# Magnitude calculation of filter

Electrical Engineering Asked by smake5730 on December 5, 2020

I am working on a problem for a given transfer function given below and am having trouble calculating the magnitude and phase of it. The problem is specifically that I do not know what method to use to seperate the imaginary value out of from the real to allow me to use their seperate values in calculating the magnitude and phase.

$$frac{Vout}{Vin}=frac{1}{1-(4*PI^2*0.04)+(j*2*PI*0.2)}$$

Ideally I’d like to end up with something like a+bi to allow me to do the srqt(a^2 + b^2) and the tan equation for the phase but I do not know how to do this.

Could anyone suggest a method to use or some mathematical identity to use?

Thanks!

So implementing what was in the first comment from ocrdu

It simplifies to

$$frac{Vout}{Vin}=frac{1}{(-0.579+j*1.25)}$$

Then multiplying top and bottom by inverse of bottom gives:

$$frac{Vout}{Vin}=frac{(0.579-j*1.25)}{(-0.579+j*1.25)*(0.579-j*1.25)}$$

But how does this help me?

So the solution was simply to take the square as follows:

$$frac{1^2}{sqrt{(1-(4*PI^2*0.04))^2+(j*2*PI*0.2)^2}}$$

j squared is j * j = -1

$$=frac{1^2}{sqrt{(1-(4*PI^2*0.04))^2+(-1*2*PI*0.2)^2}}$$

$$=0.722$$

Correct answer by smake5730 on December 5, 2020

$$frac{Vout}{Vin}=frac{-0.579-j*1.25}{(-0.579+j*1.25)*(-0.579-j*1.25)}=$$ $$frac{-0.579-j*1.25}{0.335+0.724j-0.724j+1.563} =$$ $$frac{-0.579-j*1.25}{1.898} = -0.305-0.659j$$

Please check for errors (yours and mine), but you get the idea.

Multiplying top and bottom by the denominator's complex conjugate gives a real number in the denominator.

Answered by ocrdu on December 5, 2020

$$(1 + ja)cdot(1 - ja) = 1 + a^2$$

Answered by Andy aka on December 5, 2020