# How was this equation simplified? (Fourier coefficients)

## Homework Statement

http://puu.sh/gGhdb.jpg [Broken]

Solution:[/B]
http://puu.sh/gGh3E.jpg [Broken]

## The Attempt at a Solution

How did they get that solution for the Fourier coefficient? When I evaluate the integral I can only seem to get it to:

(1/-jk2π)[2*exp(-jkπt)-exp(-jk2πt)-1]

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## Answers and Replies

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DrClaude
Mentor
When I evaluate the integral I can only seem to get it to:
(1/-jk2π)[2*exp(-jkπt)-exp(-jk2πt)-1]
It's a definite integral. You can't have a result that is a function of t.

It's a definite integral. You can't have a result that is a function of t.
True, my mistake. Take out the t's and that's what I have. I did the integration right I think (plugged in for t) but forgot to remove the t also:

(1/-jk2π)[2*exp(-jkπ)-exp(-jk2π)-1]

Still can't seem to get it in the form they have, which is necessary for the next part of the problem.

DrClaude
Mentor
What values can k take?

What values can k take?
Sorry I'm not sure what you mean by that, but I'm looking at a table of properties for coefficients and it says if x(t) is real and odd, then ak should be jIm(ak). Using that with Euler's equation, I get:

(1/kπ)[cos(kπ)-(1/2)cos(k2π)-(1/2)], which doesn't seem anywhere near right.

DrClaude
Mentor
Sorry I'm not sure what you mean by that, but I'm looking at a table of properties for coefficients and it says if x(t) is real and odd, then ak should be jIm(ak).
My question goes back to the definition of Fourier series. What does the ##k## in ##a_k## stand for? What can be its value?

Using that with Euler's equation, I get:

(1/kπ)[cos(kπ)-(1/2)cos(k2π)-(1/2)], which doesn't seem anywhere near right.
The answer in the solution is in terms of an exponential, so don't use Euler's equation.

My question goes back to the definition of Fourier series. What does the ##k## in ##a_k## stand for? What can be its value?

The answer in the solution is in terms of an exponential, so don't use Euler's equation.
The k's are the harmonics, and they can be from 0 to 8 because of the function's range?

DrClaude
Mentor
The k's are the harmonics, and they can be from 0 to 8 because of the function's range?
We need to go back to basics. Whats is the general for ##x(t)## written as a Fourier series?

We need to go back to basics. Whats is the general for ##x(t)## written as a Fourier series?
x(t) = a summation from negative infinity to positive infinity of ak*exp(jkwot)

sorry about my ignorance on this subject, it's really difficult for me to grasp this stuff just from reading my textbook

DrClaude
Mentor
x(t) = a summation from negative infinity to positive infinity of ak*exp(jkwot)
We're almost there: summation over what?

sorry about my ignorance on this subject, it's really difficult for me to grasp this stuff just from reading my textbook
Don't worry, and don't think I am trying to give you a hard time. I simply think you'll learn more if you find it by yourself with a bit of guidance.

izelkay
We're almost there: summation over what?

Don't worry, and don't think I am trying to give you a hard time. I simply think you'll learn more if you find it by yourself with a bit of guidance.
Summation over k, the harmonic components

DrClaude
Mentor
Ok, so you have
$$\frac{1}{-jk2\pi} \left[ 2 \exp(-jk\pi) - \exp(-jk2\pi)-1 \right]$$
where ##k## can take any integer value. Use this fact to simplify the equation.

Ok, so you have
$$\frac{1}{-jk2\pi} \left[ 2 \exp(-jk\pi) - \exp(-jk2\pi)-1 \right]$$
where ##k## can take any integer value. Use this fact to simplify the equation.
This is where I thought it'd be helpful to expand it with Euler's equation. The term in the brackets becomes:

2coskπ - j2sinkπ - cosk2π + jsink2π - 1

If k can be any integer value, all the sines should become 0 right? so i'm left with:

2coskπ - cosk2π - 1

in the brackets.

But like you said before the simplified equation is in exponential terms so what I have doesn't help. I don't know how to simplify the equation as it is.

DrClaude
Mentor
The term in the brackets becomes:

2coskπ - j2sinkπ - cosk2π + jsink2π - 1
You're missing a factor 1/2 here.

If k can be any integer value, all the sines should become 0 right?
Right

so i'm left with:

2coskπ - cosk2π - 1

in the brackets.

But like you said before the simplified equation is in exponential terms so what I have doesn't help.
You can use the same kind of identities directly in the exponential. In any case, you should be able to simplify ##\cos(k 2\pi)## further. You can at the end convert the cosines back into exponentials.

You're missing a factor 1/2 here.

Right

You can use the same kind of identities directly in the exponential. In any case, you should be able to simplify ##\cos(k 2\pi)## further. You can at the end convert the cosines back into exponentials.
Okay so distributing that (1/2) from the term outside the brackets, I have:

cos(kπ) - (1/2)cos(k2π) - (1/2)

Then using the identity cos(2x) = 1 - 2sin²(2x), I have:

cos(kπ) - 1, since the sine term becomes 0

Using Euler's equation, I then have:

(1/2)exp(jkπ) + (1/2)exp(-jkπ) - 1

Bump.

That's still not where it should be though.
Right now I have:

(1/jπk)[(1/2)exp(jkπ) + (1/2)exp(-jkπ) - 1]

but I'm not sure how to simplify it further.

DrClaude
Mentor
After looking in more details at your solution, it appears that you can't simply convert back to an exponential using Euler formula, even though your solution is formally equivalent to the one in the book. To obtain the one in the book, when you have
$$\frac{1}{-jk2\pi} \left[ 2 \exp(-jk\pi) - \exp(-jk2\pi)-1 \right]$$
you take that ##\exp(-jk2\pi) = 1 \forall k##, and then you get
$$\frac{1}{jk\pi} \left[ 1 - \exp(-jk\pi) \right]$$
This is formally equivalent to your answer,
$$\frac{1}{jk\pi} \left[ 1 - \cos(k\pi) \right]$$
because both ##\exp(-jk\pi)## and ##\cos(k\pi)## equal 1 when ##k## is even and -1 when ##k## is odd.

izelkay
After looking in more details at your solution, it appears that you can't simply convert back to an exponential using Euler formula, even though your solution is formally equivalent to the one in the book. To obtain the one in the book, when you have
$$\frac{1}{-jk2\pi} \left[ 2 \exp(-jk\pi) - \exp(-jk2\pi)-1 \right]$$
you take that ##\exp(-jk2\pi) = 1 \forall k##, and then you get
$$\frac{1}{jk\pi} \left[ 1 - \exp(-jk\pi) \right]$$
Okay this right here makes sense. Thanks!