Fourier Series - Asymmetric Square Wave

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DmytriE
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Good morning everyone,

I am taking a signals and systems course where we are now studying the Fourier series. I understand that this is for signals that are periodic. But I get hung up when determining the Fourier coefficients. In the video by Alan Oppenheim, he derives the equation for the Fourier series. Below is the analysis equation.

[itex]a_{k} = \frac{1}{T_{0}} ∫ x(t)*e^{jk\omega_{0}t}[/itex]

He goes through an example using an asymmetric square wave with an amplitude of 1. I understand the bounds that he chooses [itex](-T_0/2, 0)[/itex] and [itex](0, T_0/2)[/itex].

This leads to a Fourier coefficient equation of the following:
[itex]a_{k} = \frac{1}{T_{0}} [∫ (-1)*e^{jk\omega_{0}t} + ∫ (1)*e^{jk\omega_{0}t}][/itex]

To compute the general equation for [itex]a_k[/itex] should I treat the ±1 as a function and use u-substitution / integration by parts? If so, can someone at least show the first step or 2? I haven't done I.B.P. or U-substitution in some time. If I can treat the x(t) as a constant then how can I integrate to get the answer below?

General equation: [itex]a_k = \frac{1}{jkπ} (1 - (-1^{k}))[/itex]

This example is from M.I.T. Open Courseware Alan V. Oppenheim Signals and Systems course Lecture 7 approximately 20 minutes into the video.
 
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DmytriE said:
This leads to a Fourier coefficient equation of the following:
[itex]a_{k} = \frac{1}{T_{0}} [∫ (-1)*e^{jk\omega_{0}t} + ∫ (1)*e^{jk\omega_{0}t}][/itex]
To compute the general equation for [itex]a_k[/itex] should I treat the ±1 as a function and use u-substitution / integration by parts?[/itex]
You don't need to do all that. "1" and "-1" are just constants. If you are studying this, you should know ##\int e^{at}dt##, and that formula works when ##a## is a complex number as well as when ##a## is real.

You should also know ##e^{j\omega t} = \cos \omega t + j \sin \omega t##.

You also have an equation connecting ##T_0## and ##\omega_0##, which is why they both disappeared in the final equation for ##a_k## (and that's also where the ##\pi## came from).

If you are confused by the ##(-1)^k## part, just work out the first few values of ##a_1##, ##a_2##, etc and see what happens.
 
AlephZero said:
If you are confused by the ##(-1)^k## part, just work out the first few values of ##a_1##, ##a_2##, etc and see what happens.

This is precisely what I am confused about. Is there a rule that applies to the [itex](-1)^k[/itex] or do I have to input a couple numbers for [itex]a_k[/itex]?