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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?

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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.

enter image description here

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:

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

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

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