how can i plot 5 periods of this signal

hi, i have the following signal and i want to plot 5 cycles of it. but i'm getting an error

samples = 500;
T = 5;
f0 = 1/T;
t = linspace(0,T,samples+1); % 5 cycles
t(end) = [];

s2_hinf = exp(-((t-6)/4));
s2 = repmat(s2_hinf, [1 5]);
figure(1)
plot(t, s2);

but i'm getting the following error Error using plot Vectors must be the same length.

[Jan]

@OMAR ALQARNI: Please do not remove the relevant part of the question. Thanks.

Did you do this, because this is a homework and you do not want it to be found? Remember that the original version is stored in e.g. Google's cache. In addition, you did not let the forum solve your homework, but showed your own effort and asked a specific question. This is not cheating. Communicating with other programmers is a useful strategy to learn programming.

[Rukshi Sundararaj]

i just want to say thank you. <3 you are amazing

Answer 1

[Jan]

If you use repmat to concatenate the data 5 times, you have to concatenate the time t also 5 times, but add T in each block cumulatively:

samples = 500;
T       = 5;
t       = linspace(0,T,samples+1); % 5 cycles
t(end)  = [];

s2_hinf = exp((6 - t) / 4);
s2      = repmat(s2_hinf, [1 5]);
t2      = reshape(t(:) + (0:T:4*T), 1, []);  % >= Matlab R2016b
figure;
plot(t2, s2);

With older Matlab versions:

t2      = reshape(bsxfun(@plus, t(:), 0:T:4*T), 1, []);

Answer 2

[Image Analyst]

Your linspace() is wrong. You could do this:

samples = 500;
T = 5; % Period
f0 = 1/T; % Not used!
t = linspace(0, 5 * T, samples); % 5 cycles
s2_hinf = exp(-((t-6)/4));
plot(t, s2_hinf);

But I don't see where any cycles comes in. You are not using anything like sin() or cos(), so are you expecting oscillations?

Answer 3

[Image Analyst]

Perhaps you mean this:

samples = 500;
T = 5; % Period
f0 = 1/T; % Not used!
t = linspace(0, T, samples); % 5 cycles
s2_hinf = exp(-((t-6)/4));
s5 = repmat(s2_hinf, [1, 5]);
t = linspace(0, length(s5), samples*5); % 5 cycles
plot(t, s5);
grid on;

[ocsse]

how do you make the x axis from 0 to 25?

[Image Analyst]

@ocsse, how about

t = t / 100;

or

t = linspace(0, 25, samples*5);