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