physical exponential decay and growth processes

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Hi All,
I am trying to model the following exponential function for signal decay according to the equation N = N0*e^-1/tau where:
N = signal can vary from 0 to 1;
t = time from 0 to 10 (in seconds);
tau: the time required for N to decrease in size by a factor of 1/e when t = 1.
For my purpose, I hold N0 = 1 and T =1 since I am trying to model different curves at tau = 0.5, tau = 1, tau = 2, etc. to represent how the exponential decay changes as result of changes in tau.
I have tried the following code but I get an exp growth curve instead, which I don't understand why since the exp is negative. Of course, if I remove the negative sign, I get a decay curve, which is not the intended purpose of the formula:
N = linspace(0,1,11);
t = linspace(0,10,11);
tau = linspace(0.5,10,11);
N0= 1;
N = N0*exp(-1./tau);
plot(t,N,'-')
thanks for your help

采纳的回答

Walter Roberson
Walter Roberson 2021-11-22
You have two changing variables: t and tau. And you are calculating N using changes in tau, but plotting over changes in t.
If you want both t and tau to be changing, then your code needs to use a plot with two independent variables and one dependent variable, such as a surf() -- and your formula needs to include t, even if only implicitly. For example if you were to rewrite N0 to be something that depends on time:
num_t = 11;
num_tau = 12;
t = linspace(0, 10, num_t);
N0 = t ./ 10; %signal is straight line
tau = linspace(0.5, 10, num_tau);
N = N0 .* exp(-1./tau.'); %tau is vertical, time is horizontal
surf(t, tau, N, 'edgecolor', 'none');
xlabel('time'); ylabel('tau'); zlabel('N')

更多回答(1 个)

Image Analyst
Image Analyst 2021-11-22
You need to put t into the equation for N, and just plot one curve for each tau. You were varying N with tau and then plotting vs. t, which is wrong:
N = linspace(0,1,11);
t = linspace(0,10,11);
tau = linspace(0.5,10,11);
N0= 1;
for k = 1 : length(tau)
N = N0*exp(-t ./ tau(k));
plot(t,N,'-')
hold on;
end
grid on;
  2 个评论
Felix
Felix 2021-11-22
thanks, this is beautiful! it's the same curve I worked manually
Image Analyst
Image Analyst 2021-11-22
You're welcome. If it works, then could you click the "Accept this answer" link? Thanks in advance.

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