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

Thread Starter

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
Hi hitmen,

Sorry if the page didn't load up with your function entered.

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)

View attachment WIMS Home.pdf

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
 
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