How Do You Calculate the Magnitude of a Filter Response?

[jmher0403]

Homework Statement

F(w) = a / [1 - be^(-jwT)]

where a and b are constants

Find the magnitude of this filter response

Homework Equations

e^(-jX) = cos(X) - isin(X)

The Attempt at a Solution

the answer is a/ √[1-2bcos(wT) + b^2*cos^2(2wT)]

but I can't seem to get rid of sines.



F(w) = a / [1 - be^(-jwT)]
= a / [ 1 - b(cos(wT) - isin(wT) ]
= a / [1 - bcos(wT) + ibsin(wT)]
|F(w| = √{ a^2/ [1 - bcos(wT) + ibsin(wT)]^2}
= a /√[1-2bcos(wT) + b^2*cos^2(wT) + i 2bsin(wT) - i 2b^2*sin(wT)cos(wT) -b^2*sin^2(wT)]

Can anyone point me at the right direction??

 

[NascentOxygen]

I think you'll find that when determining the magnitude, you take √[/size]( Real^2 + Imag^2 ). The i operator itself does not appear in the magnitude term.

 

[milesyoung]

the answer is a/ √[1-2bcos(wT) + b^2*cos^2(2wT)]

Where did you get this from? It's not correct.

 

[rude man]

Homework Statement

|F(w| = √{ a^2/ [1 - bcos(wT) + ibsin(wT)]^2}

Where did this come from? Makes no sense, and you can't have an imaginary component in a magnitude.

In general, given (a + jb)/(c + jd), a thru d real, magnitude = √(a² + b²)/√(c² + d²).

Plus, miles is right, the given answer is incorrect.