Interpolation (interp1) in MATLAB

Introduction

Interpolation helps you find values between known points. It fills in gaps smoothly.

You have temperature readings every hour and want to estimate values every 30 minutes.

You know the speed of a car at certain times and want to find speed at times in between.

You have a few points on a curve and want to draw a smooth line through them.

You want to estimate missing data points in a measurement series.

Syntax

MATLAB

vq = interp1(x, v, xq)
vq = interp1(x, v, xq, method)
vq = interp1(x, v, xq, method, extrapolation)

x and v are vectors of known data points.

xq are the query points where you want to find values.

Examples

Finds value at 2.5 by linear interpolation between points.

MATLAB

x = [1 2 3 4 5];
v = [10 20 30 40 50];
xq = 2.5;
vq = interp1(x, v, xq);

Finds value at 3.7 using nearest neighbor method.

MATLAB

vq = interp1(x, v, 3.7, 'nearest');

Estimates value at 6 by linear extrapolation beyond known points.

MATLAB

vq = interp1(x, v, 6, 'linear', 'extrap');

Sample Program

This program finds the value at 2.5 using two methods: linear and nearest neighbor interpolation.

MATLAB

x = [0 1 2 3 4];
v = [0 2 4 6 8];
xq = 2.5;
vq_linear = interp1(x, v, xq, 'linear');
vq_nearest = interp1(x, v, xq, 'nearest');
fprintf('Linear interpolation at %.1f: %.2f\n', xq, vq_linear);
fprintf('Nearest interpolation at %.1f: %.2f\n', xq, vq_nearest);

Important Notes

Make sure x is strictly increasing.

Common methods: 'linear', 'nearest', 'spline', 'pchip'.

Use 'extrap' to allow values outside the known range.

Summary

Interpolation estimates values between known points.

interp1 is used for 1D interpolation in MATLAB.

You can choose different methods for smoothness or speed.

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