# What is the physical meaning of Bode plot in case of unstable system?

I know that from the mathematical point of view it doesn't matter if we plot Bode diagram of stable or unstable system. It's just a function of complex value.

However from the physical point of view, Bode plot shows steady response to a sinusoidal input. If the system is unstable there is no steady response, am I right? So is it only a "magic" of mathematics that although there is no meaning, it's still useful to use it for design?

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You can interpret that Bode plot as the response to a sinusoidal input even for unstable systems, if instead of "steady state" you say "with initial conditions exactly zero".

For a stable system, those two interpretations coincide. Remember that the space of all solutions of an inhomogenous system of linear differential equations $f'(t) = Af(t) + g(t)$ can always be written as $$\lambda_1f_1(t) + \ldots + \lambda_nf_n(t) + f_p(t)$$ where $f_1,\ldots,f_n$ are (linearly independent) solutions of the homogenous system $f'(t) = Af(t)$, and $f_p$ is one particular solution of the inhomogenous system. Also remember that $\lambda_1,\ldots,\lambda_n$ are determined by the initial condition of the system. Now, if the system is stable, all the $f_1,\ldots,f_n$ decay exponentially. Thus, no matter what the initial conditions are, if $t$ gets large enough, the solution will basically just be $f_p(t)$.

For an unstable system, some of the $f_1,\ldots,f_n$ grow exponentially (or are constant), so you cannot count on their influence to vanish at some point. You can, however, simply decide to set $\lambda_1=\ldots=\lambda_n=0$ in the solution, and hence look only at $f_p$.

Take, for example, the simple system $$f'(t) = f(t) + e^{i\omega t}$$ The corresponding homogenous system has the solution $f_1(t)=e^t$ and is thus unstable. A particular solution of the inhomogenous system is $$f_p(t)=-\frac{1+i\omega}{1+\omega^2}e^{i\omega t}$$ and the frequency response (i.e. what you plot in a Bode plot) is therefore $$A(\omega) = -\frac{1+i\omega}{1+\omega^2}$$

Note how it is perfectly reasonable to interpret this as the attenuation of the sinusoidal input $e^{i\omega t}$ - it is, after all, derived from an actual solution of the differential equation. You just won't be able to measure it, because in every actual experiment the initial condition (i.e. $\lambda_1$) will never be exactly zero - and once it differs from zero only the slightest, $\lambda_1f(t)=\lambda_1e^t$ will quickly dominate the result.

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"You can interpret that Bode plot as the response to a sinusoidal input even for unstable systems, if instead of "steady state" you say "with initial conditions exactly zero"." -- Note that this is not true in general - zero initial conditions doesn't necessarily correspond to $\lambda_1=0,\lambda_2=0,\dots$. In fact, it would be quite a coincidence. Try with the your own example. Also, the $\lambda$s are, in general, polynomials in $t$ (now I'm just being picky/pedantic). –  jkn Mar 6 '13 at 19:19
As far as I know, the only situation where the state of an unstable system will not diverge if the input is a sinusoid $\sin(\omega t)$ is if $i\omega$ is a zero of the system's transfer function. –  jkn Mar 6 '13 at 19:21
@jkn Yeah, I used initial condition far too sloppily there. –  fgp Mar 20 '13 at 14:03