Solution for HW02
4.1
Fourier transform calculated directly:
F [(t t0 )] =
=e
=e
(t t0 )ej2f t dt
j2f t
|t=t0
j2f t0
.
The second step holds since (t t0 ) is zeros everywhere except at t0 . Using the
time shift rule F [(t t0 )] = F [(t)ej2f t0 ] = ej2f t0
ECE 3311: Principles of Communication Systems
Homework 2: Due at start of class on 12-Nov.
B-term 2013
Problems From Text
1. 4.1. In addition, calculate the magnitude and phase response from the Fourier Transform,
and show your steps.
2. 4.11
3. 4.18
4. 4
2
Solution for HW01
2.5
Start with the equation (2.5):
R(f ) = F cfw_r(t) = F cfw_s(t) cos(2f1 t)
1
= F cfw_s(t)[ (ej2f1 t + ej2f1 t )]
2
+
1
s(t)[ (ej2f1 t + ej2f1 t )]ej2f t dt
=
2
1
1 +
s(t)ej2(f f1 )t dt +
2
2
1
1
= S(f f1 ) + S(f + f1 ).
2
2
+
=
s(t
ECE 3311: Principles of Communication Systems
Quiz #2
2:002:25 PM, November 14, 2012
Name:
Instructions:
Do not open this quiz until you are instructed to do so.
This quiz is closed book, but you are permitted to bring one two-sided 8.5 by 11 sheet o
6
Sampling with Automatic Gain
Control
6.1
According to Nyquist Sampling Theorem, the sampling rate must be highter
than or equal to twice the maximum frequency of the analog signal so that the
original signal can be reconstructed. fs 2 Bhuman = 40kHz.
44
HW04 solution
8.4
Below is the code for this problem:
% p u l s e s h a p e 0 .m: a p p l y i n g a p u l s e shape t o a t e x t
s t r= Transmit this text string ; % message t o be t r a n s m i t t e d
m etters2pam ( str ) ;
=l
N=length (m) ; % 4 l e v
ECE 3311: Principles of Communication Systems
Homework 4: Due at start of class on 26-Nov.
B-term 2013
Problems From Text
1. 8.4
2. 8.6
3. 8.7
4. 8.9
5. 8.13
6. 8.14
7. 9.6 (a) and (b)
Problem 8: Familiarity with Matlab Transmitter
This problem will begin
10 HW05 solution
10.2
Using freqz command, you will see both the frequency and phase response. From
the gure, we can read the phase response at the desired location. In our case,
it is 2fc normalized by fs /2. Notice that the phase response is given in te