Running simulations in MATLAB - Step-by-Step Execution

Concept Flow - Running simulations

Initialize parameters

↓

Set simulation time

↓

Start loop: t = 0 to end

↓

Calculate next state

↓

Store results

↓

Increment time

Back to loop check↓

Plot or analyze results

↩Back to loop check

This flow shows how a simulation runs step-by-step: setting parameters, looping over time, updating states, storing data, and finally analyzing results.

Execution Sample

MATLAB

dt = 0.1;
t_end = 0.3;
x = 0;
for t = 0:dt:t_end
    x = x + dt;
    disp(x)
end

This code runs a simple simulation increasing x by dt at each time step and displays the value.

Execution Table

Iteration	t	Condition	x before update	x after update	Output
1	0.0	0.0 <= 0.3 (True)	0	0 + 0.1 = 0.1	0.1
2	0.1	0.1 <= 0.3 (True)	0.1	0.1 + 0.1 = 0.2	0.2
3	0.2	0.2 <= 0.3 (True)	0.2	0.2 + 0.1 = 0.3	0.3
4	0.3	0.3 <= 0.3 (True)	0.3	0.3 + 0.1 = 0.4	0.4

💡 Loop ends after iteration 4 because the next t would be 0.4 which exceeds t_end = 0.3, so the loop stops.

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	Final
t	0.0	0.1	0.2	0.3	0.3	0.3
x	0	0.1	0.2	0.3	0.4	0.4

Key Moments - 3 Insights

Why does the loop run 4 times even though t_end is 0.3?

Because the loop includes t = 0.3 (condition t <= 0.3 is true), so it runs for t = 0, 0.1, 0.2, and 0.3 before stopping.

Why does x increase by 0.1 each iteration?

Because inside the loop, x is updated by adding dt (0.1) every time (see 'x after update' column in execution_table).

What stops the simulation loop from running forever?

The loop condition t <= t_end becomes false when t exceeds 0.3, so the loop exits.

Visual Quiz - 3 Questions

Look at the execution table, what is the value of x after the 3rd iteration?

A 0.3

B 0.2

C 0.4

D 0.1

💡 Hint

Check the 'x after update' column at iteration 3 in the execution_table.

At which iteration does the loop condition become false?

A Iteration 3

B Iteration 5

C Iteration 4

D Iteration 2

💡 Hint

Look at the 'Condition' column in the execution_table where it shows False.

If dt was changed to 0.2, how many iterations would the loop run?

A 3

B 4

C 2

D 5

💡 Hint

Think about how t increments by dt each iteration and compare to t_end = 0.3.

Concept Snapshot

Running simulations in MATLAB:
- Set parameters (dt, t_end)
- Use a for loop: for t = 0:dt:t_end
- Update state variables inside loop
- Store or display results
- Loop ends when t > t_end

Full Transcript

This example shows how to run a simple simulation in MATLAB by increasing a variable x over time steps from 0 to 0.3 with step size 0.1. The loop runs while the time variable t is less than or equal to the end time. Each iteration updates x by adding dt and displays the new value. The execution table traces each iteration's values and conditions, showing when the loop stops. Key points include understanding loop conditions, variable updates, and loop termination. The quiz tests understanding of variable values and loop behavior.

Practice

1. What is the main purpose of using a for loop in running simulations in MATLAB?

easy

A. To repeat a set of commands multiple times automatically

B. To create graphical user interfaces

C. To store data permanently on disk

D. To define new functions

Solution

1. Step 1: Understand the role of for loops

   A for loop repeats commands a fixed number of times, which is essential for running simulations step-by-step.

2. Step 2: Match the purpose to simulation needs

   Simulations require repeating calculations or processes many times to model changes over time, which for loops handle well.

3. Final Answer:

   To repeat a set of commands multiple times automatically -> Option A

4. Quick Check:

   Loop repeats commands = repeat steps [OK]

Hint: Loops repeat steps automatically, perfect for simulations [OK]

Common Mistakes:

• Confusing loops with data storage
• Thinking loops create GUIs
• Mixing loops with function definitions

2. Which of the following is the correct syntax to run a simulation loop 10 times in MATLAB?

easy

A. repeat 10 times
% simulation code
endrepeat

B. for i = 1:10
% simulation code
end

C. loop i = 1 to 10
% simulation code
endloop

D. while i < 10
% simulation code
end

Solution

1. Step 1: Recall MATLAB for loop syntax

   MATLAB uses for variable = start:end to run loops a fixed number of times.

2. Step 2: Compare options with correct syntax

   for i = 1:10 % simulation code end matches MATLAB syntax exactly; others use incorrect keywords or structures.

3. Final Answer:

   for i = 1:10 % simulation code end -> Option B

4. Quick Check:

   MATLAB loops use 'for i = start:end' [OK]

Hint: MATLAB loops use 'for i = start:end' syntax [OK]

Common Mistakes:

• Using non-MATLAB keywords like 'loop' or 'repeat'
• Confusing while with for loops
• Missing the end keyword

3. What will be the output of the following MATLAB code?

for t = 1:3
  disp(t^2)
end

medium

A. 1 4 9

B. 1 2 3

C. 2 4 6

D. Error: Undefined function

Solution

1. Step 1: Understand the loop iterations

   The loop runs with t = 1, 2, 3.

2. Step 2: Calculate the displayed values

   Each iteration displays t squared: 1^2=1, 2^2=4, 3^2=9.

3. Final Answer:

   1 4 9 -> Option A

4. Quick Check:

   Squares of 1,2,3 = 1 4 9 [OK]

Hint: Square each loop index to find output quickly [OK]

Common Mistakes:

• Printing t instead of t squared
• Confusing output with sums
• Expecting syntax errors

4. Identify the error in this MATLAB simulation code:

for i = 1:5
  result = i * 2
end
disp(result)

medium

A. Missing semicolon after result assignment

B. The loop variable 'i' is not initialized

C. No error; code runs and displays the last result

D. Variable 'result' is not defined before the loop

Solution

1. Step 1: Check variable initialization

   In MATLAB, variables can be created inside loops without prior definition.

2. Step 2: Understand loop and output behavior

   The loop runs 5 times, updating 'result' each time; after loop, 'result' holds last value (5*2=10), which is displayed.

3. Final Answer:

   No error; code runs and displays the last result -> Option C

4. Quick Check:

   Loop updates variable; last value shown [OK]

Hint: Variables inside loops update each iteration; last value remains [OK]

Common Mistakes:

• Thinking semicolons are mandatory to avoid errors
• Assuming variables must be pre-declared
• Confusing loop variable initialization

5. You want to simulate the growth of a population starting at 100, increasing by 10% each step, for 5 steps. Which MATLAB code correctly runs this simulation and stores all population values in an array pop?

hard

A. pop = zeros(5,1);
pop(1) = 100;
for i = 1:5
pop(i+1) = pop(i) * 1.1;
end

B. pop = 100;
for i = 1:5
pop = pop * 1.1;
end

C. pop = [];
for i = 1:5
pop(i) = 100 * 1.1^i;
end

D. pop = zeros(1,5);
pop(1) = 100;
for i = 2:5
pop(i) = pop(i-1) * 1.1;
end

Solution

1. Step 1: Understand the simulation requirements

   Start at 100, increase by 10% each step, store all 5 values in an array.

2. Step 2: Analyze each option for correctness

   pop = zeros(1,5); pop(1) = 100; for i = 2:5 pop(i) = pop(i-1) * 1.1; end initializes array, sets first value, then correctly updates subsequent values using previous step's value. pop = 100; for i = 1:5 pop = pop * 1.1; end overwrites single value, no array. pop = []; for i = 1:5 pop(i) = 100 * 1.1^i; end calculates powers but starts from step 1, missing initial 100 at index 1. pop = zeros(5,1); pop(1) = 100; for i = 1:5 pop(i+1) = pop(i) * 1.1; end tries to access index 6 in 5-element array, causing error.

3. Final Answer:

   pop = zeros(1,5); pop(1) = 100; for i = 2:5 pop(i) = pop(i-1) * 1.1; end -> Option D

4. Quick Check:

   Initialize array, set first value, update from previous [OK]

Hint: Initialize array, set first value, update from previous step [OK]

Common Mistakes:

• Not storing all values in an array
• Indexing beyond array size
• Overwriting single variable instead of array