Transients formulas

Holds whenever $i_C$ and $v_L$ contain no impulse. It can fail in two ways: a source that itself contains an impulse (such as $\delta (t)$), or a switching that creates a loop of capacitors and ideal voltage sources whose voltages do not already satisfy KVL, or a cut-set of inductors and ideal current sources whose currents do not already satisfy KCL (see the tip below). Put in terms of charge and flux linkage: $q_C=C{v}_C$ and ${\lambda }_L=L{i}_L$ are continuous.

Find $i_C(0^{+})$ and $v_L(0^{+})$ from the $t=0^{+}$ circuit, using KCL and KVL with capacitors replaced by voltage sources and inductors by current sources. A second-order circuit needs this as its second initial condition. For the derivative of a resistor quantity, differentiate its KVL or KCL equation and substitute these values.

$R_{th}$ is the resistance seen from the capacitor or inductor terminals with independent voltage sources shorted and independent current sources opened. Keep dependent sources and find $R_{th}$ with a test source, $R_{th}=v_{test}/i_{test}$. Use the circuit that exists after the switching.

Valid for any voltage or current in a first-order circuit with constant sources. $t_0$ is the switching instant (usually $0$). Use the $0^{+}$ value, not the $0^{-}$ value, for anything other than $v_C$ and $i_L$.

DC source $V_s$ switched on at $t=0$ with $v_C(0^{-})=V_0$. $i$ flows from the source into the + terminal of the capacitor. $V_0=0$ gives charging, $V_s(1-e^{-t/RC})$; $V_s=0$ gives discharge, $V_0{e}^{-t/RC}$.

DC source $V_s$ switched on at $t=0$ with $i_L(0^{-})=I_0$. $I_0=0$ gives build-up, $\frac{V_s}{R}(1-e^{-Rt/L})$; $V_s=0$ gives decay, $I_0{e}^{-Rt/L}$. $v_L$ jumps to $V_s-R{I}_0$ at $t=0^{+}$; if the inductor was in DC steady state before switching, $v_L(0^{-})=0$.

The tangent at the start reaches the final value after exactly one time constant. The second form gives the instant at which $x$ reaches a stated value between $x(t_0^{+})$ and $x(\infty )$.

For a first-order step response. The rise time is the time to go from 10 to 90 percent of the change; $t_{50}$ is the time for half the change.

For the source-free series loop (a voltage source is replaced by a short). $\alpha$ is the damping coefficient (neper frequency, in s${}^{-1}$) and ${\omega }_0$ is the undamped natural frequency (rad/s).

For the source-free parallel circuit (a current source is replaced by an open). It has the same ${\omega }_0$ as the series case but a different $\alpha$: here a larger $R$ means less damping.

Both cases have the form $s^2+2\alpha s+{\omega }_0^2=0$. ${\omega }_d$ is real only when $\alpha <{\omega }_0$. The roots multiply to $s_1{s}_2={\omega }_0^2$ and add to $-2\alpha$.

A series RLC is overdamped for $R>2\sqrt{L/C}$ and underdamped for $R<2\sqrt{L/C}$. A parallel RLC is the other way round: overdamped for $R<\frac{1}{2}\sqrt{L/C}$, underdamped for $R>\frac{1}{2}\sqrt{L/C}$.

DC source $V_s$ applied at $t=0$, with $i(0^{-})=0$ and $v_C(0^{-})=V_0$; $i$ flows from the source into the + terminal of $C$. All three cases follow from $i(0^{+})=0$ and $\frac{di}{dt}(0^{+})=\frac{V_s-V_0}{L}$. With $R=0$: $i(t)=(V_s-V_0)\sqrt{C/L}\sin {\omega }_{0}t$.

Same conditions as the current formulas above. Each form satisfies $v_C(0^{+})=V_0$ and $\frac{d{v}_C}{dt}(0^{+})=i(0^{+})/C=0$, and $i=C\frac{d{v}_C}{dt}$ gives back the matching current. With $R=0$: $v_C(t)=V_s-(V_s-V_0)\cos {\omega }_{0}t$.

Same conditions, with $V_s>V_0$, so the first extremum is the maximum (for $V_s<V_0$ the same instant gives the lowest point, below $V_s$). The extrema of $v_C$ fall at $t=k\pi /{\omega }_d$ ($k=1,2,\dots$), the instants where $i=0$. Successive maxima are one damped period $2\pi /{\omega }_d$ apart, and the overshoot above $V_s$ shrinks by the factor $e^{-2\pi \alpha /{\omega }_d}$ from one maximum to the next. The fractional overshoot is the peak-overshoot formula of the standard second-order system.

A capacitor charged to $V_0$ is connected across an uncharged inductor at $t=0$; $i$ is the current leaving the + terminal of $C$ and entering $L$. The energy swings between $\frac{1}{2}C{V}_0^2$ and $\frac{1}{2}L{I}_{\max }^2$ every quarter period. An uncharged LC switched onto a DC source $V_s$ gives $v_C=V_s(1-\cos {\omega }_{0}t)$, which peaks at $2V_s$.

Series model: impedance $sL$ in series with a voltage source $L{i}_L(0^{-})$ that drives current in the direction of $i_L$ (its + terminal is on the side where $i_L$ leaves). Parallel model: impedance $sL$ in parallel with a current source $i_L(0^{-})/s$ pointing in the direction of $i_L$.

Series model: impedance $\frac{1}{sC}$ in series with a voltage source $v_C(0^{-})/s$ of the same polarity as $v_C$. Parallel model: impedance $\frac{1}{sC}$ in parallel with a current source $C{v}_C(0^{-})$ whose arrow points toward the + terminal of $v_C$ (the charged capacitor pushes current out of its + terminal).

For a rational $X(s)$, the initial-value theorem needs $X(s)$ to be strictly proper; otherwise $x(t)$ has an impulse at $t=0$. The final-value theorem is valid only when every pole of $sX(s)$ lies in the open left half-plane. It gives a wrong answer for a sinusoid or a growing response.

Impulse and step responses are defined with every initial condition zero. The zero-state response to any input is ${\mathcal{L}}^{-1}\{H(s)X(s)\}$, the convolution of $h(t)$ with the input. The complete response adds the zero-input response produced by the initial-condition sources. Every pole of $H(s)$ is a natural frequency of the circuit; a natural frequency can be missing from a particular $H(s)$ when it cancels against a zero, but it still appears in the zero-input response.

Time-shift property, with $T>0$. A pulse of height $V$ and width $T$ applied at zero initial state gives, by linearity and time invariance, the step response minus a copy delayed by $T$ (each copy is zero before its own step). For the series RC with output $v_C$: $v_C=V(1-e^{-t/RC})$ for $0\le t\le T$, and $v_C=V(1-e^{-T/RC})e^{-(t-T)/RC}$ for $t\ge T$.

A current impulse $Q\delta (t)$ into a capacitor raises its voltage by $Q/C$. A voltage impulse $\Lambda \delta (t)$ across an inductor raises its current by $\Lambda /L$. When the impulse is the only source, the circuit is source-free after $0^{+}$, so the impulse response is the zero-input response from these new initial values.

$x_{ss}(t)$ is the steady-state response (from phasors, for a sinusoid). This generalises the DC formula, where $x_{ss}=x(\infty )$. The transient term vanishes when $x_{ss}(0^{+})=x(0^{+})$.

With $i(0^{-})=0$. The steady-state current lags the voltage by $\varphi$. The exponential (DC offset) term starts at $-\frac{V_m}{|Z|}\sin (\theta -\varphi )$.

With $v_C(0^{-})=0$. The steady-state current $\frac{V_m}{|Z|}\sin (\omega t+\theta +\varphi )$ leads the voltage by $\varphi$. The current's own transient term is $-\frac{V_m}{|Z|}\frac{\cos (\theta +\varphi )}{\omega RC}{e}^{-t/RC}$, which gives $i(0^{+})=\frac{V_m\sin \theta }{R}$.

Both come from one rule: there is no transient when the steady-state value of the state variable ($i_L$ or $v_C$) at the switching instant equals its actual initial value, which is zero here. For RL, switch when the steady-state current passes through zero; for RC, switch when the steady-state capacitor voltage does.

The transient is largest when the switch closes at a peak of the steady-state current. About half a cycle later, the total current's first peak approaches twice the steady-state amplitude, provided the time constant is long compared with the period.

These depend only on the present $v_C$ or $i_L$, which is why those two quantities are the continuous ones.

From $v_C=V_0$ to the final value $V_s$, through a resistor of any value. Starting uncharged, the source delivers $C{V}_s^2$ and exactly half of it is dissipated. If $V_0>V_s$ the capacitor discharges toward $V_s$, and the same $W_R$ holds.

The power $i^{2}R$ decays with time constant $\tau /2$. For an RL decay ($\tau =L/R$) the total is $\frac{1}{2}L{I}_0^2$, and for an RC discharge from $V_0$ it is $\frac{1}{2}C{V}_0^2$. After one time constant, $1-e^{-2}=0.865$ of the stored energy has been dissipated.

For $C_1$ at $V_1$ and $C_2$ at $V_2$ connected in parallel, through a switch or a resistor. $V_1$ and $V_2$ are measured across the same pair of joined terminals, so a capacitor connected with reversed polarity enters with a negative voltage. The loss does not depend on the connecting resistance $R$, which only sets the time constant $R\frac{C_1{C}_2}{C_1+C_2}$.