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MAE 4182November 4, 2026← Back to the lesson

After-class practice

Why calculate steady-state error?

Watch the video. Work on paper, then open each worked answer.

1

A moving gripper: is matching speed enough?

Watch the KUKA conveyor-tracking video. Focus on how the gripper follows a moving object.

KUKA · Conveyor tracking ↗Open on YouTube ↗

For this exercise, use the following simplified tracking model after the startup transient:

Object:   r(t) = t cm
Gripper:   y(t) = (t − 0.25) cm

t is measured in seconds. Required positioning accuracy: within 1 mm.

  1. Both move at 1 cm/s. Does having the same speed mean their positions match?
  2. Calculate ess = limt→∞ [r(t) − y(t)]. Express it in millimeters.
  3. Does this controller meet the accuracy requirement? Why does ess matter for controller design?
Show worked answer

  1. Compare the motion.

    Both move at 1 cm/s. Equal speeds keep their separation constant; they do not make that separation zero.

  2. Calculate the remaining error.

    e(t) = r(t) − y(t)
    = t − (t − 0.25) = 0.25 cm

    The gap is constant after the startup transient, so

    ess = 0.25 cm = 2.5 mm

  3. Check the requirement.

    2.5 mm > 1 mm. The controller does not meet the required accuracy. The gripper follows the motion but remains too far behind the object.

Why calculate ess? It tells us the remaining tracking offset. We compare that offset with the allowable error to decide whether a controller is accurate enough.

2

Repeat our classroom example

F(s) = 12 divided by s(s + 3)
  1. Rewrite F(s) into standard factors.
  2. Identify the System Type and g.
  3. Calculate Kv.
  4. Find the unit-ramp steady-state error. Explain what it means.
Show worked answer

  1. Rewrite the factor.

    s + 3 = 3(1 + s/3)
    F(s) = 12 divided by 3s(1 + s/3) = 4 divided by s × 1 divided by 1 + s/3

  2. Identify the Type and g.

    One pole at s = 0 → Type 1. Comparing with the standard factors gives g = 4.

  3. Calculate the error constant.

    Kv = lims→0 s · 12 divided by s(s + 3)

    Cancel s before taking the limit:

    = lims→0 12 divided by s + 3 = 12 divided by 3 = 4

  4. Calculate the error.

    ess = 1 divided by Kv = 1 divided by 4 = 0.25

    For the unit ramp, the output eventually follows the same slope but remains 0.25 output units behind the reference.

Kv = 4 is the intermediate result. The remaining tracking error is 0.25.

3

Change the pole factor

F(s) = 15(s + 2) divided by s(2s + 3)(s + 5)

Rewrite 2s + 3 with a constant term of 1 inside the parentheses.

  1. Rewrite F(s) into standard factors.
  2. Identify the System Type and g.
  3. Calculate Kv.
  4. Find the unit-ramp steady-state error. Explain what it means.

Is the unit-ramp error larger or smaller than in Question 2?

Show worked answer

  1. Rewrite the factor.

    s + 2 = 2(1 + s/2)
    2s + 3 = 3(1 + 2s/3)
    s + 5 = 5(1 + s/5)
    F(s) = 15 · 2(1 + s/2) divided by 3 · 5s(1 + 2s/3)(1 + s/5)
    = 2 divided by s × 1 + s/2 divided by (1 + 2s/3)(1 + s/5)

  2. Identify the Type and g.

    One pole at s = 0 → Type 1, with g = 2.

  3. Calculate the constant.

    Kv = lims→0 s · 15(s + 2) divided by s(2s + 3)(s + 5)

    Cancel the single factor s, then substitute s = 0:

    = lims→0 15(s + 2) divided by (2s + 3)(s + 5) = 15 · 2 divided by 3 · 5 = 2

  4. Calculate and compare the error.

    ess = 1 divided by 2 = 0.5

    This is larger than 0.25 in Question 2, so the remaining unit-ramp tracking error is larger.

SystemTypeKvUnit-ramp error
Classroom example140.25
Changed pole120.5

Both are Type 1, but their tracking accuracy differs. Calculating Kv tells us how large the remaining error is.