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enter image description hereI'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

My try My other try

Try

Aleksandar Simonović
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  • 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
  • True, but I'm having hard time determining them. If you could work the math out and post photo i would appriciate – Aleksandar Simonović Mar 22 '19 at 21:16
  • again, you should edit your question to express $x(t)$ as a Fourier series with complex coefficients $C_k$. then use the standard result from Fourier series that gives you the coefficients. – robert bristow-johnson Mar 22 '19 at 21:18
  • I got this as a part of an exam task, and it said using Fourier transform determine spectrum of a signal. – Aleksandar Simonović Mar 22 '19 at 21:29
  • but it's a periodic signal, with finite power and infinite energy. the Fourier Transform is essentially meant to be applied to non-periodic signals with finite energy. but with a mathematical slight-of-hand, the Fourier Transform applied to a sinusoid can meaningfully be shown to be a pair of dirac delta functions in the frequency domian. a general periodic function would have more dirac deltas *as you have indicated in your question*. ALL you need to do is determine the Fourier series coefficients $C_k$ and you have your answer. – robert bristow-johnson Mar 22 '19 at 21:46
  • Ok, I will post my results in few minutes. Please take a look and help me understand – Aleksandar Simonović Mar 22 '19 at 21:49
  • just to be clear (i don't wanna add this to your question), the Fourier Transform (using angular frequency, $\omega$):

    $$ X(j\omega) \triangleq \mathscr{F}\big{ x(t) \big} = \int\limits_{-\infty}^{\infty} x(t) e^{-j \omega t} \mathrm{d}t$$

    and the Fourier series

    $$ x(t) = \sum\limits_{k=-\infty}^{\infty} C_k e^{j k \omega_0 t} $$

    where

    $$ C_k = \frac{\omega_0}{2 \pi} \int\limits_{t_0}^{t_0+2 \pi/\omega_0} x(t) e^{-j k \omega_0 t} \mathrm{d}t $$ for any real $t_0$.

     – robert bristow-johnson Mar 22 '19 at 21:57
  • I posted my try, take a look – Aleksandar Simonović Mar 22 '19 at 22:12
  • So 2 of my methods give kinda different result in terms of sinc function. In first method, there is sinc(k*pi/4) while in other there is sinc(wT/2) – Aleksandar Simonović Mar 22 '19 at 22:36
  • remember $$\omega_0=\frac{2\pi}{4T}$$ your period of repetition is $4T$. (actually, it looks like you're doing that correctly.) and in your Fourier series summation, you are including the even $k$ terms even though we know they are zero (your computation of $C_k$ was correct). so, to get just the odds, substitute $k=2n+1$ and sum over all integer $n$. that will make sure you're only adding up the odd terms. – robert bristow-johnson Mar 23 '19 at 02:50
  • Can you take a look at all my solutions and explain why they don't match up – Aleksandar Simonović Mar 25 '19 at 16:38

2 Answers2

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Lemma:

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

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

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 )

Community
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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.

Royi
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