Fourier Transfrom of scaled periodic impulse train

In summary, the equation for the pulse train is 2*pi*sigma*delta(w - 2*pi/T), where T = 2. The original equation needed to be modified to alternate between values of 2 and 1, but this can be done by splitting the problem into two summations.
  • #1
neg_ion13
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0

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[tex]\Pi[/tex][tex]\Sigma[/tex][tex]\delta[/tex](w - [tex]\Pi[/tex]k) + [tex]\Pi[/tex][tex]\Sigma[/tex][tex]\delta[/tex](w - [tex]\Pi[/tex](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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  • #2
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[tex]\Pi[/tex][tex]\Sigma[/tex][tex]\delta[/tex](w - [tex]\Pi[/tex]k) + [tex]\Pi[/tex][tex]\Sigma[/tex][tex]\delta[/tex](w - [tex]\Pi[/tex](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.
 
  • #3
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
 
  • #4
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[tex]\Pi[/tex][tex]\Sigma[/tex][tex]\delta[/tex](w - [tex]\Pi[/tex]k) + [tex]\Pi[/tex][tex]\Sigma[/tex][tex]\delta[/tex](w - [tex]\Pi[/tex](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?
 
  • #5
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.
 
  • #6
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))]
 
  • #7
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 +∞.
 
  • #8
Cool. Thanks a lot.
 

Related to Fourier Transfrom of scaled periodic impulse train

1. What is the Fourier Transform of a scaled periodic impulse train?

The Fourier Transform of a scaled periodic impulse train is a continuous function that represents the frequency components of the original signal. It is a complex-valued function that contains information about the amplitude and phase of each frequency component.

2. How is the Fourier Transform of a scaled periodic impulse train related to the original signal?

The Fourier Transform of a scaled periodic impulse train is the mathematical representation of the original signal in the frequency domain. It allows us to analyze the frequency components of the signal and understand how different frequencies contribute to the overall signal.

3. What is the significance of scaling in the Fourier Transform of a periodic impulse train?

The scaling factor in the Fourier Transform of a periodic impulse train determines the spacing between the impulses in the frequency domain. A larger scaling factor results in a narrower spacing between the impulses, which means that the frequency components are more closely spaced and the signal contains more high-frequency components.

4. How can the Fourier Transform of a scaled periodic impulse train be used in signal processing?

The Fourier Transform of a scaled periodic impulse train is used in signal processing to remove specific frequency components from a signal. By multiplying the Fourier Transform of the signal with a filter function, we can selectively remove or attenuate certain frequencies in the signal.

5. Can the Fourier Transform of a scaled periodic impulse train be used for non-periodic signals?

Yes, the Fourier Transform of a scaled periodic impulse train can be applied to non-periodic signals as well. This is because any signal can be represented as a combination of periodic signals through the use of the Fourier series. However, the resulting Fourier Transform will have an infinite number of impulses spaced at different frequencies, making it more difficult to analyze and interpret.

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