Connect gain and time constant to a process step response and closed-loop tuning decisions.
G(s)=τs+1K
A first-order model describes a process that moves exponentially toward a new steady value. Its gain determines how far it moves and its time constant determines how quickly.
01
The 63.2 percent point
After one time constant, an ideal first-order response has completed about 63.2 percent of its total change. After roughly four time constants it is close to steady state.
TIME VIEW 01
One time constant means 63.2 percent complete
Changing τ stretches or compresses the same exponential shape without changing the final process gain.
Fast τMedium τSlow τ
READ THE PLOTτ controls speed; it does not control final value
02
Process gain
The steady change in output divided by the applied input change estimates K. Its sign also determines whether increasing the actuator raises or lowers the measured process.
GAIN VIEW 02
Process gain changes how far the output moves
The response shape stays first-order while K scales the steady change produced by the same input step.
K = 0.5K = 1.0K = 1.5
READ THE PLOTEstimate K from steady output change ÷ input change
03
Why delay changes everything
Adding dead time does not change the final gain but postpones feedback. The controller acts without seeing the result, reducing the gain and bandwidth that remain safe.
DELAY VIEW 03
Dead time adds a silent interval before the rise
These plants have the same gain and time constant. Only the start of the response moves, but that waiting period strongly limits feedback control.