How was this equation simplified? (Fourier coefficients)

In summary: RightYou 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...cos(k 2\pi) = k*cosk2π + j*coskπ
  • #1
izelkay
115
3

Homework Statement


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

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

Homework Equations

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]
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
izelkay said:
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.
 
  • #3
DrClaude said:
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.
 
  • #4
What values can k take?
 
  • #5
DrClaude said:
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.
 
  • #6
izelkay said:
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?

izelkay said:
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.
 
  • #7
DrClaude said:
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?
 
  • #8
izelkay said:
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?
 
  • #9
DrClaude said:
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
 
  • #10
izelkay said:
x(t) = a summation from negative infinity to positive infinity of ak*exp(jkwot)
We're almost there: summation over what?

izelkay said:
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.
 
  • Like
Likes izelkay
  • #11
DrClaude said:
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
 
  • #12
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.
 
  • #13
DrClaude said:
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.
 
  • #14
izelkay said:
The term in the brackets becomes:

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

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

izelkay said:
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.
 
  • #15
DrClaude said:
You're missing a factor 1/2 here.RightYou 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
 
  • #16
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.
 
  • #17
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.
 
  • Like
Likes izelkay
  • #18
DrClaude said:
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!
 

What are Fourier coefficients?

Fourier coefficients are the coefficients that represent the contribution of each frequency component in a complex wave. They are used to simplify complex equations and analyze periodic functions.

How are Fourier coefficients calculated?

Fourier coefficients are calculated using a mathematical technique called Fourier analysis. This involves breaking down a complex wave into simpler components or frequencies and determining the contribution of each component to the overall wave.

Why are Fourier coefficients important?

Fourier coefficients are important because they allow us to understand and analyze complex waveforms. They are used in various fields such as signal processing, physics, and engineering to simplify equations and study periodic functions.

How does simplifying an equation using Fourier coefficients work?

Simplifying an equation using Fourier coefficients involves breaking down the equation into simpler components or frequencies. These components are then represented by the Fourier coefficients, making the equation easier to analyze and understand.

Can Fourier coefficients be used to solve real-world problems?

Yes, Fourier coefficients can be used to solve real-world problems. They are commonly used in fields such as engineering, physics, and mathematics to analyze and understand complex waveforms. They can also be used in signal processing to filter and extract useful information from signals.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
134
  • Calculus and Beyond Homework Help
Replies
5
Views
222
  • Calculus and Beyond Homework Help
Replies
3
Views
165
  • Calculus and Beyond Homework Help
Replies
6
Views
274
  • Calculus and Beyond Homework Help
Replies
3
Views
267
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
253
  • Calculus and Beyond Homework Help
Replies
1
Views
492
  • Calculus and Beyond Homework Help
Replies
4
Views
273
Back
Top