I have a question regarding an alternating pulse train signal, $p(t)$, and a system which looks like this:

Alternating Pulse Train


Immediately, I'm supposed to be able to extrapolate that $$ P_t = P_1(t)-P_1(t-\Delta) $$

  • How do you obtain this?
  • How would you get the Fourier transform of this pulse train from there?
  • $\begingroup$ The first point you can obtain just by inspection (I'm assuming that $p_1(t)$ is the time-domain representation of the positive-going pulse train. As a hint on the second, take advantage of the linear property of the Fourier transform. $\endgroup$ – Jason R Dec 8 '13 at 13:35
  • $\begingroup$ This looks a lot like a homework problem, and if so, please add the homework or self-study tag. Hint: $\displaystyle P(t) = \sum_{n=-\infty}^\infty (-1)^n\delta(t-n\Delta)$ where $\delta(t)$ is the unit impulse (or delta function if that is what your instructor calls it, shame on him). $\endgroup$ – Dilip Sarwate Dec 8 '13 at 15:26
  • $\begingroup$ Still a bit confused - if p<sub>1</sub>(t) is the positive going pulse train, I'm just subtracting a negative pulse train delayed by Δ? $\endgroup$ – imonlysleeping Dec 8 '13 at 19:03
  • $\begingroup$ If $\displaystyle Q(t) = \sum_{n=-\infty}^\infty \delta(t-2n\Delta)$ is a periodic impulse train of period $2\Delta$, then $Q(t) - Q(t-\Delta)$ is a periodic impulse train of period $2\Delta$ in which the impulses alternate in polarity, that is, it is the $P(t)$ shown in the figure you have included. I have no idea what that $p_t = p_1(t) - p_1(t-\Delta)$ in your question means. $\endgroup$ – Dilip Sarwate Dec 8 '13 at 21:43

Since this is a periodic signal, we have to first represent the signal $x(t)$ in terms of its Fourier Series and then compute its Fourier transform

$$\text{Representing } x(t) \text{ in terms of its Fourier series }$$

Let us assume $\Delta=T$

enter image description here


$$x(t)=\sum_{n=-\infty}^\infty x_ne^{j2\pi \frac{n}{T_0}t}=\sum_{n=-\infty}^\infty x_ne^{j\pi nt}$$

$$x_n=\frac{1}{T_0}\int_{\alpha}^{\alpha +T_0} x(t)e^{-j2\pi \frac{n}{T_0}t}dt=\frac{1}{2T}\int_{0T^-}^{2T^-} x(t)e^{-j2\pi \frac{n}{2T}t}dt$$

$$\frac{1}{2T}\int_{0T^-}^{T^-} \delta (t)e^{-j\pi \frac{n}{T}t}dt+\frac{1}{2T}\int_{1T^-}^{2T^-} -\delta (t-T)e^{-j\pi \frac{n}{T}t}dt$$

$$=\frac{1}{2T}(1)-\frac{1}{2T}e^{-j\pi \frac{n}{T}T}=\frac{1}{2T}(1-e^{-j\pi n})=\frac{1}{2T}(1-(-1)^n)$$ $$x(t)=\sum_{n=-\infty}^\infty \frac{1}{2T}(1-(-1)^n)e^{j\pi\frac{n}{T}t} \tag 1$$

$$\text{Determing Fourier Transform of } x(t) \text{ from its Fourier series representation}$$

Let $n=2k+1$, then (1) becomes $$=> x(t)=\sum_{k=-\infty}^\infty \frac{1}{2T}(2)e^{j\frac{(2k+1)}{T}\pi t}$$

$$X(jw)=\int_{-\infty}^\infty x(t)e^{-jwt}dt$$ $$=\int_{-\infty}^\infty \sum_{k=-\infty}^\infty \frac{1}{2T}(2)e^{j\frac{(2k+1)}{T}\pi t}e^{-jwt}dt$$

$$\frac{1}{T}\sum_{k=-\infty}^\infty \int_{-\infty}^\infty [1.e^{j\frac{(2k+1)}{T}\pi t}]e^{-jwt}dt$$

$$\text{ Since } F[1]=2\pi\delta (w) \text{ and } F[e^{j\Omega_0t}]=2\pi \delta(\omega-\Omega_0) \text{ Therefore}$$

$$X(jw)=\frac{1}{T}\sum_{k=-\infty}^\infty 2\pi \delta(w-\frac{(2k+1)}{T}\pi)$$

$$=\frac{2\pi}{T}\sum_{k=-\infty}^\infty \delta(w-\frac{(2k+1)}{T}\pi) \tag 2$$

$$X(f)=\frac{2\pi}{T}\sum_{k=-\infty}^\infty \delta(2\pi f-\frac{2(2k+1)}{2T}\pi)$$

$$\text{Since } \delta(ax)=\frac{1}{|a|}\delta(x)$$

$$X(f)=\frac{2\pi}{|2\pi|T}\sum_{k=-\infty}^\infty \delta( f-\frac{(2k+1)}{2T})$$

$$=\frac{1}{T}\sum_{k=-\infty}^\infty \delta( f-\frac{(2k+1)}{2T})$$ Let $2k+1=m$ $$X(f)=\frac{1}{2T}\sum_{m=-\infty}^\infty (1-(-1)^m)\delta(f-\frac{m}{2T}) \tag 3$$

  • $\begingroup$ I think your variable of integration is wrong - should it not be performed over $t$, rather than $n$? I also found this really hard to follow due to your use of notation - I would expect to see $c_n$ or $X[k]$ to describe the Fourier coefficients. Using $x_n$ suggests some discrete time series to me. Do you agree? $\endgroup$ – Speedy Feb 23 '17 at 10:27
  • $\begingroup$ Yes true. I have corrected it now. I have used the conventions used in 'Communication Systems Engineering' by Proakis and salehi(chapter 2,Frequency Domain Analysis of Signals and Systems). Please note that $c_n$ is just the same as $x_n$, just the varibles are changed. $\endgroup$ – Soumee Feb 23 '17 at 13:35
  • $\begingroup$ Please note that while you are using $$x(t)=\sum_{-\infty}^\infty c_ne^{jnw_0t};c_n=\frac{1}{T_0}\int_{\alpha}^{\alpha +T_0} x(t)e^{-jnw_0t}dt;\text{ where }w_0=\frac{2\pi}{T_0};T_0\text{ being the time peroiod of the given periodic signal}$$ I am using $$x(t)=\sum_{n=-\infty}^\infty x_ne^{j2\pi \frac{n}{T_0}t};x_n=\frac{1}{T_0}\int_{\alpha}^{\alpha +T_0} x(t)e^{-j2\pi \frac{n}{T_0}t}dt;w_0=\frac{2\pi}{T_0}$$ Both are the same. Just use $c_n$ in place of $x_n$ and you are good to go. $\endgroup$ – Soumee Feb 23 '17 at 13:36
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    $\begingroup$ @Speedy I have tried to resolve your doubts. Check the comments above. Should you have any other doubts, do let me know. $\endgroup$ – Soumee Feb 23 '17 at 14:07

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