Transient response and residues
Before doing a single calculation, the location of the poles already says whether a circuit will oscillate, how long it will take to settle and whether it will overshoot. The residues only supply the numbers.
01The whole circuit, transformed into \( s \)
Steady state was solved with phasors. The transient —what happens from the moment a switch closes until the circuit settles— is solved with the Laplace transform, which does for derivatives the same thing that phasors do, but without requiring the signal to be sinusoidal or a long time to have gone by. It also incorporates, in a natural way, the initial conditions: the charge a capacitor already had or the current that was already flowing through an inductor.
| Element | In time | In \( s \), with initial condition |
|---|---|---|
| Resistor | \( v = Ri \) | \( V = RI \) |
| Inductor | \( v = L\,di/dt \) | Impedance \( sL \) in series with a source \( L\,i(0) \) |
| Capacitor | \( i = C\,dv/dt \) | Impedance \( 1/sC \) in series with a source \( v(0)/s \) |
With that substitution, the circuit is solved with the same rules as always —mesh analysis, nodal analysis, dividers— and the response comes out as a ratio of polynomials in \( s \).
02Residues: from \( Y(s) \) back to time
To go back to time, the ratio is decomposed into partial fractions, one per pole. The coefficient of each fraction is the residue of that pole, and it is calculated without solving any system of equations:
The consequence is powerful: the shape of the transient is decided by the location of the poles, before doing any calculation. The residues only adjust the amplitudes.
| Poles | Contribute |
|---|---|
| Negative real | Decaying exponentials, without oscillating |
| Complex conjugates with negative real part | An oscillation that dies out: the “ringing” |
| On the imaginary axis | An oscillation that never dies out |
| With positive real part | Something that grows without bound: instability |
03The series RLC with a step input
The case that sums everything up: a series RLC connected to a DC source \( V \). The capacitor voltage in \( s \) has three poles: one at the origin, which gives the final value, and two that depend on the components.
Series RLC with \( L = 1 \) mH and \( C = 100 \) nF, connected to 5 V. You move \( R \): on the left, the two poles in the \( s \)-plane; on the right, the capacitor voltage that results from adding up their residues.
04Where it shows up in electronics
| Case | The second-order transient |
|---|---|
| Poorly compensated oscilloscope probe | The “ringing” on the edges of a square wave is the inductance of the ground lead resonating with the input capacitance |
| Fast data lines on a circuit board | The oscillation after each edge can cross the logic threshold and produce a false bit |
| Input LC filter of a switched-mode power supply | When it is plugged in, the capacitor can overshoot the input voltage if the filter is underdamped |
| Feedback amplifier | Its step response reveals, through the overshoot, how close it is to oscillating |
05In the lab
Build the lab’s RLC with R = 100 Ω and drive it with a low-frequency square wave. Measure the overshoot of the capacitor voltage and compare it with the 16.3 % predicted by \( \zeta = 0.5 \).
Using this page’s lab, locate the value of \( R \) at which the two poles merge. Verify that it matches \( R = 2\sqrt{L/C} \).
Calculate the three residues of \( V_C(s) \) for \( R = 300 \) Ω and write \( v_C(t) \) as a sum of exponentials. Compare with the lab’s data panel.
06Common mistakes
- Forgetting the initial conditions. A charged capacitor or an inductor carrying current are sources in the transformed circuit.
- Looking for the shape of the transient in the residues. The shape is decided by the poles; the residues only set the amplitudes.
- Treating a double pole as if it were two simple ones. A double pole contributes \( t\,e^{pt} \), not two equal exponentials.
- Confusing \( \omega_n \) with the frequency at which it oscillates. The damped oscillation runs at \( \omega_d = \omega_n\sqrt{1-\zeta^2} \), somewhat lower.
- Believing that more resistance always gives a faster response. Past critical damping, the response becomes slower.
07Self-assessment
What does a charged capacitor represent in the transformed circuit?
An impedance \( 1/sC \) in series with a source \( v(0)/s \).
How is the residue of a simple pole calculated?
\( K_k = \lim_{s\to p_k}(s-p_k)Y(s) \).
What transient shape do two complex poles with negative real part give?
An oscillation that dies out exponentially.
What is the overshoot with \( \zeta = 0.5 \)?
16.3 % of the final value.
At what resistance is the damping of a series RLC of 1 mH and 100 nF critical?
\( R = 2\sqrt{L/C} = 200 \) Ω.
08Further reading
- Mac E. Van Valkenburg. Network Analysis. 3rd ed., Prentice Hall, 1974 (Spanish edition, “Análisis de redes”, Limusa). The Laplace transform applied to networks, with the residue calculation and the reading of poles that this page uses.
- Charles K. Alexander and Matthew N. O. Sadiku. Fundamentals of Electric Circuits. 7th ed., McGraw-Hill, 2021. The chapters on second-order circuits and on Laplace, with many examples of initial conditions.
- Katsuhiko Ogata. Modern Control Engineering. 5th ed., Prentice Hall, 2010 (Spanish edition, “Ingeniería de control moderna”, Pearson). The second-order system, the overshoot and the settling time, from the control point of view.