Why is a0 a value if pi when It shouldn't be?
%% Problem 17.5
display('Problem 17.5')
syms n t
Wo = 1;
T = 2*pi;
n = 1:10;
a0 = (1/T)*[int(-14,t,-pi,0) + int(7,t,0,pi)]
an = (2/T)*[int(-14*cos(n*Wo*t),-pi,0) + int(7*cos(n*Wo*t),0,pi)]
bn = (2/T)*[int(-14*sin(n*Wo*t),-pi,0) + int(7*sin(n*Wo*t),0,pi)]

3 Kommentare

Utkarsh Belwal
Utkarsh Belwal am 7 Okt. 2020
I am not understanding your issue, can you please tell your query again ?
Salik Mallick
Salik Mallick am 7 Okt. 2020
Hello Mr.Belwal,
I meant why does ao have pi in it? I did the hand calculation and there shouldn't be a pi for the DC coefficent.
Star Strider
Star Strider am 11 Okt. 2020
See also an apparent duplicate: Fourier Series of Rectified Cosine Function

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Utkarsh Belwal
Utkarsh Belwal am 8 Okt. 2020
Bearbeitet: Utkarsh Belwal am 8 Okt. 2020

1 Stimme

Hello Salik,
You are right, there should be no pi in the DC coefficient and the value should be 3.5. According to me as a0 is a symbolic variable so it has the pi symbol included in it for better precision. I calculated the value of a0 that MATLAB is displaying, and it is coming 3.5. If you want to use the numeric value, then you can do a symbolic to numeric conversion as follows:
A0_numeric = double(a0);
DISCLAIMER: These are my own views and in no way depict those of MathWorks.

1 Kommentar

Salik Mallick
Salik Mallick am 10 Okt. 2020
Hello Mr. Belwal,
Would you know why this script isn't running?
%% Problem 17.9
display('Problem 17.9')
% Determine the Fourier coefficients an and bn of the
% first three harmonic terms of the rectified cosine
% wave
syms n t
T = 8;
Wo=2*pi/T;
n = 1:3;
an = (2/T)*[int(2.5*cos(n*Wo*t)*cos(n*Wo*t),-2,2)]
AN_numeric = double(an)

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