Fourier Series Expansion

[soccer4life]

Homework Statement

There is a sawtooth function with u(t)=t-π.

Find the Fourier Series expansion in the form of
a₀ + ∑α_kcos(kt) + β_ksin(kt)

Homework Equations

a₀ = ...
α_k = ...
β_k = ...

The Attempt at a Solution

After solving for a₀, a_k, and b_k, I found that a₀=0, a_k=0, and b_k=-2/k

Therefore, my Fourier expansion was the summation from k=1 to infinity of -2*sin(kt)/k

I think this is right, but can someone cross-reference final solutions with me? If you need to see my work I can provide that also.

 

[BvU]

You can cross reference with this
some converting may be required
(I'm surprised you didn't already do that... -- the Googling I mean)

 

[soccer4life]

You can cross reference with this
some converting may be required
(I'm surprised you didn't already do that... -- the Googling I mean)

I'm a bit confused from those tables. I'm assuming that I am translating them incorrectly, but aren't they essentially stating that there must be an a₀ term by always including a single value in front of the summation? On another note, I got an answer of π/4+2Σsin((2k-1)t)/(2k-1) by going through the methods outlined in that link. The summation term is the opposite term of my solution, but the π/4 in front of the summation still confuses me. There is definitely not an a₀ term in this series. Here's a picture of my solution for β_k if it is any help. Thank you for the response!

 

[BvU]

Your sawtooth has zero average, so you did fine there. Amplitude is easy too.
Now how about the conversion of the period ? Shouldn't be too hard: they already show $\omega_0 = {2\pi\over T}$.

 

[soccer4life]

Your sawtooth has zero average, so you did fine there. Amplitude is easy too.
Now how about the conversion of the period ? Shouldn't be too hard: they already show $\omega_0 = {2\pi\over T}$.

The period is just 1. T=2π. That's how I performed all of my calculations, both in the table and by hand

 

[BvU]

No, your period is $2\pi$ That's why you integrate from 0 to $2\pi$. Your frequency is $1/T = 1/(2\pi)$ and your $\omega = 2\pi f = 1$ .
Apart from that, you're in good shape: I already see their $1/n$ in your $1/k$ !

What about the 2 ?

 

[soccer4life]

No, your period is $2\pi$ That's why you integrate from 0 to $2\pi$. Your frequency is $1/T = 1/(2\pi)$ and your $\omega = 2\pi f = 1$ .
Apart from that, you're in good shape: I already see their $1/n$ in your $1/k$ !

What about the 2 ?

Ah sorry I had my terminology mixed up (it's been a rough week). Can you clarify which 2 you are referring to? Also, just to make sure I understand where you're leading me, is the issue that I am translating the table incorrectly? Also, it it fair to assume that I have a mistake somewhere in my current answer?

 

[BvU]

You're being too modest. I think you did just fine and I want to help you so you can do the conversion between the table and your result all by yourself -- and that way you don't need to ask PF .

If we shift the table graph down by $A\over 2$ we see that your $b_0 = 0$ is correct and we see that the $A$ in the table is $2\pi$ in your case. Meaning the factor 2 is in agreement.

Now we have me puzzled about the minus sign. Took me a while to find out YOU are correct and the table I quoted is wrong !

A lot of texts (such as p 6 here) integrate from $-\pi$ to $+\pi$ -- that is a horizontal shift which introduces a $(-1)^k$ and I even needed an excel plot to convince myself that in your case your result is really correct...

(this was only for the sign, I didn't look at A anymore)

 

[soccer4life]

Wow, you've gone above and beyond for me. Thank you! However, this entire question all leads into the overarching problem I'm having... which is finding the steady state output for this input. I'm going to go ahead and post the issues I'm having here, but if I need to start a new thread let me know and I can do so.
1. Problem Statement
Find the steady state output y_ss(t) for the input u(t)=t-π in terms of an infinite sum of sinusoids.
We are given the transfer function as:

2. Homework Equations
G(i) = ...
|G(ik)| = ...
Φ(ik) = ... (this is the angle)
y_ss(t) = β_k||G(ik)|e^i(kt+Φ(ik)) ***check that this is the correct formula please***

3. Attempt at Solution
I've found the following:
G(i)=1
|G(ik)| =

(Any tips/tricks on how to input fractions/square roots into PF would be greatly appreciated...)
Φ(ik) =

I know that these values are right. However, I don't fully understand how to incorporate them into the steady state formula (assuming that my formula is correct)

 

[Mark44]

I'm going to go ahead and post the issues I'm having here, but if I need to start a new thread let me know and I can do so.

Yes, please start a new thread since this is a completely different problem.