(50 points) Consider the full-wave rectified cosine function shown in the Figure

Question

(50 points) Consider the full-wave rectified cosine function shown in the Figure below (refer to Appendix A (page 866 Table A.1 of 10h edition of textbook). For all the parts below assume A 2 volts and T-1 msec. The FSE for the signal s(t) is given by 2A 4A n-2,4,6,… The total power for this signal is given by A2/2 Watts g(t) Full-wave rectified cosine function (Textbook Appendix A-Table A.1) For the given periodic signal, it is required to perform the following: 90 a) (10 points) Plot the signal 3774 714 774 374 time, t s(t) for -T

Explanation / Answer

First take time t as vector :

t= 0.1:0.001:10 ; t is in miliseconds because T is in miliseconds.

A=2;

T=1;

According to given qsn signal is combination of DC(constant part) and AC signal(sinosoidal part) so seperate these two

s1=2*A/pi;

and

for s2 we have to run loop of n

sum=0;

for n = 2:2:100

s2=sum+(4*A/pi)*(-1)^(1+n/2)*cos(2*pi*n*t/T)/(n^2-1);

end

s=s1+s2;

Now you can plot signal from -T to T

plot(t,s)

axis(0 2 -T T);

One part is very interestng in first section i.e. fandamental frequency

you can find fandamental frquency by taking fft of given signal and take the index at which signal has maximum value

sfft= fft(s);

[maxvalue,indx]=max(abs(sfft));

fandamental frequency= Fs*length(sfft)*idx;

Fs is sampling frquency.

Also you can find out Power by applying "Parseval's Theorem"

take FFT of s the sum the square of sfft in the loop of n

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