# Frequency response of marginally stable LTI systems

The frequency response of a system is defined as: $$\int_0^\infty{h(t)e^{-j\omega t}dt}$$ where $h(t)$ is the impulse response. But in marginally stable systems, $h(t)$ does not decay so the integral doesn't converge. Does the frequency response "exist" in this case?

• that bottom limit of 0 on the integral is okay as long as everyone is on board with the stable LTI system also being causal. i think it would be better, for generality, to put the bottom limit at $-\infty$. Jun 11 '18 at 4:37

## 2 Answers

As long as the system is linear and time-invariant, it can be described using the frequency response / transfer function $H(j\omega)$. For marginally stable systems, $H(j\omega)$ does exist, but cannot be expressed in terms of the Fourier-transformed impulse response $h(t)$.

The Form $H(j\omega) = \frac{Y(j\omega)}{X(j\omega)}$ however is still valid for systems not stable in the more narrow sense simply because they are both linear and time-invariant.

• In this lecture: web.stanford.edu/~boyd/ee102/freq.pdf, the frequency response is used to solve the forced response of a system due to a sinusoidal excitation. You can see that $H(j\omega)$ was directly substituted for $\int_0^\infty{h(\tau)e^{-j\omega\tau}d\tau}$. What bugs me is that I can use this derivation to find the forced response of marginally stable systems and I get the correct answer, as if I solved it through undetermined coefficients if I instead formulated an ODE. Jun 11 '18 at 1:42
• @mjtsquared I can't find it now, but I remember there was a question(+answer) that explained why the frequency response of an unstable system is irrelevant to time-domain response, or vice-versa. Jun 11 '18 at 5:48

Same problem occurs with the Laplace transform when the system is not stable. There is a region of convergence associated with these transforms but somehow we still talk about them as if the integral is converged right?

This is possible thanks to a concept called analytical continuation of analytic functions. And more impressively if you constrain the continuation to be conformal (circles are mapped to circles or angles are preserved) then this mysterious continuation is also unique.

Hence, the concrete answer is : The transform does not converge and yes the transform does not exist with this formulation. However, we have mathematical tools to extend this definition to domains where originally the operation is not defined.