Interpolation (interp1) in MATLAB - Mini Project: Build & Apply

Interpolation with interp1 in MATLAB

📖 Scenario: You are working with temperature data collected at specific hours of the day. However, you want to estimate the temperature at times between the recorded hours.

🎯 Goal: Build a MATLAB script that uses interp1 to find estimated temperatures at new times using linear interpolation.

📋 What You'll Learn

Create vectors for recorded times and temperatures

Create a vector for new times where temperature needs to be estimated

Use interp1 with linear method to interpolate temperatures

Display the interpolated temperatures

💡 Why This Matters

🌍 Real World

Interpolation helps estimate unknown values between measured data points, useful in weather forecasting, engineering, and science.

💼 Career

Knowing how to use interpolation functions like <code>interp1</code> is important for data analysis and modeling tasks in engineering and research jobs.

1

Create recorded time and temperature vectors

Create a vector called time with values [0 3 6 9 12 15 18 21 24] representing hours of the day. Create a vector called temp with values [15 17 20 22 24 23 21 18 16] representing temperatures at those times.

MATLAB

% Create vectors 'time' and 'temp' with given values
% Your code here

Hint

Use square brackets to create vectors. Separate values with spaces.

2

Create new time points for interpolation

Create a vector called new_time with values [1 4 7 10 13 16 19 22] representing hours where you want to estimate temperature.

MATLAB

time = [0 3 6 9 12 15 18 21 24];
temp = [15 17 20 22 24 23 21 18 16];
% Create vector 'new_time' with given values
% Your code here

Hint

Use square brackets and spaces to create the vector.

3

Interpolate temperatures at new times

Use interp1 with time, temp, and new_time to create a vector called interp_temp that contains temperatures estimated by linear interpolation.

MATLAB

time = [0 3 6 9 12 15 18 21 24];
temp = [15 17 20 22 24 23 21 18 16];
new_time = [1 4 7 10 13 16 19 22];
% Use interp1 to interpolate temperatures
% Your code here

Hint

Use interp1(time, temp, new_time, 'linear') to get interpolated values.

4

Display the interpolated temperatures

Use disp to print the vector interp_temp showing the estimated temperatures at the new times.

MATLAB

time = [0 3 6 9 12 15 18 21 24];
temp = [15 17 20 22 24 23 21 18 16];
new_time = [1 4 7 10 13 16 19 22];
interp_temp = interp1(time, temp, new_time, 'linear');
% Display the interpolated temperatures
% Your code here

Hint

Use disp(interp_temp) to show the results.

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