Can we choose a sampling frequency to remove unwanted noise at a specific frequency?

Question

I am studying for my exam in signal processing. In one of the old exam papers I am told to find a sampling frequency, which will remove 80 Hz noise. The filter the exam question is based around has an impulse response $h[n]$ of

\begin{bmatrix} -3&0&-3&0 \end{bmatrix}

My approach is to just choose the sampling frequency at a multiple of the noise frequency. In this case, I would choose it to be 320 Hz, since 4 samples gives a sum of 0. When doing a DFT with the known impulse response it would also result in a net 0 frequency response.

Does this seem like a sustainable approach?

Answer 1

The general approach would be to find the zeros of the filter's transfer function. If you do things right it will turn out that there are two complex conjugate zeros on the unit circle at $\pm j$, i.e., at half the Nyquist frequency. Now you need to choose the sampling frequency such that $80$ Hertz corresponds to half the Nyquist frequency.

Answer 2

Or another way, visually, with the plot of the magnitude response of your filter shown below you see that it goes to zero at $\pi/2$.

So what you could do is use the relationship between the angular frequency $\omega$ in [radians/sample] and the relative frequency $f$ in [cycles/sample]

$$ \omega = 2\pi f = 2\pi \left(\frac FF_s\right) $$ Then compute $F_s$ for $F = 80 \ \rm Hz$ and $\omega = \pi/2$. You then get \begin{align} \frac \pi 2 &= 2\pi \left(\frac{80}{F_s}\right)\\ \implies \frac 12 &= 2\cdot \left(\frac{80}{F_s}\right) \\ \iff 80 &= \frac 12\cdot \left(\frac{F_s}{2}\right) \end{align}

Giving you the $320\ \rm Hz$.

In MATLAB

You can plot the magnitude response from the frequency response using MATLAB's freqz function where you can specify the frequency grid from $0$ to Nyquist (i.e. $\pi$). With your resulting $320\ \rm Hz$, you can run the function with this $F_s$ and see that $80\ \rm Hz$ is indeed suppressed.