Filter design with zero - pole placement method

Question

I have some question for you about Filter design.

• Α. Calculate the transfer function in order to stop the frequency of $300\textrm{ Hz}$ for sampling frequency at $12\textrm{ kHz}$. Use the zero – pole placement method
• Β. Calculate the absolute value for the designed filter at $f_1=300\textrm{ Hz}$ and at $f_2=200\textrm{ Hz}$.

Can someone show me how can i solved this problem.

Answer 1

Here let me show you a simple procedure very similar to pole zero placement which will be helpful for your notch filter design.

First, lets analyse the frequency response of a single zero and let $$ H(z) = 1 - b z^{-1} $$ be a first order system with a single zero at $z = b$ where $b$ is a complex constant with a radius $r$ and phase angle $\phi$ radians; i.e., $$ b = r e^{j\phi} $$

Lets see this zero on the z-plane and the corresponding frequency response magnitude for the values of $r = 0.9$ and $\phi = \pi/4$ radians:

Note that since there is a single zero, this is a non-symmetric frequency response (due to a complex impulse response $h[n]$). To get rid of this non-symmetric frequency response, we shall force the impulse response to be real and the way to achieve this is to add a second zero at the complex-conjugate location resulting in the filter: $$ H_{az}(z) = (1 - bz^{-1})(1 - b^{*} z^{-1}) $$ as the second figure shows:

Now we have a pair of complex-conjugate zeros, whose frequency response is conjugate symmetric (and therefore the magnitude response is even symmetric as displayed)

next, we display the same figures, for a pair of complex-conjugate poles with a radius of $r=0.85$ and the same angle $\phi = \pi/4$ radians: $$ H_{ap}(z) = \frac{1}{(1 - az^{-1})(1 - a^{*} z^{-1})} $$

So far we have displayed the pole-zero locations and corresponding frequency responses for individual pairs of poles and zeros at the same frequency. Lets combine them into a single filter and display the result: $$ H(z) = \frac{(1 - bz^{-1})(1 - b^{*} z^{-1})}{(1 - az^{-1})(1 - a^{*} z^{-1})} $$

Nice! We have reached a system whose frequency response now resembles that of a notch, but a weak one? The solution comes by recognizing that we should better put the zero on the unit circle (i.e., set its radius $r=1$) for an infinetely deep nulling at that frequency;

Now this system is what can be called as a 2nd order notch filter with a pole radius of $r=0.85$ and radian frequency of $\omega = \pi/4$ radians per sample. This is a quite satisfactory notch filter.

Finally lets display a much sharper notch by moving the pole closer to the zero (closer to the unit circle) by setting its radius to $r = 0.99$. But note that it's very dangerous for a pole to wander about unit circle, as it can easily fall over it, thereby making the system unstable...

For your convenience, choose the pole radius between 0.9 and 0.99 depending on your numerical accuracy and how sharp a notch you need. And adjust the frequency of the pole-zero pair according to which frequency you want to null.