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ELE4606 Communication Systems

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ELE4606 Communication Systems

1 Download4 Pages / 987 Words

Course Code: ELE4606

University: University Of Southern Queensland

MyAssignmentHelp.com is not sponsored or endorsed by this college or university

Country: Australia

Question:

Use the integrate-per-symbol method to demodulate the channel waveform, as depicted. Show the output of each integrator as subplots one above the other, in a similar way to that shown. The input sequence should be the same as you used in part (a).

Answer:

Given, the modulation signal is

x(n) =

Here, for 1 bit input signal

= 0.5 for 0 bit input signal.

M = 60 = samples per carrier cycle.

N = 3 cycles.

So, x(n) =

Block diagram:

Part (b):

Given that, samples per carrier cycle = M = 60, ω = 2π/M phase steps in radians,

N = 4 carrier cycles per symbol.

MATLAB program:

% 4 bit input

bit_stream = [1 0 1 1];

% Amplitude for 0 bit

Ab1 = 0.5;

% Amplitude for 1 bit

Ab2 = 1;

% samples per carrier cycle

M = 60;

% sample size

S = 600;

% sample vector

n = 1:S;

ASK_signal1 = zeros(1,S);

ASK_signal2 = zeros(1,S);

ASK_signal = zeros(1,S);

for i = 1: 1: length(bit_stream)

% The ASK signal construction

if bit_stream(1,i)==0

ASK_signal1 = Ab1*sin((2*pi/M).*n);

elseif bit_stream(1,i)==1

ASK_signal2 = Ab2*sin((2*pi/M).*n);

end

ASK_signal = ASK_signal1 + ASK_signal2;

% Final ASK signal

ASK_signal = ASK_signal + ASK_signal2;

end

% graph of the ASK Signal

plot(n,ASK_signal,’b’);

xlabel(‘sample number n’);

ylabel(‘channel Amplitude’);

title(‘ASK Signal with two Amplitudes’);

grid on;

Plot of ASK signal:

Question 2:

Part a)

Given that,

x(n) = Asin(nω + ?b)

The simple block diagram of the modulation scheme for one-bit modulation is given below.

Now, given that the carrier amplitude A = 1, M = 60 samples per carrier cycle, ω = 2π/M, ?b = ±180°, N = 4.

MATLAB code:

% 4 bit input

bit_stream = [1 0 1 1];

% sample size

S = 600;

% amplitude of modulation signal

A =1;

% phase shift for 0 bit

phi1 = zeros(1,S);

% phase sift for 1 bit

phi2 = repmat(pi,1,S);

% samples per carrier cycle

M = 60;

% sample vector

n = 1:S;

ASK_signal1 = zeros(1,S);

ASK_signal2 = zeros(1,S);

ASK_signal = zeros(1,S);

for i = 1: 1: length(bit_stream)

% The phase modulation construction

if bit_stream(1,i)==0

ASK_signal1 = A*sin((2*pi/M).*n + phi1);

elseif bit_stream(1,i)==1

ASK_signal2 = A*sin((2*pi/M).*n + phi2);

end

ASK_signal = ASK_signal1 + ASK_signal2;

% Final Phase modulated signal

ASK_signal = ASK_signal + ASK_signal2;

end

% graph of the ASK Signal

plot(n,ASK_signal,’b’);

xlabel(‘sample number n’);

ylabel(‘channel amplitude’);

title(‘Phase modulated signal’);

grid on;

Plot of the phase modulated output:

Part b)

The block diagram of demodulation of channel waveform is shown below.

MATLAB code:

M = 60; % sampling rate

t = (0:2*M+1)’/M; % time vector

x = sin((2*pi/M + pi)*t) + sin((2*pi/M)*t);

fc = 10; % carrier freq

phasedev = pi; % phase deviation

mx = pmmod(x,fc,M,phasedev); % modulation

dx = pmdemod(mx,fc,M,phasedev); % demodulation

figure; plot(t,[x dx]);

legend(‘Original signal’,’demodulated signal’);

xlabel(‘Time’)

ylabel(‘Amplitude’)

Graph:

Question 3:

The modulator for two constellation is shown below.

The input for both constellation is 00 01 11 10.

Constellation A modulation MATLAB code:

bit_stream = [0 0 0 1 1 1 1 0];

% sample size

S = 600;

% amplitude of modulation signal

A =1;

% phase shift for 0 bit

phi1 = repmat(pi/2,1,S);

% phase sift for 1 bit

phi2 = repmat(pi,1,S);

% samples per carrier cycle

M = 60;

% sample vector

n = 1:S;

ASK_signal1 = zeros(1,S);

ASK_signal2 = zeros(1,S);

ASK_signal = zeros(1,S);

for i = 1: 1: length(bit_stream)

% The phase modulation construction

if bit_stream(1,i)==0

ASK_signal1 = A*sin((2*pi/M).*n + phi1);

elseif bit_stream(1,i)==1

ASK_signal2 = A*sin((2*pi/M).*n + phi2);

end

ASK_signal = ASK_signal1 + ASK_signal2;

% Final Phase modulated signal

ASK_signal = ASK_signal + ASK_signal1 + ASK_signal2;

end

% graph of the ASK Signal

plot(n,ASK_signal,’b’);

xlabel(‘sample number n’);

ylabel(‘channel amplitude’);

title(‘ modulated by Constellation A’);

grid on;

Graph:

Similarly in case of Constellation B the phase shifts will be changed to π/4 for 0 bit and 3π/4 for 1 bit.

Part b)

The demodulation of the above signal based on the two Constellation can be done in the similar way demodulation done in question 2(b). In this case the input bits will be changed as there are a total of 8 bits [0 0 0 1 1 1 1 0] and the phases for the corresponding Constellation can be adjusted.

Part a)

MATLAB code:

% time waveform x(n)

N = 64;

n = 0:N-1;

w = 2*pi/N;

x = sin(n*w);

figure(1)

stem(x, ‘.’);

title(‘Input Waveform’);

% perform FFT

X = fft(x);

% plotting real and imaginary components

subplot(2,1,1);

stem(real(X));

title(‘Real’);

subplot(2,1,2);

stem(imag(X));

title(‘Imaginary’);

for x(n) = sin(nw)

For x(n) = cos(nw)

For x(n) = sinnω + cosnω

x(n) = −sinnω + cosnω

x(n) = −sinnω−cosnω

It is clear that the real parts of FT outputs are negative when signal x(n) is positive and when x(n) is negative then the real part is positive. The imaginary plot suggest that with negative x(n) the imaginary plot starts with positive conjugate part and ends with negative conjugate plot. For positive x(n) the reverse occurs.

Part B)

For x(n) = -sin(2*n*w) – cos(2*n*w)

For x(n) = sin(2*n*w) + cos(2*n*w)

The Orthogonal Frequency Division Multiplexing (OFDM) of a signal produces large number of narrowband carriers which are very closely associated in the frequency domain. The complex numbers in the baseband fashion is modulated by the Inverse First Fourier transform and then those are converted back to serial data for the transmission purpose.

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