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This is the solution to a fourier series problem, of the function $sin(\omega_0t)$:

enter image description here

I understand how the author has used Euler's formula to split this function into two exponential terms. However, I don't understand how the coefficients were obtained.

They have used a method called inspection. However, they have also stated we can use the formula: $$c_n=\frac{1}{T_0}\int^{T_0/2}_{-T_0/2} f(t)e^{-jn\omega_0t}\,dt$$

But what do we put in $T_0$ in this formula? And what is the method of inspection?

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without any other specification than that cited we assume that we are working over a period with $-\pi \le \omega_0t \le \pi$ so $T_0 = \frac{2\pi}{\omega_0}$

so we are looking for an expansion: $$ f(t) = \sum_{n=-\infty}^\infty c_n e^{ni\omega_0 t} $$

inspection generally means "just by looking at" - usually when we can directly guess the correct solution, perhaps with a quick bit of mental arithmetic. in applied situations this can save time by reducing formality, though there is always the danger of making an error through haste, or a false assumption etc.

however the fact that, knowing the expression given for the $\sin$ function, we immediately may write: $$ f(t) = \frac1{2j}\left( e^{j\omega_0 t} - e^{-j\omega_0 t}\right) \tag{1} $$ means inspection is quite reliable in the present case. the expression (1) is already the required expansion!

if you plug it into the integral as given you will find, from the orthogonality of the functions $e^{i\omega_0t}$ over the given range, that all coefficients except $c_{-1}$ and $c_1$ evaluate to zero.

for example: $$ c_1 = \frac{\omega_0}{2\pi}\int_{-\frac{\pi}{\omega_0}}^{\frac{\pi}{\omega_0}} \frac1{2j}\left( e^{j\omega_0 t} - e^{-j\omega_0 t}\right) e^{-j\omega_0 t}dt \\ =\frac{\omega_0}{4\pi j}\int_{-\frac{\pi}{\omega_0}}^{\frac{\pi}{\omega_0}} \left( 1- e^{-2j\omega_0 t}\right)dt \\ =\frac{\omega_0}{4\pi j} \left[ t +\frac1{2 j \omega_0} e^{-2j\omega_0 t}\right]_{-\frac{\pi}{\omega_0}}^{\frac{\pi}{\omega_0}} \\ = \frac1{2j} $$

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  • $\begingroup$ How does one know it is already the required expansion? $\endgroup$ – Hassaan Apr 20 '15 at 14:52
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    $\begingroup$ well if you don't know that, then at this stage of your learning process, inspection is not a suitable method for you in this problem, and you may proceed with the more detailed calculations as i have indicated. perhaps you could try to derive $c_{-1} $ yourself, which only requires very small adjustments to the working. good luck! $\endgroup$ – David Holden Apr 20 '15 at 15:08
  • $\begingroup$ I understood your working - thanks for being so specific! $\endgroup$ – Hassaan Apr 20 '15 at 15:13
  • $\begingroup$ glad to be of assistance! it helps me too, to go through the working, as these calculations, although not difficult, involve quite a bit of fiddly arithmetic - it is easy to make a minor error, and then have to spend time trying to find out what went wrong! ;-) $\endgroup$ – David Holden Apr 20 '15 at 15:15
  • $\begingroup$ btw, re the inspection thing wrt Fourier series, what really brought it home to me was when i wanted to expand $\sin x$ over $[-\pi,\pi]$. of course $\sin x$ is already its own Fourier expansion! however a Fourier expansion of $\sin \frac{x}2$ over the same range is a different proposition! (because this function, unlike $\sin x$ or $\sin 2x$, say, is not an element of the orthonormal basis for the Hilbert space $L^2(-\pi,\pi)$ $\endgroup$ – David Holden Apr 20 '15 at 15:20

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