Fourier Transform of Alternating Periodic Rectangular Pulse

Question

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

Answer 1

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 )

Answer 2

The answer is simple.
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.