ECE453, Fall 2014, Assignment 1
1. When designing a transmitter it is important to consider the peak envelope power (PEP) as well as
the average power of the signals to be transmitted. The envelope power of a modulated carrier signal
s(t) is a timevaryin
I used Gnu Octave to do these calculations. Here is the Octave script that I used:
# define pulse amplitude sequence a_n=[1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1];
beta=0.95;
# sampling interval for s(t)
dt = 0.01;
# time interval
tp = (0:dt:19
ECE 453: Homework Assignment 7 Solution
61 Solution
The data given in the problem statement tell us the following:
Vs  = 5 V
2.795 = Vs 
50
50 + Zs 
Vs 
= 0.1 A
Zs 
(1)
(2)
(3)
Using (1) in (3) leads to Zs  = 50 . Manipulating equation (2):
2
ECE 453: Homework Assignment 2 Solution
111 Solution
a. s(t) = 10 cos[2 108 t 2 cos(2 104 t + /4)]. This signal is in the form A cos(c t + (t), where (t) =
2 cos(2 104 t + /4). Hence, the signal is purely angle modulated. The message signal consists of a
22 Solution
a. The 3 dB bandwidth is found using the roots of
F ( ) =
F (o )
1
= .
2
2
Using the identity
1
1
=
1j
2
it is easy to see that the 3 dB frequencies are the roots of
o
Q(
) = 1.
o

Or
2
o
2
o = 0.
Q
The roots are:
o
o
1 + 4Q2
2Q 2
ECE 453: Homework Assignment 1 Solution
116 Solution
The bandwidth of a raisedcosine pulse is W = 21 (1 + ). With T = 1s and = 0.5, W = 750kHz. The
T
bandwidth occupied by a DSBSC signal is 2W, hence the bandwidth occupied by the transmitted signal is
[813]
(a)
L
in
ZL
ZL
S11
0 .6
Z in
'
S11
(b)
Av
'
S 21
Z 0 500 50
Z 0 500 50
S12 S 21 L
1 S 22 L
9
11
(8.41)
0.175 105 9 11
1 0 .1
30 9 11
107.33
100
0.4615
1 in
Z0
1 in
1 0.4615
107.33
50
1 0.4615
107.33
26.447 j 29.608
Z in 500
Z in 500
26.447 j 29.608
ECE 453
Wireless Communication Systems
Mixer
1st IF Filter
Mixer
2nd IF Filter
Preselector
Filter
RF Amplier
fIF 1
IF Amplier
fIF 2
IF Amplier
fLO2
fLO1
Course Notes
Spring 2009
Steven J. Franke
Department of Electrical and Computer Engineering
University
56 Solution
a. Oscillation will occur only if
<(Z + Z 0 ) 0
=(Z + Z 0 ) = 0
where Z =
600 and
Z0 =
1
R j!(R2 C + ! 2 L2 C L)
(R + j!L) =
.
j!C
(1 ! 2 LC)2 + ! 2 R2 C 2
The potential frequency of oscillation is the frequency where the imaginary part of Z
[121 Solution]
This amplifiers inputoutput voltage characteristic contains only first and secondorder terms. The second order terms will
generate all frequency components of the form:
 fi fj 
where i, j 2 cfw_1, 2, 3. Thus, the following frequencies
ECE453 Lab 1. Spectrum Exploration
The Spectrum Exploration is an exercise for ECE453 with the set of educational objectives outlined
below. It is designed as an engaging introduction to Wireless Communications. It also develops skills
with the Spectrum A
ECE453 Lab 2. Modeling Discrete Components
The purpose of this lab is to learn how to simulate simple passive circuits using Agilents Advanced
Design System (ADS) software, how to measure impedance using a network analyzer (NA), and to
become familiar wit
ECE 353
Power and dB Handout
Fall 1998
ECE 353
Power and dB Handout
Introduction
Power is a common quantity measured in RF and communications systems. In many measurement contexts, power is
measured in decibels referenced from milliwatts (dBm). This hando
Test & Measurement
Application Note 1501
H
Spectrum Analysis
Amplitude
&Frequency
Modulation
http:/www.hp.com/go/tmappnotes
H
Contents
Spectrum Analysis AM and FM
Spectrum Analysis 1501
Amplitude and
Frequency Modulation
Chapter 1 Modulation Methods
Cha
83 Solution
2
4
S11
S12
S21
S22
3
5=
2
1
4
N2 + 1
1
1
N2
2N
2N
N2
1
3
5
85 Solution
a. With ZL = 0 (
L
=
1),
in
= 2.220\
68.90 , so Zin = 65.92\
133.5 = ( 45.4
j47.8) .
b. From part a, we already know that the 2port is NOT unconditionally stable. Note
75 Solution
V1 = AV2
BI2
I1 = CV2
DI2
The parameter A is the inverse of the open circuit voltage gain:
A=(
V2
)
V1
1
I2 =0
The parameter B is the negative inverse of the forward shortcircuit transfer admittance:
B=
(
I2
)
V1
1
V2 =0
The parameter C is
43 Solution
1
a. Current Ii ows through a parallel RLC circuit with impedance Zp = Rj!L j!C . Denote the reactance
of the inductor and capacitor at resonance by jXo . The Q of this parallel resonant circuit is
Qp =
R
.
Xo
At the resonant frequency, Zp
211 Solution
a. The possible LO frequency ranges are:
Low LO: 20602085 MHz
High LO: 25802605 MHz
b. For fc = 2335 MHz:
Low LO: fLO = 2075 MHz, fIM = 1815 MHz.
High LO: fLO = 2595 MHz, fIM = 2855 MHz.
c. For an interferer at frequency fi to cause interf
a. The
dB bandwidth is found using the roots of
F (!) =
F (!o )
1
p
=p .
r
r
where r = 10/10 . A little algebraic manipulation can be used to show that the
roots of
p
!
!o
Q(
) = r 1.
!o
!
Or
!2
!o p
r
Q
1!
dB frequencies are the
2
!o = 0.
There ar
11 Solution
Signal 1 contains components at f1 = 0.5 kHz and f2 = 0.9 kHz and signal 2 contains components at
f3 = 3.5 kHz and f4 = 10 kHz. The output of the multiplier is
[cos 1 t + cos 2 t][cos 3 t + cos 4 t].
When expanded, the product includes 4 term
12 Solution
a. The output of a halfwave rectier can be written as s(t)p(t) where p(t) is a square wave that alternates between 0 and 1.
The Fourier series representation of p(t) is
p(t) =
1
1
2X1
n
+
sin
cos(n!c t).
2 n=1 n
2
The rst few terms of the se
15 Solution
a. (t) = 5 cos m t + 15 cos 3m t, so (t) = 5m sin m t 45m sin 3m t, or f (t) = 10 sin m t 90 sin 3m t (kHz). I
found the peak value of this function using gnu octave, as follows:
t=(0:9999)/10000;
f=10*sin(2*pi*t)90*sin(6*pi*t);
max(abs(f)
ECE 453: Homework Assignment 3 Solution
22 Solution
a. The 3 dB bandwidth is found using the roots of
F ( ) =
1
F (o )
= .
2
2
Using the identity
1
1
=
1j
2
it is easy to see that the 3 dB frequencies are the roots of

Q(
Or
o
) = 1.
o
2
The ro
ECE 453: Homework Assignment 4 Solution
214 Solution
a. Downconversion with highside LO: fLO = fc + fIF = (1930 to 1975) + 70 = 2000 to 2045 MHz. lowside
LO: fLO = fc fIF = (1930 to 1975) 70 = 1860 to 1905 MHz.
b. highside LO: fIM = fc + 2fIF = 1940
ECE 453: Homework Assignment 5 Solution
413 Solution
At 10 MHz, the reactance of the capacitors is j 5 . The branch consisting of C1 in parallel with R = 50
has Q = 50/5 = 10, and transforms into j 5 in series with 50/102 = 0.5 , as shown in Figure a.
j
ECE 453: Homework Assignment 8 Solution
74 Solution
[Z ] =
Z1 + Z3
Z3
Z3
Z2 + Z3
710 Solution
a.
[Y ] =
b.
Av = gm
R=
1
Rf
gm r
Re +r +gm Re r
1
Rf
1
Rf
R L Rf
1
RL + Rf 1 + Re (gm +
1
r )
+
RL
R L + Rf
(R + Rf )(Re + r + gm Re r )
R + Rf + Re + r + g
[56]
(a)
Oscillation will occur only if
Recfw_Z + Recfw_Zin < 0
Imcfw_Z + Imcfw_Zin = 0
In the problem
Zin = C/(L+R)
=
If oscillation would happen, then
At oscillation frequency f0 , Recfw_Zin =
Therefore, the circuit will NOT oscillate.
(b)
If the negat