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In the book Schaum's Outlines of Analog and Digital Communications solved problem 1.2, the author calculates the fourier coeffecient $C_0$ for the rectangular pulse train: enter image description here

where $a$ is assumed to be $(1/4)T$ as such:

enter image description here

Usually, I've seen around the web that we have to integrate over the whole period, which in this case would be $0$ to $T$. Why is this wrong and the author's approach correct?

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  • $\begingroup$ did you mean why the integration is taken on $ [-a, a]$ instead of $[0, T]$? $\endgroup$ Apr 20, 2015 at 2:33
  • $\begingroup$ yes, I meant that $\endgroup$
    – Hassaan
    Apr 20, 2015 at 2:34
  • $\begingroup$ because function is zero on both intervals $[-T/2, -a]$ and $[a, T/2]$. $\endgroup$ Apr 20, 2015 at 2:35
  • $\begingroup$ oh, how trivial - thanks! $\endgroup$
    – Hassaan
    Apr 20, 2015 at 2:36
  • $\begingroup$ You are welcome $\endgroup$ Apr 20, 2015 at 2:37

2 Answers 2

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Firstly, the function is zero on both intervals $[-T/2, -a]$ and $[a, T/2]$;

Also, Integral kernel of Fourier series, the $e^{inω}$, is periodic for each $n∈ \mathbb N$ that means integral can be taken on $[-T/2, T/2]$;

Since $C_0$ is a DC component which means $n$ = 0, whole above make the integration of $C_0$ stand on $[-a, a]$.

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I haven't contributed anything, just spelling it all out.

We have the assumption that $a=T/4$ which implies $T/2=2a$. So using the formular for the $0$'th coefficient, we get:

$$c_0=\frac{1}{T}\int_{0}^{T}s(t)\;\mathrm{d}t=\frac{1}{T}\int_{-T/2}^{T/2}s(t)\;\mathrm{d}t.$$

If we define $\Pi_k(t)$ to be $1$, if $|t|<k/2$ and $0$, if $|t|\geq k/2$, then we get that the signal $s(t)$ on the interval $[-T/2,T/2]=[-2a,2a]$ is just $s(t)=\Pi_{2a}(t)$.

$$c_0=\frac{1}{T}\int_{-T/2}^{T/2}\Pi_{2a}(t)\;\mathrm{d}t= \frac{1}{T}\int_{-2a}^{2a}\Pi_{2a}(t)\;\mathrm{d}t= \frac{1}{T}\left(\int_{-2a}^{-a}\Pi_{2a}(t)\;\mathrm{d}t+\int_{-a}^{a}\Pi_{2a}(t)\;\mathrm{d}t+\int_{a}^{2a}\Pi_{2a}(t)\;\mathrm{d}t\right)$$

now since $\Pi_{2a}(t)=0$ on $[-2a,-a]$ and $[a,2a]$ and $\Pi_{2a}(t)=1$ on $[-a,a]$, we get:

$$c_0=\frac{1}{T}\int_{-a}^{a}\mathrm{d}t=\frac{2a}{T}=\frac{2a}{4a}=\frac{1}{2}.$$

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  • $\begingroup$ Thank you for such a detailed explanation! $\endgroup$
    – Hassaan
    Apr 20, 2015 at 12:42
  • $\begingroup$ No problem. You're welcome $\endgroup$ Apr 20, 2015 at 19:58

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