# Understanding s-domain poles, negative frequencies and bode plots

The transfer function of a low-pass filter, $H(s)=\frac{1}{s+1}$ has a pole at $s=-1$. Setting the real part of s to 0, we have a pole at $jw=-1$. In the bode plot of this transfer function, at $\omega=1$ the gain has dropped to -3db.

What I don't understand is why the effect of the pole $jw=-1$ appears on the bode plot at $\omega=1$.

I guess I am confused about the significance and relationship between the s-domain, the jw axis, and real frequencies, and how a pole in the s-domain affects the other domains.

• With s=(sigma+jw)=-1 we have sigma=-1 (and NOT jw=-1). – LvW Oct 15 '16 at 7:42
• Setting $s\rightarrow j\omega$ means that we are ignoring the exponential transients, $e^{-\alpha t}$, leaving only the steady state sinusoidal terms, $e^{j\omega t}$ – Chu Oct 15 '16 at 12:42

I guess I am confused about the significance and relationship between the s-domain, the jw axis, and real frequencies, and how a pole in the s-domain affects the other domains.

This might help: -

The three pictures along the top show a 2nd order low pass filter magnitude response i.e. amplitude versus frequency. This is more conventionally called the bode plot and is the sort of thing you would see on a spectrum analyser.

Bottom left takes the bode plot into a 3D image - as you can see, behind the bode plot there is the $\sigma$ axis and there is a pole drawn the corresponds with the values shown.

The bottom right picture is the plan view of the 3D picture and is the standard view of the pole zero diagram.

The transfer function of a low-pass filter, has a pole at s=−1. Setting the real part of s to 0, we have a pole at jw=−1.

No, you have a pole at co-ordinates $\sigma$ = -1, jw = 0.

The Laplace transform evaluated along the jw-axis gives the frequency response of the system. The "real" frequencies are therefore mapped onto the imaginary axis!

For a low-pass transfer function the corner-frequency is defined as the frequency where the absolute value of the transfer function drops by 3dB which is equivalent to a reduction by the factor of $1/\sqrt{2}$.

The transfer function is given by $$H(s) = \frac{1}{s+1}$$ Now the question is for which $\omega$ is |H(s)| reduced by $1/\sqrt{2}$ which is answered by looking at the denominator and solving $$\sqrt{2} = |j\omega + 1| = \sqrt{\omega^2 + 1}$$ which results in $\omega = 1$.

• $\omega = -1$ also solves this equation – jagjordi Apr 2 '17 at 9:50