Analog Intuition

Divider vs RC

Scope at one \(f\) · Bode 10 Hz–10 MHz · solid = with \(R_s\) · dashed = ideal

fp —
fz —
H(s) = 1 + s 2π · ∞ 1 + s 2π · —
Input signal multiplied by H(s) gives output signal IN Vin(s) H(s) OUT Vout(s)

With ESR = 0 the zero is at infinity — H(s) is a single pole. Increase ESR to add a zero. Whether you see it depends on fz vs the test frequency.

Resistor divider

Resistive voltage divider Vin Rtop 90 Ω Vout Rbot 10 Ω
\(H = R_{\mathrm{bot}}/(R_{\mathrm{top}}+R_{\mathrm{bot}})\) — magnitude only, 0° phase.

RC low-pass

RC low-pass with capacitor ESR Vin R 10 Ω Vout Rs 0 Ω C 7.9 µF
Series \(R\), then shunt \(R_s\) in series with \(C\).

Scope — 20 kHz sine

V(in) divider RC phase lag outputs ×5 vs 1 V in

Bode · fp red · fz green · f_scope violet

Magnitude Phase Mag · Rs=0 Phase · Rs=0 f_scope
f — div — RC — Δ|H| — ESR Δφ — Hdiv —
Writes the schematic with these R, C, Rs, and sine frequency. Run is .tran (scope). Uncomment .ac on the sheet for the Bode.
Writing \(H(s)\) the right way

The schematic is a divider. That algebra is correct — and it hides the corners. Factor so each \(1+s/\omega\) is a Bode breakpoint you can point at (Basso’s low-entropy form). Drag \(R_s\); the live fractions below track it.

\[ Z_C(s)=R_s+\frac{1}{sC},\qquad H(s)=\frac{Z_C}{R+Z_C} =\frac{1+s R_s C}{1+s C(R+R_s)} =\frac{1+s/\omega_z}{1+s/\omega_p} \] \[ \omega_z=\frac{1}{R_s C},\qquad \omega_p=\frac{1}{C(R+R_s)} \]

If \(R_s=0\), \(\omega_z\to\infty\) and \(\omega_p=1/RC\) — a single pole. Finite ESR adds the zero and moves \(f_p\) a little, because the time constant is \(C(R+R_s)\), not \(RC\).

High entropy — parts hide the corners
H(s) = 1 —
Low entropy — each factor is a breakpoint
H(s) = 1 + s 2π · ∞ 1 + s 2π · —

\(f_p=\) — · \(f_z=\) ∞ · if \(R_s=0\), \(f_p=\) —