# problem of fourier series

Discussion in 'Homework Help' started by bhuvanesh, Mar 6, 2015.

1. ### bhuvanesh Thread Starter Member

Aug 10, 2013
268
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Choose the function f(t) ; −∞ < < +∞, for which a Fourier series cannot be defined.
(a) 3 sin(25t) (b) 4 cos(20t + 3) + 2 sin (10t)
(c) e-|t| sin(25t) (d) 1

fourier series are available for all periodic signal .its obvious that all the four are periodic.But somewhere i hear that fourier series is not possible for constant .could someone give me hint for the answer ,so try to come up. Thank you

Last edited: Mar 6, 2015
2. ### MrAl Well-Known Member

Jun 17, 2014
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Hi,

How is '1' periodic?
That's example (d).

3. ### bhuvanesh Thread Starter Member

Aug 10, 2013
268
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yes 1 is periodic with period infinity (or we can make period as we need and its extending from minus infinity to plus infinity )

4. ### WBahn Moderator

Mar 31, 2012
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What are the requirements for the Fourier series to exist?

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5. ### bhuvanesh Thread Starter Member

Aug 10, 2013
268
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it should be periodic ,
it should finite discontinuous over one period
it should absolutely integrable over any period
it should not have no more that finite number of maxima and minima over single period
thats all i think

6. ### anhnha Active Member

Apr 19, 2012
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7. ### WBahn Moderator

Mar 31, 2012
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Where is this somewhere that you heard this?

Take the basic equation for x(t) in terms of it's Fourier series components and set all of them equal to zero except Ao (the coefficient of the cosine term at zero frequency) and what do you get?

But if they are being nitpicky, then ask yourself if a constant is absolutely integrable over one period.

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8. ### WBahn Moderator

Mar 31, 2012
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Do you understand what each of these mean?

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9. ### bhuvanesh Thread Starter Member

Aug 10, 2013
268
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except this "it should not have no more that finite number of maxima and minima over single period"i know all

10. ### WBahn Moderator

Mar 31, 2012
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The simplest example would be a sine wave at an infinite frequency. In lay and imprecise terms, the problem is that such a waveform effectively becomes multivalued.

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11. ### WBahn Moderator

Mar 31, 2012
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Okay. So is a constant absolutely integrable over one period?

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12. ### bhuvanesh Thread Starter Member

Aug 10, 2013
268
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no, it is not. Integrating constant over one period leads to infintiy. At the same time look my waveform for e-|t| sin(25t). There at infinite time signal amplitude leading infinite so it would not be periodic ,would't be?

13. ### WBahn Moderator

Mar 31, 2012
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Sketch that waveform and consider the effect of the dying exponential on the integral. Notice that it is important that the exponent be |t| and not just t. Why?

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14. ### bhuvanesh Thread Starter Member

Aug 10, 2013
268
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e-|t| sin(25t).waveform attached

15. ### WBahn Moderator

Mar 31, 2012
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Not attached.

But that's okay.

Consider a simpler waveform.

x(t) = e^-|t|

Is this function absolutely integrable?

If so, then any function y(t) for which |y(t)| <= |x(t)| must also be absolutely integrable. Why?

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16. ### bhuvanesh Thread Starter Member

Aug 10, 2013
268
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yes it integrable over -infinity to infinty .it gives zero

but it not integrable for x(t) = e^|t|,right?

if x(t) is integrable ,sure then y(t) is integrable since y(t) is less than or equal to x(t).

x(t)= e-|t| sin(25t)

here how would you say |y(t)| <= |x(t).why it may not like |y(t)| => |x(t)?

Mar 31, 2012
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18. ### bhuvanesh Thread Starter Member

Aug 10, 2013
268
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sorry i was doing wrong calculation
x(t)= e-|t| sin(25t)
Because of magnitude we are going to have only e^-x ,where x= positive value for (minus infinity to plus infinty )
yes,its giving both positive and negative values,example e^ -1 it gives positive value but e^-10 gives negative value,but for any value its giving bounded value ,i mean its not giving infinity so its integrable,right?

if |x(t)| gives bounded value(not infinity ),then |y(t)| gives less than the bounded values of |x(t)|,right?

19. ### WBahn Moderator

Mar 31, 2012
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What is the numerical value of e^-10 ?

20. ### bhuvanesh Thread Starter Member

Aug 10, 2013
268
2
-7.28171817154