First-order control systems: feedback, time constant, and steady-state error

Who this is for: Beginners who have seen Laplace transforms and transfer functions, or want a guided first example.

What you will learn: form a unity negative-feedback transfer function, identify its time constant, and calculate its final output and steady-state error for a unit step.

Prerequisites: Basic algebra, the Laplace transform of a unit step, and familiarity with poles and transfer functions are helpful.

Model assumptions: This is a continuous-time, linear time-invariant mathematical model with zero initial conditions, an ideal proportional controller, unity negative feedback, and a stable closed loop. It is not a physical measurement or a design guarantee.

Read the feedback relationship

Let the reference input be R(s), the output be Y(s), the plant be G(s), and the controller gain be Kp. With unity negative feedback, the error is the reference minus the output:

E(s) = R(s) - Y(s)
Y(s) = Kp G(s) E(s)
T(s) = Y(s) / R(s) = Kp G(s) / (1 + Kp G(s))

The plus sign in the denominator follows from the negative-feedback relationship. A non-unity feedback path would require its transfer function in the denominator.

Worked example: close the loop

Use a first-order plant and a proportional controller:

G(s) = 1 / (2s + 1)
Kp   = 3

Substitute them into the unity negative-feedback formula and simplify:

T(s) = 3G(s) / (1 + 3G(s))
     = 3 / (2s + 1 + 3)
     = 3 / (2s + 4)
     = 0.75 / (0.5s + 1)

The closed-loop pole is at s = -2, so the response is stable. In the standard first-order form K / (tau s + 1), the closed-loop gain is 0.75 and the time constant tau is 0.5 seconds.

Find the unit-step response

A unit step has R(s) = 1/s. The output is:

Y(s) = T(s)R(s)
     = 3 / (s(2s + 4))

y(t) = 0.75(1 - e^(-2t)), for t >= 0

The final value is 0.75, not 1. The steady-state error for a unit-step reference is therefore 1 - 0.75 = 0.25. For this system, the same result follows from the closed-loop DC gain: T(0) = Kp / (1 + Kp) = 3/4.

At one time constant, t = 0.5 seconds, y(t) is approximately 0.474, or 63.2% of its final value. At five time constants, t = 2.5 seconds, it is approximately 0.745, about 99.3% of its final value. These are rounded values from the stated equation.

Practice: change the controller gain

Keep G(s) = 1/(2s + 1), but set Kp = 1. Find the closed-loop transfer function, time constant, final output for a unit step, and steady-state error.

What this example does not tell you

• A transfer function is a model; actual systems can have delays, nonlinearities, noise, disturbances, actuator limits, and parameter changes that are not represented here.
• The calculation checks the response of this one ideal model. It does not choose a safe gain for hardware or certify stability when the model changes.
• Higher proportional gain can affect stability and practical behavior in ways this simple example does not cover. Verify a real system with an appropriate model, analysis, and safety process.

References

• University of Michigan Control Tutorials for MATLAB and Simulink: System Modeling
• University of Michigan Control Tutorials for MATLAB and Simulink: Introduction