Inverse Laplace transformation using reside method of transfer function that contains time delay I'm having a problem trying to inverse laplace transform the following equation
$$
  h_0 = K_p * \frac{1 - T s}{1 + T s} e ^ { - \tau s}
$$
I've tried to solve this equation using the residue method and got the following.
$$
  y(t) = 2 K_p  e ^ {- \tau s} e ^ {-t/T}
$$
$$
  y(t) = 2 K_p e ^ {\frac{\tau}{T}} e ^ {-t/T}
$$  
And that is clearly wrong.
Is it so that you can't use the residue method on functions that contains time delay, or is it possible to do a "work around" or something to get to the right answer?

First do polynomial division to simplify the fraction:
$$\frac{1-Ts}{1+Ts}=-1+\frac{2}{1+Ts}$$
Now expand $h_0$:
$$h_0=-K_pe^{-\tau{s}}+2K_p\frac{1}{Ts+1}e^{-\tau{s}}$$
Recall the time-domain shift property:
$$\scr{L}(f(t-\tau))=f(s)e^{-\tau{s}}$$
$$\scr{L}^{-1}h_0=-k_p\delta{(t-\tau)}+2k_p g(t-\tau)$$
Where $g(t)=\scr{L}^{-1}\frac{1}{ts+1}$.
To take the inverse Laplace transform of this term, recall the frequency domain shift property:
$$\scr{L}^{-1}f(s-a)=f(t)e^{at}$$
$$\frac{1}{Ts+1}=\frac{1}{T(s+\frac{1}{T})}=\frac{1}{T}\frac{1}{s+\frac{1}{T}}$$
Therefore the inverse Laplace transform is:
$$\frac{1}{T}e^{-\frac{t}{T}}$$
Finally, putting all of it together, the full inverse Laplace transform of the original expression is:
$$-K_p\delta(t-\tau)+\frac{2K_p}{T}e^{-\frac{t-\tau}{T}}$$
