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**Home Assignment 6 / Re: Problem 2**

« **on:**November 07, 2012, 09:30:35 PM »

Solution is attached!

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Solution is attached!

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I was wondering (1/2 pi) in Fourier transform formula is an optional? (I mean, can we just ignore it?)

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Part (b) solution is attached!

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Part (a) solution is attached!

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WTH you are bringing me from textbook the plots of the partial sums of the FS on the given interval rather than the complete sum on $(-\infty,\infty)$ which is due to continuation?

Professor Ivrii,

The interval is [0,Ï€]!

But F.s. converges everywhere!

Yes, I got it.

I am sorry,

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WTH you are bringing me from textbook the plots of the partial sums of the FS on the given interval rather than the complete sum on $(-\infty,\infty)$ which is due to continuation?

Professor Ivrii,

The interval is [0,Ï€]!

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Part (a) solution is attached!

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Part (c) solution is attached!

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Part (a) solution is attached!

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Solution is attached,

Aida, I think your solution is not correct - the integral from -1 to 0 will only be 0 if x < y in H(x-y), i.e. x < -1.

Also, consider extreme cases - for C(-100), H(-100-y) can never be greater than 0, so a solution with C(x) = 1 cannot be right (unless I've totally misunderstood something).

I believe your solution is not correct; H(x) and H(x-y) has the same value for these two different domain because it is a constant function.

Shouldn't they be different though? For H(x), it's 1 for x>0 and 0 for x<=0, but H(x-y) is 1 for x-y>0, or x>y, and 0 for x<=y.

In any case, how can C(-100) be nonzero? x-y=-100-y>0 is impossible for -1<=y<=1, which is the only area where Q(y) is nonzero.

Sorry Ian, You are right

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My solution. Please check there might be mistakes. Sorry for quality of print and hand writing.

Djirar, I posted my solution 10 seconds after you

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Check the attachment please,