laplace of dirac delta function

조회 수: 27 (최근 30일)
sahar
sahar 2017년 11월 13일
답변: Walter Roberson 2017년 11월 13일
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

답변 (1개)

Walter Roberson
Walter Roberson 2017년 11월 13일
dirac(t,1) is the first derivative of dirac(), not dirac(t-1)

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