physical exponential decay and growth processes

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

Answer 1

[Walter Roberson]

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')

Answer 2

[Image Analyst]

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;

[Felix]

thanks, this is beautiful! it's the same curve I worked manually

[Image Analyst]

You're welcome. If it works, then could you click the "Accept this answer" link? Thanks in advance.