Interpolation (interp1) in MATLAB - Step-by-Step Execution

Concept Flow - Interpolation (interp1)

Define known points x, y

↓

Choose query points xi

↓

Call interp1(x, y, xi)

↓

interp1 finds position of xi in x

↓

Calculate interpolated yi values

↓

Return yi as output

This flow shows how interp1 takes known data points and query points, then calculates interpolated values step-by-step.

Execution Sample

MATLAB

x = [1 2 3 4];
y = [10 20 30 40];
xi = 2.5;
yi = interp1(x, y, xi);

This code finds the interpolated y value at x=2.5 using linear interpolation.

Execution Table

Step	Action	Input/Condition	Result/Output
1	Define x and y	x=[1 2 3 4], y=[10 20 30 40]	Known data points set
2	Define xi	xi=2.5	Query point set
3	Find interval for xi	2 <= 2.5 <= 3	xi lies between x(2)=2 and x(3)=3
4	Calculate interpolation fraction	fraction = (2.5-2)/(3-2) = 0.5	fraction=0.5
5	Calculate yi	yi = y(2) + fraction*(y(3)-y(2))	yi = 20 + 0.5*(30-20) = 25
6	Return yi	yi=25	Interpolated value at xi=2.5 is 25

💡 Interpolation complete for xi=2.5 within x range

Variable Tracker

Variable	Start	After Step 2	After Step 4	Final
x	[1 2 3 4]	[1 2 3 4]	[1 2 3 4]	[1 2 3 4]
y	[10 20 30 40]	[10 20 30 40]	[10 20 30 40]	[10 20 30 40]
xi	undefined	2.5	2.5	2.5
fraction	undefined	undefined	0.5	0.5
yi	undefined	undefined	undefined	25

Key Moments - 3 Insights

Why does interp1 look for the interval where xi fits between x values?

Because interpolation estimates yi by using the two known points around xi, as shown in step 3 of the execution_table.

What happens if xi is exactly one of the x points?

interp1 returns the corresponding y value directly without interpolation, since the point is known exactly.

Why do we calculate the fraction (step 4) before finding yi?

The fraction tells us how far xi is between the two x points, so we can scale the difference in y values accordingly.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of 'fraction' at step 4?

A 25

B 2.5

C 0.5

D 1

💡 Hint

Check the 'fraction' value in the 'Result/Output' column at step 4.

At which step does interp1 find the interval where xi fits between x values?

A Step 2

B Step 3

C Step 5

D Step 6

💡 Hint

Look for the step describing 'Find interval for xi' in the execution_table.

If xi was 3 instead of 2.5, what would yi be according to the variable_tracker?

A 30

B 20

C 25

D 35

💡 Hint

Recall that if xi matches an x point exactly, yi equals the corresponding y value.

Concept Snapshot

interp1(x, y, xi) finds yi by linear interpolation.
It locates xi between two x points,
calculates fraction of distance,
and scales y difference accordingly.
Returns yi as interpolated value.

Full Transcript

This visual execution traces MATLAB's interp1 function. First, known points x and y are defined. Then, a query point xi is chosen. interp1 finds where xi fits between x values, calculates how far xi is between those points as a fraction, and uses that fraction to interpolate yi between the corresponding y values. The final yi is returned as the estimated value at xi. This step-by-step shows how interpolation works by estimating values between known data points.

Practice

1. What does the MATLAB function interp1 do?

easy

A. It calculates the derivative of a function.

B. It sorts a vector in ascending order.

C. It estimates intermediate values between known data points.

D. It plots data points on a graph.

Solution

1. Step 1: Understand the purpose of interpolation

   Interpolation is used to estimate values between known data points.

2. Step 2: Identify the role of interp1

   interp1 in MATLAB performs 1D interpolation to find intermediate values.

3. Final Answer:

   It estimates intermediate values between known data points. -> Option C

4. Quick Check:

   Interpolation = estimating values [OK]

Hint: Remember: interp1 fills gaps between points [OK]

Common Mistakes:

• Confusing interpolation with sorting
• Thinking interp1 calculates derivatives
• Assuming interp1 creates plots

2. Which of the following is the correct syntax to interpolate the value at x = 2.5 using vectors x and y in MATLAB?

easy

A. interp1(x, y, 2.5)

B. interp1(y, x, 2.5)

C. interp1(2.5, x, y)

D. interp1(x, 2.5, y)

Solution

1. Step 1: Recall interp1 function syntax

   The syntax is interp1(x, y, xi) where x and y are known points, xi is query point.

2. Step 2: Match the correct order of arguments

   interp1(x, y, 2.5) uses interp1(x, y, 2.5), which matches the syntax correctly.

3. Final Answer:

   interp1(x, y, 2.5) -> Option A

4. Quick Check:

   interp1(x, y, xi) = correct syntax [OK]

Hint: Remember: interp1(x, y, xi) order matters [OK]

Common Mistakes:

• Swapping x and y vectors
• Putting query point before vectors
• Using wrong argument order

3. Given x = [1 2 3 4] and y = [10 20 30 40], what is the output of interp1(x, y, 2.5) in MATLAB?

medium

A. Error

B. 20

C. 30

D. 25

Solution

1. Step 1: Understand linear interpolation between points

   At x=2, y=20 and at x=3, y=30. For x=2.5, interpolate linearly.

2. Step 2: Calculate interpolated value

   Value = 20 + (2.5-2)*(30-20)/(3-2) = 20 + 0.5*10 = 25.

3. Final Answer:

   25 -> Option D

4. Quick Check:

   Linear interp at 2.5 = 25 [OK]

Hint: Linear interp averages between neighbors [OK]

Common Mistakes:

• Choosing nearest point value instead of interpolating
• Confusing x and y values
• Expecting an error for valid input

4. What is wrong with this MATLAB code snippet?

x = [1 2 3];
y = [10 20];
interp1(x, y, 2)

medium

A. Vectors x and y have different lengths causing an error.

B. The query point 2 is outside the range of x.

C. interp1 requires y to be a matrix, not a vector.

D. The syntax of interp1 is incorrect.

Solution

1. Step 1: Check vector lengths

   Vector x has length 3, y has length 2; they must be equal for interp1.

2. Step 2: Identify error cause

   Different lengths cause MATLAB to throw an error when calling interp1.

3. Final Answer:

   Vectors x and y have different lengths causing an error. -> Option A

4. Quick Check:

   Equal vector lengths needed [OK]

Hint: Check x and y lengths match before interp1 [OK]

Common Mistakes:

• Ignoring vector length mismatch
• Thinking query point must be outside range
• Assuming interp1 needs matrix y

5. You have temperature data temp = [15 18 21 24] measured at times time = [0 3 6 9] hours. How would you find the temperature at 4.5 hours using cubic interpolation in MATLAB?

hard

A. interp1(time, temp, 4.5)

B. interp1(time, temp, 4.5, 'cubic')

C. interp1(time, temp, 4.5, 'linear')

D. interp1(temp, time, 4.5, 'cubic')

Solution

1. Step 1: Identify correct argument order and method

   Syntax is interp1(x, y, xi, method). Here x=time, y=temp, xi=4.5, method='cubic'.

2. Step 2: Choose cubic interpolation

   interp1(time, temp, 4.5, 'cubic') uses 'cubic' method correctly for smooth interpolation.

3. Final Answer:

   interp1(time, temp, 4.5, 'cubic') -> Option B

4. Quick Check:

   Use 'cubic' for smooth interp [OK]

Hint: Use 'cubic' as 4th argument for smooth interp [OK]

Common Mistakes:

• Swapping time and temp vectors
• Using default linear instead of cubic
• Omitting interpolation method argument