I want to perform inverse Laplace transform but my code does not seem to work

I am trying to perform inverse Laplace transform the following way, but I am getting an answer that is not satisfying.

syms s K tau a

fun2 = ((60*K*(exp(-a*s)))/tau)/(s*(s + 1/tau));
y2 = ilaplace(fun2);

pretty(y2)

          /   exp(-a s)         \
K ilaplace| -------------, s, t | 60
          |   /      1  \       |
          | s | s + --- |       |
          \   \     tau /       /
------------------------------------
                 tau

I am getting the answer that code shows, but it is not helpful as it does not perform the Laplace inverse transform. I do not know how to solve the problem.

Answer 1

[Paul]

Hi Carlos,

ilaplace assumes that all signals are causal, which wouldn't be true in this case if a < 0.

Add the appropriate assumption, and ilaplace returns a closed-form expression.

syms s K tau a
assume(a >= 0)
fun2 = ((60*K*(exp(-a*s)))/tau)/(s*(s + 1/tau))

fun2 = 

y2 = ilaplace(fun2)

y2 = 

[Carlos Puente]

Indeed a>=0, but I still can't understand what "heaviside" means in that expression. I am asking just for means of getting to understand the syntax, my code worked very well with your solution!

[Walter Roberson]

See heaviside

heaviside() is 0 when the argument is negative.

heaviside() is 1 when the argument is positive

sympref 'HeavisideAtOrigin' permits you to configure the value when the argument is exactly 0. The default is 1/2 but 0 and 1 are also common. There is no "standard" for what the heaviside function should be when the parameter is 0: 0, 1/2, or 1 are all useful in different circumstances.

In the context of that y2 result, it is saying that the y2 value should be 0 if t < a, and that it should be 60K/tau times the (tau minus exponential) expression for t > a ... and it is fuzzy about what should happen at time a exactly.

[Paul]

@Carlos Puente

You may be more familiar with the term unit step, or step, function, rather than Heaviside function.

It shows up in y2 as a result of the time shift property of the unilateral Laplace transform:

f(t-a)*u(t-a) -> exp(-a*s)*F(s) , a > = 0

u(t) is the same as heaviside(t)

@Walter Roberson

y2(a) = 0 regardless of the value of heaviside(0).

[Paul]

Also, you have to be careful with Matlab if you want to apply the time shift property by hand because ilaplace does not always tack on the heaviside to the result like it should (others may have different opinion).

For example

syms s K tau t a
assume(a >= 0)
fun2(s) = ((60*K)/tau)/(s*(s + 1/tau))

fun2(s) = 

y2(t) = ilaplace(fun2)

y2(t) = 

Now, if we want to find the ilapalce of exp(-a*s)*fun2(s), it's tempting to just shift y2(t) to the right by a

y2(t-a)

ans = 

whch is incorrect. Instead, we have to tack on the heaviside to the y2(t) first and then shfit

y2(t) = y2(t)*heaviside(t);
y2(t-a)

ans = 

which is the same result as above (after canceling tau in result above).