Fourier Transfrom of scaled periodic impulse train

[neg_ion13]

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

 

[Redbelly98]

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.

 

[neg_ion13]

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

 

[Redbelly98]

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.

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?

 

[neg_ion13]

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.

 

[neg_ion13]

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))]

 

[Redbelly98]

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 +∞.

 

[neg_ion13]

Cool. Thanks a lot.