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Question

The fourier series for a periodic signal is given as
x(t)=cos(1.2πt)+cos(2πt)+cos(2.8πt).
The fundamental frequency of the signal is

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Solution

Given signal is
x(t)=cos(1.2πt)+cos(2πt)+cos(2.8πt).
Step-1:

ω1=1.2π

2πT1=1.2π

T1=2π1.2π

=2012=53;ω2=2π

2πT2=2π

T2=1

ω3=2.8π

2πT3=2.8π

T3=22.8=57

Step - 2:
T1T2=53,T1T3=73

Step-3:
L.C.M. of denominators of step 2 is
L.C.M. = 3.

Step-4:
T = (L.C.M.)× T1

=3×53=5sec

Fundamental period = 5 sec
Fundamental frequency of the signal
=1T=15=0.2Hz


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