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Proportional control effect in Simulink - Step-by-Step Execution

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Concept Flow - Proportional control effect
Input Error (e)
↓
Multiply by Kp
↓
Control Output (u = Kp * e)
↓
System Response
↓
Feedback to Input
The proportional controller multiplies the error by a gain Kp to produce the control output, which affects the system and feeds back to reduce error.
Execution Sample
Simulink
e = setpoint - measurement;
u = Kp * e;
apply u to system;
measure new output;
repeat
This code calculates the error, multiplies it by the proportional gain Kp to get control output, applies it, and updates the measurement.
Execution Table
StepSetpointMeasurementError (e)KpControl Output (u)System Output
1100100.55.05.0
2105.05.00.52.57.5
3107.52.50.51.258.75
4108.751.250.50.6259.375
5109.3750.6250.50.31259.6875
6109.68750.31250.50.156259.84375
7109.843750.156250.50.0781259.921875
8109.9218750.0781250.50.03906259.9609375
9109.96093750.03906250.50.019531259.98046875
10109.980468750.019531250.50.0097656259.990234375
Exit109.9902343750.0097656250.50.00488281259.9951171875
💡 Error becomes very small, control output approaches zero, system output nears setpoint.
Variable Tracker
VariableStartAfter 1After 2After 3After 4After 5After 6After 7After 8After 9After 10Final
Setpoint101010101010101010101010
Measurement05.07.58.759.3759.68759.843759.9218759.96093759.980468759.9902343759.9951171875
Error (e)105.02.51.250.6250.31250.156250.0781250.03906250.019531250.0097656250.0048828125
Control Output (u)5.02.51.250.6250.31250.156250.0781250.03906250.019531250.0097656250.00488281250.00244140625
System Output05.07.58.759.3759.68759.843759.9218759.96093759.980468759.9902343759.9951171875
Key Moments - 3 Insights
Why does the system output never exactly reach the setpoint?
Because the proportional controller output is proportional to the error, as error gets very small, control output also becomes very small, causing the system to approach but not exactly reach the setpoint (see execution_table rows 9-10).
What happens if Kp is too large?
If Kp is too large, the control output can overshoot the setpoint causing oscillations or instability, unlike the smooth approach shown here with Kp=0.5.
Why does the error decrease each step?
Because the control output moves the system output closer to the setpoint, reducing the difference (error) each iteration as shown in the error column of execution_table.
Visual Quiz - 3 Questions
Test your understanding
Look at the execution table, what is the control output (u) at step 4?
A0.625
B1.25
C2.5
D0.3125
💡 Hint
Check the 'Control Output (u)' column at step 4 in the execution_table.
At which step does the error first become less than 1?
AStep 3
BStep 4
CStep 5
DStep 6
💡 Hint
Look at the 'Error (e)' column in execution_table and find when it drops below 1.
If Kp was doubled to 1.0, how would the control output at step 2 change?
AIt would be 5.0
BIt would be 1.25
CIt would be 10.0
DIt would be 2.5
💡 Hint
Control output u = Kp * error; check error at step 2 and multiply by new Kp.
Concept Snapshot
Proportional Control:
- Output u = Kp * error (e)
- Error = setpoint - measurement
- Larger Kp means stronger correction
- System output approaches setpoint smoothly
- Does not eliminate steady-state error alone
Full Transcript
This visual trace shows how a proportional controller works step-by-step. We start with an error between the desired setpoint and the current measurement. The controller multiplies this error by a gain Kp to get the control output. This output adjusts the system, moving its output closer to the setpoint. Each step reduces the error and control output until the system output nearly matches the setpoint. The trace highlights how the error shrinks and the control output decreases accordingly. It also shows that the system output approaches but does not exactly reach the setpoint due to the proportional nature of the control. This helps beginners see the effect of proportional control clearly.

Practice

(1/5)
1. What does the proportional gain (Kp) in a proportional controller primarily affect?
easy
A. The integral of the error over time
B. The delay time before the system reacts
C. The strength of the system's response to the error
D. The derivative of the error

Solution

  1. Step 1: Understand the role of proportional gain

    The proportional gain (Kp) multiplies the error to determine the output response strength.
  2. Step 2: Identify what Kp controls

    Kp controls how strongly the system reacts to the current error, not delay or integral/derivative terms.
  3. Final Answer:

    The strength of the system's response to the error -> Option C
  4. Quick Check:

    Proportional gain = response strength [OK]
Hint: Kp scales error to output strength [OK]
Common Mistakes:
  • Confusing Kp with integral or derivative gains
  • Thinking Kp controls delay or timing
  • Mixing proportional control with PID control
2. Which of the following Simulink blocks is typically used to implement proportional control?
easy
A. Integrator block
B. Derivative block
C. Sum block
D. Gain block

Solution

  1. Step 1: Identify the block that multiplies input by a constant

    The Gain block multiplies the input signal by a constant value, which is the proportional gain Kp.
  2. Step 2: Match block function to proportional control

    Proportional control output = Kp * error, so Gain block is the correct choice.
  3. Final Answer:

    Gain block -> Option D
  4. Quick Check:

    Proportional control uses Gain block [OK]
Hint: Gain block multiplies error by Kp [OK]
Common Mistakes:
  • Choosing Integrator or Derivative blocks which are for other control types
  • Confusing Sum block as proportional gain
  • Not knowing block functions
3. Given a proportional controller with gain Kp=3 and an error signal of 2, what is the controller output?
medium
A. 6
B. 5
C. 3
D. 2

Solution

  1. Step 1: Calculate output using proportional control formula

    Output = Kp * error = 3 * 2 = 6.
  2. Step 2: Verify calculation

    Multiplying gain 3 by error 2 gives 6, which matches 6.
  3. Final Answer:

    6 -> Option A
  4. Quick Check:

    3 * 2 = 6 [OK]
Hint: Multiply Kp by error to get output [OK]
Common Mistakes:
  • Adding instead of multiplying gain and error
  • Using wrong gain value
  • Confusing output with error
4. A Simulink proportional controller uses a Gain block with Kp set to -4. The error signal is positive. What is the likely effect on the system output?
medium
A. Output will be zero regardless of error
B. Output will be negative, causing inverse response
C. Output will be positive, increasing error
D. Output will oscillate randomly

Solution

  1. Step 1: Analyze effect of negative gain on positive error

    Output = Kp * error = -4 * positive value = negative output.
  2. Step 2: Understand system response

    Negative output means control acts opposite to error, causing inverse or destabilizing response.
  3. Final Answer:

    Output will be negative, causing inverse response -> Option B
  4. Quick Check:

    Negative gain flips output sign [OK]
Hint: Negative Kp reverses output sign [OK]
Common Mistakes:
  • Assuming output stays positive with negative gain
  • Thinking output is zero without error
  • Confusing oscillation with negative gain effect
5. In a Simulink model, increasing the proportional gain Kp too much causes the system to oscillate. What is the best way to reduce oscillations while keeping proportional control?
hard
A. Decrease Kp to a moderate value
B. Increase Kp further to stabilize
C. Remove the Gain block entirely
D. Set Kp to zero

Solution

  1. Step 1: Understand effect of high Kp

    High Kp causes strong reactions, leading to overshoot and oscillations.
  2. Step 2: Adjust Kp to reduce oscillations

    Lowering Kp reduces reaction strength, smoothing output and reducing oscillations.
  3. Final Answer:

    Decrease Kp to a moderate value -> Option A
  4. Quick Check:

    Lower Kp reduces oscillations [OK]
Hint: Lower Kp to smooth control [OK]
Common Mistakes:
  • Increasing Kp worsens oscillations
  • Removing Gain block stops control
  • Setting Kp to zero disables proportional control
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