frequency response in control systems

ericgibbs

Joined Jan 29, 2010
13,256
hi avyan.
This explains
E
The reason why S=jω is chosen to evaluate AC signals is that it allows to convert the Laplace transform into Fourier transform. The reason is that while S is a complex variable, what's used in the Fourier representation is just the rotational (imaginary) component, hence σ=0.
 

Papabravo

Joined Feb 24, 2006
16,498
Setting s=jω allows us to evaluate the magnitude of a transfer function along the jω axis as a function of frequency alone. We could also evaluate it along another path in the left half plane which would then have an exponential decay as well as a dependence on frequency. We can of course do the same thing with phase. When you do this you get the familiar Bode plots.
 

Thread Starter

avyan reddy

Joined Oct 30, 2020
7
hi avyan.
This explains
E
The reason why S=jω is chosen to evaluate AC signals is that it allows to convert the Laplace transform into Fourier transform. The reason is that while S is a complex variable, what's used in the Fourier representation is just the rotational (imaginary) component, hence σ=0.
thank you
 

Thread Starter

avyan reddy

Joined Oct 30, 2020
7
Setting s=jω allows us to evaluate the magnitude of a transfer function along the jω axis as a function of frequency alone. We could also evaluate it along another path in the left half plane which would then have an exponential decay as well as a dependence on frequency. We can of course do the same thing with phase. When you do this you get the familiar Bode plots.
thank you
 
Top