Nyquist/Shannon Theorem MATLAB Code suffering from minor issues, don't know where I went wrong?

A few months ago I asked a similar question as to how I could properly create nyquist/shannon graphs on MATLAB, and fortunately I got help from someone here. However, in those few months I managed to learn a lot and tried this task again with this new knowledge and manipulation of matlab tools. When I gave it a try, I only ended up getting a half/small sampled graph instead the whole thing.

What I did was what I learned about the Nyquist theorem, which was that:

Nyquist Frequency is the highest frequency, therefore this nyquist frequency gets doubled to get the nyquist rate. This nyquist rate then gets multiplied by 5 to get the sampling frequency and finally is divided over one to get the sampling period(1/F = T). The code below shows perfectly how I followed this process.

%Time Base
t = 0:0.001:2;

%Nyquist Frequencies
Fn1 = 1;
Fn2 = 6;

%Nyquist Rates 
Fnr2 = 2*(Fn2);

%Sampling Period
Sp2 = 5*(Fnr2);


Ts2 = 1/(Sp2);
T2 = 1/(Fn2);

%Number of Samples
N2 = (T2/Ts2); 
n2 = 0:1:N2-1; 

nTs2 = n2 * Ts2; 

x_c = sin(2*pi*Fn1*nTs2);
x_c2 = sin(2*pi*Fn2*nTs2);

signal = sin(2*pi*Fn1*t) + sin(2*pi*Fn2*t);
ct = nTs2;
nqsignal = x_c + x_c2;

%Second Part
h = stem(ct, nqsignal, 'linewidth', 2);
hold
plot(t, signal, 'linewidth', 2)
lgd = legend('Discrete Data', 'Continuous Data');
set (lgd, "fontsize", 12)
set(gca,'XTick',[0:0.2:1.8]);
set(gca,'YTick',[-2:0.5:2]);
title('Time vs Magnitude','fontweight','bold','fontsize',16);
xlabel('Time(s)','fontweight','bold','fontsize',14)
ylabel('Magnitude','fontweight','bold','fontsize',14)
grid

My Output from this code:

What I'm supposed to get:

I later found out that if I added more to the Number of samples (which is the variable N2), I could actually get the same output depending on whatever random number I add to it(e.g. I could do N2+50 rather than N2-1 and get the same output as this graph above) instead of doing "N2-1". Why that works, i don't know, but is there any other way I could get this same answer/output without having to make random and unexplainable modifcations to the Number of Samples? I have tried reducing the number of steps to get to the number of samples but that only increases the number of samples in the given area, without extending it.

Thank you in advance, and please leave a comment if I was not clear enough in this question

[dpb]

Well, what IS the question?

The Nyquist frequency is determined by the sampling rate, not the other way around in general.

Sampling at 5X Nyquist is a rational approach to ensure adequate sample rate to reproduce the signal over a relatively short number of samples, but anything over is adequate mathematically

The number of samples is pretty-much independent of anything else once you've set the sampling frequency; all it does is determine how long the subsequent sample will be; it will have no bearing whatsoever on the frequencies; only the resolution in the frequency domain is determined by the size of the FFT if you use baseband DFT and the full-length time series.

[Joel Okanta]

What I was asking is how could I increasing the sampling to the point I get the same output as below?

OH, because from what I had seen previously I thought you had to stick with the 0:N-1 rule thing, at least that's how I learnt it. So overall it doesn't matter with the N-1 or N+1 or whatever as I could add as many samples to it just to increase its sampling range.

[dpb]

I dunno still that I understand what you're asking about.

Once again, once the sampling rate (samples/second) is set, then the Fmax is set as

Fmax=1/(2*dt)

The duration of the signal then follows as

T = N*dt

where N is the number of samples. The (N-1) number comes about in MATLAB that in order for the sample time vector, t, to be 0-based, it must be adjusted to be [0:(N-1)]*dt as [1:N]*dt would start at dt, not 0. This is immaterial for all practical purposes, but it's just generally nicer to have the time trace start at the origen.

The frequency resolution of the FFT is given by

df = Fmax/(N/2)

so if you double N you double the number of bins in the FFT and halve the distance between frequency bins. But with the given sample rate, dt, Fmax is sitll the same and so nothing changes in the resulting frequency plot of a PSD except one has twice the resolution; the frequency range is still 0:Fmax

For the time trace, changing N just keeps adding more and more points on to the end of the trace, doubling means a 1-second time trace would instead be 2-seconds, but nothing else changes; if it is a deterministic signal one will just get more cycles of the same deterministic signal.

But, you can set N arbitrarily to whatever you want, you just have to have a long-enough time record to capture at least one complete cycle of the lowest frequency in the signal you wish to detect; otherwise it will look mostly like DC offset and trend. That's determined and seen as df, the resolution of the FFT, can only capture something that is at least 0+df Hz in minimum frequency.

If you choose some combination that is a 1 Hz resolution, then you won't be able to find a frequency component in the signal that is less than that 1 Hz bin. To do better with baseband FFT you would have to increase the length of the sample, increasing the resolution and thus the ability to see something below 1 Hz.

[Joel Okanta]

So from what you're saying, that N-1 rule thing was just to start off nicely on MATLAB right? So overall it does not matter what you do with N, due to the fact that adding larger sums of numbers to it will extend the range for sampling the signal.

[dpb]

Basically, yes.

numel(0:(N-1)) == numel(1:N) = N