Why s is substitute with jw in transfer function?

[hilman]

Can anybody explain to me briefly on why in a transfer function like G(s), we substitute s=jw such that it become G(jw) in order to find its gain value?

Thanks

 

[milesyoung]

You can explain it in a variety of ways, but I prefer this one, since it's quite general:

The only difference between the bilateral Laplace transform:
$$
G(s) = \mathcal{L}\{f(t)\} = \int_{-\infty}^\infty e^{-st} f(t)dt\\
$$
and the Fourier transform:
$$
H(\omega) = \mathcal{F}\{f(t)\} = \int_{-\infty}^\infty e^{-j\omega t} f(t) dt
$$
is the variable substitution $s = j\omega$, i.e. $H(\omega) = G(s)\left.\right|_{s = j\omega} = G(j\omega)$.

For some system with transfer function $G(s)$, $G(s)$ is also its impulse response, which makes $G(j\omega)$ the Fourier transform of its impulse response.

More intuitively:
To find the gain of a linear system at some frequency $\omega$, you could apply a sinusoidal input with frequency $\omega$ and unity amplitude, and observe the amplitude of the output signal, which directly gives you the gain of the system at $\omega$.

An ideal impulse has components with unity amplitude at all frequencies, and thus $G(j\omega)$ "picks out" the gain and phase of the system at $\omega$ from its impulse response.

Makes sense?

 

[hilman]

Wow, I like your answer. I am still a beginner in this kind of things and I can understand most of your answer. But one question. What is exactly a linear system?
Thanks in advance

 

[donpacino]

Wow, I like your answer. I am still a beginner in this kind of things and I can understand most of your answer. But one question. What is exactly a linear system?
Thanks in advance

a big thing in looking at systems like this is knowing if they are linear and if they are time invarient.

you should know these

Linear:: more or less to be a linear system the sum of two inputs have to equal the sum of the two outputs
x1(n)=y1(n)
x2(n)=y2(n)
so in order to be linear
x1(n)+x2(n)=y1(n)+Y2(n)

then continue that out to inf and -inf

for a system that is time invariant, delaying the input by a constant delays the output by the same amount.

given the system x(t)=y(t)

x(t-d)=y(t-d) for all values of t and d

 

[donpacino]

a big thing in looking at systems like this is knowing if they are linear and if they are time invarient.

you should know these

Linear:: more or less to be a linear system the sum of two inputs have to equal the sum of the two outputs
x1(n)=y1(n)
x2(n)=y2(n)
so in order to be linear
x1(n)+x2(n)=y1(n)+Y2(n)

then continue that out to inf and -inf
basically, this means if you graph the system, it will be a straight line.

for a system that is time invariant, delaying the input by a constant delays the output by the same amount.

given the system x(t)=y(t)

x(t-d)=y(t-d) for all values of t and d

 

[EM_Guy]

Generally,

$s = \sigma + j \omega$

When $\sigma = 0$, $s = j \omega$

$\sigma$ is the Neper frequency. It tells us the rate at which the function decays (when it is negative).

$\omega$ is the radial frequency. It tells us the rate at which the function oscillates.