# Homework Help: Find the fourier transform

1. Aug 26, 2011

### Jncik

1. The problem statement, all variables and given/known data
find the fourier transform of the function

$$x(t)=\left\{\begin{matrix} &25 - \frac{25}{8}|t-10| &for &|t-10|<8 \\ &0 &for& |t-10|>8 \end{matrix}\right.$$

2. Relevant equations

3. The attempt at a solution

we know that

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

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?

Last edited: Aug 26, 2011
2. Aug 26, 2011

### vela

Staff Emeritus
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.

Last edited: Aug 26, 2011
3. Aug 26, 2011

### 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 $$|\frac{t}{8} - \frac{10}{8}|<1$$

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 $$|t-10|<8$$

and the second remains 0 but with interval $$|t-10|>8$$

which is essentially the x(t)

Last edited: Aug 26, 2011
4. Aug 26, 2011

### vela

Staff Emeritus
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.

5. Aug 26, 2011

### Jncik

thanks I have one more question

I think I'm wrong with my FT calculations

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

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

Last edited: Aug 26, 2011
6. Aug 26, 2011

### vela

Staff Emeritus
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.

7. Aug 27, 2011

### Jncik

that was a nice alternative

thanks a lot for your help :)