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?


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.

| improve this answer | |
  • \$\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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.