Comparing the curves of the graphs

Comparing the curves

>> NFFT = 2^nextpow2(L); % Next power of 2 from length of y
Y = fft(y,NFFT)/L;
f = Fs/2*linspace(0,1,NFFT/2+1);
% Plot single-sided amplitude spectrum.
plot(f,2*abs(Y(1:NFFT/2+1)))
title('Single-Sided Amplitude Spectrum of y(t)')
xlabel('Frequency (Hz)')
ylabel('|Y(f)|')



>> Fs = 1000; % Sampling frequency
T = 1/Fs; % Sample time
L = 1000; % Length of signal
t = (0:L-1)*T; % Time vector
% Sum of a 50 Hz sinusoid and a 120 Hz sinusoid
x = 0.7*sin(2*pi*50*t) + sin(2*pi*120*t);
y = x + 2*randn(size(t)); % Sinusoids plus noise
plot(Fs*t(1:50),y(1:50))
title('Signal Corrupted with Zero-Mean Random Noise')
xlabel('time (milliseconds)')
>>

Answer 1

[Image Analyst]

The spectrum (left plot) shows a spike at 50 and 120 Hz. I'd never know that there is monotone noise at those exact frequencies just by looking at the time domain plot on the right. So I'd say the frequency domain plot is most informative in this particular case.

[ivko ivkovic]

hi thanks for your reply,

have you got any idea whats missing in the code above to get the pure signal? I managed to get the main code but I struggle with this bit

the graph should look similar to this:

[Image Analyst]

Try setting the spectrum to zero where there are the spikes. Then inverse transform.

[ivko ivkovic]

basically I need to add function that the pure signal and is seen on the graphs

tried modiyfing the code and adding x = 0.7*sawtooth(2*pi*50*t);%Function to calculate Sawtooth wave but I get an error. Any ideas?

[Image Analyst]

I don't know why you did that instead of what I suggested. Anyway, attach your y in a .mat file with the paperclip icon.