note that the complex discrete-time exponentials e
jkw
D
n
are all periodic with period
N =2π/w
D
= 20 (although N is not the fundamental period for all of them),Hence
x[n] must also be periodic with period N = 20,
1
(b) To?nd the Fourier series representation for x[n],we rewrite Equation (1) in the form
of a Fourier series synthesis equation,
k10
1
| |
jkw
D
n jkw
D
n
(2)e = a
k
e
2
k=?10 k=<20>
Note that there are 21 terms (k =?10...10) in the left-hand side of Equation (2),but
there are only 20 terms in the Fourier series on the right-hand side (as x[n]isperiodic
with N = 20),Looking at the k =10 and k =?10 terms,
k =10,
1
10
e
j10(0.1π)n
1
10
(?1)
n
=
1
10
(e
jπ
)
n
=
2 2 2
k =?10,
1
10
e
j(?10)(0.1π)n
1
10
=
1
10
(e
jπ
)
n
= (?1)
n
2 2 2
We can add these two terms since they involve the same complex exponential,(?1)
n
,
Now we have the full set of Fourier series coe?cients for x[n],The coe?cients are
periodic with period N = 20,and over the period?9 ≤k ≤ 10 have values
k
1
| |
,|k|≤9
a
k
=?
2
1
10
2,k =10.
2
O&W 8.23
(a) The block diagram of modulation and demodulation for this problem is shown below,
-w
co
w
co
x(t)
y(t) w(t)
r(t)
2
ω
cos(ω
c
t) cos(ω
d
t)
We want to show that the output of the lowpass?lter in the demodulator is proportional
to x(t)cos(?ωt),where?ω = ω
d
ω
c
,We show this algebraically as well as graphically
for the example in Figure P8.23,y(t)isjust x(t) multiplied by cos(ω
c
t),Using the
multiplication property,Y (jω) consists of two shifted copies of X(jω),centered at
ω = ω
c
and ω =?ω
c
,with amplitude scaled by 1/2,
1
Y (jω)=
2
[X(j(ω?ω
c
)) + X(j(ω + ω
c
))]
Y (jω)issketchedbelow,
2
ωY(j )
ω
c
+ω
m
ω
c
ω
m
ω
c c
ω
m
ω
c
ω c+ω
m
ω
ω
1/2 1/2
Similarly,W(jω) will consist of two shifted and scaled copies of Y (jω),
1
W(jω)= [Y (j(ω? ω
d
)) + Y (j(ω + ω
d
))]
2
1
= [X(j(ω? ω
d
ω
c
)) + X(j(ω? ω
d
+ ω
c
)) + X(j(ω + ω
d
ω
c
))
4
+X(j(ω + ω
d
+ ω
c
))]
1
= [X(j(ω? ω
c
ω
d
)) + X(j(ωω)) + X(j(ω +?ω)) + X(j(ω + ω
c
+ ω
d
))]
4
If we assume that the two copies centered at?ω andω do not overlap,then W(jω)
will look like the following,
ωW(j )
ω?ω
m
ω+ω
m
ω
c
+ ω
dc d
ω?ω
1/4 1/4 1/4
1/4
ω
ω?ω
Finally,R(jω)is W(jω) passed through a lowpass?lter with cuto? frequency between
ω
m
+?ω and 2w
c
+?w? w
m
= ω
c
+ ω
d
ω
m
,This?ltering process removes the high
frequency content of W(jω) so that only the two triangles at low frequencies remain,
Their amplitudes are scaled by 2,since the gain of the lowpass?lter is 2,Therefore,
R(jω),the output of the demodulator,will be
1
R(jω)= [X(j(ωω)) + X(j(ω +?ω))]
2
which in the time domain gives us
r(t)= x(t)cos(?ωt)
(b) From part (a),the spectrum of the output of the demodulator will be
1
R(jω)= [X(j(ωω)) + X(j(ω +?ω))]
2
If the two copies of X(jω) do not overlap,as shown in the?gures above,this will look
like,
3
ω?ω
m
ω+ω
m
ωR(j )
1/2 1/2
ω?ω
ω
This will happen if?ω ≥ ω
M
,If this is not true,then the two copies will overlap,and
the output will look like,
R(jω)
1
2
a0?
a0
a0
a0?
a0
a0
a0 ω
ω? ω
mω?ω
ω + ω
m
4
Problem 1
(a) We are given x(t)=cos(10t),Here,ω
o
= 10 rad/sec,Taking the Fourier transform of
x(t),
X(jw)
10 10
(π) (π)
w
2π
The sampling function,s(t)=
+∞
δ(t?kT),with T =
90
.
k=?∞
s(t)
T T?2T 2T0
(1)
··· ···
t
Taking the Fourier transform of s(t) (note that ω
s
=
2
T
π
= 90),
S(jw)
(90)
··· ···
w
180?90 0 90 180
Using the multiplication property,z(t)= x(t)s(t) in frequency domain is Z(jw)=
2
1
π
(X(jw)?S(jw)),i.e,we need to convolve X(jw) with the periodic impulse train in
S(jw) and scale the amplitude by
2
1
π
(see section 7.1.1 in O&W),
+∞
z(t)= x(nT)δ(t?nT)
n=?∞
1
+∞
Z(jw)= X(jθ)S(j(w?θ))dθ
2π
∞
5
Therefore,Z(jw) is as follows:
Z(jw)
(45)
··· ···
190?170?100?80?10 10 80 100 170 190
w
(b) y(t) is the output from the band-pass?lter,H(jw),with input z(t) as derived in part
(a),We know,
Y (jw)= H(jw)Z(jw)
Let us consider |Y (jw)| and ∠Y (jw) separately,|Y (jw)| is the band-pass?ltered
version of |Z(jw)| with frequency components between 90 to 180 and -180 to -90
rad/sec,
|Y (jw)|
(45)
170?100
0
100 170
w
∠Y (jw)= ∠H(jw)+ ∠Z(jw)
πw πw
=? +0 =?
200 200
Combining the magnitude and angle,Y (jw)= |Y (jw)|e
j∠Y (jw)
.
Consider Y (jw) as the Fourier transform of the sum of two sinusoidal signals; one with
w
o
= 100 and another with w
o
= 170,Using the time-shifting property of Fourier
←→ e
jwt
o
X(jw),transform,x(t?t
o
)
FT
45 π 45 π
y(t)= cos(100(t? )) + cos(170(t? ))
π 200 π 200
45 π 45 17π
= cos(100t? )+ cos(170t? )
π 2 π 20
(c) Now the sampling function s(t) is changed with T =
2π
90
,
6
s(t)
0
(1)
(?1)
T
2
T
2
T T
3T
2
3T
2
2T 2T
··· ···
t
T
s(t)= δ(t?kT)? δ(t?kT? )
∞ ∞
2
k= k=?∞?∞
Taking the Fourier transform,
∞ ∞
2π 2π 2π 2π
e
jw
T
2
S(jw)=
δ(w?k )?
T
T
δ(w?k )
T T
k= k=?∞?∞
2π
∞
2π 2π
∞
2πT
δ( k
2
w
π
T
e
jk
2
T
δ(w?k )?
T
T
)
=
T
k= k=?∞?∞
2π 2π
∞
=
2π
∞
δ(w?k )? (e
jπ
)
k
δ(w?k
2π
)
T T T T
k= k=?∞?∞
2π 2π
∞
2π
∞
2π
= δ(w?k )? (?1)
k
δ(w?k )
T T T T
k= k=?∞?∞
Seperating the odd and even terms of k,
2π
2π 2π
2π
S(jw)= δ(w?k )? δ(w?k )
T T T T
k=even k=even
2π
2π 2π
2π
+ δ(w?k )+ δ(w?k )
T T T T
k=odd k=odd
4π
2π
= δ(w?k )
T T
k=odd
x(t)=cos(10t) as before,To?nd Z(jw),we need to convolve X(jw) with the impulse
train in S(jw) and scale the result by
2
1
π
,
S(jw)isassketchedbelow,
7
S(jw)
0
(180)
90 90?270 270
··· ···
w
The convolution will place two scaled impulses (from X(jw)) centered at each impulse
in the impulse train of S(jw),Finally,H(jw) will only pass impulses that exist between
90 to 180 and?180 to?90 radians,We plot |Y (jw)| (output from H(jw)) as follows,
|Y (jw)|
(90)
w
100 0 100
As derived in part (b),∠Y (jw)= ∠H(jw)=?
πw
.
200
From the plot of |Y (jw)| and the ∠Y (jw),we can view y(t) as a time-shifted cos
functions,Therefore,
90 π
y(t)= cos(100(t? ))
π 200
90 π
= cos(100t? )
π 2
8
Problem 2 (O&W 7.30 except let x
c
(t)= δ(t?
T
2
))
(a) We are given x
c
(t)
T
x
c
(t)= δ(t? )
2
X
c
(jw)=e
jw
2
T
We take the Fourier transform of the system’s di?erential equation and?nd the fre-
quency response,H(jw),of the system,
dy
c
(t)
+ y
c
(t)= x
c
(t)
dt
jwY
c
(jw)+ Y
c
(jw)= X
c
(jw)
H(jw)=
Y
c
(jw)
X
c
(jw)
=
1
1+ jw
Now,we can write,
Y
c
(jw)= X
c
(jw)H(jw)=e
jw
2
T 1
1+ jw
y
c
(t)= e
(t?
2
T
)
u(t?
T
)
2
(b) y[n]= y
c
(nT)where y
c
(t) is as de?ned in part (a),Therefore,y
c
(nT) will pick-up
values from y
c
(t)at nT time values with n =0,1,2,..,
2
T
u[
e
nT +
n? 1]y[n]= y
c
(nT)=
2
T
)(e
T
)(e
T
)
n?1
u[n? 1]=(e
Using the time-shifting property of DTFT and basic DTFT table,
jw
)= e
2
T
e
jw
1
e
jw
1? e
T
Y (e
Now we choose H(e
jw
) such that,
y[n]? h[n]= w[n]= δ[n]
jw
)H(eY (e
jw
)=1
1
H(e
jw
)= e
jw
)(1? e
T
2
T
e
jw
e
T T
2
H(e
jw
)=
jw
e
2
e e
Taking the inverse FT,
T T
e
h[n]= e δ[n +1]
δ[n]
9
2 2
Problem 3
First,we need to?nd frequency response of the DT?lter,y[n]=
3
4
y[n?2]+x[n]+
1
x[n?1]
4
When x[n]= δ[n],y[n]= h[n],Therefore,
3 1
h[n]= h[n?2] + δ[n]+ δ[n?1]
4 4
3 1
e
j2?
H(e
j?
H(e
j?
)= )+1+ e
j?
4 4
1+
1
e
j?
4
H(e
j?
)=
1?
3
e
j2?
,|?|<π
4
2π
It is given that X(jω) = 0 for |ω|≥
π
and we have a sampling frequency,ω
s
=,So
T T
there will be no aliasing,
Therefore,the e?ective frequency response of the entire CT system,H
c
(jw),is related to
the frequency response of the DT system,H(e
j?
),by (assume? = wT and?nd appropriate
range of w),
ω
s
H(e
jωT
),|w|≤
π
=
T 2
H
c
(jω)=
0,|w|>
π
T
1
4
jωT
e1+
ω
s
|w|≤
π
=
T
π
j2ωT
,
3
4
2
H
c
(jω)=
1? e
0,|w|>
T
10
Problem 4,O&W 8.22
Let us de?ne v(t),r(t),and z(t) as shown below in the system diagram,
1
1
a12
×
v(t)
w
a12
×
r(t) z(t)
w
5w?3w 3w 5w
.
3w 3w
cos(5wt) cos(3wt)
v(t) is the output of sinusoidal amplitude modulation,To?nd V (jω),
v(t)=
V (jω)=
=
x(t) cos(5wt)
1
2π
(X(jω)? π [δ(ω? 5w) + δ(ω +5w)])
1
2
[X(j(ω? 5w)) + X(j(ω + 5w))]
Graphically,
V (jω)
y(t)
x(t)
ω
-7w -5w -3w 7w5w3w
1
2
1
2
To?nd R(jω),we see that the bandpass?lter will only passes frequency components that
are between of magnitude 3w to 5w,
R(jω)
1
2
1
2
-7w -5w -3w 3w 5w 7w
ω
11
Now z(t) is a sinusoidal amplitude demodulation with a di?erent frequency carrier signal,
cos(3wt),We can derive Z(jω),
z(t)= r(t) cos(3wt)
1
Z(jω)=
2π
(R(jω)? π [δ(ω? 3w) + δ(ω + 3w)])
1
=
2
[R(j(ω? 3w)) + R(j(ω + 3w))]
Z(jω)
ω
-8w -3w 0 6w3w2w
1
4
-6w -2w 8w
Finally,y(t)is z(t) lowpass-?ltered,only allowing frequency content from?3w to 3w to go
through,Therefore,Y (jω) is as follows,
Y (jω)
1
4
-3w -2w 0 2w 3w
ω
12
Problem 5 (O&W 8.49)
(a) Let us assume the frequency of the sampling function s(t)is w
c
,s(t)isaperiodic
2π
pulse-train with period T and pulse width,? =
T
,Therefore,w
c
=
T
and we can
2
write the Fourier series coe?cients,a
k
,as:
sin(kw
c
/2)
a
k
=
πk
sin(k(2π/T)(T/4))
=
πk
sin(k
π
2
)
=
πk
a
k
is a periodic sample train with frequency w
c
,
Let us assume the input signal,x(t),with Fourier transform,X(jw),as shown below,
X(jw)
1
w
w
m 0
w
m
Let us de?ne the input of the?lter,H
1
(jw),to be z
1
(t),Therefore,z
1
(t)= x(t)s(t)
and as discussed in section 8.5.1,Z
1
(jw) is the convolution of X(jw) with impulse
train with area a
k
,
+∞
Z
1
(jw)= a
k
X(j(w? kw
c
))
k=?∞
In order to prevent aliasing,we need to make sure X(jw) centered at kw
c
’s in Z
1
(jw)
does not overlap with another one,w
c
> 2w
m
will ensure that there is no aliasing or
overlap,i.e,
w
c
w
m
<
2
2π 1 π
highest allowable w
m
= =
T 2 T
If x(t) has frequency greater than
π
T
,there will be aliasing and y(t) will not be propor-
tional to x(t),
13
(b) Let us de?ne the output of the?lter,H
1
(jw),to be z
2
(t),z
2
(t) is the bandpass-?ltered
version of z
1
(t),From the pass-band of H
1
(jw),we see that z
2
(t) will only contain
frequency components centered at ±(w
c
=
2
T
π
) which corresponds to k = ±1in Z
1
(jw),
1
The magnitude of Z
1
(jw)at k = ±1is a
k=1
=
sin(
π
π/2)
=, Therefore,Z
2
(jw)isas
π
follows,
Z
2
(jw)
0
A
π
w
c
w
m
w
c
+ w
m
w
c
A
π
(w
c
w
m
)?(w
c
+ w
m
)
w
c
w
The second modulation with s(t) will result in convolution of Z
2
(jw) with the impulse
train with area a
k
again,We are only interested in low frequency components as
the result is low-pass?ltered with H
2
(jw),The convolution will cause two images
1
of
A
X(jw)from Z
2
(jw)tobescaledby
π
and superimposed centered at w =0,
π
The low-pass?lter,H
2
(jw),will only pass frequency components between?
π
to?
π
.
T T
Therefore,Y (jw)isasasfollows,
Y (jw)
1 2A
2
A
=
ππ π
2
w
w
m 0
w
m
From the plot,we can see Y (jw) has the same shape and bandwidth as X(jw),except
2A
that the amplitude is multiplied by
2A
π
2
,Therefore,gain of the overall system =
π
2
,
14
Problem 6
(a) Let’s analyze the system graphically in the frequency domain,Consider one step at a
time,
(i) p(t)= x(t)cos(ω
c
t)
Multiplying by cos(ω
c
t) in the time domain means that the Fourier transform of
p(t) will be the sum of two X(jw)shifted by ω
c
and?ω
c
,and magnitude scaled
by
1
2
.
(ii) q(t)= p(t)+ z(t)
The Fourier transform of q(t) is just the sum of the Fourier transforms of p(t)
and z(t),
Q(jω)
ω
w
c
w
x
w
c
w
c
+w
x
w
z
w
z
w
c
+w
x
w
c
w
c
w
x
A
2
A
2
B
(iii) r(t)= q(t)cos(ω
c
t)
i.e,q(t) is modulated by cos(ω
c
t),Therefore,repeating the procedure in (i),
R(jω)
ω?2w
c
w
c
2w
c
w
c
w
x?w
x
A
4
A
4
A
2
B
2
B
2
(iv) Filter r(t)with H
LP
(jw),
The baseband copy of X(jω) is nonzero only between?ω
x
and ω
x
.We can
recover it if it doesn’t overlap with the shifted Z(jω) on the sides,Therefore,we
can write,
ω
c
ω
z
>ω
x
ω
c
>ω
x
+ ω
z
15
(b) We want to?nd the parameters C and ω
f
for the low-pass?lter,From the?gure in
part (a),we can see that for ω ∈ [?ω
x
,ω
x
],R(jω)=
1
X(jω),so to get back X(jω),
2
C =2,
The cuto? frequency ω
f
just needs be between the baseband copy of X(jω)and the
shifted Z(jω),Therefore,
ω
x
≤ ω
f
≤ ω
c
ω
z
(c) This is the same system as above,except that we need to account for the delay in the
channel,We can account for this delay with h
d
(t)= δ(t? T),Following is the system
diagram with the delay element,
Channel
x(t) p(t)
×
+ δ(t? T)
q(t)
r(t)
× HLP (jω)
y(t)
H
LP
(jw)
C
z(t)
ω
ω
f
0 ω
f
cos(ω
c
t)
Let us de?ne s(t) to be the signal right before the delay,Then,
S(jω)= P(jω)+ Z(jω)
S(jω)H
d
(jω)= e
jωT
S(jω).Q(jω)=
We are essentially repeating what we did in (a) and (b),but this time Q(jω)isscaled
by e
jωT
,
Q(jω)
ω
w
c
w
x
w
c
w
c
+w
x
w
z
w
z
w
c
+w
x
w
c
w
c
w
x
A
2
e
A
2
e
Be
jωT jωT
jωT
Carrying out the the convolution,we see that the new R(jω)is,
16
R(jω)
ω
2w
c
w
c
A
4
e
j(ω+ω
c
)T A
4
e
j(ω?ω
c
)TA
4
(e
j(ω+ω
c
)T
+ e
j(ω?ω
c
)T
)
B
2
e
j(ω+ω
c
)T B
2
e
j(ω?ω
c
)T
w
x
w
c 2w
c
As we use the parameters chosen in parts (a) and (b),only the baseband (centered
around origin) component will pass through the?lter,It will be the same as in part
(b),except that we use the new,height” factor from the graph,If we let Y
0
(jω)be
the output of the system in part (b),then
1
(e
j(ω+ω
c
)T
+ e
j(ω?ω
c
)T
Y (jω)= Y
0
(jω) )
2
= Y
0
(jω)e
jωT
1
(e
jω
c
T
+ e
jω
c
T
)
2
= Y
0
(jω)e
jωT
cos(ω
c
T)
Recalling that the system without delay gave an output Y
0
(jω)= X(jω),we can?nd
the new system’s frequency response as following,
Y (jw)= X(jω)e
jωT
cos(ω
c
T)
Y (jw)
H(jω)= = e
jωT
cos(ω
c
T),
X(jw)
17
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Electrical Engineering and Computer Science
6.003,Signals and Systems—Fall 2003
Problem Set 7 Solution
Exercise for home study
O&W 7.28
(a) Using the Fourier series coe?cients of x(t),we can write,
k∞
1
| |
x(t)= e
jkw
0
t
,
2
k=?∞
where
2π
w
0
= =20π rad/sec,
0.1sec
The lowpass?lter H(jw) has a cuto? frequency w
c
= 205π rad/sec,Thus,x
c
(t)is x(t)
where all terms with frequency above w
c
are removed by the lowpass?lter,The terms
which are kept have |kw
0
|≤205π rad/sec =?|k|≤10.25,so the output,x
c
(t),is
k10
1
| |
x
c
(t)= e
jkw
0
t
2
k=?10
To obtain x[n],we sample x
c
(t)every T =5 × 10
3
seconds with an impulse train,
The sampling frequency is
2
T
π
= 400π =2 × maximum frequency in x
c
(t),Therefore,
we can write,
x[n]= x
c
(nT )
k10
1
| |
jkw
0
(nT )
= e
2
k=?10
k10
1
| |
= e
jkw
D
n
,where w
D
= w
0
T =0.1π rad,(1)
2
k=?10
Note that the complex discrete-time exponentials e
jkw
D
n
are all periodic with period
N =2π/w
D
= 20 (although N is not the fundamental period for all of them),Hence
x[n] must also be periodic with period N = 20,
1
(b) To?nd the Fourier series representation for x[n],we rewrite Equation (1) in the form
of a Fourier series synthesis equation,
k10
1
| |
jkw
D
n jkw
D
n
(2)e = a
k
e
2
k=?10 k=<20>
Note that there are 21 terms (k =?10...10) in the left-hand side of Equation (2),but
there are only 20 terms in the Fourier series on the right-hand side (as x[n]isperiodic
with N = 20),Looking at the k =10 and k =?10 terms,
k =10,
1
10
e
j10(0.1π)n
1
10
(?1)
n
=
1
10
(e
jπ
)
n
=
2 2 2
k =?10,
1
10
e
j(?10)(0.1π)n
1
10
=
1
10
(e
jπ
)
n
= (?1)
n
2 2 2
We can add these two terms since they involve the same complex exponential,(?1)
n
,
Now we have the full set of Fourier series coe?cients for x[n],The coe?cients are
periodic with period N = 20,and over the period?9 ≤k ≤ 10 have values
k
1
| |
,|k|≤9
a
k
=?
2
1
10
2,k =10.
2
O&W 8.23
(a) The block diagram of modulation and demodulation for this problem is shown below,
-w
co
w
co
x(t)
y(t) w(t)
r(t)
2
ω
cos(ω
c
t) cos(ω
d
t)
We want to show that the output of the lowpass?lter in the demodulator is proportional
to x(t)cos(?ωt),where?ω = ω
d
ω
c
,We show this algebraically as well as graphically
for the example in Figure P8.23,y(t)isjust x(t) multiplied by cos(ω
c
t),Using the
multiplication property,Y (jω) consists of two shifted copies of X(jω),centered at
ω = ω
c
and ω =?ω
c
,with amplitude scaled by 1/2,
1
Y (jω)=
2
[X(j(ω?ω
c
)) + X(j(ω + ω
c
))]
Y (jω)issketchedbelow,
2
ωY(j )
ω
c
+ω
m
ω
c
ω
m
ω
c c
ω
m
ω
c
ω c+ω
m
ω
ω
1/2 1/2
Similarly,W(jω) will consist of two shifted and scaled copies of Y (jω),
1
W(jω)= [Y (j(ω? ω
d
)) + Y (j(ω + ω
d
))]
2
1
= [X(j(ω? ω
d
ω
c
)) + X(j(ω? ω
d
+ ω
c
)) + X(j(ω + ω
d
ω
c
))
4
+X(j(ω + ω
d
+ ω
c
))]
1
= [X(j(ω? ω
c
ω
d
)) + X(j(ωω)) + X(j(ω +?ω)) + X(j(ω + ω
c
+ ω
d
))]
4
If we assume that the two copies centered at?ω andω do not overlap,then W(jω)
will look like the following,
ωW(j )
ω?ω
m
ω+ω
m
ω
c
+ ω
dc d
ω?ω
1/4 1/4 1/4
1/4
ω
ω?ω
Finally,R(jω)is W(jω) passed through a lowpass?lter with cuto? frequency between
ω
m
+?ω and 2w
c
+?w? w
m
= ω
c
+ ω
d
ω
m
,This?ltering process removes the high
frequency content of W(jω) so that only the two triangles at low frequencies remain,
Their amplitudes are scaled by 2,since the gain of the lowpass?lter is 2,Therefore,
R(jω),the output of the demodulator,will be
1
R(jω)= [X(j(ωω)) + X(j(ω +?ω))]
2
which in the time domain gives us
r(t)= x(t)cos(?ωt)
(b) From part (a),the spectrum of the output of the demodulator will be
1
R(jω)= [X(j(ωω)) + X(j(ω +?ω))]
2
If the two copies of X(jω) do not overlap,as shown in the?gures above,this will look
like,
3
ω?ω
m
ω+ω
m
ωR(j )
1/2 1/2
ω?ω
ω
This will happen if?ω ≥ ω
M
,If this is not true,then the two copies will overlap,and
the output will look like,
R(jω)
1
2
a0?
a0
a0
a0?
a0
a0
a0 ω
ω? ω
mω?ω
ω + ω
m
4
Problem 1
(a) We are given x(t)=cos(10t),Here,ω
o
= 10 rad/sec,Taking the Fourier transform of
x(t),
X(jw)
10 10
(π) (π)
w
2π
The sampling function,s(t)=
+∞
δ(t?kT),with T =
90
.
k=?∞
s(t)
T T?2T 2T0
(1)
··· ···
t
Taking the Fourier transform of s(t) (note that ω
s
=
2
T
π
= 90),
S(jw)
(90)
··· ···
w
180?90 0 90 180
Using the multiplication property,z(t)= x(t)s(t) in frequency domain is Z(jw)=
2
1
π
(X(jw)?S(jw)),i.e,we need to convolve X(jw) with the periodic impulse train in
S(jw) and scale the amplitude by
2
1
π
(see section 7.1.1 in O&W),
+∞
z(t)= x(nT)δ(t?nT)
n=?∞
1
+∞
Z(jw)= X(jθ)S(j(w?θ))dθ
2π
∞
5
Therefore,Z(jw) is as follows:
Z(jw)
(45)
··· ···
190?170?100?80?10 10 80 100 170 190
w
(b) y(t) is the output from the band-pass?lter,H(jw),with input z(t) as derived in part
(a),We know,
Y (jw)= H(jw)Z(jw)
Let us consider |Y (jw)| and ∠Y (jw) separately,|Y (jw)| is the band-pass?ltered
version of |Z(jw)| with frequency components between 90 to 180 and -180 to -90
rad/sec,
|Y (jw)|
(45)
170?100
0
100 170
w
∠Y (jw)= ∠H(jw)+ ∠Z(jw)
πw πw
=? +0 =?
200 200
Combining the magnitude and angle,Y (jw)= |Y (jw)|e
j∠Y (jw)
.
Consider Y (jw) as the Fourier transform of the sum of two sinusoidal signals; one with
w
o
= 100 and another with w
o
= 170,Using the time-shifting property of Fourier
←→ e
jwt
o
X(jw),transform,x(t?t
o
)
FT
45 π 45 π
y(t)= cos(100(t? )) + cos(170(t? ))
π 200 π 200
45 π 45 17π
= cos(100t? )+ cos(170t? )
π 2 π 20
(c) Now the sampling function s(t) is changed with T =
2π
90
,
6
s(t)
0
(1)
(?1)
T
2
T
2
T T
3T
2
3T
2
2T 2T
··· ···
t
T
s(t)= δ(t?kT)? δ(t?kT? )
∞ ∞
2
k= k=?∞?∞
Taking the Fourier transform,
∞ ∞
2π 2π 2π 2π
e
jw
T
2
S(jw)=
δ(w?k )?
T
T
δ(w?k )
T T
k= k=?∞?∞
2π
∞
2π 2π
∞
2πT
δ( k
2
w
π
T
e
jk
2
T
δ(w?k )?
T
T
)
=
T
k= k=?∞?∞
2π 2π
∞
=
2π
∞
δ(w?k )? (e
jπ
)
k
δ(w?k
2π
)
T T T T
k= k=?∞?∞
2π 2π
∞
2π
∞
2π
= δ(w?k )? (?1)
k
δ(w?k )
T T T T
k= k=?∞?∞
Seperating the odd and even terms of k,
2π
2π 2π
2π
S(jw)= δ(w?k )? δ(w?k )
T T T T
k=even k=even
2π
2π 2π
2π
+ δ(w?k )+ δ(w?k )
T T T T
k=odd k=odd
4π
2π
= δ(w?k )
T T
k=odd
x(t)=cos(10t) as before,To?nd Z(jw),we need to convolve X(jw) with the impulse
train in S(jw) and scale the result by
2
1
π
,
S(jw)isassketchedbelow,
7
S(jw)
0
(180)
90 90?270 270
··· ···
w
The convolution will place two scaled impulses (from X(jw)) centered at each impulse
in the impulse train of S(jw),Finally,H(jw) will only pass impulses that exist between
90 to 180 and?180 to?90 radians,We plot |Y (jw)| (output from H(jw)) as follows,
|Y (jw)|
(90)
w
100 0 100
As derived in part (b),∠Y (jw)= ∠H(jw)=?
πw
.
200
From the plot of |Y (jw)| and the ∠Y (jw),we can view y(t) as a time-shifted cos
functions,Therefore,
90 π
y(t)= cos(100(t? ))
π 200
90 π
= cos(100t? )
π 2
8
Problem 2 (O&W 7.30 except let x
c
(t)= δ(t?
T
2
))
(a) We are given x
c
(t)
T
x
c
(t)= δ(t? )
2
X
c
(jw)=e
jw
2
T
We take the Fourier transform of the system’s di?erential equation and?nd the fre-
quency response,H(jw),of the system,
dy
c
(t)
+ y
c
(t)= x
c
(t)
dt
jwY
c
(jw)+ Y
c
(jw)= X
c
(jw)
H(jw)=
Y
c
(jw)
X
c
(jw)
=
1
1+ jw
Now,we can write,
Y
c
(jw)= X
c
(jw)H(jw)=e
jw
2
T 1
1+ jw
y
c
(t)= e
(t?
2
T
)
u(t?
T
)
2
(b) y[n]= y
c
(nT)where y
c
(t) is as de?ned in part (a),Therefore,y
c
(nT) will pick-up
values from y
c
(t)at nT time values with n =0,1,2,..,
2
T
u[
e
nT +
n? 1]y[n]= y
c
(nT)=
2
T
)(e
T
)(e
T
)
n?1
u[n? 1]=(e
Using the time-shifting property of DTFT and basic DTFT table,
jw
)= e
2
T
e
jw
1
e
jw
1? e
T
Y (e
Now we choose H(e
jw
) such that,
y[n]? h[n]= w[n]= δ[n]
jw
)H(eY (e
jw
)=1
1
H(e
jw
)= e
jw
)(1? e
T
2
T
e
jw
e
T T
2
H(e
jw
)=
jw
e
2
e e
Taking the inverse FT,
T T
e
h[n]= e δ[n +1]
δ[n]
9
2 2
Problem 3
First,we need to?nd frequency response of the DT?lter,y[n]=
3
4
y[n?2]+x[n]+
1
x[n?1]
4
When x[n]= δ[n],y[n]= h[n],Therefore,
3 1
h[n]= h[n?2] + δ[n]+ δ[n?1]
4 4
3 1
e
j2?
H(e
j?
H(e
j?
)= )+1+ e
j?
4 4
1+
1
e
j?
4
H(e
j?
)=
1?
3
e
j2?
,|?|<π
4
2π
It is given that X(jω) = 0 for |ω|≥
π
and we have a sampling frequency,ω
s
=,So
T T
there will be no aliasing,
Therefore,the e?ective frequency response of the entire CT system,H
c
(jw),is related to
the frequency response of the DT system,H(e
j?
),by (assume? = wT and?nd appropriate
range of w),
ω
s
H(e
jωT
),|w|≤
π
=
T 2
H
c
(jω)=
0,|w|>
π
T
1
4
jωT
e1+
ω
s
|w|≤
π
=
T
π
j2ωT
,
3
4
2
H
c
(jω)=
1? e
0,|w|>
T
10
Problem 4,O&W 8.22
Let us de?ne v(t),r(t),and z(t) as shown below in the system diagram,
1
1
a12
×
v(t)
w
a12
×
r(t) z(t)
w
5w?3w 3w 5w
.
3w 3w
cos(5wt) cos(3wt)
v(t) is the output of sinusoidal amplitude modulation,To?nd V (jω),
v(t)=
V (jω)=
=
x(t) cos(5wt)
1
2π
(X(jω)? π [δ(ω? 5w) + δ(ω +5w)])
1
2
[X(j(ω? 5w)) + X(j(ω + 5w))]
Graphically,
V (jω)
y(t)
x(t)
ω
-7w -5w -3w 7w5w3w
1
2
1
2
To?nd R(jω),we see that the bandpass?lter will only passes frequency components that
are between of magnitude 3w to 5w,
R(jω)
1
2
1
2
-7w -5w -3w 3w 5w 7w
ω
11
Now z(t) is a sinusoidal amplitude demodulation with a di?erent frequency carrier signal,
cos(3wt),We can derive Z(jω),
z(t)= r(t) cos(3wt)
1
Z(jω)=
2π
(R(jω)? π [δ(ω? 3w) + δ(ω + 3w)])
1
=
2
[R(j(ω? 3w)) + R(j(ω + 3w))]
Z(jω)
ω
-8w -3w 0 6w3w2w
1
4
-6w -2w 8w
Finally,y(t)is z(t) lowpass-?ltered,only allowing frequency content from?3w to 3w to go
through,Therefore,Y (jω) is as follows,
Y (jω)
1
4
-3w -2w 0 2w 3w
ω
12
Problem 5 (O&W 8.49)
(a) Let us assume the frequency of the sampling function s(t)is w
c
,s(t)isaperiodic
2π
pulse-train with period T and pulse width,? =
T
,Therefore,w
c
=
T
and we can
2
write the Fourier series coe?cients,a
k
,as:
sin(kw
c
/2)
a
k
=
πk
sin(k(2π/T)(T/4))
=
πk
sin(k
π
2
)
=
πk
a
k
is a periodic sample train with frequency w
c
,
Let us assume the input signal,x(t),with Fourier transform,X(jw),as shown below,
X(jw)
1
w
w
m 0
w
m
Let us de?ne the input of the?lter,H
1
(jw),to be z
1
(t),Therefore,z
1
(t)= x(t)s(t)
and as discussed in section 8.5.1,Z
1
(jw) is the convolution of X(jw) with impulse
train with area a
k
,
+∞
Z
1
(jw)= a
k
X(j(w? kw
c
))
k=?∞
In order to prevent aliasing,we need to make sure X(jw) centered at kw
c
’s in Z
1
(jw)
does not overlap with another one,w
c
> 2w
m
will ensure that there is no aliasing or
overlap,i.e,
w
c
w
m
<
2
2π 1 π
highest allowable w
m
= =
T 2 T
If x(t) has frequency greater than
π
T
,there will be aliasing and y(t) will not be propor-
tional to x(t),
13
(b) Let us de?ne the output of the?lter,H
1
(jw),to be z
2
(t),z
2
(t) is the bandpass-?ltered
version of z
1
(t),From the pass-band of H
1
(jw),we see that z
2
(t) will only contain
frequency components centered at ±(w
c
=
2
T
π
) which corresponds to k = ±1in Z
1
(jw),
1
The magnitude of Z
1
(jw)at k = ±1is a
k=1
=
sin(
π
π/2)
=, Therefore,Z
2
(jw)isas
π
follows,
Z
2
(jw)
0
A
π
w
c
w
m
w
c
+ w
m
w
c
A
π
(w
c
w
m
)?(w
c
+ w
m
)
w
c
w
The second modulation with s(t) will result in convolution of Z
2
(jw) with the impulse
train with area a
k
again,We are only interested in low frequency components as
the result is low-pass?ltered with H
2
(jw),The convolution will cause two images
1
of
A
X(jw)from Z
2
(jw)tobescaledby
π
and superimposed centered at w =0,
π
The low-pass?lter,H
2
(jw),will only pass frequency components between?
π
to?
π
.
T T
Therefore,Y (jw)isasasfollows,
Y (jw)
1 2A
2
A
=
ππ π
2
w
w
m 0
w
m
From the plot,we can see Y (jw) has the same shape and bandwidth as X(jw),except
2A
that the amplitude is multiplied by
2A
π
2
,Therefore,gain of the overall system =
π
2
,
14
Problem 6
(a) Let’s analyze the system graphically in the frequency domain,Consider one step at a
time,
(i) p(t)= x(t)cos(ω
c
t)
Multiplying by cos(ω
c
t) in the time domain means that the Fourier transform of
p(t) will be the sum of two X(jw)shifted by ω
c
and?ω
c
,and magnitude scaled
by
1
2
.
(ii) q(t)= p(t)+ z(t)
The Fourier transform of q(t) is just the sum of the Fourier transforms of p(t)
and z(t),
Q(jω)
ω
w
c
w
x
w
c
w
c
+w
x
w
z
w
z
w
c
+w
x
w
c
w
c
w
x
A
2
A
2
B
(iii) r(t)= q(t)cos(ω
c
t)
i.e,q(t) is modulated by cos(ω
c
t),Therefore,repeating the procedure in (i),
R(jω)
ω?2w
c
w
c
2w
c
w
c
w
x?w
x
A
4
A
4
A
2
B
2
B
2
(iv) Filter r(t)with H
LP
(jw),
The baseband copy of X(jω) is nonzero only between?ω
x
and ω
x
.We can
recover it if it doesn’t overlap with the shifted Z(jω) on the sides,Therefore,we
can write,
ω
c
ω
z
>ω
x
ω
c
>ω
x
+ ω
z
15
(b) We want to?nd the parameters C and ω
f
for the low-pass?lter,From the?gure in
part (a),we can see that for ω ∈ [?ω
x
,ω
x
],R(jω)=
1
X(jω),so to get back X(jω),
2
C =2,
The cuto? frequency ω
f
just needs be between the baseband copy of X(jω)and the
shifted Z(jω),Therefore,
ω
x
≤ ω
f
≤ ω
c
ω
z
(c) This is the same system as above,except that we need to account for the delay in the
channel,We can account for this delay with h
d
(t)= δ(t? T),Following is the system
diagram with the delay element,
Channel
x(t) p(t)
×
+ δ(t? T)
q(t)
r(t)
× HLP (jω)
y(t)
H
LP
(jw)
C
z(t)
ω
ω
f
0 ω
f
cos(ω
c
t)
Let us de?ne s(t) to be the signal right before the delay,Then,
S(jω)= P(jω)+ Z(jω)
S(jω)H
d
(jω)= e
jωT
S(jω).Q(jω)=
We are essentially repeating what we did in (a) and (b),but this time Q(jω)isscaled
by e
jωT
,
Q(jω)
ω
w
c
w
x
w
c
w
c
+w
x
w
z
w
z
w
c
+w
x
w
c
w
c
w
x
A
2
e
A
2
e
Be
jωT jωT
jωT
Carrying out the the convolution,we see that the new R(jω)is,
16
R(jω)
ω
2w
c
w
c
A
4
e
j(ω+ω
c
)T A
4
e
j(ω?ω
c
)TA
4
(e
j(ω+ω
c
)T
+ e
j(ω?ω
c
)T
)
B
2
e
j(ω+ω
c
)T B
2
e
j(ω?ω
c
)T
w
x
w
c 2w
c
As we use the parameters chosen in parts (a) and (b),only the baseband (centered
around origin) component will pass through the?lter,It will be the same as in part
(b),except that we use the new,height” factor from the graph,If we let Y
0
(jω)be
the output of the system in part (b),then
1
(e
j(ω+ω
c
)T
+ e
j(ω?ω
c
)T
Y (jω)= Y
0
(jω) )
2
= Y
0
(jω)e
jωT
1
(e
jω
c
T
+ e
jω
c
T
)
2
= Y
0
(jω)e
jωT
cos(ω
c
T)
Recalling that the system without delay gave an output Y
0
(jω)= X(jω),we can?nd
the new system’s frequency response as following,
Y (jw)= X(jω)e
jωT
cos(ω
c
T)
Y (jw)
H(jω)= = e
jωT
cos(ω
c
T),
X(jw)
17
