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Quiz: Steady-State Error Analysis

Test your understanding of steady-state error, error constants, and system type classification.

1. Steady-state error is defined as:

  1. The maximum overshoot during transient response
  2. The difference between reference input and output as time approaches infinity
  3. The initial error at $t = 0$
  4. The error at the peak time
Show Answer

The correct answer is B. Steady-state error is $e_{ss} = \lim_{t \to \infty}[r(t) - y(t)]$, the difference between the desired reference and actual output after all transients have decayed. It measures how accurately the system tracks inputs in the long term.

Concept Tested: Steady-State Error

2. The position error constant $K_p$ is used to calculate steady-state error for:

  1. Ramp inputs
  2. Step inputs
  3. Parabolic inputs
  4. Impulse inputs
Show Answer

The correct answer is B. The position error constant $K_p = \lim_{s \to 0} G(s)H(s)$ determines steady-state error to a step input: $e_{ss} = 1/(1 + K_p)$. For ramp inputs use $K_v$, for parabolic use $K_a$.

Concept Tested: Position Error Constant, Error Constants

3. A Type 1 system has:

  1. No integrators in the open-loop transfer function
  2. Exactly one integrator (pole at origin) in the open-loop transfer function
  3. Two or more integrators in the open-loop transfer function
  4. Only proportional control
Show Answer

The correct answer is B. System type equals the number of pure integrators (poles at $s = 0$) in the open-loop transfer function. Type 1 has exactly one integrator: $GH = K/[s(s+a)...]$. Type 0 has none; Type 2 has two.

Concept Tested: Type 1 System, System Type

4. For a Type 0 system responding to a unit step input, the steady-state error is:

  1. Zero
  2. $1/(1 + K_p)$ where $K_p$ is finite
  3. Infinite
  4. Equal to the DC gain
Show Answer

The correct answer is B. Type 0 systems have finite $K_p$, so steady-state error to a step is $e_{ss} = 1/(1 + K_p)$. Increasing gain reduces but never eliminates this error. Only Type 1 or higher systems can track steps with zero steady-state error.

Concept Tested: Type 0 System, Position Error Constant

5. The velocity error constant $K_v$ is calculated as:

  1. $\lim_{s \to 0} G(s)H(s)$
  2. $\lim_{s \to 0} sG(s)H(s)$
  3. $\lim_{s \to 0} s^2G(s)H(s)$
  4. $\lim_{s \to \infty} G(s)H(s)$
Show Answer

The correct answer is B. The velocity error constant is $K_v = \lim_{s \to 0} sG(s)H(s)$. For a ramp input, steady-state error is $e_{ss} = 1/K_v$. Type 0 systems have $K_v = 0$ (infinite error to ramp); Type 1+ systems have finite positive $K_v$.

Concept Tested: Velocity Error Constant

6. A Type 2 system responding to a ramp input has steady-state error of:

  1. Infinite
  2. $1/K_v$
  3. Zero
  4. Equal to the ramp slope
Show Answer

The correct answer is C. Type 2 systems have two integrators, so $K_v = \infty$, making $e_{ss} = 1/K_v = 0$ for ramp inputs. Type 2 systems can track both steps and ramps with zero steady-state error, but have finite error to parabolic inputs.

Concept Tested: Type 2 System, Steady-State Error

7. Adding an integrator to a control system:

  1. Always improves stability margins
  2. Increases the system type by one and reduces steady-state error
  3. Decreases the system order
  4. Has no effect on steady-state performance
Show Answer

The correct answer is B. Adding an integrator ($1/s$) increases the system type by one, improving steady-state accuracy (e.g., converting a Type 0 with step error into a Type 1 with zero step error). However, it adds phase lag of -90°, which can reduce stability margins.

Concept Tested: System Type, Steady-State Error

8. The steady-state error to a unit ramp input for a Type 1 system is:

  1. Zero
  2. $1/K_v$
  3. Infinite
  4. $1/(1 + K_p)$
Show Answer

The correct answer is B. Type 1 systems have finite velocity error constant $K_v$, so steady-state error to a ramp is $e_{ss} = 1/K_v$. This error can be reduced by increasing gain but cannot be eliminated without adding another integrator (making it Type 2).

Concept Tested: Type 1 System, Velocity Error Constant

9. Disturbance errors can be reduced by:

  1. Decreasing the loop gain
  2. Increasing the loop gain at frequencies where disturbances occur
  3. Removing all integrators
  4. Reducing the system bandwidth
Show Answer

The correct answer is B. High loop gain suppresses disturbances. For a disturbance entering at the plant input, the transfer function from disturbance to output is approximately $1/L(s)$ where $L(s)$ is loop gain. Higher $|L(j\omega)|$ at the disturbance frequency reduces its effect.

Concept Tested: Disturbance Error

10. The acceleration error constant $K_a$ determines steady-state error for:

  1. Step inputs
  2. Ramp inputs
  3. Parabolic inputs
  4. Sinusoidal inputs
Show Answer

The correct answer is C. The acceleration error constant $K_a = \lim_{s \to 0} s^2G(s)H(s)$ determines steady-state error to a parabolic input (constant acceleration): $e_{ss} = 1/K_a$. Only Type 2 or higher systems have finite $K_a$.

Concept Tested: Acceleration Error Constant