2
$\begingroup$

Assume that you are giving an arbitrary amplitude frequency response $A(\omega)=|H(j\omega)|$

Is there a characterization that ensures that $A$ is monotone? i.e, $A$ has a global maximum at $\omega=0$ and does not have a local peak at some other frequency?

thanks

$\endgroup$
1
$\begingroup$

As far as I know there is no simple characterization of filters with a monotonic magnitude response. I could only come up with a weak time domain characterization of LTI systems with monotonically decreasing amplitude for positive frequencies (cf. Eq. $(6)$ below).

Let $h(t)$ be the real-valued impulse response of an LTI system, and let $H(\omega)=A(\omega)e^{j\phi(\omega)}$ be its Fourier transform. Furthermore, define $A(\omega)$ and $\phi(\omega)$ such that $A(\omega)\ge 0$ is satisfied. Now we require

$$\frac{dA(\omega)}{d\omega}\le 0,\qquad\omega>0\tag{1}$$

Let $r_h(t)$ be the deterministic auto-correlation function of $h(t)$:

$$r_h(t)=\int_{-\infty}^{\infty}h(\tau)h(\tau+t)d\tau\tag{2}$$

Note that the Fourier transform of $r_h(t)$ is given by $A^2(\omega)$. By a basic property of the Fourier transform we have

$$\begin{align}t\,r_h(t)&=\mathcal{F}^{-1}\left\{j\frac{dA^2(\omega)}{d\omega}\right\}\\&=\frac{j}{2\pi}\int_{-\infty}^{\infty}\frac{dA^2(\omega)}{d\omega}e^{j\omega t}d\omega\\&=-\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{dA^2(\omega)}{d\omega}\sin(\omega t)d\omega\\&=-\frac{1}{\pi}\int_{0}^{\infty}\frac{dA^2(\omega)}{d\omega}\sin(\omega t)d\omega\tag{3}\end{align}$$

Note that since $A(\omega)\ge 0$, condition $(1)$ implies

$$\frac{dA^2(\omega)}{d\omega}\le 0,\qquad\omega>0\tag{4}$$

From $(3)$ and $(4)$ we get the following inequality:

$$|t\, r_h(t)|\le -\frac{1}{\pi}\int_{0}^{\infty}\frac{dA^2(\omega)}{d\omega}d\omega=\frac{A^2(0)-A^2(\infty)}{\pi}\tag{5}$$

which finally results in

$$|r_h(t)|\le \frac{A^2(0)-A^2(\infty)}{\pi |t|},\qquad t\neq 0\tag{6}$$

In sum, condition $(1)$ implies that the deterministic auto-correlation of the corresponding impulse response must satisfy inequality $(6)$.

$\endgroup$
  • $\begingroup$ that is very interesting. thanks $\endgroup$ – M.A Jul 30 at 18:18
0
$\begingroup$

The obvious solution would be $$\frac{\partial A(\omega)}{\partial \omega } < 0$$

If you just have a table of numbers, that's probably the best you can do. Are you looking for any relationships to the poles and zeros or some other parametric description of the transfer function ?

$\endgroup$
0
$\begingroup$

If A($\omega$) has a global maximum at $\omega=0$, you might consider pre-emphasizing the signal, that removes the DC components and then you might be able to see some peaks in the spectra. Also, consider smoothing the spectra before finding the gradient, this will help eliminate the local noisy peaks. Gradient will give you peaks in the spectrum.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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