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The question is as follows:

Design a simple digital filter of the form $$y[n] = a_0 x[n] + a_1 x[n-1] + a_2 x[n-2]$$ to remove a narrowband disturbance with frequency $f_0 = 70 \,\rm{Hz}$. Let $f_s = 280 \, \rm{Hz}$ be the sampling frequency.

a) Determine the desired zeros and poles of the filter.

So in order to find the zeroes and poles; First I find the transfer function:

$$ H(z) = K \frac{(z-z_1)(z-z_2)}{(z-p_1)(z-p_2)} $$

However, since we want to remove the disturbance, we must have $H(e^{i\omega_0})$ such that $ \omega_0 = 2 \pi \left( \frac{f_0}{f_s} \right) = \frac{\pi}{2}$. This means that we have $z_1 = i$ and $z_2 = -i$. I assume I've worked out the zeroes correctly. However, I'm not sure how to calculate the poles. Any guidance would be appreciated.

EDIT: Follow up question:

What would be the process in calculating the transform function for the filter. Where

$$y[n] = a_0x[n] + a_1x[n-1] + a_2x[n-2]+b_1y[n-1]+b_2y[n-2]$$

Rodrigo de Azevedo
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Priit
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1 Answers1

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Since this is a FIR filter, there is no pole. The Z-transform writes $$ Y(z)= b_0 X(z)+b_1 z^{-1}X(z)+b_2 z^{-2}X(z) $$

The transfer function is $$ H(z)= \frac{Y(z)}{X(z)} = b_0(1-z_1z^{-1})(1-z_1^*z^{-1}) $$ So $z_1=\rho e^{i2\pi f_N/f_s}$ with $f_N=70$ Hz, $f_s=280$ Hz from which $b_2=b_0\rho^2,b_1=-2b_0\rho \cos(2\pi f_N/f_s)$.

The gain is set to unity if $b_0=1/|1-z_1|^2$.

Steph
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