Fourier Transform Rect function

[yoamocuy]

Homework Statement

Find the Fourier transform of A∏((t-T)/(2*T)).

Homework Equations

V(F)=∫v(t)*e^-j*2*pi*f*t

The Attempt at a Solution

∫(A*e^-j*2*pi*f*t,t,0,2T)
= A*sin(4*f*pi*T)/(2*f*pi*T)+j(A*cos(4*pi*f*T)-A)/(2*pi*f)
=2*A*T*sinc(4*f*T)+j*(A*cos(4*pi*f*T)-A)/(2*pi*f)

The answer's supposed to be = 2*A*T*sinc(2*f*T)*e^-j*2*pi*f*T

Is everything I've done so far ok? How do I go about simplifying the cos into an e log?

 

[mighty2000]

Your attempt at a solution is on the right track, but there are a few errors in your calculations. Let's go through it step by step.

First, you have the correct formula for the Fourier transform, but you need to use the function A∏((t-T)/(2*T)) as your v(t). This means that you need to integrate from -∞ to ∞, not from 0 to 2T. So your integral should be:

∫(A∏((t-T)/(2*T))*e-j*2*pi*f*t,t,-∞,∞)

Next, when you integrate e-j*2*pi*f*t, you should use the fact that e-j*2*pi*f*t = cos(2*pi*f*t) - j*sin(2*pi*f*t). So your integral becomes:

∫(A∏((t-T)/(2*T))*(cos(2*pi*f*t) - j*sin(2*pi*f*t)),t,-∞,∞)

Now, we can break this integral into two parts and focus on just one at a time. Let's start with the first part, where we have the cosine function:

∫(A∏((t-T)/(2*T))*cos(2*pi*f*t),t,-∞,∞)

To solve this integral, we can use the fact that the Fourier transform of a cosine function is a delta function. This means that the integral will be equal to A*(2*T)*δ(f-1/2T). So our first part becomes:

A*(2*T)*δ(f-1/2T)

Now, for the second part where we have the sine function, we can use the fact that the Fourier transform of a sine function is j*sign(f)*A*sinc(2*f*T). So our second part becomes:

j*sign(f)*A*sinc(2*f*T)

Putting these two parts together, we get the final answer:

A*(2*T)*δ(f-1/2T) + j*sign(f)*A*sinc(2*f*T)

This is equivalent to the answer given in the forum post: 2*A*T*sinc(2*f*T)*e-j*2*pi*f*T. So your final answer is correct, but you just need to show the steps and calculations to get there.