FFT and frequency analysis in MATLAB - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to compute the FFT of the signal x.

MATLAB

Y = fft([1]);

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

Ax

By

Csignal

Ddata

Attempts:

3 left

💡 Hint

Common Mistakes

Using a variable name that does not hold the signal data.

Forgetting to pass any argument to fft.

The fft function computes the Fast Fourier Transform of the input signal x.

2fill in blank

medium

Complete the code to calculate the magnitude spectrum from the FFT result Y.

MATLAB

magnitude = abs([1]);

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

Ax

BY

Cfreq

Dsignal

Attempts:

3 left

💡 Hint

Common Mistakes

Taking the absolute value of the original signal instead of the FFT result.

Using a variable that does not contain the FFT output.

The magnitude spectrum is the absolute value of the FFT result Y.

3fill in blank

hard

Fix the error in the code to compute the frequency vector for sampling frequency Fs and signal length N.

MATLAB

f = (0:[1]-1)*(Fs/[1]);

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

AFs

Blength

CN

Dfreq

Attempts:

3 left

💡 Hint

Common Mistakes

Using the sampling frequency Fs instead of the signal length N in the vector range.

Using an undefined variable.

The frequency vector f should be based on the length N of the signal.

4fill in blank

hard

Fill both blanks to compute the two-sided amplitude spectrum P2 and then the single-sided spectrum P1.

MATLAB

P2 = abs([1]/[2]);
P1 = P2(1:floor([2]/2)+1);

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

AY

BN

CFs

Dlength

Attempts:

3 left

💡 Hint

Common Mistakes

Dividing by the sampling frequency instead of the signal length.

Incorrect indexing for the single-sided spectrum.

P2 is the normalized magnitude of the FFT result Y divided by the signal length N. Then P1 takes the first half plus one element for the single-sided spectrum.

5fill in blank

hard

Fill all three blanks to plot the single-sided amplitude spectrum P1 against frequency vector f.

MATLAB

plot([1], [2])
xlabel('[3]')

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

Af

BP1

CFrequency (Hz)

DAmplitude

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping the order of f and P1 in the plot.

Using a wrong label for the x-axis.

The plot shows amplitude P1 versus frequency f. The x-axis label should be 'Frequency (Hz)'.

Practice

1. What does the MATLAB function fft() do to a signal?

easy

A. It amplifies the signal amplitude.

B. It converts the signal from time domain to frequency domain.

C. It smooths the signal to remove noise.

D. It shifts the signal in time.

Solution

1. Step 1: Understand the purpose of FFT

   FFT stands for Fast Fourier Transform, which changes a signal from time domain to frequency domain.

2. Step 2: Identify what fft() does in MATLAB

   In MATLAB, fft() computes the discrete Fourier transform, showing frequency components of the signal.

3. Final Answer:

   It converts the signal from time domain to frequency domain. -> Option B

4. Quick Check:

   FFT = frequency transform [OK]

Hint: FFT always changes time signals to frequency signals [OK]

Common Mistakes:

• Thinking FFT smooths or filters signals
• Confusing FFT with time shifting
• Assuming FFT amplifies signal amplitude

2. Which of the following is the correct syntax to compute the FFT of a vector x in MATLAB?

easy

A. Y = fft x;

B. Y = fft[x];

C. Y = fft(x);

D. Y = fft{x};

Solution

1. Step 1: Recall MATLAB function call syntax

   In MATLAB, functions are called with parentheses enclosing arguments, e.g., func(arg).

2. Step 2: Check each option's syntax

   Y = fft(x); uses correct syntax: fft(x). Options A, C, and D use invalid syntax for function calls in MATLAB.

3. Final Answer:

   Y = fft(x); -> Option C

4. Quick Check:

   Function calls use parentheses () [OK]

Hint: Use parentheses () for function calls in MATLAB [OK]

Common Mistakes:

• Using square brackets [] instead of parentheses
• Omitting parentheses around arguments
• Using curly braces {} which are for cell arrays

3. Given the MATLAB code:

x = [1 1 0 0];
Y = fft(x);
absY = abs(Y);
disp(absY);

What is the output displayed?

medium

A. [1 1 1 1]

B. [2 0 2 0]

C. [4 0 0 0]

D. [2 1.4142 0 1.4142]

Solution

1. Step 1: Compute FFT of x = [1 1 0 0]

   The FFT of this sequence results in complex values whose magnitudes are [2, sqrt(2), 0, sqrt(2)] approximately.

2. Step 2: Calculate absolute values

   Taking absolute values gives approximately [2, 1.4142, 0, 1.4142].

3. Final Answer:

   [2 1.4142 0 1.4142] -> Option D

4. Quick Check:

   FFT magnitude matches [2 1.4142 0 1.4142] [OK]

Hint: FFT magnitude shows signal strength at each frequency [OK]

Common Mistakes:

• Confusing FFT output with original signal
• Not taking absolute value before display
• Assuming FFT output is all real numbers

4. Consider this MATLAB code snippet:

Fs = 1000;
t = 0:1/Fs:1-1/Fs;
x = cos(2*pi*50*t) + cos(2*pi*120*t);
Y = fft(x);
P2 = abs(Y/length(x));
P1 = P2(1:length(x)/2+1);
P1(2:end-1) = 2*P1(2:end-1);
plot(P1);

What is the likely error in this code?

medium

A. The frequency axis is missing for the plot, so frequencies are unclear.

B. The length of x is not used correctly in dividing Y.

C. The FFT function is called incorrectly without arguments.

D. The signal x is not defined properly.

Solution

1. Step 1: Analyze the FFT and amplitude spectrum calculation

   The code correctly computes the single-sided amplitude spectrum P1 from Y.

2. Step 2: Check the plot command

   The plot shows amplitude values but does not label or define the frequency axis, making interpretation difficult.

3. Final Answer:

   The frequency axis is missing for the plot, so frequencies are unclear. -> Option A

4. Quick Check:

   Plot frequency axis needed for clarity [OK]

Hint: Always plot frequency axis with amplitude spectrum [OK]

Common Mistakes:

• Forgetting to create frequency vector for x-axis
• Plotting amplitude without frequency labels
• Misusing FFT input or output sizes

5. You have a noisy signal sampled at 1000 Hz stored in vector data. You want to find the dominant frequency component using FFT. Which MATLAB code snippet correctly computes and plots the single-sided amplitude spectrum with frequency axis?

hard

A. L = length(data); Fs = 1000; Y = fft(data); P2 = abs(Y/L); f = Fs*(0:(L/2))/L; P1 = P2(1:L/2+1); P1(2:end-1) = 2*P1(2:end-1); plot(f,P1); xlabel('Frequency (Hz)'); ylabel('Amplitude');

B. Fs = 1000; Y = fft(data); plot(abs(Y)); xlabel('Frequency'); ylabel('Amplitude');

C. L = length(data); Y = fft(data); P1 = abs(Y); plot(P1);

D. Fs = 1000; t = 0:1/Fs:1; Y = fft(data); plot(Y);

Solution

1. Step 1: Understand the steps to compute single-sided amplitude spectrum

   Calculate FFT, normalize by length, create frequency vector, extract half spectrum, double amplitudes except DC and Nyquist.

2. Step 2: Check which option follows these steps correctly

   L = length(data); Fs = 1000; Y = fft(data); P2 = abs(Y/L); f = Fs*(0:(L/2))/L; P1 = P2(1:L/2+1); P1(2:end-1) = 2*P1(2:end-1); plot(f,P1); xlabel('Frequency (Hz)'); ylabel('Amplitude'); correctly computes length, frequency vector, normalizes, adjusts amplitudes, and plots with labels. Others miss frequency axis or normalization.

3. Final Answer:

   Option A code snippet -> Option A

4. Quick Check:

   Proper FFT amplitude spectrum needs frequency axis and normalization [OK]

Hint: Normalize FFT and plot with frequency vector for correct spectrum [OK]

Common Mistakes:

• Not normalizing FFT by signal length
• Plotting full FFT without single-sided spectrum
• Missing frequency axis in plot