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I found an example at the following URL that asks for the Fourier Transform of $f(t) = 2t-3$ for 0 < t < 3/2, for all other values of t, f(t) = 0 : http://www.usciences.edu/~lvas/MathMethods/fourier.pdf

Their solution approximates this as an even function because it uses the cosine transform but it is obviously a piece wise function that looks like a triangle. When i approached this solution i did not think of doing this because it isnt even at first glance to me.

If they can just use the cosine transform then why couldnt they just use the sine transform since we have no other information as to what happens with x less than zero.

Why are they able to do this?

Thank you.

Dr. Dan
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2 Answers2

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Firstly, their solution does not approximate anything. As said in http://www.usciences.edu/~lvas/MathMethods/fourier.pdf, page 13, I quote from it " suppose that $f(t)$ is defined just for $t>0$. We can extend $f(t)$ so that it is even.

Please note the word "extend" (not approximate). Since $f(t)$ is defined for $t>0$, we extend to the negative side and think of it as an even function. Since $f(t)$ is even, then

$F(\omega)=\frac{1}{\sqrt{2 \pi}}\int_{-3/2}^{3/2} f(t) e^{-jwt}dt= 2\frac{1}{\sqrt{2 \pi}}\int_{0}^{3/2} f(t) cos(wt)dt$.

We could use sine transform if we extend the function $f(t)$ as an odd function. I hope you can try that yourself.

Oliver
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  • thanks @Oliver for the answer! Sorry i'm quite new to this subject so please forgive for any naive questions If I extend to the other side then should I not divide by two. The range for f(t) is given as 0<t<3/2 Meaning if $I = \int_{0}^{3/2} \frac{1}{\sqrt{2\pi}}f(t)e^{-j\omega t}dt = \frac{1}{2} \int_{-3/2}^{3/2} \frac{1}{\sqrt{2\pi}}f(t)e^{-j\omega t}dt$ – Dr. Dan Apr 18 '14 at 17:47
  • did you mean to type more after "Then,"? – Dr. Dan Apr 19 '14 at 14:40
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I appreciate your question and no need to say sorry. Fourier cosine transform (FCT) is basically the real part of the complex Fourier transform (CFT).

That is, FCT[$f(t)$]=$Re[\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{+\infty} f(t) e^{-jwt} dt]=\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{+\infty} f(t) \cos(wt) dt$. Please note that cosine transform is defined from $-\infty$ to $+\infty$.

If $f(t)$ is even, then the FCT $\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{+\infty} f(t) \cos(wt) dt=\sqrt{\frac{2}{\pi}} \int_{0}^{+\infty} f(t) \cos(wt) dt$.

The question in the file http://www.usciences.edu/~lvas/MathMethods/fourier.pdf, is to find the FCT. Thus, you can directly use $\sqrt{\frac{2}{\pi}} \int_{0}^{+\infty} f(t) \cos(wt) dt$ to find the FCT. I hope this helps you.

Oliver
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