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
Messages
118
Reaction score
1
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)
========
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) ?
 
Engineering news on Phys.org
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
 
Hi all I have some confusion about piezoelectrical sensors combination. If i have three acoustic piezoelectrical sensors (with same receive sensitivity in dB ref V/1uPa) placed at specific distance, these sensors receive acoustic signal from a sound source placed at far field distance (Plane Wave) and from broadside. I receive output of these sensors through individual preamplifiers, add them through hardware like summer circuit adder or in software after digitization and in this way got an...
I have recently moved into a new (rather ancient) house and had a few trips of my Residual Current breaker. I dug out my old Socket tester which tell me the three pins are correct. But then the Red warning light tells me my socket(s) fail the loop test. I never had this before but my last house had an overhead supply with no Earth from the company. The tester said "get this checked" and the man said the (high but not ridiculous) earth resistance was acceptable. I stuck a new copper earth...
I am not an electrical engineering student, but a lowly apprentice electrician. I learn both on the job and also take classes for my apprenticeship. I recently wired my first transformer and I understand that the neutral and ground are bonded together in the transformer or in the service. What I don't understand is, if the neutral is a current carrying conductor, which is then bonded to the ground conductor, why does current only flow back to its source and not on the ground path...
Back
Top