# Calculating the main lobe width of Dirichlet kernel

The Dirichlet kernel of order $$N$$ is defined as: $$D(\theta,N)=\frac{\sin(N\theta /2)}{\sin(\theta /2)}$$

We know that the main lobe widths between zero crossings is: $$4\pi/N$$

I was wondering if there is a closed form mathematical expression for the FWHM?

Meaning I'm looking for $$\theta_s$$ s.t.: $$D(\theta_s,N)=\frac{\sin(N\theta_s /2)}{\sin(\theta_s /2)} = N/2$$

If so, how does one go about calculating this?

There's no analytical solution for the general case. However, it's quite straightforward to compute that value using a few iterations of Newton's method. I've done that and the plot below shows the result for Dirichlet kernels of orders $$n=2$$ to $$n=100$$. As is to be expected, the curve converges to zero.

Now I remember that there's a very related thread in which some of us competed in computing ridiculously accurate approximations to the $$3$$dB cut-off frequency of a moving average filter of arbitrary order.

The same type of approximation also works for this problem, because the only difference is that we're now looking for the $$6$$dB cut-off instead of the $$3$$dB cut-off.

Using the methods explained in the answers to the question mentioned above, we can derive a simple yet accurate approximation for the half-maximum width $$W(n)$$ of a Dirichlet kernel of order $$n$$:

$$W(n)=\frac{7.581977068135924}{\sqrt{n^2-0.7236694265416452}},\qquad n>1\tag{1}$$

This formula is asymptotically correct, i.e., its error converges to zero for large orders $$n$$. The error decreases monotonically for $$n>2$$, with its maximum value $$7.2892\cdot 10^{-4}$$ at $$n=3$$.

For completeness I add a brief explanation of the derivation of Eq. $$(1)$$. Starting from the normalized Dirichlet kernel

$$D_n(\omega)=\frac{\sin(n\omega /2)}{n\sin(\omega/2)}\tag{2}$$

we compute a second-order Taylor (Maclaurin) series:

$$D_n(\omega)\approx 1-\frac{n^2-1}{24}\omega^2\tag{3}$$

Using $$(3)$$ to obtain an estimate for $$\omega_c$$ satisfying $$D_n(\omega_c)=\frac12$$ gives

$$\hat{\omega}_c(n)=\frac{2\sqrt{3}}{\sqrt{n^2-1}}\tag{4}$$

This estimate is biased, i.e., its error doesn't converge to zero for $$n\to\infty$$. This can be corrected by using the estimate

$$\hat{\omega}_c(n)=\frac{a}{\sqrt{n^2-1}}\tag{5}$$

with $$a$$ optimized such that

$$\lim_{n\to\infty}D_n(\hat{\omega}_c(n))=\frac12\tag{6}$$

Plugging $$(5)$$ into $$(2)$$ and computing the limit $$n\to\infty$$ results in the following equation for $$a$$:

$$\sin(a/2)=a/4\tag{7}$$

This equation must be solved numerically. The result is

$$a=3.790988534067962\tag{8}$$

With this value of $$a$$, the estimate $$\hat{\omega}_c(n)$$ in $$(5)$$ is unbiased, i.e., the error converges to zero for large values of $$n$$. The error for small values of $$n$$ can be further decreased by a slight modification of the estimate:

$$\hat{\omega}_c(n)=\frac{a}{\sqrt{n^2-b}}\tag{9}$$

where $$b$$ must be optimized. Note that the estimate $$(9)$$ is still unbiased, regardless of the value of $$b$$.

I chose $$b$$ to make the error zero for $$n=2$$. That's where the maximum error occurred. For $$n=2$$ it's straightforward to show that $$\omega_c=2\pi/3$$. Hence, the constant $$b$$ is obtained by solving

$$\hat{\omega}_c(2)=\frac{a}{\sqrt{2^2-b}}=\frac{2\pi}{3}\tag{10}$$

for $$b$$, yielding

$$b=4-\left(\frac{3a}{2\pi}\right)^2=0.7236694265416452\tag{11}$$

Equation $$(1)$$ is obtained from $$(9)$$ with $$a$$ and $$b$$ given by $$(8)$$ and $$(11)$$, respectively:

$$W(n)=2\hat{\omega}_c(n)=\frac{2a}{\sqrt{n^2-b}}\tag{12}$$

• Wow, wasn't expecting to get such a detailed answer! Thank you so much! Commented Nov 16, 2023 at 14:01