EE435 HW#1
Due: 24 October 2011
Problem-1) Let
s (t ) A 1 cos m t 0.3cos 2m t cos c t
where is chosen such that there is no over modulation. a) Plot s(t) assuming f c f m for 0.5 . b) What is the maximum value of ? c) Consider the following definitions:
EE352Communication Systems 1
Laboratory Report
Frequency Modulation and Demodulation using MATLAB
Abstract
The principal objectives of the experiment is understand frequency modulation and
demodulation processes. In order to understand well, we analyzed f
EE451Communication Systems 2
Laboratory Report
Pulse Code Modulation (PCM) and Quantization
Abstract
The principal objective of the experiment is to understand Pulse Code Modulation
(PCM) and Quantization. In order to understand well, we write the MATLAB
EE352Communication Systems 1
Laboratory Report
Introduction to Probability and Random Processes
Abstract
The principal objectives of the experiment is understand the basics of probability and
random processes. In order to understand well, we studied on pr
EE352Communication Systems 1
Laboratory Report
Effect of Noise on Frequency Modulation
Abstract
The principal objectives of the experiment is understand the effect of noise on FM
modulation. In order to understand well, we studied on prelaboratory. Prelab
EE352Communication Systems 1
Laboratory Report
Effect of Noise on Amplitude Modulation
Abstract
The principal objectives of the experiment is understand the effect of noise on AM
modulation. In order to understand well, we studied on prelaboratory. Prelab
EE352Communication Systems 1
Laboratory Report
Sampling
Abstract
The principal objective of the experiment is to understand sampling process. In order
to understand well, we analyzed sampling of a continuous time signal and the effect of the
sampling rate
EE435 HW #1
Due: 30 October 2013, 17:30 (Submit to D-217 Telecomm. Lab. EE435 HW box)
Problem-1)
We have a voice file TSM.wav which we use to make an AM wave (you can listen to it by double
clicking the file). To process it, well use MATLAB and the follow
EE435 HW #2
Due: 18 November 2013, 17:30 (Submit to D-217 Telecomm. Lab. EE435 HW box)
Problem-1)
An upper sideband SSB signal is obtained from a message signal
m(t) = a(t) cos(2 fm t) where a(t) is a waveform bandlimited to Wa=600 Hz, and fm=1000 Hz.
The
EE 435 Fall 2016-2017
HW 3
Group Number :8
Group Members : Ahmed Sleyman Kl 1876457
zlem Batzel 1875814
1) a)
b)
c)
d)
2) a)
b)
c)
3) a)
b)
c)
d)
KVmin=2000*pi
Frequencies of two signals are equal and they have a little phase difference as expected.
e)
Th
EE 435 Fall 2016-2017
HW 3
Group Number :8
Group Members : Ahmed Sleyman Kl 1876457
zlem Batzel 1875814
1) a)
b)
c)
d)
2) a)
b)
c)
3) a)
b)
c)
d)
KVmin=2000*pi
Frequencies of two signals are equal and they have a little phase difference as expected.
e)
Th
EE 435 Fall 2016-2017
HW 1
Group Number : 8
Group Members : zlem Batzel, Ahmed Sleyman Kl
1) a)
b)
2) a)
b) We use bandpass filter. Because, we want to get DSB-SC dignal.
c)
d) We use low pass filter. Because, we want to get message signal.
3)
Spectrum sh
EE352Communication Systems 1
Laboratory Report
Amplitude Modulation / Demodulation
Abstract
The principal objectives of the experiment is understand modulation and demodulation
processes. In order to understand well, we analyzed amplitude modulation / dem
EE421Communication Systems 2
Laboratory Report
Delta Modulation
Abstract
The principal objective of the experiment is to understand Delta Modulation (DM). In
order to understand well, we built an integrated circuit. In modulation part of experiment,
Flip-
EE451Communication Systems 2
Laboratory Report
Delta Modulation
Abstract
The principal objective of the experiment is to understand Delta Modulation (DM). In
order to understand well, we built an integrated circuit. In modulation part of experiment,
Flip-
EE435 HW#1 Fall 2011 Solutions
Problem-1)
s(t ) A 1 cos mt 0.3cos 2mt cos ct
where is chosen such that there is no over modulation. For A=1, fc =20, fm=1 and =0.5:
1.5
a)
1
0.5
s(t)
0
-0.5
-1
-1.5
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
t (sec)
b)
For maxi
EE435 HW #2
Due: 04 November 2011, 17:30
Problem-1)
We have a voice file chimes.wav which we use to make an AM wave (Listen to it by double
clicking the file). The file contains stereo voice, but well use only one channel of it. To process it
well use MAT
EE435 HW#2 Fall 2011 Solutions
Problem-1)
a)
For an AM modulated signal s(t ) A1 mt cos c t
A2
1 2 Pm
2
A2
PC
2
A2 2
Pm
PS
2
PT
2 Pm
PS
PT 1 2 Pm
For maximum power efficiency, we have to maximize 2 and choose 2 such that no
overmodulation occurs for
EE435 HW #3 Due: 01 December 2011, 17:30
Problem-1) Consider a NBFM generator for which the modulation index () is restricted to be less than 0.1. The message signal has two equal amplitude frequency components, the lower one being at 50 Hz and the higher
AMPLITUDE
MODULATION
A. zgr Ylmaz & E. Uysal-Bykolu
METU
EE 435 Fall 2016
Modulation
Translation of a baseband signal to high frequencies in
various forms
Why modulation?
Accomodation of multiple transmissions
Smaller devices by EM theory
Sometimes b
EE 435 Fall 2016/2017
HW Set #1 (Part-I and Part-II)
Due: 23:55, November 17
1) Single tone signals are often transmitted for providing synchronization to various devices that need
accurate time information. A 1kHz, 3 V peak amplitude sync signal () = 3si
EE 435 Fall 2016
HW Set #3
Due: 23:55, November 13, 2016
Note: In one of the problems of this homework, you will need the pdf of a random sinusoidal wave. The
related document is included in HW3 support files.
1) Consider the frequency modulation of a car
Q3)
Figure 1: Spectrum analyzer window snapshots for Q3.
As can be observed in the above figure, we see many radio stations in the FM frequency range between
88.5 MHz and 101.5 MHz. The spacing between the FM channels is about 200 kHz, which can also be
o
3) a)
Fig.1: The spectrum of the DSB-SC signal (with 3kHz sinusoidal signal as message signal) demodulated
with an envelope detector
As can be observed from Fig.1, we have a spectrum with impulses at multiples of 6 kHz (0, 6, 12, 18 kHz,).
The reason to o
EE 435 Fall 2016-2017
HW 3
Group Number : 30
Group Members : Bertan Kurun , Trker Dolap
1) a)
c(t)=10 cos (2pi* 103e6 t)
m(t)=1e3 cos (2pi*2e3 t)
u(t) = 10 cos(2pi*103e6*t+2pi*kf*10e3
= cos(2pi*103e6*t+(10e3* kf/ fm)*sin
fi(t) = fc + kf*1000* cos
= 103e6+