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

 

Hi,

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

 

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

 

What are the requirements for the Fourier series to exist?

 

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

 

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

http://www.encyclopediaofmath.org/index.php/Absolutely_integrable_function

 

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

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.

 

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

Do you understand what each of these mean?

 

Do you understand what each of these mean?

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

 

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

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.

 

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

Okay. So is a constant absolutely integrable over one period?

 

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?

 

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?

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?

 

e-|t| sin(25t).waveform attached

 

e-|t| sin(25t).waveform attached

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?

 

x(t) = e^-|t|

Is this function absolutely integrable?

yes it integrable over -infinity to infinty .it gives zero

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

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

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)?

 

yes it integrable over -infinity to infinty .it gives zero

How can if give zero?

Can you find any value of t for which e^-|t| is strictly positive?

Can you find any value of t for which e^-|t| is strictly negative?

If you can find one but not the other, then it can't integrate to zero.

 

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?

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

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

 

What is the numerical value of e^-10 ?

 

-7.28171817154