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QUESTION: Dear Randy,

I am attaching this image because it's the easiest way for me to sum up the task all my problems.

Once I get at the end of what I wrote on the image, I don’t know how to carry on solving this problem. I’d like to simplify my expression, getting rid of the b term, so that I can calculate more easily magnitude and phase.

Could you please help me?

Thank you very much!

Best regards

Stefano

ANSWER: I went about doing the integral differently and used techniques you should find very useful in solving these types of problems. BTW, just to get nomenclature straight, h(t) = impulse response function and H(ω) = Fourier Transform of h(t) (= Laplace transform if we need ω = complex variable; typically only when analyzing transient responses).

My derivation is in the attached image but the main "trick" is to write

sin(bt) = = Im( cos(bt)+isin(bt) ) = Im(e^ibt), where Im(*) means the imaginary part of *,

and then do the simple exponential integral. A couple of points;

1) I use i = sqrt(-1) instead of j just out of habit, recognizing that EEs ("sparkies") use j because i = current

2) note that in the evaluation of the limits that e^(-a+i(anything))t = 0 as t -> ∞ because of the negative real exponent.

Thanks for the problem and especially thanks for trying to solve it first and showing your work. Please send me any future problems you run across.

Good luck.

Randy

---------- FOLLOW-UP ----------

QUESTION: Dear Randy,

first of all thank you so much!

I couldn't believe to what I was reading, your answer is absolutely brilliant! Thank you very much!!!

I just don't get a few points:

1) We are asked to find H(jω), whereas you are writing H(ω). Is it the same because of the Im in front of the expression?

2)The last result you get, is my transfer function, isn't it? From there, how can I find magnitude and phase? (I am attaching a screenshot of one of the examples we did in the lecture) I mean, to me the IM in front of it sounds like "the phase" of the transfer function.

I am sorry for this follow up question,

I look forward to hearing from you

Thanks in advance,

all the best!

Stefano

I made a mistake in my previous derivation in that I shouldn't have taken the imaginary part of the whole integrand. A corrected derivation is shown in the attached image. The final expression is

H(iω) = b/[ (a+iω)^2 + b^2 ].

This has an imaginary component in it, which you can rearrange as per the notes in the more recent image you sent. I kept it in this form since it corresponds to the Laplace transform with the transform variable p = iω, i.e., with its real part = 0 so that it corresponds to a Fourier Transform.

Sorry for the confusion. I am impressed though that you are working hard to understand this stuff. I will be more careful in the future.

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