In Laplace transform, we defined 's' to be a complex number and wrote it as s= sigma+jw. But in frequency response, why do we zero-out the real part(sigma)?
Few sites on the internet told it to be the 'removal of transients', but I'm not fully satisfied with the answer. It would be great if someone explains it in detail. Thanks in advance
Hi Sanjeev C Achar
The real component in the complex variable s = σ + jω represents the decay effect e–σ·t. When we talk about the frequency response of a system, we want to study the steady-state response of the system to a sinusoidal input signal. Since there is no damping in the sinusoid, naturally σ = 0, and we substitute s = jω in the closed-loop transfer function.
Hi Sanjee,
Yew-Chung's answer is correct and accurate. Putting it in a different way: take any Table of Laplace transforms and you will see that a pure sine wave is represented by purely imaginary poles, that is, poles on the jw axis. The value of the frequency of the sinusoid coincides with the position of the pole. So, as we evaluate the Laplace transform on the jw axis, we are actually, evaluating the Fourier transform. Remember that the frequency response is defined as the Fourier transform of the impulse response. The frequency response is how a system responds to pure frequencies, that is, pure sinusoids in steady state. The same is valid for discrete-time systems if we replace the jw axis for the circumference of radius equal to one.
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