# 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) &= % \frac{z-1}{z} \mathscr{Z} \left\{ \mathscr{L}^{-1} \left\{ \frac{F(s)}{s} \right\} \right\} \\ &= \frac{z-1}{z} \mathscr{Z} \left\{ \mathscr{L}^{-1} \left\{ \frac{K}{s \left( Ts+1 \right) } \right\} \right\} \\ &= \frac{z-1}{z} \mathscr{Z} \left\{ K \left( 1 - e^{-\frac{1}{T} T_s n} \right) \right\} \\ &= \frac{z-1}{z} K \left( \frac{z}{z-1} - \frac{z}{z - e^{-\frac{T_s}{T}}} \right) \\ &= \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)} \\ &= 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.

## 1 Answer

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) &= K \frac {\left( 1 - e^{-\frac{T_s}{T}} \right)} {\left( z - e^{-\frac{T_s}{T}} \right)}. \\ &= 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] &= \mathscr{Z}^{-1} \left\{ G(z) \right\} \\ &= 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\} \\ &= 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\} \\ &= K \left( 1 - e^{-\frac{T_s}{T}} \right) \delta[n-1] * \left( e^{-\frac{T_s}{T}n} u[n] \right) \\ &= 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.