A particular waveform for Fourier Analysis is given by 5 cos

A particular waveform for Fourier Analysis is given by 5 cos 3x 2 cos 2x. What is the period of this periodic function- (a) (b) (c) (d) (e) 5

Solution

The fundamental period of sin function is 2. Now,Add the period of 2 in each cos function of given equation.

y=5cos(3x)+2cos(2x)

y=5cos(3x+2)+2cos(2x+2)

taking 3 common from 1st cos fun. and 2 from 2nd cos fun.

y=5cos[3(x+2/3)]+2cos[2(x+2/2)]

so, the period of 1st cos fun. is 2/3

and the peiod of 2nd cos fun. is

One has period and the other has period 2/3. What you want now is to see when they \"match up\". This is obtained in 2. Basically, this is 3×2/3 and 2×. We\'re just cross multiplying periods.