Interpolation (interp1) in MATLAB - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to interpolate the value at x = 2.5 using linear interpolation.

MATLAB

x = [1 2 3 4 5];
y = [10 20 30 40 50];
result = interp1(x, y, [1]);

Drag options to blanks, or click blank then click option'

A3.5

B2.5

C1.5

D4.5

Attempts:

3 left

💡 Hint

Common Mistakes

Using a value outside the range of x, which causes NaN result.

Using an x value that is not numeric.

The value 2.5 is between 2 and 3, so interp1 will interpolate at this point.

2fill in blank

medium

Complete the code to use 'nearest' method for interpolation at x = 3.7.

MATLAB

x = [1 2 3 4 5];
y = [10 20 30 40 50];
result = interp1(x, y, 3.7, [1]);

Drag options to blanks, or click blank then click option'

A'nearest'

B'linear'

C'spline'

D'pchip'

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'linear' instead of 'nearest' changes the interpolation behavior.

Forgetting the quotes around the method name.

The 'nearest' method picks the nearest data point for interpolation.

3fill in blank

hard

Fix the error in the code to interpolate at x = 6 using extrapolation.

MATLAB

x = [1 2 3 4 5];
y = [10 20 30 40 50];
result = interp1(x, y, 6, 'linear', [1]);

Drag options to blanks, or click blank then click option'

A'spline'

B'linear'

C'extrap'

D'nearest'

Attempts:

3 left

💡 Hint

Common Mistakes

Not specifying 'extrap' causes NaN for out-of-range points.

Using 'nearest' or 'linear' without extrapolation option.

To interpolate outside the range of x, use 'extrap' to allow extrapolation.

4fill in blank

hard

Fill both blanks to create a vector of interpolated values at points 1.5 and 4.5 using 'pchip' method.

MATLAB

x = [1 2 3 4 5];
y = [10 20 30 40 50];
xi = [[1] [2]];
result = interp1(x, y, xi, 'pchip');

Drag options to blanks, or click blank then click option'

A1.5

B4.5

C2.5

D3.5

Attempts:

3 left

💡 Hint

Common Mistakes

Choosing points outside the x range without extrapolation.

Using points that are not numeric.

The points 1.5 and 4.5 lie between the x values and are used for interpolation.

5fill in blank

hard

Fill all three blanks to interpolate values at points 2, 3.3, and 5 using 'spline' method.

MATLAB

x = [1 2 3 4 5];
y = [10 20 30 40 50];
xi = [[1] [2] [3]];
result = interp1(x, y, xi, 'spline');

Drag options to blanks, or click blank then click option'

A2

B3.3

C5

D'spline'

Attempts:

3 left

💡 Hint

Common Mistakes

Using points outside the x range without extrapolation.

Not using the correct method string for interpolation.

The points 2, 3.3, and 5 are used for interpolation with the 'spline' method.

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