The circuit in the transformed domain
In the Laplace domain, a charged capacitor is an impedance 1/sC in series with a source v(0)/s. The initial conditions stop being a separate problem and become part of the circuit.
01Why transform
A circuit with inductors and capacitors is described by differential equations, and solving them in the time domain requires setting up the homogeneous and particular solutions and fitting constants to the initial conditions. The Laplace transform turns derivatives into products by \( s \) and differential equations into algebraic ones. The entire circuit can be transformed element by element and solved with the same tools as a resistive circuit: meshes, nodes, Thévenin. In Circuit Theory I the s-plane was used in transient response; here the method is completed.
The foundation is that the circuit be linear and time-invariant: its components do not change value and the response to a sum of excitations is the sum of the responses. That is what the theory of LTI systems requires.
02The transformed elements
| Element | In time | In the s domain |
|---|---|---|
| Resistor | \( v = R\,i \) | \( V = R\,I \) |
| Inductor | \( v = L\, di/dt \) | \( V = sL\,I - L\,i(0) \): impedance \( sL \) in series with a source \( L\,i(0) \) |
| Capacitor | \( i = C\, dv/dt \) | \( V = I/(sC) + v(0)/s \): impedance \( 1/sC \) in series with a source \( v(0)/s \) |
The energy stored at the start appears as additional sources, and the transformed circuit is solved in one go with all the sources. The complete response breaks down into two parts that can be calculated separately and added:
03The poles decide the shape
Each pole \( p \) of the response contributes a term \( K e^{pt} \). A negative real pole gives a decaying exponential; a complex conjugate pair \( -\alpha \pm j\omega_d \), a damped oscillation; a pole in the right half-plane, a growing response: an unstable circuit. In the series RLC:
A 5 V step at \( t = 0 \) on a series RLC with \( L = 10 \) mH and \( C = 1\ \mu \)F (resonance at 1.59 kHz; critical damping with 200 Ω). The capacitor can start charged and the inductor with current. At the top, the capacitor voltage; on the right, the poles in the s-plane.
04Transfer function and frequency response
With zero initial conditions, the ratio of the transformed output to the transformed input is the transfer function \( H(s) \). Evaluated on the imaginary axis, \( s = j\omega \), it gives the frequency response; its inverse transform, the impulse response. A single function then describes both the transient and the steady-state behavior: the same poles that make the step response oscillate produce the resonance peak in the Bode plot.
05Where it appears in electronics
| Circuit | What is analyzed in the s domain |
|---|---|
| LC output filter of a switching power supply | The voltage overshoot in response to a load step |
| Motor start-up circuit | The inrush current with the capacitor discharged |
| Snubber of a switching transistor | The damping of the oscillation between the parasitic inductance and the capacitance of the transistor |
| Control loop | The closed-loop poles and stability |
06In the lab
Build a series RLC with 10 mH and 1 µF, drive it with a low-frequency square wave and observe the voltage across the capacitor with 40 Ω, 200 Ω and 600 Ω. Measure the overshoot and the oscillation frequency.
Charge the capacitor to 3 V with a source and a switch, switch over to the circuit and compare the response with that of the lab.
Measure the frequency response of the same circuit and relate the resonance peak to the overshoot of the step response.
07Common mistakes
- Forgetting the initial-condition sources. The transform of the derivative brings in the initial value.
- Getting the sign of the inductor source wrong. The direction of the initial current has to be respected.
- Looking for poles in the numerator. The poles are the roots of the denominator; those of the numerator are zeros.
- Applying the final value theorem to an unstable system. It only holds if all the poles, except perhaps one at the origin, are in the left half-plane.
08Self-assessment
What resistance gives critical damping with 10 mH and 1 µF?
\( 2\sqrt{L/C} = 2\sqrt{10^4} = 200\ \Omega \).
How is a capacitor charged to 3 V represented in s?
An impedance \( 1/sC \) in series with a source of \( 3/s \).
What overshoot does the step response have with ζ = 0.2?
\( e^{-\pi\zeta/\sqrt{1-\zeta^2}} \approx 0.53 \): 53 %.
What shape does the response have if the poles are −1000 and −4000 s⁻¹?
The sum of two decaying exponentials, with no oscillation: overdamped.
09Further reading
- Mac E. Van Valkenburg. Network Analysis. 3rd ed., Prentice Hall, 1974 (Spanish edition, “Análisis de redes”, Limusa). The classic of network analysis with the Laplace transform, network functions and poles.
- Charles K. Alexander and Matthew N. O. Sadiku. Fundamentals of Electric Circuits. 6th ed., McGraw-Hill, 2017 (Spanish edition, “Fundamentos de circuitos eléctricos”, 2018). The transformed circuit step by step, with many worked examples.
- William H. Hayt, Jack E. Kemmerly and Steven M. Durbin. Engineering Circuit Analysis. 8th ed., McGraw-Hill, 2012 (Spanish edition, “Análisis de circuitos en ingeniería”). The complete response, complex frequency and the transfer function.