Islamic University of GazaElectrical Engineering Department

Communication I laboratory

Submitted by:

Adham Abu-Shamla Mohammed Hajjaj

120063320 120063640

Submitted to:

Eng. Mohammed kamel Abu-Foul

Amplitude modulation DSB-LC

(full AM)

1

Contents:The objective of this experiment.Quick review about the AM modulation and

(DSB-LC)Part 1 code and its comments and results

(step by step)Part 2 code and its comment and resultsConclusion.

2

Objective:Understanding AM modulation, Double Side Band-Large Carrier (DSB-LC) “known as Full AM”.

Using MATLAB to plot the modulated signal.

To simulate coherent demodulator and an envelope detector to obtain the real signal using MATLAB.

3

Quick review about the AM modulation and (DSB-LC)

Type of AM Modulation1. DSB-SC2. DSB-LC3. SSB4. VSB

(AM) and (FM) :

6

Part 1Use MATLAB to simulate this block (AM

block) (f(t)=cos(2π2000t),Ac=4,m=0.25,fc=20Khz)

X +

f(t)

mS(t)=Ac[1+mf(t)] .cos(2.π.fct)Ac.cos(2.π.fct)

Part 1 (a) code: the input signal fc=20000; % Carrier frequency ts=1/(10*fc);t=[0:2000]*ts;fs=1/ts; % Sampling frequencym=.25; % Modulation indexAc=4; % DC shiftx=cos(2*pi*2000*t); % the original signalfigure(1)subplot(211)plot(t,x)title('plot of baseband signal x(t)')xlabel('time (t)')ylabel('x(t)')

Carrier freqyancy Sampling period Sampling frequancy Modulation index

µ=mp/ADC shift

plot the real signal

Comment:In the above code we chose ts =1/(10*fs) to avoid overlapping in signal, then we make Fourier transform to plot the magnitude spectrum.

Code cont.Xf=fftshift(fft(x));Xf=Xf/length(Xf);deltax=fs/length(Xf);fx=-fs/2:deltax:fs/2-deltax;subplot(212)plot(fx,abs(Xf))title('the fourier transform of x(t)')xlabel('frequency (f)')ylabel('X(f)')

Fourier Transform of a real signal

Plot the magnitude spectrum

The results (part 1a)

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01-1

-0.5

0

0.5

1plot of baseband signal x(t)

time (t)

x(t)

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

x 105

0

0.2

0.4

0.6

0.8the fourier transform of x(t)

frequency (f)

X(f

)

Two Sym. pulses on the signal frequency

Code (part 1b): modulated signaly=(1+m*x)*Ac.*cos(2*pi*fc*t); % (Modulated

signal)figure(2)subplot(211)plot(t,y)title('the modulated signal

y(t)=(1+m*x)*Ac.*cos(2*pi*fc*t)')xlabel('time (t)')ylabel('y(t)')yf=fftshift(fft(y));

Code (part 1b): modulated signal cont.yf=yf/length(yf);delta=fs/length(yf);f=-fs/2:delta:fs/2-delta;subplot(212)plot(f,abs(yf))title('the fourier transform of the modulated

signal Y(f)')xlabel('frequency (f)')ylabel('Y(f)')

Plot the magnitude spectrum

The results (part 1b)

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01-5

0

5the modulated signal y(t)=(1+m*x)*Ac.*cos(2*pi*fc*t)

time (t)

y(t)

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

x 105

0

0.5

1

1.5

2the fourier transform of the modulated signal Y(f)

frequency (f)

Y(f

)

The same signal shifted at fc and the magnitude divided

by 2

Coherent detector

LPHS(t) w(t)

)v(t)

cos(2.π.fc.t)

Code (Part 1c-1) coherent demodulator “before filtering”. w=y.*cos(2*pi*fc*t); % Coherent

demodulated signal figure(3)subplot(211)plot(t,w)title('plot of demodulated signal w(t) before

LPF')xlabel('time (t)')ylabel('w(t)')wf=fftshift(fft(w));

Code (Part 1c-1) coherent demodulator “before filtering”. wf=wf/length(wf);delta=fs/length(wf);f=-fs/2:delta:fs/2-delta;subplot(212)plot(f,abs(wf))title('fourier transform of the demodulated

signal W(f)')xlabel('frequency (f)')ylabel('W(f)')

The results (part 1c-1)

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

2

4

6plot of demodulated signal w(t) before LPF

time (t)

w(t

)

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

x 105

0

1

2

3fourier transform of the demodulated signal W(f)

frequency (f)

W(f

)

The modulated signal shifted at 2*fc and the magnitude

divided by 2

There are a signal in 0 as real signal

Code (Part 1c-2) coherent demodulator “after filtering”. Wp=5000/fs; Ws=20000/fs; Rp=-1; Rs=-100;[N, Wn] = BUTTORD(Wp, Ws, Rp, Rs); %[num,den]=butter(N,Wn); %

the lower frequency of the transient region ( must be

between 0 and 1 )

The Upper frequency of the transient region ( must be

between 0 and 1 ) losses due to rippels

Fn. that return the order of the filter and the cutt off frequency

Fn. that return the transfer function of the Butterworth

filter

Code (Part 1c-2) coherent demodulator “after filtering”. v=filter(num,den,w); figure(4)subplot(211)plot(t,v)title('the demodulated signal after LPF v(t)')xlabel('time (t)')ylabel('v(t)')

Filtering process

Code (Part 1c-2) coherent demodulator “after filtering”. Vf=fftshift(fft(v));Vf=Vf/length(Vf);deltav=fs/length(Vf);fv=-fs/2:deltav:fs/2-deltav;subplot(212)plot(fv,abs(Vf))title('the fourier transform of v(t)')xlabel('frequency (f)')ylabel('V(f)')

The results (part 1c-2)

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

1

2

3the demodulated signal after LPF v(t)

time (t)

v(t)

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

x 105

0

0.5

1

1.5

2the fourier transform of v(t)

frequency (f)

V(f

)

There are a signal in 0 as real signal only

Envelope detector

Code (part 1d) %C=0.1e-6F,R=3.2e2%%%%%c=0.1e-6;r=3.2e2;RC=r*c;Vc=ones*(1:length(y));Vc(1)=y(1);for i=2:length(y) if y(i)>=Vc(i-1)

Code (part 1d) Vc(i)=y(i); else Vc(i)=Vc(i-1).*exp(-ts/RC); endendfigure(5)plot(t,y,t,Vc)%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%We can explain this code by the following flow chart

start

Vc(i)=y(i)

Vc(1)=y(1)i=2

Y(i)>=Vc(i-1)

Vc(i)=Vc(i-1).*exp(-ts/RC);

i=length(Y)

start

YES NO

YES

NO

The results (C=0.1e-6F,R=3.2e2)

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01-5

-4

-3

-2

-1

0

1

2

3

4

5

If we change R

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01-5

-4

-3

-2

-1

0

1

2

3

4

5

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01-5

-4

-3

-2

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0

1

2

3

4

5

R=3.2e(3)

R=3.2e(4)

Part 2Repeat part 1 with Ac=1, m = 2

After we make the simulation, the result is the same in part 1, but we

saw some difference in the envelope detector .

we will show the results of this part and comment the reason

Result of envelope

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01-1

0

1

2

3

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

x 105

0

2

4

6x 10

-3

overlap

Result of envelope

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01-3

-2

-1

0

1

2

3

Not as real signal

Comment in part 2 The reason of part 2 become like this becouse

the Ac is not enough to alternate the signal up to zero so envelope detector can’t get the real signal

Conclusion The experiment is a good simulate for AM

signals.We must make sure of the code because any

error causes fail in compiling