Solving the amplitude response for a real tricky expression.

Aedrha2

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Jun 14, 2021
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14
Hi there!
I am soon to have an exam in signal processing. I am practicing on old exams and there is this one qeustion which I am really stuck at:

f0=0,05f1=0,045f_0=0,05 \quad f_1=0,045

G12cos(2πf0)+112rcos(2πf0)+r2=1G\left|\frac{1-2\cos(2\pi f_0)+1}{1-2r\cos(2\pi f_0)+r^2}\right|=1
G12cos(2πf0)ej2πf1+ej4πf112rcos(2πf0)ej2πf1+r2ej4πf1>0,95G12cos(2πf0)+112rcos(2πf0)+r2G\left|\frac{1-2\cos(2\pi f_0)e^{-j2\pi f_1}+e^{-j4\pi f_1}}{1-2r\cos(2\pi f_0)e^{-j2\pi f_1}+r^2e^{-j4\pi f_1}}\right|>0,95G\left|\frac{1-2\cos(2\pi f_0)+1}{1-2r\cos(2\pi f_0)+r^2}\right|
12cos(2πf0)ej2πf1+ej4πf112rcos(2πf0)ej2πf1+r2ej4πf112cos(2πf0)+112rcos(2πf0)+r2>0.95\left|\frac{\frac{1-2\cos(2\pi f_0)e^{-j2\pi f_1}+e^{-j4\pi f_1}}{1-2r\cos(2\pi f_0)e^{-j2\pi f_1}+r^2e^{-j4\pi f_1}}}{\frac{1-2\cos(2\pi f_0)+1}{1-2r\cos(2\pi f_0)+r^2}}\right|>0.95
for some r.

I was thinking i could set r to a value 0.99 for example. calculate 12cos(2πf0)+112rcos(2πf0)+r2\left|\frac{1-2\cos(2\pi f_0)+1}{1-2r\cos(2\pi f_0)+r^2}\right| for that value and then find a G that makes it work.

I can't seem to get it right!

Am I going about this the right way?
 
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