Formula for designing Lynn's low pass filter

Question

For the second order Lynn's low pass filter, the general form of the transfer function is: $$ H(z)=\frac{(1−z^{-m})^2}{ (1−z^{-1})^2 } $$ where $m$ is a positive integer. The gain for this is $m^2$, time delay is $m-1$ samples and the nominal frequency corresponding to the location of the lowest frequency zero is $f_l = f_s /m$. The frequency response is given by: $$ |H(e^{j\omega T})|=\frac{\sin^2(m\omega /2f_s)}{\sin^2(\omega /2f_s)} $$ At the 3 dB cutoff frequency, for unit gain we have: $$ |H(e^{j\omega_c T})|=\frac{\sin^2(m\omega_c /2f_s)}{m^2\sin^2(\omega_c /2f_s)} \\ \therefore \frac{1}{\sqrt2}=\frac{\sin^2(m\pi f_c /f_s)}{m^2\sin^2(\pi f_c /f_s)} $$

Now, to design the filter for our required cutoff frequency $f_c$ and sampling frequency $f_s$, we need to find the value of $m$ corresponding to these inputs. So how can we get this equation to the form $m = f(f_c, f_s)$, so that we can avoid getting the $m$ value by trial-and-error or numerical methods? Or is there some other quick approximation formula that we can use?

Answer 1

Just analyzing your equation, let $a = f_c / f_s$

Also we know that for $m = k/(2a)$ it reduces to $1/(m^2 sin(\pi a))$ for odd $k$ and $0$ for even $k$.

$$\lim_{m \to 0} \frac{sin^2( \pi m a)}{m^2 sin(\pi a)} = \frac{a^2 \pi^2}{sin(\pi a)} = \frac{1}{\sqrt{2}}$$

Even for this case, determining $a$ is not an easy task. I computed numerically the curve defined by this equation.

Answer 2

Meanwhile, we have got an approximation solution for this. Consider the equation: $$ \frac{1}{\sqrt2}=\frac{\sin^2(m\pi f_c /f_s)}{m^2\sin^2(\pi f_c /f_s)} $$

In the right-hand side denominator, since $f_c << f_s$, we can write: $$sin(\pi f_c /f_s) \simeq \pi f_c /f_s$$ Also, in the numerator, since $f_l = f_s/m$, we have $m\pi f_c /f_s = \pi f_c /f_l$. Now, since $f_c < f_l$, we see that $0 < \pi f_c /f_l < \pi$. This means we can use Bhaskara I's sine approximation formula and write: $$sin(m\pi f_c /f_s) \simeq \frac{16 (\pi -(m\pi f_c /f_s)) (m\pi f_c /f_s)}{5 \pi ^2-4 (\pi -(m\pi f_c /f_s)) (m\pi f_c /f_s)}\qquad $$ Plugging these back into the main equation, simplifying and solving the resulting quadratic equation for $m$, we get: $$ m \simeq 0.3f_s/f_c$$ This gives quite accurate results. For example, if we need to design the filter for a sampling frequency of 200 Hz and cutoff frequency of 15 Hz, we get $m$ as 4. This gives us $|H(e^{j\omega_c T})|= 0.75$