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From the pole-zero plot, you can compute the system frequency response by assuming a locus of test points along the \$j\omega\$ axis.

enter image description here
Figure from: http://web.mit.edu/2.14/www/Handouts/PoleZero.pdf

\begin{align} |H(j\omega)| &= K \frac{r_1\ldots r_m}{q_1\ldots q_n}\\ \angle H(j\omega) &= (\phi_1 + \ldots + \phi_m) - (\theta_1 + \ldots + \theta_n) \end{align}

This means that if I stimulate \$H(s)\$ with a steady-state sinusoidal input,

$$A\sin\omega_1t$$

at the output I'll get

$$A|H(j\omega_1)|\sin(\omega_1t + \angle H(j\omega_1))$$

Question

Evaluating \$H(j\omega)\$ means I'll get its magnitude and phase response when it is stimulated with a steady-state sinusoidal input

$$A\sin \omega t$$

If I evaluate \$H(\sigma + j\omega)\$, what kind of input does that imply? Does that mean I would stimulate the system with a decaying sinusoid?

$$Ae^{-\sigma t} \sin (\omega t + \phi)$$

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  • \$\begingroup\$ \$\sigma\$ can be either positive or negative (or zero.) If zero, you just have rotation but the vector length (magnitude) stays the same. If negative, the vector length is declining with time and is "decaying" or "spiraling inward" as the vector also rotates. If positive, the vector length is increasing with time and is "spiraling outward" as the vector also rotates. (Part of why "right-half-plane" usually isn't such a good thing.) \$\endgroup\$
    – jonk
    Commented Dec 11, 2018 at 0:18
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    \$\begingroup\$ Where have you seen \$H(\sigma + j\omega)\$ used? It's meaningless. \$\endgroup\$
    – Chu
    Commented Dec 11, 2018 at 1:59
  • \$\begingroup\$ H(σ+jω) = H(s) >> complex transfer function (meaningless?). \$\endgroup\$
    – LvW
    Commented Dec 13, 2018 at 10:08

2 Answers 2

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To answer your question: Yes - evaluating the function H(σ+jω) with the described method would - theoretically !! - mean that the input stimulus is a decaying or rising sinus. However, in practice this is not possible because you never will get a "steady-state" condition. Hence, you would remain under transient conditions - and the definition of a transfer function is not valid anymore.

Therefore, in order to allow the definition/calculation of a transfer function we need steady-state conditions which can be created with a continuous sinus stimulus only.

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The "real world" is along the \$j\omega\$ axis and, ignoring such things as negative frequencies, measurable reality is from the origin and upwards (frequency rising). The pole zero diagram vertical axis embodies the bode plot amplitude like this: -

enter image description here

The above (and below) is just an example for a 2nd order low pass filter. Looking vertically down on the diagram above gives the traditional pole-zero view: -

enter image description here

Picture source.

Basically it's meaningless to analyse \$H\$ in any other place than the axis that embodies the bode plot.

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