EE315 Exam 2 Solutions
1)
a) Take Fourier Transforms of both sides and use derivative property to get that
Y (ω)((jω)2 + 5jω + 4) = −3X(ω)
yielding H(ω) =
−3
.
(jω)2 +5jω+4
b) Use partial fraction expansion to get that H(ω) = 1/(4 + jω) − 1/(1 + jω) and
take inverse Fourier Transform to get that h(t) = (exp(−4t) − exp(−t))u(t).
4+jω
,
2+jω
c) Note X(ω) =
then
Y (ω) = X(ω)H(ω) =
−3
= −3/(1 + jω) + 3/(2 + jω)
(1 + jω)(2 + jω)
Take inverse Fourier Transform to get that y(t) = −3(exp(−t) − exp(−2t))u(t).
2)
a) Let V (ω) = X(ω + 1). Therefore x(t) = v(t) exp(jt)/(2π). Note that V (ω) is
real and even therefore v(t) will be real and even. We then have that h(x(t) =
t + u(−v(t))π.
R∞
b) x(0) = 1/(2π)
−∞ X(ω)dω
R∞
c) EX = 1/(2π)
d)
R∞
−∞ x(t)dt
−∞ |X(ω)|
2
= 3/(2π).
dω = 4/(3π).
= X(0) = 1.
P∞
e) Y (ω) = 1/2
k=−∞ X(ω
− 2k).
f ) Z(ω) = .5(X(ω) + X(−ω)).
3)
b) Distortionless bands from -1000π to 1000π radians.
c) y(t) = |H(500π)| cos(500πt+h(H(500π)))−2|H(3000π)| sin(3000πt+h(H(3000π))) =
cos(500πt − π/4) + cos(3000πt) .
4)
a) BX = 4B. No, minimal bandwidth needed is 4B.
1
0.8
Y
0.6
0.4
0.2
0
−5
−4
−3
−2
−1
0
w
1
2
3
4
5
1
0.8
Z
0.6
0.4
0.2
0
−3
−2
−1
0
w
1
2
3
Figure 1: 2e,2f: Plots of Y (ω) and Z(ω)
0
−0.2
1
−0.4
−0.6
Arg(H(w))
|H(w)|
0.8
0.6
0.4
−0.8
−1
−1.2
−1.4
0.2
−1.6
0
0
5000
10000
w
15000
−1.8
0
5000
10000
15000
w
Figure 2: 3a): Plots of magnitude and phase of H(ω)
X(w)
M3
M1
M2
w0−2B w0
w0+2B w
BPF
w0−B
X
m1(t)
LPF
cos((w0−B)t)
x(t)
BPF
w0+B
X
m2(t)
LPF
cos((w0+B)t)
BPF
w0
X
m3(t)
LPF
sin(w0t)
Figure 3: 4) Spectrum of X and DMUX