Manually trying to calculate output of an transfer function.

Question

I am trying to calculate the output of an transfer function due to the input of an step, But some weird reason, I am only getting the inverse output, what Matlab says it should be.

My transfer function is $ \frac{1}{s+1}$ which i Apply an step to $1/s$ which in time domain gives an function $e^{-t}$... which is inverse of what matlab gives me..

Why am i getting this???

Answer 1

If your transfer function is $1/(s+1)$ then the Laplace transform of the step response is

$$Y(s)=\frac{1}{s(s+1)}=\frac{1}{s}-\frac{1}{s+1}\tag{1}$$

Using basic Laplace transform relations, the inverse of (1) is

$$y(t)=(1-e^{-t})u(t)$$

EDIT:

If you want to use convolution then you must compute the following integral

$$y(t)=\mathcal{L}^{-1}\{\frac{1}{s}\}*\mathcal{L}^{-1}\{\frac{1}{s+1}\}=\int_0^te^{-\tau}d\tau,\quad t>0$$

which of course results in the same solution.