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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. enter image description here

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!

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Since I gave you a wrong hint first, let me try to make it up. You have the correct transfer function and the rest is (I think) just to slog through the algebra. Here is an outline how this can be approached.

We can use the geometric sum property to simply things a bit

$$4\cdot H(\omega) = 1 + e^{-j\omega} + e^{-j2\omega} e^{-j2\omega} = \frac{1-e^{-4j\omega} }{1-e^{-j\omega} } = \frac{H_4(\omega)}{H_1\omega)} $$

So we can write is as the ratio of two transfer functions and the total magnitude is the ratio of the the individual magnitudes and the total phase is the difference of the individual phases.

Let's do the magnitude squared of the first one: $$|H_4(\omega)|^2 = (1-e^{-4j\omega}) \cdot (1-e^{+4j\omega}) = (1 - 2cos(4\omega)+1) = 2+ 2cos(4\omega) = sin^2(2\omega)$$

The last step uses a trig identity for $sin^2(x)$.

For the phase we do $$H_4(\omega) = 1-e^{-4j\omega} = 1 - cos(4\omega) + j \cdot sin(4\omega)$$

$$ \phi_4 = tan^{-1}\frac{sin(4ω)}{1−cos(4ω)}$$

We leverage the identity $\frac{sin(x)}{1−cos(x)} = cot(x/2)$ and we get

$$ \phi_4 = tan^{-1} \left ( cot(2\omega) \right ) = \pi/2-2\omega$$

Repeat for the other transfer function and combine. There may be an easier way to do this, but I'm not seeing it right now.

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  • $\begingroup$ Ohhh now that you showed the geometric sum all the pieces fell into place! Thanks a bunch!! $\endgroup$
    – Aedrha
    Jul 20 at 16:22

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