Transient response of second-order circuits [Jaeyoung Wiki]

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====== Transient response of second-order circuits ======

The solutions of all time-dependent circuits look like this:

$$ x(t) = x_p + x_h(t) $$

where $x_p$ is the steady state or particular solution, and $x_h(t)$ is the transient or homogeneous solution. We can get the steady-state solution by replacing capacitors with opens and replacing inductors with shorts, just as in the first-order case. Finding the transient solution takes a bit more effort and is the subject of these notes.

All second-order circuits (in fact, all second-order systems) have characteristic polynomials of the following form:

$$ s^2 + 2 \alpha s + \omega_0^2 $$

  * $\alpha$ is the rate of exponential decay.
  * $\omega_0$ is the **undamped** oscillation frequency.

^  ^ Series RLC circuit ^ Parallel RLC circuit ^
| $\alpha$ | $\frac{R}{2L}$ | $\frac{1}{2RC}$ |
| $\omega_0$ | $\frac{1}{\sqrt{LC}}$ | $\frac{1}{\sqrt{LC}}$ |

The roots of this characteristic polynomial are:

$$ s = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2} $$

From the roots, it follows that:

  * If $\alpha > \omega_0$, the system is overdamped - there will be no oscillations.
  * If $\alpha = \omega_0$, the system is critically damped - this is the smallest amount of damping that results in no oscillations.
  * If $\alpha < \omega_0$, the system is underdamped, and the output will oscillate while decaying.

^ Damping type ^ Condition ^ Form of transient solution ^ 
| Overdamped | $\alpha \gt \omega_0$ | $ A e^{s_1 t} + B e^{s_2 t} $ where $s_1 = -\alpha + \sqrt{\alpha^2 - \omega_0^2}$ and $s_1 = -\alpha - \sqrt{\alpha^2 - \omega_0^2}$ |
| Critically damped | $\alpha = \omega_0$ | $ A e^{-\alpha t} + B t e^{-\alpha t} $ |
| Underdamped | $\alpha \lt \omega_0$ | $ e^{-\alpha t} [ A \mathrm{cos} (\omega_d t) + B \mathrm{sin} (\omega_d t) ] $ where $ \omega_d = \sqrt{\omega_0^2 - \alpha^2} $ |

===== Characteristic impedance =====

The characteristic impedance $Z_0$ is the ratio of the peak capacitor voltage to the peak inductor current:

$$ Z_0 = \frac{v_{peak,C}}{i_{peak,L}} $$

**It should not be confused with impedance.**

In terms of component values, the characteristic impedance is given by:

$$ Z_0 = \sqrt{\frac{L}{C}} $$

===== Quality factor =====

In the time domain, the quality factor is approximately the number of times an underdamped circuit will oscillate before the peaks decay to 4% of the starting values. In terms of component values, it can be expressed as:

$$ Q = \frac{\omega_0}{2\alpha} $$

===== Finding the transient response =====

In one common type of problems, we are given a second-order circuit and are asked to find an expression for the transient response of the output of that circuit to some change in the input.

  - The first step is to identify the circuit topology and choose the correct expressions for exponential decay rate $\alpha$ and undamped oscillation frequency $\omega_0$
  - Then, determine whether the circuit is overdamped, critically damped, or underdamped. We can then write the form of solutions for the output. This form will have 2 unknowns that we need to solve for with 2 initial conditions.
  - Find 2 initial conditions. The initial value (e.g. $v(0^+)$ or $i(0^+)$) and the initial first time-derivative (e.g. $\left.\frac{dv}{dt}\right|_{t=0^+}$ or $\left.\frac{di}{dt}\right|_{t=0^+}$) of the output will work.
  - Finally, use the initial conditions to find the coefficients in the solution.

==== Example (from Final Exam Practice Set 1) ====

The figure below shows a parallel RLC circuit driven by a current source. The circuit parameters are $L = 1 \mu H$, $C = 1 \mu F$, $R = 1 \Omega$, $I = 1 A$. The switch is closed for $t < 0$, and it opens permanently at $t = 0$. (You may assume that the inductor current is zero at $t = 0_–$.) Find $v(t)$ for $t > 0$.

{{:kb:ee:parallel_rlc.png?600|}}

First, let's find the steady-state or particular solution. We can see that if we replace the capacitor with an open and the inductor with a short, the voltage $v(t)$ will be $0$. Therefore, the steady-state solution is:

$$v_p = 0$$

Now let's find the transient solution. Clearly, this is a parallel RLC circuit. That means that for this circuit:

  * $\alpha = \frac{1}{2RC} = \frac{1}{2(1 \Omega)(1 \mu F)} = 5 \times 10^5 rad/s $
  * $\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{(1 \mu H)(1 \mu F)}} = 10^6 rad/s$

Since $\alpha < \omega_0$, this circuit is **underdamped**.

That means our homogeneous solution will be of the form:

$$ v_h(t) = e^{-\alpha t} [ A \mathrm{cos} (\omega_d t) + B \mathrm{sin} (\omega_d t) ] $$

And the total solution, including the particular solution, is:

$$ v(t) = v_p(t) + v_h(t) = 0 + e^{-\alpha t} [ A \mathrm{cos} (\omega_d t) + B \mathrm{sin} (\omega_d t) ] $$

where

  * $\alpha = 5 \times 10^5 rad/s$
  * $\omega_d = \sqrt{\omega_0^2 - \alpha^2} = \sqrt{(10^6)^2 - (5 \times 10^5)^2} = 8.66 \times 10^5 rad/s$

Let's find the initial conditions. First, we can find the initial value of the solution. Since $v_c(0^+) = v_c(0^-)$ and $v_c(0^-) = 0$ (because it was shorted), 

$$v(0^+) = v_c(0^+) = 0$$

Next, we can find the initial value of the time derivative of the solution $\left.\frac{dv}{dt}\right|_{t = 0^+}$.

The constitutive relation of the capacitor, $i_C = C \frac{dv_C}{dt}$, can give us this time derivative.

Because $v(0^+) = 0$, $i_R(0^+) = \frac{v(0^+)}{R} = 0$. And because $i_L(0^+) = i_L(0^-)$ and $i_L(0^-) = 0$ (given),

$$i_L(0^+) = 0$$

Therefore, all of the current from the current source is flowing into the capacitor at time $t = 0^+$:

$$i_C(0^+) = I$$

So,

$$\left.\frac{dv}{dt}\right|_{t = 0^+} = \left.\frac{dv_C}{dt}\right|_{t = 0^+} = \frac{i_C(0^+)}{C} = \frac{I}{C}$$

Let's now find coefficients $A$ and $B$ that satisfy these initial conditions.

Starting with the $v(0^+) = 0$ condition:

$$ e^{-\alpha (0)} [ A \mathrm{cos} (\omega_d (0)) + B \mathrm{sin} (\omega_d (0)) ] = 0 $$

$$ A = 0 $$

Now we know that:

$$ v(t) = B e^{-\alpha t} \mathrm{sin} (\omega_d t) $$

Let's now use the condition $\left.\frac{dv}{dt}\right|_{t = 0^+} = \frac{I}{C}$ condition. We have to take the derivative of $v(t)$:

$$ \frac{dv}{dt} = -B \alpha e^{-\alpha t} \mathrm{sin} (\omega_d t) + B e^{-\alpha t} \omega_d \mathrm{cos} (\omega_d t) $$

Plug in $t = 0^+$:

$$ \left.\frac{dv}{dt}\right|_{t = 0^+} = -B \alpha e^{-\alpha (0)} \mathrm{sin} (\omega_d (0)) + B e^{-\alpha (0)} \omega_d \mathrm{cos} (\omega_d (0)) $$

$$ \left.\frac{dv}{dt}\right|_{t = 0^+} = B \omega_d $$

Equate the expression for $\left.\frac{dv}{dt}\right|_{t = 0^+}$ to the initial condition we found earlier.

$$ B \omega_d = \frac{I}{C} $$

$$ B = \frac{I}{C \omega_d} $$

Therefore, our solution is:

$$ v(t) = \frac{I}{C \omega_d} e^{-\alpha t} \mathrm{sin} (\omega_d t) $$

We can plug in numbers:

$$ v(t) = \frac{1 A}{(1 \mu F)(8.66 \times 10^5 rad/s)} e^{-(5 \times 10^5 rad/s) t} \mathrm{sin} ((8.66 \times 10^5 rad/s)t) $$

===== Finding component values given a plot =====

Another common type of problems gives us a plot of a transient response of an RLC circuit and asks us to find the values of components in the circuit.

If we are given a plot of a transient response, we can extract the following information:

  * Decay rate $\alpha$
  * Damped oscillation frequency $\omega_d$

$\omega_d$ is straightforward to extract. We can just consider the time between two adjacent peaks (or troughs) as the "period" $T$ (in scare quotes because this signal is technically not periodic) and then divide $2\pi$ by the period:

$$ \omega_d = \frac{2\pi}{T} $$

$\alpha$ can also be extracted from a plot. Consider two adjacent peaks, which occur at times $t_1$ and $t_2$ and have magnitudes $x_1$ and $x_2$. Then, the exponential decay can be calculated by:

$$ \frac{x_2}{x_1} = \frac{e^{-\alpha t_2}}{e^{-\alpha t_1}} $$

$$ \frac{x_2}{x_1} = e^{-\alpha (t_2 - t_1)} $$
$$ \ln{\frac{x_2}{x_1}} = -\alpha (t_2 - t_1) $$
$$ \alpha = -\frac{\ln{\frac{x_2}{x_1}}}{t_2 - t_1} $$

From $\alpha$ and $\omega_d$ we can also find $\omega_0$:

$$ \omega_0 = \sqrt{\omega_d^2 + \alpha^2} $$

With this information, we can equate $\alpha$ and $\omega_0$ to the expressions in terms of component values, which are shown in the table above.

==== Example (adapted from Final Exam Practice Set 1) ====

This problem concerns the second-order network shown below. To begin, the connection between the voltage source and the capacitor is broken, and the capacitor is connected to the RL circuit in parallel. A graph of the subsequent capacitor voltage v in volts and inductor current i in milliamps is also shown below as a function of time in microseconds. What are the approximate values of R, L, C, and the quality factor Q?

{{:kb:ee:rlc_plot_circuit.png?400|}}

{{:kb:ee:rlc_plot.png|}}

Let's start with the easiest value to extract, which is the damped frequency $\omega_d$. If we look at the capacitor voltage, we can see that there is a peak at $(0us, 2V)$ and another one at $(12.5us, 1.5V)$. So the damped frequency is:

$$ \omega_d = \frac{2\pi}{12.5us - 0us} = 5.027 \times 10^5 rad/s $$

Now onto damping $\alpha$. We can again look at the two peaks $(0us, 2V)$ and $(12.5us, 1.5V)$. The formula for calculating $\alpha$ is:

$$ \alpha = -\frac{\ln{\frac{x_2}{x_1}}}{t_2 - t_1} $$

We can plug in our values:

$$ \alpha = -\frac{\ln{\frac{1.5}{2}}}{12.5us - 0} = 2.3 \times 10^4 rad/s$$

Undamped frequency is therefore:

$$ \omega_0 = \sqrt{\omega_d^2 + \alpha^2} = 5.032 \times 10^5 rad/s $$

Because this is a parallel RLC circuit, we know that:

$$ \alpha = \frac{1}{2RC} = 2.3 \times 10^4 rad/s $$
$$ \omega_0 = \frac{1}{\sqrt{LC}} = 5.032 \times 10^5 rad/s $$

We now have two equations for three unknowns. The third equation will come from solving for and using the characteristic impedance.

$$ Z_0 = \frac{v_{peak,C}}{i_{peak,L}} \approx \frac{2V}{3.75mA} = 533.3 \Omega $$

$$ Z_0 = \sqrt{\frac{L}{C}} = 533.3 \Omega $$

Using these three equations, we can solve for R, L, and C:

$$ R = 6.702 \times 10^3 \Omega $$
$$ L = 1.061 \times 10^{-3} H $$
$$ C = 3.730 \times 10^{-9} F $$

And quality factor Q is:

$$ Q = \frac{\omega_0}{2\alpha} = 12.57 $$