- #1

seanbow

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Say you have the coefficients [itex]a_k[/itex] of a Fourier series representation of some function [itex]x(t)[/itex]. You can easily then give [itex]x(t)[/itex] as

$$x(t) = \sum_{k = -\infty}^{\infty} a_k e^{i k \omega_0 t}$$

But this doesn't do much good in telling you what the actual function looks like. For example, if we have

$$a_k = \frac{ \sin ( k \pi / 2)} {k \pi}$$

we can write [itex] x(t) [/itex] as

[itex]x(t) = \sum_{k \neq 0} \frac{ \sin (k \pi / 2)} {k \pi} e^{i k \omega_0 t}[/itex]

but you would have a hard time telling that this is a square wave with a duty cycle of 50% unless you've previously derived the series coefficients for that exact function.

Basically, my question is: is there a way to derive a more intuitive representation of a function given its Fourier series representation?

$$x(t) = \sum_{k = -\infty}^{\infty} a_k e^{i k \omega_0 t}$$

But this doesn't do much good in telling you what the actual function looks like. For example, if we have

$$a_k = \frac{ \sin ( k \pi / 2)} {k \pi}$$

we can write [itex] x(t) [/itex] as

[itex]x(t) = \sum_{k \neq 0} \frac{ \sin (k \pi / 2)} {k \pi} e^{i k \omega_0 t}[/itex]

but you would have a hard time telling that this is a square wave with a duty cycle of 50% unless you've previously derived the series coefficients for that exact function.

Basically, my question is: is there a way to derive a more intuitive representation of a function given its Fourier series representation?

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