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Consider the following transfer function:

$$\frac{Y}{X} = \frac{A_0}{\omega_o^2 s^2 + 1} $$

or something similar that is supposed to represent an undamped block on a spring. I encountered the following question about it which has me flummoxed:

Linear systems, when given a pure oscillating input, are supposed to produce an output signal that is also a pure oscillating output and at the same frequency. However, an oscillating input at resonance produces an output in the form $t \cos \omega_ot$. That signal is not a single frequency. What is going on? Is this still a linear system?

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  • $\begingroup$ In this case linearity means linear in the inputs and initial conditions. It does not mean that the response is a linear function. Also, I suspect that the oscillating input signal is real, hence has two frequencies. $\endgroup$ – copper.hat Jun 11 '12 at 22:51
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It is still linear. As you say, you would expect that the system response to a "complex oscillating" input $e^{st}$ is a function of the same form, $H(s)e^{st}$, where $H$ is the transfer function.

But as you notice, there are usually a number of exceptional frequencies, or eigenfrequencies, for the system, where this is not quite true. This will happen at frequencies $s$ for which $H(s)$ blows up. In your example, the transfer function has poles at $s = \pm i/\omega_0$. (The transfer function in your example should probably have been $H(s) = \dfrac{A_0}{s^2+\omega_0^2}$ with poles at $\pm i\omega_0$.)

Note that a complex oscillation of frequency $i\omega_0$ corresponds to a cosine term: $$e^{i\omega_0 t} = \cos \omega_0t + i\sin\omega_0t.$$

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