如何在Matlab中对离散输入信号执行数值Laplace和拉普拉斯逆变换？

Question

I want to evaluate the Laplace transform of a discrete stochastic signal I sample from a communication device.

As a simple test case I am trying to use the ilaplace to get the original signal, but I am not sure if my approach is plausible.

x = sym('x','real');
y = sym('y','real');
t=linspace(0,1000,1000);
f=sin(t);
s = x+i*y;               
F_s=sum(f.*exp(-s*t));   
ilaplace(F_s)

The above might seem silly, though in my real problem I am trying to estimate the medium Green's function which is of the form ilaplace(2*F_s/(-s*F_s +f(0))) .

I have also tried to use a signal symbolic variable s and it gives me a train of deltas which I am not sure it's correct and what is the error estimation.

syms s;                                
F_s = symfun(sum(sin(t).*exp(-s*t)), s);
ilaplace(F_s)

Answer 1

Try increasing the number of elements in t to more accurately sample the sine function, or smoothing the signal before calculations. Another problem you face is that the inverse Laplace transform expects a function to be defined for s>0 , ie up to infinity. You truncate your signal at t=1000 , thus the Laplace transform is not going to infinity either.

Judging the documentation of ilaplace it tries to transform each individual term in your array F_s . This has a sum of a thousand entries of some constant times some exponential of the form exp(-s*t) , whose inverse transform is a delta pulse, see this Wikipedia page, second entry of the table . Hence, your inverse transform of F_s is a series of a thousand delta pulses, which will presumably be exactly the thousand input points you generated with your sin(t) , up to some numerical error which emerged in the forward and backward Laplace transform.

Finally: MATLAB expects, as stated in the documentation, a single symbolic variable s . You made s consist of two symbolic variables. Whilst mathematically correct, MATLAB only takes the first symbolic variable to be the transform variable, thus you get an answer with delta pulses multiplied by some exp(1i*y) function. Using the following code, you at least get rid of that problem (but obviously keep the summation of delta pulses):

t=linspace(0,1000,1000);
f=sin(t);
syms s
F_s=sum(f.*exp(-s*t));   
ilaplace(F_s)

Answer 2

Ok so that's the way I used matlab to do laplace transform for a discrete signal and recovered it back using ilaplace for validation purposes:

t=linspace(0,10,500);
f=exp(-t/0.2);   
syms s;                                
F_s = sum(f.*exp(-s*t));
f_t = (ilaplace(F_s)); 
F_t = (int(f_t));

y=subs(F_t,t);
Ft_recovered = diff(double(y));


subplot(2,2,1)
plot(t,f)
title('numerical input exp(-t/0.2)')


subplot(2,2,2)
ezplot(F_s)
title('numerical laplace')


subplot(2,2,3)
plot(t(1:end-1), Ft_recovered)
title('recovered signal')


subplot(2,2,4)
syms x;
fx = symfun(exp(-x/0.2),x);
ezplot(laplace(fx))
title('symbolic laplace transform')

The big difficulty here is to make matlab to evaluate the sum of the dirac delta train, so I did this trick: integrated the expression to convert it to train of heaviside functions, evaluate it and then plot the derivative:

F_t = (int(f_t));
y=subs(F_t,t);
Ft_recovered = diff(double(y));