I am studying a course in signalanalysis and have run into som trouble with a exercise.
I am to dimension the circuit below in such a way that the DC-amplification is 1 and that the frequencies $\omega=\frac
{\pi}{2}$ and $\omega=\pi$ is completely filtered out.
So i figured some of this out I think. I wrote this system as:
$y(n)=b_0x(n)+b_1x(n-1)+b_2x(n-2)+b_3x(n-3)$
I was not sure what to make of the DC-amplification criteria but I figured that it simply means x(n)=y(n). Meaning that all the coefficient should equal 1/4 With a z-transform this gave me:
$y(z)=b_0(x(z)z^0+x(z)z^{-1}+x(z)z^{-2}+x(z)z^{-3})$
$H(z)=b_0(1+z^{-1}+z^{-2}+z^{-3})$
If you make the substition $z=e^{j\omega}$ this transferfucton does actually filter out the requested frequencies.
$H(\omega)=b_0(1+e^{-j\omega}+e^{-j2\omega}+e^{-j3\omega})$
This is where I get stuck.
To get the amplitude- and phase functions respectively I was thinking:
$|H(\omega)|=\sqrt{\Re\{H(\omega)\}^2+\Im\{H(\omega)\}^2}$
and
$\theta(\omega)=\arctan\left(\frac{\Im\{H(\omega)\}}{\Re\{H(\omega)\}}\right)$
I have tried and tried but i can't get i right.
The answer sheet gives:
$|H(\omega)|=\frac{1}{4}|\frac{\sin(2\omega)}{\sin(\frac{\omega}{2})}|=\frac{1}{2}|\cos(\frac{3}{2}\omega)+\cos(\frac{1}{2\omega})|$
and
$\arg\{H(\omega)\}=-\frac{3}{2}\omega-\pi \quad for \quad \pi/2<\omega<\pi$
If anyone could help me with this last bit or give any hints i would be eternaly grateful! Please and thank you!