Time Domain Behavior of First Order Systems to a Step Input
It is usually straight forward to find the step response for any first order system. The analysis is based upon two facts:
The time constant for a first order system is either t=RC (for a system with resistors and capacitors) or t=L/R (for a circuit with inductors).
The response of a first order system (to a constant
input, for t>0) is given by:
where v(0) is the quantity (typically current or voltage) at t=0+ (i.e., just after t-0), and v(¥) is the steady state value of the quantity.
Typically v(0), v(¥) and t are all easily determined.
Consider the circuit shown. The switch is closed for a long time, and opens at t=0. Find vx(t), and ix(t).
Clearly, the time constant after the switch is thrown is R2C and:
Now consider the situation where the switch has been open for a long time and closes at t=0. Clearly this reverses the initial and final voltages and currents.
The difficulty lies in finding the time constant. With the switch closed, the resistance "seen" by the capacitor is simply R1 and R2 in parallel (the voltage source has an equivalent impedance of zero -- the voltage is unchanged no matter what the current does). So:
Consider the circuit shown. The switch is open for a long time, and closes at t=0. Find ix(t).
The time constant after the switch is thrown is
This is simple L/Req where Req is
is the resistance seen from the terminals of the inductor. Also:
and ix(t) can be found as in the previous example.