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DFT of periodic impulse train!

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purnapragna

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periodic impulse train

Hi i have a problem like how to calculate the DFT of a dicrete time periodic impulse train.

please help me in this regard

thank you

purna
 

knmaheshy2k

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dft of impulse train

hi purnapragna,
The question u have asked is What is the DFT of periodic impulse train?

Solution: Periodic impulse train can be represented as say
x[n] = {1,1,1,1,1,1,1,.........,1]
then DFT of x[n] is given by

X{k} = N*del(k)

if u still have doubt then let me know.
Happy learning....
 

purnapragna

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discrete impulse train

i mean periodic impulse train in DT domain is not just

x[n] = {1,1,1,1,1,1,1,.........,1]


which u have written as it extends up to infinity

so how to tackle this with DFT ?

thanx

purna
 

knmaheshy2k

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dtft impulse train

Hi purna,
As u know that we determine Fourier Transform only for aperiodic signals, the case that Fourier transform of a periodic signal doesnt come into picture.
DFT of an aperiodic Discrete time signal x[n] is the sampled version of Discrete Time Fourier Transform {DTFT} .So here too the periodicity of x[n] doesnt come into picture. For discrete time periodic signals we have something called as Discrete Time Fourier Series {DTFS}.
I hope this convinces u, if u still have doubt plz be free to ask
Happy learning.........
 

purnapragna

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dft of impulse

HI

I think DFT will result in periodicity in Time Domain. pls check out.

thnax

regards

purna
 

Gandharva

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fourier transform of periodic impulse train

hi purna,

applying DFT doesnt result in periodicity in the time domain,
but, DFT assumes the time domain signal to be periodic so that the
frequency domain representation becomes discrete and hence easy
to compute.

i hope this should help you..
 

purnapragna

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periodic impulse fft

hey i have one more doubt that does DFT of a periodic signal for exaple let the continuous time signal be x(t) = cos(ω0t). So will the DFT of the Discretized version of this signal exists?

thanx

regards

purna
 

knmaheshy2k

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dft impulse train

hi purnapragna,
We can represent the discretized version of the signal x(t) = cos(ω0t) as
x[n] = cos[ω0n].Now letz calculate the N point DFT of this signal, as it is periodic with 2Π consider 1 period where w0 = (2Π/N)k0.
Now express the cos[ω0n] in terms of exponentials and find out DFT of the sequence.I cant represent all the symbols here, try findin out urself if u still cant find out tell me i'll edit it in MS-Paint and post here.
Anyhow answer should be
X(k) = [{δ(k-k0)} + {δ(k-(N-k0))}]N/2
i hope this will help u.....Happy learning............
 

purnapragna

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impulse train fft

hi

i think the frequencies in DT case and CT time cases are not equal. i mean ω0 is the CT frequency of the sinsoid and it is not the same for DT case. Then how can u find the DFT?

My actual doubt is DFT is the discretized version of DTFT i.e. i sample the FT of the sampled sinusoid and and take the strengths to represent the DFT. So i mean to get DFT i have to first multipy the DTFT of the sinusoid with periodic impulse train and take the strengths of the resulting signal. Now DTFT of the sinusoid is two impulses. So how can i get DFT as the multiplication of impulses is not defined? Hope u understood my problem. I want to attack the problem in frequency domain rather than time domain.

thank u

purna
 

knmaheshy2k

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n-point dft u(k)-u(k-k0)

Hi purnapragna,
u told " i think the frequencies in DT case and CT time cases are not equal. i mean ω0 is the CT frequency of the sinsoid and it is not the same for DT case. Then how can u find the DFT? "........i dont understand wat u mean by this.w0 is a constant and will be the same for both CT and DT signals at sampled spaces
Consider for a CT...Θ = w0*t1
for DT....φ = w0*n1
Θ = φ at places where t1 = n1,ie at places where u sample the signal...,its as simple as that....

comming to ur next question....." to get DFT i have to first multipy the DTFT of the sinusoid with periodic impulse train and take the strengths of the resulting signal " Its not as u say....To get DFT from DTFT wat we do is sample the DTFT only at certain points ie obtain the value of DTFT only at certain points....It is not equivalent of multiplying the DTFT by a periodic impulse train...
i hope this convinces u....
happy learning..............
 

purnapragna

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impulse train signal

hi mahesh,

first one is ok but i have reservations with the second answer how can u get the samples from DTFT without multiplying it with impulse train? I think there is no ther way thatn this. If there is something like this pls tell me how can we do that mathematically

thnx

purna
 

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periodic unit impulse train

hi purnapragna,
as i told earlier sampling of a signal is not multiplying it by an impulse train...
If u want to get the sample of a signal by multiplyin it with an impulse train then u have to take care that both the signals start at same time so that if u want to measure the sample value of the signal at some point then multiply it with the unit impulse function at that point.I had forgotten to mention this condition earlier..
i hope this should satisfy u....
happy learning...........
 

purnapragna

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dft impulse function

dear mahesh,

but i have read in DTSP by oppenheim and schafer that sampling is nothing but multiplying the signal with periodic impulse train and if u want to get DT version of the sampled signal u have to take the strengths of the corresponding samples(CT impulses). I got the same idea from multirate signal processing by Rabiner and Crocherie. But now i m getting different idea from u. I dont know which one to follow.!!!!!!!!!!

pls clarify me

regards

purna!
 

Gandharva

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multiplication with impulse train

hi purna,

if i have understood you right you want to first get the DTFT of a signal and then sample that in the frequency domain to get the DFT of the same signal assuming it to be periodic.

please read the following link, this gives very good mathematical analysis on how to obtain DFT from a DTFT.

https://en.wikipedia.org/wiki/Discrete-time_Fourier_transform

i hope this should help you...
 
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