Does the frequency response $H(j\omega)$ exist for system with unstable poles?

This question arose from a discussion between me and TimWescott in the comments here. The question is:

Does $$\H(j\omega) \$$ exist for systems with poles in the RHP (unstable poles)?

My own answer to this question is no. If there is even one pole in the RHP, then the Laplace integral doesn't converge for $$\s=j\omega \$$, the system is unstable and $$\H(j\omega) \$$ is meaningless. A quote from signal processing and linear systems by Lathi seems to support me.

Math example

Say we have a system $$\F(s)=\frac{1}{2-s} \$$ with a pole at $$\s=2 \$$. In the time domain this function is $$\f(t) = -e^{2t} \$$. Let's try to evaluate the Laplace integral for this function.

$$F(s) = \int_{0}^{\infty}f(t)e^{-st} \: dt= \int_{0}^{\infty} -e^{2t}e^{(-\sigma-j\omega)t} \: dt$$ $$F(s) = \int_{0}^{\infty}-e^{(2-\sigma-j\omega)t} \: dt$$ If we want to find the frequency response for the system, we need to evaluate the integral on the imaginary axis, where $$\\sigma = 0 \Rightarrow s=j\omega \$$. $$F(j\omega) = \int_{0}^{\infty}-e^{(2-j\omega)t} \: dt$$ As can be seen, we end up with a growing exponential term inside the integral, and $$\-e^{2t}\rightarrow -\infty \: \text{for} \: t\rightarrow \infty \$$ which causes the integral to not converge, and therefore $$\F(j\omega) \$$ is meaningless.

It might be that my reasoning is wrong, and if so, I would appreciate being corrected. I just want to understand the premise and prerequisites of the Laplace transformation.

• @ThePhoton I think he was referring to the ROC of the definite integral from t=0 to t=+∞ -- the imaginary axis s=0 is outside the ROC, so the FT wouldn't converge (?), but of course the transfer function still exists and behaves just fine, and I've been doing that for years. I also find this very confusing May 15, 2021 at 17:28
• @PeteW I guess I should instead of "$H(j\omega)$ doesn't exist" say "$H(j\omega)$ is meaningless", because, as you say, the imaginary axis outside the region of convergence, but the transfer function still exists.
– Carl
May 15, 2021 at 17:32
• Except I'm having a hard time even saying it's meaningless! It is fairly clear to me (not in a formal sense) that there is a finite gain to finite sinusoidal inputs in time domain (despite the superimposed exponential not converging!), as the code in AJN's answer demonstrates. (just ran it) May 15, 2021 at 17:33
• @PeteW And I'm having a hard time seeing how this can be :). How can $H(j\omega)$ have any meaning, if the integral used to derive doesn't spit out a proper function/number? If one has the transfer function $H(s) = \frac{1}{1-s}$, one can't simply substitute s=jw and convert $H(j\omega) = \frac{1}{1-j\omega}$, because the prerequisites of the Laplace transformation isn't met. It's kinda like using the concept of energy conservation for a free fall in physics, even though there is air resistance. The result would be wrong, because the prerequisites aren't met. I hope you understand me.
– Carl
May 15, 2021 at 17:41
• yes, I understand the contradiction completely, but my practical experience tells me something is missing. I think AJN has demonstrated it. So let's say you have a sine input - the output is a convolution of the sine, with a gain & phase, and the exponential. It is the exponential that blows up, the transformation of the sine behaves itself. Thus by superimposition, the same is true for any periodic input... May 15, 2021 at 17:44

The book you quote is a little bit misleading:

such systems (unstable) are useless for any signal processing application

It is true, but that refers to an application, or a practical usage for those filters, which is never practical (outside an educational, or amusing scope). But that doesn't mean they can't exist, because if they didn't then evaluating those transfer functions would be impossible, therefore nobody could tell you what's happening in there. It wouldn't even be possible to say that its impulse response blows up. But it is possible. Take your transfer function, for example:

$$H(s)=\dfrac{1}{2-s}\rightarrow H(j\omega)=\dfrac{1}{2-j\omega} \tag{1}$$

This can be evaluated anywhere in the Laplace domain, not just the frequency axis. The textbook only says that the integral doesn't converge, not that it's not possible to evaluate the transfer function. Your claim:

If we want to find the frequency response for the system, we need to evaluate the integral on the imaginary axis

is not the only way to do it. Just replace $$\j\omega\$$ with any frequency value and you have a result:

$$H(j)=\dfrac{1}{2-j}=\dfrac{2+j}{5}\tag{2}$$

That's a result as good as any other. Is the phase positive? Yes, but so what? Positive phase is still phase. And if all these don't convince you, then why not take time to build such a filter and see how the output explodes to ±Vcc, or gets clamped to within numerical limits. It may not be possible to use a frequency analyzer because of that, but that's only because they do not function the way a SPICE engine does (for example) -- they actually inject signals in time domain. But you actually built that thing, it exists, and it functions exactly the way it's meant to function. A software package can show you a clear frequency analysis of such a system without blinking (see AJN's answer, or use some flavour of SPICE simulator):

I forgot to label the output, but that's what's plotted, and the small window at the top is the IFFT. Surely the program would have complained when I tried to do that! It didn't, and you can see that the output blows up.

In the end, you have to draw a line between theory and practise: in theory you can have whatever transfer function you want and do pretty much anything to it. In practise, such a filter would, indeed, be useless, and evaluating its frequency response, practically, would be impossible because we don't have the luxury to test in frequency domain -- everything we do starts in the time domain, one way or another.

Say we have a system $$\F(s)=1/(2−s)\$$ with a pole at $$\s=2\$$. In the time domain this function is $$\f(t)=−e^{2t}\$$.

No. The sentence started with system, and suddenly switched to function. To get the (time domain) function, an input was assumed to be applied to the system; viz, the impulse input.

What is done next, is the Laplace transform of the impulse response (and not the system). Of course, for this unstable system, the impulse response diverges faster than any signal represented on the imaginary axis (pure sinusoids) will converge. So the Laplace transform doesn't exist. From this, you cannot conclude that the evaluation of $$\H\$$ at $$\s=j\omega\$$ is meaningless.

Even for unstable system, the response to an input has two parts; a natural response (zero input response) and a forced response (zero state response). Evaluation of $$\H\$$ at $$\s=j\omega\$$ still represents the forced response part IIRC.

A confirmatory simulation

sys = tf([1], [1, -0.1]);

time = [0:0.001:5]'; u = sin(2*pi*10*time);

% evaluate expected forced response at w=2*pi*10 rad/s
[m, p] = bode(sys, 2*pi*10)
% m =  0.015915

% simulate
% since our input sinusoid isn't
% really starting from t = -infinity
% give an adhoc initial condition
% to prevent the natural response from
% running away.
lsim (sys, u, time, [-.016]);

% observe the actual response
ylim([-0.02, 0.02]);
• Hey @AJN. I know it's late but I just re-read your good answer and just wanted to clear up a misconception for future reference. It is indeed true that the total response $y_{\text{total}} = y_{\text{natural}}+y_{\text{forced}} = y_{\text{zs}}+y_{\text{zi}}$ where $y$ is the response. But you cannot set $y_{\text{natural}} = y_{\text{zi}}$ or $y_{\text{forced}} = y_{\text{zs.}}$. Their sum are equal, but the components are not.
– Carl
Aug 27, 2021 at 19:51