EE351–Spectrum Analysis and Discrete Time Systems (Fall 2005)
SOLUTIONS
Solutions to Assignment 7
1. (Frequency Response or Transfer Function of an LTI System) Consider two signals
x(t) and y(t) shown below.
x(t )
y (t )
1
−0.5 0
1
t (sec)
0.5
0
1
2
t (sec)
3
(a) The function z(t) = dy(t)/d(t) (first derivative of y(t)) is shown in Fig. 27-(a).
Using the differentiation property
x (t ) of FT, one has:
[6]
Z(jω) = (jω)Y (jω)
(29)
1
x(t )
Furthermore, using the time-shifting property and the given FT of x(t), Z(jω)
can be found as:
−0.5
1.5
t (sec)
5 2
1 0.5
−2
−
1
0
1
X(jω) − exp −jω ·
X(jω)
Z(jω) = exp −jω ·
2
2
ω
5ω
1
= exp −−j
− exp −j
X(jω)
2
2
3ω
[exp (jω) − exp (−jω)] X(jω)
= exp −j
2
Ideal LPF
3ω
= exp −j
2j sin(ω)X(jω)
(30)
H ( jω )
2
LTI System
y (t )
Finally, 1
FT
h(t ) ←

→ H ( jω )
y (t )
1
3ω
Z(jω)
2 sin(ω/2)
=
exp −j
2j sin(ω)
jω
jω
2
ω
4 sin(ω) sin(ω/2)
3ω
exp −j
=
2
ω
2
ω=
(jω)
−3.5π 0 Y3.5
π
M 1 ( jω )
x (t )
(31)
A
[4]
(b) Using the convolution property, the transfer function H(jω) is simply given by:
Y (jω)
2 sin(ω)
3ω
ω
H(jω)
=m1 (t )
=
exp −j
(32)
−5000 0 5000
X(jω)
ω
2
M 2 ( response
jω )
y (t )
The impulse
h(t) can be found by taking the inverse FT of H(jω)
above. Howz (t )
∑
ever, simple inspection of h(jω) reveals that H(jω) is precisely the FT of a rectangular
A by 1.5 as shown in Fig. 27(b).
pulse delayed
The above result is also expectedmbecause
convolution of
pulses
2 (t )
x (ttwo
) rectangular
2 cos(20,
000t )x(t) ∗
h(t) gives a trapezoidal waveform of y(t).
−5000 0
5000
ω
2 cos(10, 000t )
Electrical Engineering, University of Saskatchewan
Page 31
EE351–Spectrum Analysis and Discrete Time Systems (Fall 2005)
x(t )
SOLUTIONS
y (t )
z (t )
1
1
1
−0.5 0
t (sec)
0.5
0
0
0.5
2
1
1
2.5
2
3
t
t (sec)
-1
x (t )
(a)
h(t )
1
1
−0.5
−2
−1
0
0
1.5
0.5
0.5
1.5
−1
t (sec)
t2
1
2.5
(b)
Figure 27: Plots of z(t) and y(t) in the time domain.
Ideal LPF
x(t )
2. (Frequency-Shifting Property of FT ) The manager of your division asked you to design a
H ( jω ) system to transmit two signals m1 (t)LTI
communication
andSystem
m2 (t) simultaneously over the
same channel. It is also required
that thex (design
makes use of the company’s
available
y (t )
t)
y (t )
FT
1
h(t ) cos(20,
←

→ H000t).
( jω ) After examining the
carrier generators (i.e., oscillators) cos(10, 000t) and
spectra of the two signals,
you present the following block diagram for the transmitter:
ω
−3.5π 0 3.5π
M 1 ( jω )
A
−5000 0
5000
ω
m1 (t )
M 2 ( jω )
∑
y (t )
z (t )
A
m2 (t )
−5000 0
5000
x (t )
2 cos(20, 000t )
ω
2 cos(10, 000t )
[4]
(a) The spectra of three signals x(t), y(t), and z(t) are shown in Fig. 28.
Electrical Engineering, University of Saskatchewan
Page 32
EE351–Spectrum Analysis and Discrete Time Systems (Fall 2005)
SOLUTIONS
X ( jω )
A
ω (rad/sec)
-10k
0
5k 10k 15k
Y ( jω )
A
ω (rad/sec)
-10k
0
5k 10k 15k
Z ( jω )
A
ω (rad/sec)
-30k
-20k
-10k
0
5k
15k
20k
30k 35k
Figure 28: Spectra of x(t), y(t) and z(t).
[1]
(b) The minimum bandwidth of the radio channel is computed as:
W = 35k − 5k = 30k (rad/sec)
(33)
Note that the physical meaning (representing how many radians per second) of a
negative frequency is the same as that of the positive frequency of the same magnitude. Thus one should only look at the positive frequency portion to determine
the physical bandwidth.
[5]
(c) The diagram of one possible receiver that can perfectly recover the signals m 1 (t)
and m2 (t) from the modulated signal z(t) is shown in Fig. 29. Note that you
might come up with a different structure and it also works.
Electrical Engineering, University of Saskatchewan
Page 33
EE351–Spectrum Analysis and Discrete Time Systems (Fall 2005)
SOLUTIONS
H 1 ( jω )
H 2 ( jω )
1
1
z (t )
m1 (t )
ω
15k 20k
ω
25k
-5k
0
cos(20000t )
5k
H 4 ( jω )
H 3 ( jω )
1
1
ω
25k 30k
ω
35k
5k 10k
2 cos(20000t )
15k
H 2 ( jω )
1
m2 (t )
ω
-5k
cos(10000t )
0
5k
Figure 29: Block diagram to recover m1 (t) and m2 (t).
3. (Sampling) A block diagram of impulse sampling is shown below.
X ( jω )
A
−ωx
[2]
−
ωx 0
2
xψ (t )
x (t )
ωx
2
H ( jω )
y (t )
ω (rad/sec)
ωx
ψ (t ) = Ts
+∞
n =−∞
δ (t − nTs )
xψ (t ) to prevent aliasing is:
(a) The minimum sampling frequency
2
(ωs )min = 2ωx
h (t )
1.5
1
1 Engineering, University of Saskatchewan
Electrical
1.5
1
(34)
Page 34
EE351–Spectrum Analysis and Discrete Time Systems (Fall 2005)
[5]
SOLUTIONS
(b) The spectra of xψ (t) and y(t) is shown in Fig. 30.
X ψ ( jω )
A
− ωx
ω x 3ωx
0
ω
3ωx
2
Y ( jω )
A
− ωx
ω
ωx
0
Figure 30: Spectra of xψ (t) and y(t).
[3]
(c) With the impulse response h(t) shown in Fig. 31-(a), the interpolation is piecewise constant. The reconstructed signal y(t) is shown in Fig. 31-(b).
xψ (t )
h(t )
y (t )
2
1.5
1
0
1.5
1
Ts
t
0
1
Ts
(a)
2Ts
3Ts
4Ts
t
(b)
Figure 31: Piece-wise interpolation.
Electrical Engineering, University of Saskatchewan
Page 35