# Fourier transform of given signal This is the signal whose FT i need to find, at first i thought that i could solve this as a convolution of two rectangular pulses, but i could not find pulses that fit into this (it turns out that parts from -T/2 to 0 and T to 3T/2 are always somehow narrow). I could also try by definition but that's just too much work and very time consuming, the third option is derivative theorem, but i don't know how to use it, since i never actually used it. Any help appreciated!

• Maybe that is not "one" signal. Maybe it can be decomposed into elementary signals whose FT are known... – A_A Feb 7 '19 at 11:48
• @A_A as i said in my question, i believe it is a convoultion of two rectangular pulses, but i am having hard time finding them – cdummie Feb 7 '19 at 11:57
• Maybe that is not "one" signal. Maybe it can be decomposed into elementary signals whose FT are known... – A_A Feb 7 '19 at 12:34

HINT:

It can be solved by convolution. Take two rectangles, one that is non-zero in $$[-T/2,T]$$ and the other being non-zero in $$[0,T/2]$$. Now you just have to choose the amplitudes right and the convolution of the two signals will look like the one in your question.

Hint:

If you can get away from your insistence of finding the FT via expressing the waveform as the convolution of two rectangular pulses, consider that the waveform can be expressed as a triangular waveform (base $$\left[-\frac T2, \frac{3T}{2} \right]$$ and peak $$2A$$) minus another triangular waveform (base $$[0, T]$$ and peak $$A$$). Both of these have FTs that you can look up in a table (or easily derive a tabulated FT such as that of a triangular waveform of base $$\left[-\frac T2, \frac T2\right]$$ and peak $$1$$) and you are done.

Or, each of the two triangular waveforms that I have described above can be more easily expressed as the convolution of rectangular pulses of equal width (different pulse widths for the two triangles, of course) and so the FTs of each triangular waveform is a $$\operatorname{sinc}^2$$ function. Now, whether your homework grader will accept a difference of $$\operatorname{sinc}^2$$ functions as a correct answer is a different matter.

• What is wrong with a difference of sinc^2? As a grader, you have to live with the existence of equivalent answers. – MSalters Feb 7 '19 at 15:07
• @DilipSarwate This is really neat way to find FT for this signal, i must say, it's great idea, however, when i was solving this by using convolution theorem i got product of two $$sinc^2$$ functions, i suppose that it could be equivalent to the difference of $$sinc^2$$ found this way. – cdummie Feb 7 '19 at 15:15

You could see it as an overlapping of $$3$$ Triangular functions, if the Triangular function is defined as follows:

$$\Delta(t) = \begin{cases} 1 - \frac{2}{T}|t| \qquad & |t| < \frac{T}{2} \\ 0 \qquad & \mathrm{otherwise} \\ \end{cases}$$

Notice that we have one grid period equal to $$\frac{T}{2}$$. If we look closely at the figure, we can see that we have two ramp-up's (each on the edge), where each corresponds to the left/right parts of $$\Delta(t)$$. Now, another important thing to notice is the plateau region, which contains two grid periods. Each grid period could be see as two overlapping triangles. With that being said, we can model your function as $$x(t) =A \Big(\Delta(t) + \Delta(t- \frac{T}{2} ) + \Delta(t- T ) \Big)$$

Making use of the linearity property and the the fact that a time shift translates to a phase shift in frequency domain, we get

$$X(w) = A \big(1 + e^{-jw \frac{T}{2}} + e^{-jw T} \big)\Delta(w )$$ where $$\Delta(w) = \mathcal{F} \Big(\Delta(t) \Big)$$, which is computed using the definition,

$$X(w) = \int_{-\infty}^{\infty} x(t) e^{-j w t} \ dt = A\int_{-\frac{T}{2}}^{0} (1 + \frac{2}{T} t) e^{-j w t} \ dt + A\int_{0}^{\frac{T}{2}} (1 - \frac{2}{T} t) e^{-j w t} \ dt$$ Using $$\int e^{-jwt} = \frac{j}{w} e^{-jwt}$$ and $$\int t e^{-jwt} = \frac{j}{w} te^{-jwt} - \frac{1}{w^2}e^{-jwt}$$, we get $$\begin{equation} \begin{split} X(w) &= A\Big[ \frac{j}{w} e^{-jwt} \Big]_{-\frac{T}{2}}^{0} + \frac{2A}{T}\Big[ \frac{j}{w} te^{-jwt} - \frac{1}{w^2}e^{-jwt} \Big]_{-\frac{T}{2}}^{0} \\ &+ A\Big[ \frac{j}{w} e^{-jwt} \Big]_{0}^{\frac{T}{2}} - \frac{2A}{T}\Big[ \frac{j}{w} te^{-jwt} - \frac{1}{w^2}e^{-jwt} \Big]_{0}^{\frac{T}{2}} \end{split} \end{equation}$$ Replacing and a bit of arrangements will give you $$\begin{equation} \begin{split} X(w) &= \frac{4A}{w^2 T}\Big( \cos(\frac{T}{2} w) - 1 \Big) \end{split} \end{equation}$$