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 √( 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.

 

[rezeew]

Your solution is correct. The magnitude of the filter response is given by the absolute value of the complex expression, which is equal to √(a^2 + b^2 - 2abcos(wT)). This can also be written as a/√[1-2bcos(wT) + b^2*cos^2(wT)]. The presence of the sine terms in your solution is due to the imaginary component in the original expression. However, since the magnitude is only concerned with the real component, the sine terms can be ignored.