Questions: System Type and Steady-State Error Constants

Question 1

A Type 1 closed-loop control system tracks a ramp input r(t) = 2t. The open-loop velocity error constant is Kᵥ = 4. What is the steady-state tracking error?

• A. 0 — Type 1 systems track all ramp inputs with zero steady-state error
• B. 0.5 — from e∞ = R/Kᵥ = 2/4
• C. 2 — the steady-state error equals the ramp slope
• D. Infinite — Type 1 systems cannot track ramp inputs at all

Answer: (B)

Type 1 systems track ramps with a constant finite error, not zero error. The formula is e∞ = R/Kᵥ, where R is the ramp slope and Kᵥ = lim(s→0) s·G(s). Here, e∞ = 2/4 = 0.5. Zero ramp error requires Type 2. A Type 0 system would have infinite ramp error (grows without bound). The common misconception is that 'Type 1 tracks ramps' means zero error — it means bounded, finite error.

Question 2

A designer adds an integral term to a proportional controller, converting a Type 0 system into a Type 1 system. What is the primary benefit of this change?

• A. Faster transient response — the integrator speeds up rise time
• B. Zero steady-state error for constant (step) reference inputs
• C. Increased phase margin — the integral term improves stability
• D. The ability to track parabolic inputs with finite error

Answer: (B)

Adding one integrator upgrades the system from Type 0 to Type 1. For a Type 0 system, the step error is e∞ = 1/(1+Kₚ) — finite and nonzero regardless of gain. The integrator makes Kₚ = ∞, reducing step error to zero. The tradeoff is that the integrator adds 90° of phase lag, which typically reduces phase margin and can destabilize the loop — the opposite of option C. Parabolic tracking with finite error requires Type 2 (two integrators).

Question 3

A Type 0 feedback control system will typically achieve zero steady-state error for a constant step reference input if the open-loop gain is made sufficiently large.

• T. True
• F. False

Answer: False

For a Type 0 system, the position error constant Kₚ = lim(s→0) G(s) is finite, giving e∞ = 1/(1+Kₚ). Increasing the gain increases Kₚ and reduces the error, but e∞ approaches zero only as Kₚ → ∞, which is never achieved with finite gain. No finite gain makes e∞ exactly zero for a Type 0 system. Zero step error requires a free integrator (Type 1 or higher), which makes Kₚ truly infinite.

Question 4

Each free integrator added to the open-loop transfer function increases the system type by 1, allowing the system to track the next higher-order reference input (step, ramp, parabola) with zero steady-state error.

• T. True
• F. False

Answer: True

This is the fundamental structure of system type theory. Type 0: zero integrators → finite step error. Type 1: one integrator → zero step error, finite ramp error. Type 2: two integrators → zero step error, zero ramp error, finite parabolic error. Each integrator cancels one power of s in the error transfer function's denominator, eliminating the error for one class of inputs. The cost is increasing phase lag with each added integrator, which must be managed through compensator design.

Question 5

Why does a free integrator in the open-loop transfer function allow a closed-loop system to track a constant reference input with zero steady-state error, when a system with no integrators cannot?

Model answer: An integrator in the forward path generates an output that grows without bound in response to any persistent nonzero input (its output is the integral of its input). If a constant reference tracking error persists, the integrator accumulates that error over time, increasing the control signal until the error is driven to zero. At steady state, the only way the integrator's output can be constant (not growing) is if its input — the error — is exactly zero. A proportional-only (Type 0) system has no such accumulation mechanism: it produces a control signal proportional to the current error and can settle at a nonzero steady-state error where the control signal exactly balances the plant's needs.

Mathematically, the integrator term 1/s in G(s) means the position error constant Kₚ = lim(s→0) G(s) → ∞, giving e∞ = 1/(1+Kₚ) → 0. The tradeoff is that the integrator adds 90° of phase lag at all frequencies, reducing phase margin and potentially causing instability — which is why integral terms in PID controllers must be paired with careful gain tuning.