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Why are the poles of the transfer function of the sinusoidal input signal always on the left side? I know that it makes the system bounded, but is there a way to understand it conceptually without going through the maths?

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The input signal is irrelevant. All the poles of a TF must have a negative real part because each contributes \$e^{\alpha t}\$ to the response, where \$\alpha\$ is the real value of a pole, and \$\alpha\$ must be negative to give a decaying exponential. A positive value means the response goes to infinity.

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  • \$\begingroup\$ thanks, but what about poles that are zero. Doesn't it give a sinusoidal response? Why they are not acceptable? Is it because physically the response should always decay? \$\endgroup\$ – Jack Apr 15 '16 at 6:51
  • \$\begingroup\$ If so how would we explain the steady state sinusoidal response of a system if the response has to decay? \$\endgroup\$ – Jack Apr 15 '16 at 6:55
  • \$\begingroup\$ A pole at zero represents an integrator, which can, arguably, be regarded as unstable since the response to a step input will be a ramp that clearly goes to infinity. However, the response to, say, an impulse is a constant value, which is clearly not infinite. Safest thing to say is that it's marginally stable. \$\endgroup\$ – Chu Apr 15 '16 at 7:12
  • \$\begingroup\$ An imaginary pair of poles, i.e. zero real part, will give a steady state sinusoidal response, \$e^{\pm j \omega t}=cos(\omega t)\pm j\:sin(\omega t)\$ \$\endgroup\$ – Chu Apr 15 '16 at 7:19
  • \$\begingroup\$ Well, without doing any maths, the steady state response of a stable, linear system to a sinusoidal input will be sinusoidal. And the output sinusoid will have the same frequency as the input sinusoid. \$\endgroup\$ – Chu Apr 15 '16 at 7:32

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