Side-by-side comparison
| Parameter | Type 0 | Type 1 Control System |
|---|---|---|
| Definition | Zero poles at s = 0 in G(s)H(s) | One pole at s = 0 in G(s)H(s) |
| Position error constant Kp | Finite (= K, the open-loop DC gain) | Infinite |
| Velocity error constant Kv | Zero | Finite (= K) |
| Acceleration error constant Ka | Zero | Zero |
| Steady-state error, step input | 1/(1+Kp) — finite | Zero |
| Steady-state error, ramp input | Infinite | 1/Kv — finite |
| Steady-state error, parabolic input | Infinite | Infinite |
| Typical example | Proportional controller with purely resistive plant | PI controller, or plant with motor integrator |
| Stability concern | Easier to stabilise | Integrator reduces phase margin by 90° |
Key differences
Type 0 systems have no integrator in the forward path; they eliminate position error only if loop gain K is infinitely large — impractical. Type 1 systems have exactly one integrator, giving Kp = ∞ and zero position error for step inputs, but Kv = K (finite) so a ramp input still produces error = 1/K. Each additional integrator increases the system type by one, buys zero error for one more class of input, but costs 90° of phase margin per integrator, making stability harder to maintain.
When to use Type 0
Use a Type 0 system (proportional control only) when tracking a constant setpoint with a small offset tolerable — a simple lamp dimmer circuit with manual setpoint adjustment is adequately served by Type 0.
When to use Type 1 Control System
Use a Type 1 system when the reference is a step signal that must be tracked with zero steady-state error — an elevator position controller uses an integrating drive (Type 1) to park exactly at floor level without residual error.
Recommendation
For university exams, choose Type 1 as the default answer whenever zero position error for step input is required. Remember: one integrator = one free error class eliminated. Adding more integrators without compensators will likely destabilise the system.
Exam tip: GATE directly asks: "What is the steady-state error for a ramp input to a Type 1 system with gain K?" — the answer is 1/K, and you must also state that the velocity error constant Kv = K.
Interview tip: Interviewers expect you to instantly classify a given G(s)H(s) by counting poles at the origin and to state the corresponding steady-state error for step, ramp, and parabolic inputs from memory.