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I have to derive the Fourier transform of a half triangle which is shown here:

this image

So far I got the equation of the line as 1-t/T, and now I think that I have to substitute into the Fourier transform definition with the limits set as 0 to T, but I'm not entirely sure...

(Sorry if I posted in the wrong stackexchange, this question is from my electronics course and thought here would be most appropriate)

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up vote 3 down vote accepted

You are absolutely right. As your function has a value of zero everywhere except from the [0, T[ interval, you don't need to set the limits any wider. Those 'outer' intervals will add nothing to the result. The integral below will give you the answer.

$$X_3(j\omega)=\int_0^Tx_3(t) \cdot e^{-j\omega t} dt=\int_0^T(1-{{t} \over {T}})\cdot e^{-j\omega t} dt$$

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Applying the definition of Fourier transform always work, but it's not always the best idea. There are some functions that have already been transformed and are listed in tables.

Sometimes it's better to think: how can I write my function as a sum of this known functions, so that then using linearity I can easily get the answer? In your case unfortunately there's no smart sum, but you can write your function as the product of a triangle of center 0, height 1 and width 2T, and a rectangle starting in 0, width T and height 1.

Since the F-transforms of both the triangle and the rectangle are known you can obtain your result calculating the convolution of the two functions, that may be easier than directly solving the integral as per the F transform definition.

What you state is correct, meaning that if you use the definition your interval is [0,T], infact your function is not \$x_3(t)=1-\frac{t}{T}\$, but rather \$x_3(t)=(1-\frac{t}{T})\cdot rect(\frac{t-\frac{T}{2}}{T})\$. The effect of that rect is exactly limiting the integration interval:

$$ X_3(j\omega)=\mathcal{F}[x_3(t)]= \int_{-\infty}^{+\infty}x_3(t)\cdot e^{-j\omega t}dt= \int_{-\infty}^{+\infty}\bigg(1-\frac{t}{T}\bigg)\cdot rect\bigg(\frac{t-\frac{T}{2}}{T}\bigg)\cdot e^{-j\omega t}dt=\\ =\int_0^T\bigg(1-\frac{t}{T}\bigg)\cdot e^{-j\omega t}dt=\\= \int_0^Te^{-j\omega t}dt-\frac{1}{T}\int_0^T te^{-j\omega t}dt=\Bigg[\frac{e^{-j\omega t}}{-j\omega}\Bigg]_0^T-\frac{1}{T}\cdot\Bigg[\frac{e^{-j\omega t}(-j\omega t-1)}{(j\omega)^2}\Bigg]_0^T=\\= \frac{e^{-j\omega T}-1}{-j\omega}+\frac{1}{T\omega^2}\cdot\bigg[e^{-j\omega T}(-j\omega T-1)-1\bigg]=\dots=\\=-\frac{j\omega T+e^{-j\omega T}-1}{T\omega^2} $$

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Honestly, I don't see the point of this answer. First you recommend to use some 'smart' method, which you believe to be the computation of the convolution of the transforms of a triangle and a rectangle (try it for yourself if you have free weekend :), but then you - against your recommendation - use the definition of the F-transform, you give us 4 not so neat looking lines of math but no final result. So what did this answer add to the existing answer by @hryghr? – Matt L. Aug 16 '14 at 11:54
It adds a suggestion on how this kind of problems are usually tackled, I don't understand the use of "believe", do you believe it's different, I clearly state there's no smart method to do this one, I like math but that's subjective, and I couldn't find the sign error. Thanks for your feedback @MattL. – Vladimir Cravero Aug 16 '14 at 11:58
In this case it is objectively much easier to directly evaluate the integral from the definition of the Fourier transform (as you started to do towards the end of your answer). But this path was already recommended by hryghr's earlier answer. The first part of your answer is probably more confusing than helpful, because - as you mentioned - there is no smart way in this case. Furthermore, the convolution integral that you seem to suggest is very messy. As for the sign error: link Anyway, this is just my opinion. – Matt L. Aug 16 '14 at 12:09
Thanks, I fixed the sign error, now it looks good to me – Vladimir Cravero Aug 16 '14 at 12:35
About the first part, it seems to me that OP is just beginning to explore the Fourier transform, and knowing that basic trick is a must. I understand people can find it confusing, I'm open to suggestions to improve it. – Vladimir Cravero Aug 16 '14 at 12:37

You convolve two Rect() functions to get a triangle function. See this answer to get the derivation.

Now multiply the two sided ramp function with a rect function that extends from 0 to a positive direction.

enter image description here

A multiplication in the time domain is a convolution in the frequency domain.

And finally since the red rect is shifted in time you need to invoke the time shift theorem: \$ F_t[f(t-a)] = F(t) e^{-j2\pi fa} \$

\$ F_t\$ means Fourier transform.

You will be convolving a \$ sinc * e^{-j2\pi fa} \$ with a \$sinc^2\$ in the frequency domain.

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