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Download full Solutions Manual Digital & Analog Communication Systems 8th Edition version Script + Answers at: https://getbooksolutions.com/ https://getbooksolutions.com/download/soluti

Trang 1

Download full Solutions Manual Digital & Analog

Communication Systems (8th Edition) version (Script + Answers) at: https://getbooksolutions.com/

https://getbooksolutions.com/download/solutions-manual-digital-analog-communication-systems-8th-edition-answer-and-script

Example 1_01:

% File: Example1_1.m for Example 1-1

clear;

% hr is the receiving antenna height in feet

% ht is antenna height in feet

% dr is distance to the horizon for the receiving antenna

% dt is distance to the horizon for the transmitting antenna

% d is LOS distance between the receiving and transmitting antennas

% Select a value for the height(ft) of the Receiving Antenna, hr

hr = 5;

dr = sqrt(2*hr);

ht = 0:10:1000;

dt = sqrt(2*ht);

d = dr +dt;

plot(ht,d);

text(710,22,'hr=')

text(760,22,num2str(hr))

xlabel('Transmitting Antenna Height in feet');

ylabel('LOS distance in miles');

title('Distance for LOS Propagation');

grid;

% Select a print-out value for the transmitting antenna height, htt

htt = 612.5;

dtt = sqrt(2*htt);

dfix = dr +dtt;

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fprintf('\n\nFor a Receiving Antenna Height of %6.2f',hr); fprintf(' ft\n');

fprintf('and a Transmitting Antenna Height of %6.2f',htt); fprintf(' ft\n');

fprintf('\nThe LOS distance is %6.2f',dfix);

fprintf(' miles\n');

fprintf('\nSee the Window for a plot of the LOS as a\n'); fprintf('function of the Transmitting Antenna Height for a'); fprintf('\nReceiving Antenna Height of %6.2f',hr);

fprintf(' ft\n\n');

SUB

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Example: 2.04

% File: Example2_04.m for Example 2-4

% This example plots Eq (2-44) where

% the second term is negligible for positive

% frequences if fo is sufficiently large

% The Magnitude-Phase Spectral Functions

% will be plotted for the case of positive frequencies

% The Magnitude function will be plotted in dB units

% The Phase function will be plotted in degree units

clear;

T = 2;

fo = 500;

f = (fo-50):1:(fo+50);

for (k = 1:101)

W(k) = (T/2j)*1/(1+2j*pi*T*(f(k)-fo));

end;

B = log(W);

WdB = (20/log(10))*real(B);

Theta = (180/pi)*imag(B);

subplot(211);

plot(f,WdB);

xlabel('f');

ylabel('W(f) in dB');

grid;

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plot(f,Theta);

xlabel('f');

ylabel('Angle of W(f) in degrees');

grid;

SUB

P2_25

% File: P2_25.m

clear;

M = 8;

N = 2^M;

n = 0:N-1;

T = 300;

dt = T/N;

tk = n*dt-100;

% Creating time waveform

w = u_step(tk,5)-u_step(tk,75);

for (i = 1:length(w))

w(i) = w(i) * (sin(2*pi/512*tk(i)) + sin(70*pi/512*tk(i))); end;

% Approximating the Fourier Integral using the FFT

W = dt*fft(w);

fn = n/T;

fs = 1/dt;

fprintf('\nSee Window for plot.\n');

subplot(211);

axis([0 100 -2 2]);

plot(tk,w);

title('Time waveform');

xlabel('t in sec >');

ylabel('w(t)');

axis;

subplot(212);

plot(fn,abs(W));

title('MAGNITUDE SPECTRUM from 0 to fs');

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ylabel('Magnitude of W(f)');

subplot(111);

SUB

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