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The Transfer Function and Generalized Impedance

Widen the jω you used for sinusoids into the complex frequency s, and an inductor takes the impedance sL, a capacitor 1/sC. Then any circuit’s input-output relation becomes a single line of voltage division, the transfer function H(s). Learn how this one expression holds the DC gain, the frequency response, and the transient all at once.

One pole sets both the transient and the cutoff

Seen through generalized impedance, an RC low-pass is H(s) = (1/sC)/(R + 1/sC) = 1/(1+sRC). The pole that makes the denominator zero is s = −1/(RC) = −1/τ, sitting on the negative real axis of the s-plane. Slide the time constant τ to place the pole at the target s = −2. That pole is at once the speed of the transient e^(−t/τ) and the frequency cutoff ω_c = 1/τ.

Time constant τ = RCτ = 1.00
Transfer function H(s) and its pole
H(s) = 1/(1+sτ) · s = -1.00 · ω_c = 1.00
Almost −2

Widen jω into s

In sinusoidal steady state the impedance was written with jω: an inductor as jωL, a capacitor as 1/jωC. Generalize this jω into the complex frequency s = σ + jω and the inductor becomes sL, the capacitor 1/sC. Set s = jω (σ=0) and you return exactly to sinusoidal steady state; let s carry a real part σ and the growing or decaying transient is held within the same impedance. The resistor is independent of frequency, so it stays R. This one widening binds steady state and transient in the same language.

The circuit is a voltage divider in s

Through generalized impedance, two elements in series are still a voltage divider. Taking the output across the capacitor in an RC low-pass, H(s) = Vout/Vin = Z_C/(Z_R + Z_C) = (1/sC)/(R + 1/sC). Multiply top and bottom by sC and tidy up to get H(s) = 1/(1 + sRC). A single line of voltage division is the circuit’s transfer function. Drop in an inductor as sL, or take a more involved circuit, and the same division and combination of impedances build H(s) just the same.

One expression, three faces

The transfer function H(s) = 1/(1+sRC) reads three ways. First, at s = 0 (DC), H = 1, so the DC gain is 1 and slowly varying signals pass through unchanged. Second, set s = jω and H(jω) = 1/(1+jωRC), a low-pass frequency response whose magnitude bends down to 1/√2 at ω = 1/RC. Third, the pole that makes the denominator zero, s = −1/RC, sets the transient, so closing a switch lets the output settle as e^(−t/RC). One and the same expression carries the DC gain, the frequency cutoff, and the transient time constant at once. A single pole location compresses the circuit’s entire behavior.

ObserveZC = 1sC
A capacitor is 1/sC.
ChooseH(s) = ZC(ZR+ZC) = ?
The circuit is a voltage divider in s.
Fill in1+sRC = 0 → s = ?
The pole is the root of the denominator.
On your owns = jω → ωc = ?
The cutoff is the same 1/RC.

Back to the first screen

As you slid the time constant τ, the pole s = −1/τ moved along the negative real axis and reached the target s = −2 at τ = 0.5. That single pole fixed two things at once: in time, the speed at which the output settles as e^(−t/τ); in frequency, the cutoff bending at ω = 1/τ. Because widening jω into s let the capacitor be seen as 1/sC, the RC low-pass was written as one line of voltage division H(s) = 1/(1+sτ), and that one expression held the DC gain, the frequency response, and the transient together.

A transfer function H(s) = Vout/Vin widens jω into the complex frequency s, taking an inductor as sL and a capacitor as 1/sC, then writes the circuit as voltage division in s. An RC low-pass is H(s) = 1/(1+sRC), where the pole at the denominator root, s = −1/RC, sets the transient e−t/RC, s = jω gives the frequency response with cutoff ω = 1/RC, and s = 0 gives the DC gain. One expression holds all three faces — transient, frequency response, and DC gain — at once.