# Meaning of transfer function in s evaluated at a complex number

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

\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?

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

• $\sigma$ 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.)
– jonk
Commented Dec 11, 2018 at 0:18
• Where have you seen $H(\sigma + j\omega)$ used? It's meaningless.
– Chu
Commented Dec 11, 2018 at 1:59
• H(σ+jω) = H(s) >> complex transfer function (meaningless?).
– LvW
Commented Dec 13, 2018 at 10:08

The "real world" is along the $$\j\omega\$$ axis and, ignoring such things as negative frequencies, measurable reality is from the origin and upwards (frequency rising). The pole zero diagram vertical axis embodies the bode plot amplitude like this: -
Basically it's meaningless to analyse $$\H\$$ in any other place than the axis that embodies the bode plot.