What this "t=0:0.0001:0.1;" statement represent and how this syntax work in the given program?

%FM generation
clc;
clear all;
close all;
fc=input('Enter the carrier signal freq in hz,fc=');
fm=input('Enter the modulating signal freq in hz,fm =');
m=input('Modulation index,m= ');
t=0:0.0001:0.1;
c=cos(2*pi*fc*t);%carrier signal
M=sin(2*pi*fm*t);% modulating signal
subplot(3,1,1);plot(t,c);
ylabel('amplitude');xlabel('time index');title('Carrier signal');
subplot(3,1,2);plot(t,M);
ylabel('amplitude');xlabel('time index');title('Modulating signal');
y=cos(2*pi*fc*t-(m.*cos(2*pi*fm*t)));
subplot(3,1,3);plot(t,y);
ylabel('amplitude');xlabel('time index');title('Frequency Modulated signal');
fs=10000;
p=fmdemod(y,fc,fs,(fc-fm));
figure;
subplot(1,1,1);plot(p);

[Stephen23]

https://www.mathworks.com/help/matlab/ref/colon.html

Answer 1

[KSSV]

t=0:0.0001:0.1

The above line generates an array with first element as 0 and last element as 0.1, where each sucessive elements have a common difference of 0.0001. It is a Arithmetic preogression.

Answer 2

[Idin Motedayen-Aval]

It is creating a time vector that is used in the next two lines to generate the sine and cosine functions, and then in plotting those functions. As the previous answer said: start at time 0, go up to 0.1, in steps on 0.0001.

Another way to think of it: it's time vector with a sampling time of 0.0001s (0.1ms), or a sampling frequency of 10kHz. This corresponds to "fs = 10000;" near the end of the program.

[Walter Roberson]

Quibble:

time vector with a sampling time of 0.0001 could imply that the vector is like (0:1000)*0.0001 -- time 0/10000, 1/10000, 2/10000 and so on .

But that is not actually the case. The actual vector is 0, 0+0.0001, 0+0.0001+0.0001, 0+0.0001+0.0001+0.0001, and so on

actual = 0:0.0001:0.1;
nominal = (0:1000)*0.0001;
mismatches = find(actual~=nominal);
format long g
am5 = actual(mismatches(1:5))

am5 = 1×5

0.0506 0.0511 0.0523 0.0528 0.054

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nm5 = nominal(mismatches(1:5))

nm5 = 1×5

0.0506 0.0511 0.0523 0.0528 0.054

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am5-nm5

ans = 1×5

1.0e+00 * 6.93889390390723e-18 6.93889390390723e-18 6.93889390390723e-18 6.93889390390723e-18 6.93889390390723e-18

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The actual vector has accumulated floating point round-off; the nominal vector does not have accumulated floating point round-off

@KSSV's description of arithmetic progression was correct; describing it in terms of sampling times is not completely accurate.