Fourier Transform of x(t)= 8 sin(8(pi)t + pi/8)

hitmen

Joined Sep 21, 2008
159
Using the formula sin (A+B) = sinAcosB+cosAsinB I am able to reduce the equation to

(8 cos (pi/8) ) sin( 8(pi)t ) + (8 sin (pi)/8 )cos( 8(pi)t )

Now I do not know how to do the fourier transform or what equation to use. pls help

t_n_k

Joined Mar 6, 2009
5,448

t_n_k

Joined Mar 6, 2009
5,448
Hi hitmen,

The result is complex but presumably explainable in physical terms.

Since the time domain function transforms {maps?} as single frequency value into the frequency {fourier} domain one just gets a single fourier frequency value - mathematically expressed as a dirac or delta based function....?

I've attached a printout of the "valid" entry for your function on the WIMS site

(I used s as the default variable - maybe ω would be better so as not to confuse with Laplace)

How you derive this from a knowledge of the transform I'll to leave to you to discover - the website has some information which may be useful ...?

GirishC

Joined Jan 23, 2009
58
Using the formula sin (A+B) = sinAcosB+cosAsinB I am able to reduce the equation to

(8 cos (pi/8) ) sin( 8(pi)t ) + (8 sin (pi)/8 )cos( 8(pi)t )

Now I do not know how to do the fourier transform or what equation to use. pls help
Now 8Cos(pi/8) and 8sin(pi/8) is a constant term added to your time varying signal...its a DC..so you do not need to worry about it much....

Now you have two terms as

7.39*sin(8*PI*t) + 3.06*cos(8*PI*t) = 7.39*sin(4t) + 3.06*cos(4t)

i.e. you have two components at 4t