# Fourier Transform of Alternating Periodic Rectangular Pulse

I'm having trouble determining Fourier transform of signal. I have 2 ideas on how to solve this problem. Given the signal is periodic I could use formula for Fourier transform of periodic signals:

$$X(j\omega) = \sum\limits_{k=-\infty}^{\infty} C_k \cdot 2\pi \delta(\omega-k\omega_{0})$$ where $$\omega_0 = \frac{2 \pi}{4T}$$.

Also, I could make signal $$g(t)$$ non periodic rectangular signal, and $$h(t)$$ alternating pulse train, and then $$g(t)*h(t)$$ should give me signal I needed. Although I have some ideas I'm stuck solving this problem. Please help. Thanks

• Hi! Welcome. Not quite sure what an "alternating periodic rectangular pulse" is, specifically. Can you add a formula, and/or a drawing? – Marcus Müller Mar 22 '19 at 20:32
• Hey, i had some trouble posting photo, but I managed to do it. Take a look please – Aleksandar Simonović Mar 22 '19 at 20:34
• appears to me that you want the Fourier Series. (the Fourier Transform will be a collection of impulses in the frequency domain.) – robert bristow-johnson Mar 22 '19 at 20:52
• I need spectrum of signal x(t) showed on picture above. So, I'm looking for analytical form of Fourier transform of signal x(t). – Aleksandar Simonović Mar 22 '19 at 20:56
• what you need to know are the Fourier series coefficients, $C_k$. from those you have your analytic expression. – robert bristow-johnson Mar 22 '19 at 21:13

Lemma:

for $$x_{1}(t)$$ fourier coefficient is given by $$C_{n_{1}}$$

$$C_{n_{1}}=\dfrac{\text{amplitude}\times \text{ON duration}}{\text{Time-period}}\times Sa\left(n\ .\omega_{0}. \frac{\text{ON duration}}{2}\right)=\dfrac{\tau}{T_{0}}\times Sa\left(n\ .\omega_{0}. \frac{{\tau}}{2}\right)=\dfrac{\sin\left(\dfrac{\pi n \tau}{T_{0}}\right)}{n\pi}$$

where

$$Sa(\lambda x)=\dfrac{\sin \lambda x}{\lambda x}$$

for your question : $$\tau=T ; T_{0}=4T$$

also,

fourier coefficient of $$x(t)$$ is $$C_{n}=C_{n_{1}}(1-e^{-jn\pi})=C_{n_{1}}[1-(-1)^n]$$ (by use of linearity +time shifting as stated in @ royi's answer)

$$C_{n}=\dfrac{\sin\left(\dfrac{\pi n }{4}\right)}{n\pi}[1-(-1)^n]$$

and now you can find fourier transform by using formula of periodic function's F.T. i.e., $$X(j\omega)$$

so, your first method is correct . but your second method(using convolution) is awesome ( i haven't checked it yet though )

I will give 3 points to solve it:

1. The Fourier transform is linear. Hence $$\mathcal{ F } \left\{ \alpha f \left( x \right) + \beta g \left( x \right) \right\} = \alpha \mathcal{ F } \left\{ f \left( x \right) \right\} + \beta \mathcal{ F } \left\{ g \left( x \right) \right\}$$.
2. Shift in time $$f \left( x - {x}_{0} \right)$$ equals multiplication by $${e}^{-j \omega {x}_{0}}$$ in Fourier domain.
3. Instead of solving for the case above, think of the case you have 2 rectangular signals with twice the period with one multiplied by $$-1$$ and shifted.

In your case, just have a look on the signal of the Positive Pulses. It has a period of $$4T$$.
You have another signal. The signal of the negative pulses.
It is basically the same signal as the positive one (It also has a period of $$4T$$) yet it is shifted by $$2T$$ and as it is multiplied by $$-1$$.
So if you know the transform of the positive one, follow my above points and you have the transform of the negative one and their sum.

• As I know, shift in time equals multiplication by exponential function e^(jwto) – Aleksandar Simonović Mar 22 '19 at 20:57
• You're correct and I clarified my bad choice of words. Thank You. – Royi Mar 22 '19 at 21:26
• Still not sure how would you solve this problem. If you could work it out and take a photo it would mean a lot. I'm new to this field so I'm finding these problems difficult – Aleksandar Simonović Mar 22 '19 at 21:34
• Do you know the Fourier Transform of a rectangular pulse with a period of ${T}_{0}$? – Royi Mar 23 '19 at 10:25
• Of course, Rectangular pulse with period $T$ and amplitude $A$ is $A$$T$$sinc(\frac{wT}{2})$ – Aleksandar Simonović Mar 23 '19 at 10:28