laplace of dirac delta function

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sahar
sahar el 13 de Nov. de 2017
Respondida: Walter Roberson el 13 de Nov. de 2017
my question is about the difference between this two definition of dirac delta function in Matlab:
dirac(t,1)
and
dirac(t-1)
I tried to apply the laplace transform and inverse laplace transform to this 2 functions and they gives me very different results:
syms s t
>> F_s=s+s^2; % definition of the function in s domain
>> f_t=ilaplace(F_s)
f_t =
dirac(t, 1) + dirac(t, 2)
>> F_s=laplace(f_t)
F_s =
s^2 + s
>> F_s=laplace(dirac(t-1)+dirac(t-2))
% but if I use this definition for dirac delta function it gives a very diffrent answer...
F_s =
exp(-s) + exp(-2*s)
and my problem is that when i want to plot the function as:
t=1:0.1:4;
>> f_t1=dirac(t, 1) + dirac(t, 2);
Error using double.dirac
Too many input arguments.
It can't plot the function so I used the definition as :
dirac(t-1)+dirac(t-2)
but it gives a very different answer in Laplace transform... could you please explain the reason to me?
Thank you all
sahar

Respuestas (1)

Walter Roberson
Walter Roberson el 13 de Nov. de 2017
dirac(t,1) is the first derivative of dirac(), not dirac(t-1)

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