pole-zeroplottransfer function
Pole is defined as the value at which value of a function becomes infinity.
Let's say I have one LTI system with transfer function H(s)=1/s+1.
How the system will behave when s=-1.
From Bode plot I can see at pole the slope of Bode plot changes at pole frequency.
Can someone clear this thing?
Bode plot is not a graph that plots the transfer function (\$H(s)\$) against \$s\$. \$H(s)\$ is a complex function and its magnitude plot actually represents a surface in Cartesian coordinate system. And this surface will have peaks going to infinity at each poles as shown in figure:
Bode plot is obtained by first substituting \$s= j\omega\$ in \$H(s)\$ and then representing it in polar form \$H(j\omega) = |H(\omega)|\angle\phi(\omega)\$. \$H(\omega)\$ gives the magnitude bode plot and \$\phi(\omega)\$ gives the phase bode plot.
Bode magnitude plot is the asymptotic approximation of the magnitude of transfer function (\$|H(\omega)|\$) vs logarithm of frequency in radians/sec (\$\log_{10}|\omega|\$) with \$|H(s)|\$ (expressed in dB) on y-axis and \$\log_{10}|\omega|\$ on x-axis.
Coming to the questions:
At poles, the complex surface of \$|H(s)|\$ peaks to infinity not \$|H(\omega)|\$.
When a system is fed with pole frequency, the cosponsoring output will be having the same frequency but amplitude and phase will be changing. The value can be determined by substituting the frequency in radians/sec in \$|H(\omega)|\$ and \$\phi(\omega)\$ respectively.
A pole at -2 rad/sec and 2 rad/sec have the same effect on \$|H(\omega)|\$. And our interest is in the frequency response. So we need only positive part of it.
Given that your first equation has the correct signs, \$ a\$ must be negative otherwise it's an unstable pole and \$ s\rightarrow j\omega\$ is not a valid operation, i.e. there is no steady state, therefore a frequency response does not exist.
Hence it would be better (less confusing) to let the complex pole be: $$ H(s)=\dfrac{1}{s+(a-jb)}\:, \:a>0$$
Now, determining the amplitude Bode plot for this isolated complex pole is mathematically simple, but in practice it's meaningless. Rather, you need to consider the frequency response of the complex conjugate pair.
To illustrate the problem, the DC gain of the single pole (obtained by taking \$\small s=0)\$ would be \$\small \dfrac{1}{(a-jb)}\$. A complex gain is not something that is physically realisable.
However, including the complex conjugate pole in the formulation would give a real DC gain of \$\small \dfrac{1}{(a^2+b^2)}\$, which complies with the 2nd order TF: \$\small \dfrac{1}{s+(a-jb)}.\dfrac{1}{s+(a+jb)}=\dfrac{1}{s^2+2as+(a^2+b^2)}\$
Best Answer
This picture might help: -
Along the top of the picture there are three bode plot examples of the magnitude response for a 2nd order low pass filter. These are just examples that show how the damping ratio (\$\zeta\$) affects the peak of the response.
Bottom left shows the fuller picture where you can see the bode plot and pole zero plot together. Finally, bottom right is the conventional pole zero diagram (as viewed from above in the previous diagram).
So, if you have a pole at -1, that pole exists along the \$\sigma\$ axis and is at a frequency where jw = 0. Because the \$\sigma\$ axis is concerned with damping (al la \$\zeta\$ or its inverse Q/2) the further to the left you travel the more damping there is.
A lot of systems will begin to turn oscillatory as the pole advances towards and aligns with the jw axis. If the pole advances further then almost certainly, a system will become unstable and oscillate.
Hopefully this will help or maybe trigger a related question that I should be able to clarify.