Fourier Transfrom of scaled periodic impulse train

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Discussion Overview

The discussion revolves around the Fourier transform of a scaled periodic impulse train, specifically addressing the representation of the impulse train and the corresponding frequency domain expression. Participants explore the characteristics of the pulse train, including its periodic nature and the scaling of impulses at various time points.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents a Fourier transform expression for the scaled pulse train but questions its correctness.
  • Another participant expresses confusion about the pattern of the pulse train and notes the absence of pulses at certain time points.
  • A participant clarifies the periodic nature of the pulse train, providing specific values for impulses at various time points.
  • One participant suggests that the Fourier transform for a regular pulse train needs modification to account for the alternating values of the impulses.
  • A later reply corrects an earlier expression, indicating that impulses were being overwritten and provides a revised Fourier transform expression.
  • Another participant agrees with the revised solution while offering an alternative expression that they believe is equivalent.

Areas of Agreement / Disagreement

There is some agreement on the revised Fourier transform expression, but earlier disagreements about the representation of the pulse train and its characteristics remain unresolved. Multiple competing views on the correct formulation exist.

Contextual Notes

Participants express uncertainty regarding the pattern of the pulse train and the implications for the Fourier transform. The discussion includes various assumptions about the nature of the impulses and their scaling, which are not fully resolved.

neg_ion13
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Homework Statement


Fourier transform of scaled pulse train from -inf to +inf. Starting at 0 the first impulse is scaled at 2, the second impulse at 1 is scaled as 1, third at three scaled at 2, etc...

Homework Equations


Does my solution look correct?

The Attempt at a Solution



X(jw) = 2\Pi\Sigma\delta(w - \Pik) + \Pi\Sigma\delta(w - \Pi(k + 1))
the sigma notation is from k = -inf to inf. My thinking behind this is to represent a pulse for each scaled delta funciton in freq domain
 
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neg_ion13 said:

Homework Statement


Fourier transform of scaled pulse train from -inf to +inf. Starting at 0 the first impulse is scaled at 2, the second impulse at 1 is scaled as 1, third at three scaled at 2, etc...
I'm having trouble seeing any pattern in the pulse train you describe. Is there no pulse at t=2? What about the pulses at negative times?

Homework Equations

The transform of a simple pulse train might come in useful here.

Does my solution look correct?

The Attempt at a Solution



X(jw) = 2\Pi\Sigma\delta(w - \Pik) + \Pi\Sigma\delta(w - \Pi(k + 1))
the sigma notation is from k = -inf to inf. My thinking behind this is to represent a pulse for each scaled delta funciton in freq domain
If you can describe the input pulse train better, it will be easier for us to see what is going on.
 
The pulse train is periodic so it runs from plus -inf to +inf.
At t = -2 the impulse = 2
At t = -1 the impulse = 1
At t = 0 the impulse = 2
At t = 1 the impulse = 1
At t = 2 the impulse = 2
At t = 3 the impulse = 1
At t = 4 the impulse = 2
Etc... from -inf to +inf
 
neg_ion13 said:
The pulse train is periodic so it runs from plus -inf to +inf.
At t = -2 the impulse = 2
At t = -1 the impulse = 1
At t = 0 the impulse = 2
At t = 1 the impulse = 1
At t = 2 the impulse = 2
At t = 3 the impulse = 1
At t = 4 the impulse = 2
Etc... from -inf to +inf
Thank you for clarifying that.
neg_ion13 said:
Does my solution look correct?

The Attempt at a Solution



X(jw) = 2\Pi\Sigma\delta(w - \Pik) + \Pi\Sigma\delta(w - \Pi(k + 1))
the sigma notation is from k = -inf to inf. My thinking behind this is to represent a pulse for each scaled delta funciton in freq domain
No, it does not look correct. Can you show your work?
 
The Fourier transform for a pulse train is 2*pi*sigma*delta(w - 2*pi/T). Here T = 2. the problem is that a regular pulse train has a constant value of 1 for the impulse while this one does not. This means the above equation needs to be modified to alternate between value of 2 and a value of 1. That is how I arrived at the above equation. The first part of the equation is for value of 2. When k = 0 the first part 2*pi. The second part = pi when w = pi, etc... This was my thinking behind this X(jw). Because it's linear I could split the problem into two summations to represent each of the 2 values of the delta function.
 
Oops! I am overwriting my impulses. It should be:

X(jw) = 2*pi*Sigma[delta(w - pi*(2k))] + pi*Sigma[delta(w - pi*(2k + 1))]
 
Okay, I agree with your solution now. I had in mind expressing it a different way, but it is equivalent to what you have.

What I had in mind:
X(ω) = π(∑δ(ω-πk) + ∑δ(ω-2πk))

where k runs from -∞ to +∞.
 
Cool. Thanks a lot.
 

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