Transient Analysis

Quick Review

First-Order Transient Analysis

How do recognize it?

A transient problem is one that asks you to find voltage (or current) vs. time. It also includes some instant event: a switch opens or closes, the power supply turns on, a part is suddenly pulled out the circuit, a fuse blows. Your answer can either be a formula for v(t) or i(t) or it might be a graph of v or i on the vertical axis and time on the horizontal axis.

First-order problems have only one energy storage element (either a capacitor or inductor, but not both). Note that two capacitors in parallel or series only count as one capacitor because you can find the equivalent capacitance (same with inductors).

How do you do it?

1. Find the inital condition (t=0+)
   1. Find the current for each inductor just before the switch is flipped (t=0-). Current through an inductor will remain the same for the instant just after the switch is flipped.
   2. Find the voltage over each capacitor just before the switch is flipped (t=0-). Voltage over a capacitor will remain the same the instant just after the switch is flipped.
   3. In order to find the currents and voltages suggested above, first determine if you are given the values in the problem or you have a formula for the current/voltage from a previous step. If you are not given any other method for finding the current/voltage, then you can usually assume that steady-state conditions exist up to the time the switch is flipped. This will simply the circuit considerably, since inductors look like shorts and capacitors look like opens in steady-state.
   4. Once you have the currents through each inductor and the voltage over any capacitors for the time just before the switch is flipped, then draw the circuit for just after the switch is flipped (t=0+). Each inductor will look like a current source (at its initial current), and each capacitor will look like a voltage source (at its initial current). Note that if the initial current on an inductor is 0 amps, then it will look like an open for that instant, and if the initial voltage on a capacitor is 0 volts, then it will look like a short for that instant. If your variable of interest is not the cap voltage or the inductor current, then use your normal circuit analysis techniques to find the variable of interest at the initial instant after the switch is flipped.
2. Find the final condition (t=infinity)
   1. A long time after the switch is flipped, the inductors will look like shorts and the capacitors will look like opens. Find the variable of interest (voltage/current) for that simplified circuit.
3. Find the time constant for the time between initial and final conditions ( 0 <= t <= infinity)
   1. The time constant for a circuit with a single capacitor is RC, where R is the Thevenin Resistance seen by the capacitor after the switch has flipped. The time constant for a circuit with a single inductor is L/R where R is the Thevenin Resistance seen by the inductor after the switch has flipped.
   2. For a circuit with more than one (equivalent) capacitor and/or inductor does not have a simple exponential response, but the time constant is still part of the generalized second order response. For a circuit with energy storage elements in series, the time constant is L/R, and for a parallel combination the time constant is RC. In both cases, R is the Thevenin Resistance seen by the energy storage elements.
4. Write down the final answer.
   1. For a circuit with a single energy storage element (first order transient), the formula is:
      
      where the "unknown" is the desired voltage or current (as a function of time), "final" is the final value of the unknown, "initial" is the initial value of the unknown, "t" is the time variable, and tau is the time constant. Note that for t=0, unknown = intial and for t=infinity, unknown = final.

What are some difficulties I might run into?

1. If the desired value is not the current through an inductor or the voltage over a cap, then you can still use this procedure, but the initial and final values must be derived based on what the inductor and capacitors do under instantaneous change and under steady-state. That is, you must "back up" to your desired value after seeing what the caps and inductors do in your circuit.
2. If there is more than one switch, treat each switch one at a time in order of the time they change. You can probably assume steady-state conditions before the first switch flip (unless you are told otherwise). Ignore the following switch flips when computing the final value (since the circuit doesn't know another switch flip is coming). For all the other switch flips, you must use your formula to compute what the initial conditions will be. The time constant will be different for each segment of time in between switch flips, since the cap and/or inductor will see a different Thevenin resistance depending on the switch positions.

Second Order Transient Analysis

How do recognize it?

A transient problem is one that asks you to find voltage (or current) vs. time. It also includes some instant event: a switch opens or closes, the power supply turns on, a part is suddenly pulled out the circuit, a fuse blows. Your answer can either be a formula for v(t) or i(t) or it might be a graph of v or i on the vertical axis and time on the horizontal axis.

Second-order problems have two independent energy storage elements (capacitors and/or inductors). Note that two capacitors in parallel or series only count as one capacitor because you can find the equivalent capacitance (same with inductors).

How do you do it?

1. For first-order, the only form possible was an exponential response. For second-order there are three possible forms. You can compute which one based on the following three constants that are coefficients of the differential equation that describes these types of circuits:

   Constant	Series RLC	Parallel RLC
   a	1	1
   b	R/L	1/(RC)
   c	1/(LC)	1/(LC)

   Note that in each case, R is the Thevenin resistance seen by the LC pair. If the LC pair is in parallel but the Thevenin R is in series, you may need to convert to a Norton so that all the elements are in parallel.
2. Now compare three constants to determine which of the three response forms you have:
   • if b^2 > 4*a*c then you have overdamped.
   • if b^2 = 4*a*c then you have critically damped.
   • if b^2 < 4*a*c then you have underdamped.

Here's what they look like:

Notice that underdamped looks very similar to the first-order response (it is hard to tell the difference, but in this case, it is actually two exponentials added together). The critically damped response looks like an exponential except it overshoots the target and falls back. The complex response is a combination of a sine wave and an exponential -- sometimes this is called a damped oscillation.

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© Steven H. VanderLeest, 2008
This page was written and is maintained by Steve VanderLeest.
It was last modified on 21-Apr-2008 .