What a transfer function is, and why s → jω

A six-chapter visual story, not a calculator. Why eˢᵗ is the one signal a system cannot reshape, how a sinusoid splits into two spinning arrows at ±jω, why Fourier is the Laplace map read along one line, why sin ωt for all time has no Laplace transform (its ROC is empty), and how switching it on adds a transient that dies to leave the frequency-response answer. A sixth chapter shows why “take FT both sides” turns L di/dt into jωL·I(jω): impedance.

The textbook recipe

Put $s = j2$ : $G(j2) = \dfrac{1}{2 + j2} = \dfrac{1}{2\sqrt2}\angle{-45^\circ}$ , so $y(t) = \dfrac{1}{2\sqrt2}\sin(2t - 45^\circ)$ .

The recipe works, but why? This lab answers it one question per chapter:

1
What is $G(s)$ actually? Not the formula: what does it do to a signal?
2
Why do we put $s = j\omega$ into $G(s)$ when the input is $\sin\omega t$ ?
3
How are the Laplace transform and the Fourier transform related?
4
Why does the Laplace transform exist for $\sin\omega t\,u(t)$ but not for $\sin\omega t$ ?
5
Is $|G(j\omega)|\sin(\omega t + \angle G)$ the whole output?
6
Why does $v = L\,\dfrac{di}{dt}$ become $V(j\omega) = j\omega L\, I(j\omega)$ after “take FT both sides”?

Chapter 1 of 6

The one signal a system cannot reshape

This chapter answers: What is $G(s)$ actually? Not the formula: what does it do to a signal?

Pick the input's s: click or drag in the plane

System pole at s = −aa = 2.0

input

\mathrm{Re}\,e^{st}

prediction

\mathrm{Re}\,G(s)e^{st}

the system, actually run

input s: $-0.5$
the input is: a decaying exponential
G(s) = 1/(s + a): $0.667$
what happens to it: × 0.667

Start with a real circuit. An RC low-pass filter obeys $y' + 2\,y = x(t)$ : $x$ is the voltage you put in, $y$ the voltage that comes out. The “Where 1/(s + a) comes from” box shows how $G(s) = \dfrac{1}{s+2}$ comes from this equation. The question is always the same: put a signal in, what comes out?

Most signals come out a different shape. Click “square wave” above the plot: sharp corners go in, rounded curves come out. To find that output you really have to solve the differential equation.

But some signals keep their shape. Click “s = −0.5: dies away”: the input is $e^{-0.5t}$ (grey dashes) and the output (blue) is exactly the same curve, only 0.667 times as tall. The circuit did not reshape it at all; it only multiplied it by a number.

Which signals do this? Every signal of the form $e^{st}$ , and that family is bigger than it sounds. A real $s$ gives a decay ( $s < 0$ ), a constant ( $s = 0$ ) or growth ( $s > 0$ ). An imaginary $s = j\omega$ gives a sinusoid, because $e^{j2t} = \cos 2t + j\sin 2t$ . A complex $s$ gives a wave that dies or grows. The buttons above the plot show each one.

That number is the transfer function. It depends only on which $s$ you fed in, so we call it $G(s)$ : $G(s)$ is what the system multiplies $e^{st}$ by. For this circuit $G(s) = \dfrac{1}{s+2}$ , and at $s = -0.5$ that is $\dfrac{1}{1.5} = 0.667$ , the scaling you just saw.

Why does this happen? Circuits are made of derivatives ( $i = C\,dv/dt$ , $v = L\,di/dt$ ), and the derivative of $e^{st}$ is $s\,e^{st}$ : the same shape again. Adding up scaled copies of one shape cannot create a new shape.

When $s$ is complex, $G(s)$ is a complex number, and multiplying by it does two things: it scales the wave by $|G(s)|$ and shifts it by $\angle G(s)$ . That is the magnitude and the angle in the $s \to j\omega$ method; chapter 2 shows why.

The hollow dots on the plot are the circuit equation solved step by step by the computer, not the formula, so you can see the formula is not a trick.

Try this

Drag $s$ along the real axis: dying, constant, growing. The output is always the input times one number.
Put $s$ on the vertical $j\omega$ axis: a pure sinusoid in, the same-frequency sinusoid out, smaller and shifted. Chapter 2 is about exactly this.
Try to drag $s$ left of the pole. You can't: there the input, traced back into the past, grows faster than the system forgets it, and the output would be infinite. That boundary returns in chapter 4.

Where 1/(s + a) comes from, in three lines

The system: $y' + a\,y = x(t)$ (an RC low-pass is this equation with $a = 1/RC$ ).

Feed $x = e^{st}$ and guess the output is the same shape, $y = K e^{st}$ . Then $y' = sK e^{st}$ :

sK e^{st} + aK e^{st} = e^{st}\ \Rightarrow\ K = \dfrac{1}{s+a} = G(s)

The guess works for every $s$ , so the multiplier is a function of $s$ : that function is the transfer function. Replace $y'$ with $s$ in the differential equation and you have read $G(s)$ straight off it.

The answer, in one line

$G(s)$ is the number the system multiplies $e^{st}$ by. Sinusoids are $e^{st}$ signals too (with $s = j\omega$ ), so a sinusoid also just gets multiplied.

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