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?


1 Answer 1


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


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


with $a$ optimized such that


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


This equation must be solved numerically. The result is


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:


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


for $b$, yielding


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


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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.