Impulse response of a continuous system sampled with zero-order hold I've a continuous system $$F(s) = \frac{K}{Ts+1}.$$

I sample it with zero-order hold with sampling period $T_s$. The discrete system transfer function is $$
\begin{aligned}
G(z) &amp;=
%
\frac{z-1}{z} \mathscr{Z} \left\{
  \mathscr{L}^{-1} \left\{ \frac{F(s)}{s} \right\}
\right\}
\\
&amp;= \frac{z-1}{z} \mathscr{Z} \left\{
  \mathscr{L}^{-1} \left\{ \frac{K}{s \left( Ts+1 \right) } \right\}
\right\}
\\
&amp;= \frac{z-1}{z} \mathscr{Z} \left\{
K \left( 1 - e^{-\frac{1}{T} T_s n} \right)
\right\}
\\
&amp;= \frac{z-1}{z}
K \left( \frac{z}{z-1} - \frac{z}{z - e^{-\frac{T_s}{T}}} \right)
\\
&amp;= \frac{z-1}{z} K \frac
  {z \left( 1 - e^{-\frac{T_s}{T}} \right)}
  {(z-1) \left( z - e^{-\frac{T_s}{T}} \right)}
\\
&amp;= K \frac
  {\left( 1 - e^{-\frac{T_s}{T}} \right)}
  {\left( z - e^{-\frac{T_s}{T}} \right)}.
\end{aligned}
$$

From this I can get difference equation of the sampled system: $$
y[n] = e^{-\frac{T_s}{T}} y[n-1] + K \left( 1 - e^{-\frac{T_s}{T}} \right) x[n-1].
$$

I'd like to get  impulse response function of the system in time domain $h[n]$: $$
y[n] = h[n] * x[n].
$$

To get $h[n]$, I'd have to find inverse z-transform of $G(z)$, but I can't figure out what it is. I can't find it in any tables. What is it?

Thank you.
People in ##dsp on Freenode helped me realize that the impuse response is a combination of impulse responses of two simple systems connected in series, that is, convolution of two impulse responses in the time domain.

$$
\begin{aligned}
G(z) &amp;=
K \frac
  {\left( 1 - e^{-\frac{T_s}{T}} \right)}
  {\left( z - e^{-\frac{T_s}{T}} \right)}.
\\ &amp;=
K \left( 1 - e^{-\frac{T_s}{T}} \right)
z^{-1}
\frac{1}{\left( 1 - e^{-\frac{T_s}{T}} z^{-1}\right)}
\end{aligned}
$$

$$
\begin{aligned}
h[n] &amp;= \mathscr{Z}^{-1} \left\{ G(z) \right\}
\\ &amp;=
K \left( 1 - e^{-\frac{T_s}{T}} \right)
\mathscr{Z}^{-1} \left\{
z^{-1}
\frac{1}{\left( 1 - e^{-\frac{T_s}{T}} z^{-1}\right)}
\right\}
\\ &amp;=
K \left( 1 - e^{-\frac{T_s}{T}} \right)
\mathscr{Z}^{-1} \left\{
z^{-1}
\right\}
*
\mathscr{Z}^{-1} \left\{
\frac{1}{\left( 1 - e^{-\frac{T_s}{T}} z^{-1}\right)}
\right\}
\\ &amp;=
K \left( 1 - e^{-\frac{T_s}{T}} \right)
\delta[n-1] * \left( e^{-\frac{T_s}{T}n} u[n] \right)
\\ &amp;=
K \left( 1 - e^{-\frac{T_s}{T}} \right)
e^{-\frac{T_s}{T}(n-1)} u[n-1]
\end{aligned}
$$

The impulse response is exponential decay shifted to the right by 1.