# Time Domain Behavior of First Order Systems to a Step Change at t=0

The important thing to remember is that the speed of a first order system is determined by its time constant (which determines the location of the pole of the transfer function).  The final value to a step input is determined by the value of the transfer function at s=0.

It is usually straight forward to find the step response for any first order system.  The analysis is based upon two facts:

1. The time constant for a first order system is either τ=RC (for a system with resistors and capacitors), τ=L/R (for a circuit with inductors).  For mechanical systems the time constants are τ=M/B or τ=B/K.
2. 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, voltage, position, velocity...) at t=0+ (i.e., just after t=0), and v(∞) is the steady state value of the quantity.

Typically v(0), v(∞) and τ are all easily determined either physically (see below) or from the transfer function.

First order transfer functions are of the form:

or

where N(s) is either a first order polynomial or a constant.  The time constant is determined by the location of the pole of the transfer function (pole location is at -1/τ.  The further the pole is from the origin, the shorter the time constant and the faster the response.

Consider a unit step input.  The initial and final values of the response can be obtained by the initial and final value theorems.

and if the system has a steady state value (systems with negative time constants grow exponentially with time and have no steady state value) then

In particular, the steady state value of the output is equal to the value of the transfer function at s=0.  So if there is a zero at the origin, the steady state value is zero...

### First Example

 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: Therefore 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:

### Second Example

Consider the circuit shown.  The input is a step of magnitude F1 at t=0, and the mass is initially at rest.  Velocity is positive to the right.

The time constant is M/B.

The velocity at steady state is F1/B (the mass has no effect).

The velocity at t=0+ is equal to zero (the velocity of the mass cannot change instantaneously).

So

#### Using Transfer Functions

Assuming fi(t)=u(t)=unit step (F1=1).  The transfer function of the system is given by:

### Third Example

 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 because is the resistance seen from the terminals of the inductor.  Also: and ix(t) can be found as in the previous example.