Discovering the Fourier Transform to Solving for X(jω)

[Jncik]

Homework Statement

find the Fourier transform of the function

<br /> x(t)=\left\{\begin{matrix}<br /> &amp;25 - \frac{25}{8}|t-10| &amp;for &amp;|t-10|&lt;8 \\ <br /> &amp;0 &amp;for&amp; |t-10|&gt;8<br /> \end{matrix}\right.<br />

Homework Equations

The Attempt at a Solution

we know that

<br /> g(t)=\left\{\begin{matrix}<br /> &amp;1-|t| &amp;for &amp;|t|&lt;1 \\ <br /> &amp;0 &amp;for&amp; |t|&gt;1<br /> \end{matrix}\right.\leftrightarrow X(j\omega) =\left\{\begin{matrix}<br /> &amp;\begin{bmatrix}<br /> {\frac{\frac{sin(\omega)}{2}}{\frac{\omega}{2}}}<br /> \end{bmatrix}^{2} &amp;for &amp;|\omega|&lt;1 \\<br /> &amp;0 &amp;for&amp; |\omega|&gt;1<br /> \end{matrix}\right. <br />

we can see that x(t) = 25g(\frac{1}{8} (t-10))

now

25g(t-10) \leftrightarrow 25 X(j \omega) e^{-j 10 \omega}

and

25g(1/8(t-10)) \leftrightarrow \frac{25}{8} X(j\frac{\omega}{8}) e^{-j 10 \frac{\omega}{8}}

is this correct?

thanks in advance

 

[vela]

Your approach looks fine, but I think you got the factors of 8 in the wrong place. Since it divides t, it should be multiplying ω. Also, I'm not sure where you got the overall factor of 1/8 in front.

There's an error in what you wrote for X(ω), but I assume it's just a typo. It should be sin(ω/2), not (sin ω)/2.

 

[Jncik]

Yes you're right

25g(1/8(t-10)) \leftrightarrow 200 X(j 8 \omega) e^{-j 80 \omega}

"Also, I'm not sure where you got the overall factor of 1/8 in front."

isn't 25g(1/8(t-10)) = x(t)?

for g(1/8 t - 10/8) and the first interval I will have

1 - |\frac{t}{8} - \frac{10}{8}| for |\frac{t}{8} - \frac{10}{8}|&lt;1

for the second interval I will get zero as a result

now if I multiply this by 25 I will get the following result for the first interval

25-\frac{25}{8}|t-10| for |t-10|&lt;8

and the second remains 0 but with interval |t-10|&gt;8

which is essentially the x(t)

 

[vela]

I meant the 8 that you multiplied into the 25 to get 200. I didn't think it should be there, but I just checked the tables and found you were right.

 

[Jncik]

thanks I have one more question

I think I'm wrong with my FT calculations25g(t-10) \leftrightarrow 25 X(j \omega) e^{-j 10 \omega}

but I have 25g(\frac{1}{8}(t-10))

hence
25g(1/8(t-10)) \leftrightarrow 200 X(j 8 \omega) e^{-j 10 \omega}

I mean I should first find 25 g(1/8 t)

and then

find the final result due to the shifting by 10...but I'm not sure about

 

[vela]

Good catch. As you suggested, you can look at it as a shift by 10 of g(t/8). I think that's the simplest way.

You could, however, perform the shift first and then the scaling, but it's a bit tricky. The shift would have to be by 10/8, giving you g(t - 10/8). Then when you scale time, you replace t by t/8 to get g(t/8 - 10/8) = g[(t-10)/8]. In the frequency domain, the shift factor would originally be e^-j(10/8)ω, and then the scaling would replace ω by 8ω, giving you e^-j10ω again.

 

[Jncik]

that was a nice alternative

thanks a lot for your help :)