How Is x(t) Determined from Its Fourier Series Coefficients and Given Integrals?

AI Thread Summary
The discussion focuses on determining the periodic signal x(t) from its Fourier series coefficients and given integrals. It is established that x(t) consists of impulses at t=0 and t=1.5, with a repetition period of 3. The integral conditions imply that the impulse at t=1.5 must have a strength of 2, while the impulse at t=0 is a unit impulse. Participants clarify the terminology around impulse strength versus height, emphasizing the importance of correctly calculating these values. Recommendations for signal processing textbooks are also sought, with "Signals and Systems" by Simon Haykin and Barry Van Veen mentioned as a current resource.
reddvoid
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x(t) is periodic with T=3.
X(k) is FS coeff.
X(k)=X(-k) and X(k)=X(k+2)
also
integral from t=-0.5 to 0.5 of x(t)dt is 1
integral from 0.5 to 1.5 of x(t) dt is 2.
Find x(t)
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i found that the signal x(t) is even and X(0) is 5/3 (if I'm correct)
and integral from t=-1.5 to 1.5 of x(t)dt = 5
what will be x(t) ?
 
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That's a nice tricky little problem reddvoid. The trick is that there's a FS property that says if the FS is to be periodic then the function must be discrete (that is, impulses). Furthermore the spacing of the impulses is reciprocally related to the repetition period of the FS (2 f0 in this case). That is, dt = 1/(2 f_0) = \frac{1}{2} T, meaning that x(t) can only consist of an impulse at t=0 and another at t=1.5 (both repeated period 3 of course).
 
uart said:
That's a nice tricky little problem reddvoid. The trick is that there's a FS property that says if the FS is to be periodic then the function must be discrete (that is, impulses).


Furthermore the spacing of the impulses is reciprocally related to the repetition period of the FS (2 f0 in this case). That is, dt = 1/(2 f_0) = \frac{1}{2} T, meaning that x(t) can only consist of an impulse at t=0 and another at t=1.5 (both repeated period 3 of course).

thanks alot, I understood it now
but
if x(t) is a signal with impulse at t=0 and 1.5 repeating with period 3
then it doesn't satisfy the condition that
integration from 0.5 to 1.5 x(t) dt = 2
because according to you from 0.5 to 1.5 there is only one impulse
which gives integration 0.5 to 1.5 x(t)dt =1
I guess its impulse at t=0 and impulse of hight 2 at t= 1.5 repeating with period=3, right ?
 
Last edited:
I guess its impulse at t=0 and impulse of hight 2 at t= 1.5 repeating with period=3, right ?
Yes when I said "an impulse" I wasn't implying that it was a "unit impulse". I had to leave some of the work, calculating the strength of the impulses, for you. :)

BTW. We usually refer to the "strength" of an impulse rather than the "height", since technically the height of an impulse is always infinity.
 
uart said:
Yes when I said "an impulse" I wasn't implying that it was a "unit impulse". I had to leave some of the work, calculating the strength of the impulses, for you. :)

BTW. We usually refer to the "strength" of an impulse rather than the "height", since technically the height of an impulse is always infinity.
Ok, gotit :)
can you please suggest me some good book for signals
or one which you refer,
I'm using Signals and systems by Simon Haykin and Barry Van Veen
I feel quite difficult to follow this
 
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