I understand a version of this question has been asked numerous times before but I could not find an answer to this specific one.

This is how I understood replacing s with jw. From the impulse response,

\$y(t)=\int_0^\infty h(\tau) u(t-\tau)d\tau\$ where \$y,u\$ are output, input.

Taking \$u\$ as \$e^{st}\$, \$y(t)=u(t)*H(s)\$ most of the sources Ive found say that it is easier to analyze the system for constant amplitude signals so we replace \$s\$ with \$j\omega\$ and I understand till this part.

The question is if \$H(s)\$ = \$\frac{1}{s^2+2s+2}\$ which has a stable pole at \$-1+i\$ and \$u(t)=e^{(-1+i)t}\$ wouldn't that mean the output should be unstable since the denominator of \$H(s)=0\$. Ive plotted the results on simulink and mathematica and the outputs are stable.

t vs y

Can someone tell me what I'm doing wrong.

EDIT(28-oct-19)- Figured it out, I MADE A STUPID MISTAKE \$y(t) \ne \int_0^\infty h(\tau) u(t-\tau)d\tau\$ but \$y(t) = \int_0^t h(\tau) u(t-\tau)d\tau\$

  • \$\begingroup\$ Replacing \$s\$ with \$j \omega\$ only models the case where the input is a continuous sinusoid. For any other input, you need to stay in the Laplace (\$s\$) domain. \$\endgroup\$ – TimWescott Oct 23 '19 at 23:57
  • \$\begingroup\$ The roots of the TF denominator (i.e.the poles), by definition, make the TF infinite, but that, in itself, doesn't mean that the TF is unstable. It's the location of the poles in the s-plane that determine stability. The transformation \$ s\rightarrow j\omega\$ gives the steady state response to a sinusoidal input, so that's a frequency domain representation, not time domain. \$\endgroup\$ – Chu Oct 24 '19 at 0:21
  • \$\begingroup\$ @Chu if \$H(s)=\frac{1}{s^2+1}\$ and input is \$cos(t)\$ replacing s with \$j\omega\$ makes \$H(j*1)\$ infinity. So when theres no damping the output at steady state with a sinosoidal waveform of resonance frequency takes the output of the system to infinity. Why doesn't the same hold when roots are not on the \$j\omega\$ line. \$\endgroup\$ – Absolut Oct 24 '19 at 0:30
  • \$\begingroup\$ With your H(s) the poles are on the imaginary axis, therefore critically stable and the system will oscillate at 1 rad/sec with no input. It is not unstable, and nothing you apply can make it unstable. Stability has nothing to do with the input signal.If the system is linear and stable, applying a sinusoid will give a steady state output sinusoid at the same frequency. That sinusoid is not of infinite amplitude. \$\endgroup\$ – Chu Oct 24 '19 at 0:36
  • \$\begingroup\$ What would be the magnitude of output with respect to input magnitude when I apply an input sinusoid of resonance frequency be in this case? given \$H(j\omega)=\frac{1}{0}\$ \$\endgroup\$ – Absolut Oct 24 '19 at 0:43

\$y(t)=\int_0^\infty h(\tau) u(t-\tau)d\tau <=> Y(s) = H(s)U(s)\$,

\$H(s)\$ = \$\frac{1}{s^2+2s+2} = \frac{1}{(s+1+i)(s+1-i)}\$

Modifing the input \$u(t)\$ so that it is actually a real function (as suggested by Chu).

$$ u(t) = \frac{e^{(-1+i)t} + e^{(-1-i)t}}{2}, $$

$$U(s)= \frac{1/2}{s+1-i} + \frac{1/2}{s+1+i}$$

$$Y(s) = \frac{1}{(s+1+i)(s+1-i)} \left( \frac{1}{s+1+i}+\frac{1}{s+1-i} \right), $$

$$ = \frac{1/2}{(s+1+i)(s+1-i)^2} + \frac{1/2}{(s+1+i)^2(s+1-i)}, $$

$$ = \frac{ \tfrac{1}{2} (s+1+i) }{(s+1+i)^2(s+1-i)^2} + \frac{ \tfrac{1}{2}(s+1-i)}{(s+1+i)^2(s+1-i)^2}, $$

$$ = \frac{\tfrac{1}{2}/2 (s+1+i+s+1-i) }{(s+1+i)^2(s+1-i)^2}, $$

$$ = \frac{\tfrac{1}{2} (2s+2) }{(s+1+i)^2(s+1-i)^2}, $$

$$ = \frac{(s+1) }{(s+1+i)^2(s+1-i)^2}, $$

The denominator of H(s) is \$ (s+1+i)^2(s+1-i)^2 \$, it is not zero.

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  • \$\begingroup\$ by the way, I plotted my result in Octave and got the same curve as the one you got. \$\endgroup\$ – jDAQ Oct 24 '19 at 0:14
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    \$\begingroup\$ U(s), as stated, does not represent a realisable function of time. \$\endgroup\$ – Chu Oct 24 '19 at 0:28

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