- #1
roam
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Homework Statement
What is the Fourier transform of the function graphed below?
According to some textbooks the Fourier transform for this function must be:
$$ab \left( \frac{sin(\omega b/2)}{\omega b /2} \right)^2$$
Homework Equations
The Attempt at a Solution
I believe this triangular pulse is given by:
$$x(t)= \left\{\begin{matrix} a- \frac{|t|}{b} \ \ \ if |t|<b \\ 0 \ \ \ if |t|>b \end{matrix}\right.$$
So we need to find the sum of the two integrals:
$$x_1(\omega)= \int^0_{-b} \left(a+ \frac{|t|}{b} \right) e^{-j \omega t} dt$$
$$x_2(\omega)= \int^b_0 \left(a- \frac{|t|}{b} \right) e^{-j \omega t} dt$$
So using integration by parts we evaluate the two integrals:
##u=a + \frac{t}{b}##, ##\frac{du}{dt}=\frac{1}{b}##, ##dv=e^{-j\omega t}##, ##v=\frac{j}{\omega} e^{-j\omega t}##
$$x_1 (\omega)=[\frac{j}{\omega} (a+ \frac{t}{b} e^{-j\omega t}) ]^0_{-b} - \int^0_{-b} \frac{j}{\omega b} e^{-j \omega t}$$
##\therefore x_1(\omega)= \frac{ja}{\omega}-\frac{ja}{\omega}+\frac{j}{\omega}e^{+j\omega b} + \frac{1}{\omega^2b} - \frac{1}{\omega^2 b} e^{+j \omega b}## (1)
Second integral:
$$x_2 (\omega)=[\frac{j}{\omega} (a- \frac{t}{b}) e^{-j \omega t} ]^b_0 - \int^b_0 (-\frac{1}{b}) . \frac{j}{\omega} e^{-j \omega t}$$
##\therefore \ x_2(\omega)= \frac{ja}{\omega} e^{-j\omega b} - \frac{j}{\omega} e^{-j \omega b} - \frac{j}{\omega} a - \frac{1}{\omega^2 b} e^{-j \omega b} + \frac{1}{\omega^2 b}## (2)
Combining (1) and (2):
$$x(\omega)=\frac{-ja}{\omega}+\frac{2}{\omega^2 b} + \frac{j}{\omega} (e^{+j \omega b} - e^{-j\omega b}) - \frac{1}{\omega^2 b} (e^{+j \omega b} - e^{-j\omega b}) + \frac{ja}{\omega} e^{-j\omega b}$$
Now writing this in terms of sines and cosines using Euler's formula:
$$x(\omega)=\frac{-ja}{\omega}+\frac{2}{\omega^2 b} + \frac{j}{\omega} (2j sin (\omega b)) - \frac{1}{\omega^2 b} (2 j sin (\omega b)) + \frac{ja}{\omega} (cos (- \omega b)+ j sin (- \omega b))$$
$$x(\omega)=\frac{-ja}{\omega}+\frac{2}{\omega^2 b} + \frac{ja}{\omega} cos (\omega b) + \frac{a}{\omega} sin (\omega b) - \frac{2}{\omega} sin (\omega b) - \frac{2j}{\omega^2 b} sin (\omega b)$$
I'm stuck here. So how can I get from here to ##ab (\frac{sin(\omega b/2)}{\omega b /2})^2##?
Where did I go wrong?
Any help would be greatly appreciated.