# Realization of a filter based on its transfer function

How can we check whether the filter is realizable given its transfer function and What are the parameters the realization depends on?

Here is an example:

Show that a filter with transfer function

$$H(ω)=\frac{2(10^5)}{ω^2 + 10^{10}} e^{-jωt_0 }$$

is unrealizable. Can this filter be made approximately realizable by choosing a sufficiently large t0.

--Modern digital and analog communication systems (3rd edition) - B.P. Lathi

Note: The question is of general nature. The example quoted above serves only as to make it clear, what is being asked. No one has to provide a solution for this problem.

• how do you define "realizable"? is a realizable filter one that you can build? does your realizable filter have to be stable? what might be another property you need for your filter to be "realizable"? Commented Sep 17, 2018 at 23:30
• By realizable, I mean theoretically functional. We can ignore practical complications. Commented Sep 17, 2018 at 23:42
• then ya gotta define whatever the heck ya mean by "theoretically functional". Commented Sep 17, 2018 at 23:50
• hay, can any of you IEEE folks get this paper: A Note on Stable, Physically Realizable, Linear, Time Invariant Systems? i don't wanna pay money for it. Commented Sep 17, 2018 at 23:55
• maybe this segment from Rao: Signals & Systems is what you mean? Commented Sep 18, 2018 at 0:06

A transfer function is called realizable if it can be implemented by a causal and stable system. The given frequency response is continuous and doesn't have any impulses, so the corresponding system is stable.

The transfer function (as a function of $$s$$) is given by

$$H(s)=\frac{a}{b-s^2}e^{-st_0}\tag{1}$$

The given frequency response is obtained form $$(1)$$ by substituting $$s=j\omega$$. Since $$b>0$$, $$H(s)$$ has two real-valued poles at $$\pm\sqrt{b}$$, and, consequently, the corresponding impulse response is two-sided (i.e., non-causal).

Partial fraction expansion of $$(1)$$ gives

$$H(s)=\frac{a}{2\sqrt{b}}\left[\frac{1}{\sqrt{b}-s}+\frac{1}{\sqrt{b}+s}\right]e^{-st_0},\qquad -\sqrt{b}<\textrm{Re}\{s\}<\sqrt{b}\tag{2}$$

From $$(2)$$ it is straightforward to obtain the impulse response

\begin{align}h(t)=\mathcal{L}^{-1}\{H(s)\}&=\frac{a}{2\sqrt{b}}\left[e^{\sqrt{b}(t-t_0)}u(-(t-t_0))+e^{-\sqrt{b}(t-t_0)}u(t-t_0)\right]\\&=\frac{a}{2\sqrt{b}}e^{-\sqrt{b}|t-t_0|}\tag{3}\end{align}

No matter how large you choose $$t_0$$, the impulse response will never be causal. However, since the system is stable, the impulse response decays for $$|t|\to\infty$$, so you can approximate the actual impulse response well by sufficiently shifting it to the right (i.e., by choosing some large $$t_0$$), and truncating the non-causal part.

Also take a look at this related question and its answer.

The filter realization can be checked by paley-wiener criterion, it should not have any discontinuities in frequency response and it should be absolutely square integrable .if you see the text book you might find it.

• Note that the Paley-Wiener criterion only takes the magnitude into account, not the phase. So if the criterion is satisfied for a given magnitude response, we generally still can't say anything about the causality of the corresponding complex frequency response. The only conclusion we can draw in that case is that it is possible to find a phase such that a system with the given magnitude response is causal. Commented Sep 18, 2018 at 9:18
• Can you state any examples? Commented Sep 18, 2018 at 11:02
• Just take the magnitude of the frequency response of any causal and stable filter, and add a phase term $e^{j\omega \tau}$, $\tau>0$. The resulting frequency response satisfies the Paley-Wiener criterion, but it is not causal. Commented Sep 18, 2018 at 12:40
• i would modify that, @MattL. to say "add a phase term $e^{j\omega\tau}$, $\tau > \tau_\text{delay}$", in case your causal and stable filter has a delay in it such that the impulse response $h(t)=0$ for all $t<\tau_\text{delay}$. Commented Sep 18, 2018 at 20:44