# Fourier Series calculations

Discussion in 'Homework Help' started by jstrike21, Feb 1, 2010.

1. ### jstrike21 Thread Starter Member

Sep 24, 2009
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x(t)= 6+3cos(4t)+2sin(5t)
i have worked out that this equals
6+3/2e^(j4t)+3/2e^-(j4t)+e^(-j*pi/2)e^(j5t)-e^(-j*pi/2)e^(j5t)

Is this right so far?
I have attached the actual assignment for reference.

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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The first part is asking for several Fourier coefficients.

You need to use the relationship for

$a_{k}$

where

$a_{k}=\frac{1}{T}\int_{0}^{T}{f(t)e^{-jk\omega_{0}t}}dt$

to obtain the required values.

If you've used the alternate (non-complex) form to find Fourier coefficients in the past, you may be aware of the "orthogonality" properties of sinusoidal waves which help to simplify the determination of individual coefficients.

Such as

$\int_{0}^{2\pi}{cos{mt} sin{nt}dt}=0$

$\int_{0}^{2\pi}{cos{mt} cos{nt}dt}=0$

$\int_{0}^{2\pi}{sin{mt} sin{nt}dt}=0$

for n≠m

3. ### jstrike21 Thread Starter Member

Sep 24, 2009
104
0
That integral works for A0 but we arent supposed to use calculators and solving the integrals when k doesnt equal 0 is pretty taxing without one. Isn't there an easier way like through using fourier series graphs?

Mar 8, 2009
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5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
That's true and the reason why one might look for short cuts. Integration can be laborious and error prone.

One short cut worth reflecting upon ...

We know that the function is periodic over the integration interval T=2∏ (pity they didn't give us a nice Pi symbol in the RH box!).

So we would expect that the only frequency components in the series would be those with ω= 4 or 5 {x(t)= 6+3cos(4t)+2sin(5t)}.

The a0 term is readily found as you say - even by inspection! So you may reasonably assume that a1, a2, a3, a6 etc are (i.e. k ≠ 0, 4 or 5) zero and only a0, a4 and a5 are non-zero. It should be fairly obvious that a4=3 and a5=2. But I guess you may be required to show that by long hand integration - depends what reasoning the examiner would accept as sufficient in having demonstrated your appreciation of the problem's requirements.

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Sorry - I forgot it's the complex form. Since it's the complex form, the ak's (positive and negative parts) will not be the same as the real component frequency magnitudes. I was thinking non-complex form solution.

But you can still use the knowledge of the component amplitudes to extract the complex terms for the ak's. And hence avoid the integration if allowed.

Last edited: Feb 1, 2010
7. ### GirishC Active Member

Jan 23, 2009
58
0
you can Euler's identity and convert cosine and sine terms in the exponential form. You do not need to go through labor of Fourier conversion.

With this you will get positive and negative frequency component and if there is no such frequency component available implies signal isn't there.

8. ### jstrike21 Thread Starter Member

Sep 24, 2009
104
0
Can someone give me some quick help on number three on the attached pdf?
or help me start the problem I kkeep running into walls

9. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
Is this the problem you mean?

I would start by putting everything over a common denominator - such as in the second equation line below ...

Then apply Euler's Identities to the complex parts and so on ....

$f=\frac{1}{2\pi}\frac{1-e^{j(1-2k)\pi}}{(1-2k)}+\frac{1}{2\pi}\frac{1-e^{-j(1+2k)\pi}}{(1+2k)}$

$f=\frac{(1+2k)(1-e^{j(1-2k)\pi})+(1-2k)(1-e^{-j(1+2k)\pi})}{2\pi(1-2k)(1+2k)}$