Hello, i studied this fft code in mathworks but i have some challenge trying to use it. please, you can be of help. below is the code and my challenge

%Use Fourier transforms to find the frequency components of a signal buried in noise.
% Specify the parameters of a signal with a sampling frequency of 1 kHz and a signal duration of 1 second.
Fs = 1000;            % Sampling frequency                    
T = 1/Fs;             % Sampling period       
L = 1500;             % Length of signal
t = (0:L-1)*T;        % Time vector
%Form a signal containing a 50 Hz sinusoid of amplitude 0.7 and a 120 Hz sinusoid of amplitude 1.
S = 0.7*sin(2*pi*50*t) + sin(2*pi*120*t);
%Corrupt the signal with zero-mean white noise with a variance of 4.
X = S + 2*randn(size(t));
%Plot the noisy signal in the time domain. It is difficult to identify the frequency components by looking at the signal X(t). 
plot(1000*t(1:50),X(1:50))
title('Signal Corrupted with Zero-Mean Random Noise')
xlabel('t (milliseconds)')
ylabel('X(t)')
%Compute the Fourier transform of the signal. 
Y = fft(X);
%Compute the two-sided spectrum P2. Then compute the single-sided spectrum 
%P1 based on P2 and the even-valued signal length L.
P2 = abs(Y/L);
P1 = P2(1:L/2+1);
P1(2:end-1) = 2*P1(2:end-1);
%Define the frequency domain f and plot the single-sided amplitude spectrum P1. 
%The amplitudes are not exactly at 0.7 and 1, as expected, because of the added noise. 
%On average, longer signals produce better frequency approximations.
f = Fs*(0:(L/2))/L;
plot(f,P1) 
title('Single-Sided Amplitude Spectrum of X(t)')
xlabel('f (Hz)')
ylabel('|P1(f)|')
%Now, take the Fourier transform of the original, uncorrupted signal and retrieve the exact amplitudes, 0.7 and 1.0.
Y = fft(S);
P2 = abs(Y/L);
P1 = P2(1:L/2+1);
P1(2:end-1) = 2*P1(2:end-1);
plot(f,P1) 
title('Single-Sided Amplitude Spectrum of S(t)')
xlabel('f (Hz)')
ylabel('|P1(f)|')
MY CHALLENGE IN THE CODE
Fs = 1350;            % Sampling frequency                    
T = 1/Fs;             % Sampling period       
L = 5000;             % Length of signal
t = (0:L-1)*T;        % Time vector

S50 = Iabc(:,2);     %signal containing 50Hz sinusoid
X = Ia(:,2);         %sampled signal containing a 1350Hz sinusoid

%Plot the sampled signal in the time domain 
figure, plot(1350*t(1:100),X(1:100))
title('sampled Signal at 1350Hz')
xlabel('t (milliseconds)')
ylabel('X(t)')

% compute the fourier transform
 Y = fft(X);         %sampled signal
P2 = abs(Y/L);    %%Why (Y/L)?
P1 = P2(1:L/2+1);   %%Why and how come (1:L/2+1))?

i recieved an error message saying...

Index exceeds matrix dimensions."Error using plot. 
Vectors must be the same lengths.
Error in FFT_trail (line 18)
P1(2:end-1) = 2*P1(2:end-1);     %%how come P1(2:end-1) = 2*P1(2:end-1)?
f = Fs*(0:(L/2))/L;     %%how come? 
figure, plot(f,P1)

[Image Analyst]

Can you give us code that reproduces the error, not an error like this:

Undefined function or variable 'Iabc'.
Error in test2 (line 56)
S50 = Iabc(:,2);     %signal containing 50Hz sinusoid

You need to provide Iabc so I can run the code.