Digital Control Systems

Relationship between the Laplace and Z transforms
Convolution and Z Transforms
- Z Transforms and Convolution
- Here is an animation of convolution with continuous signals. Look at it so you understand what is happening with the mechanics of convolution. To convolve $x(t)$ $x(t)$ with $h(t)$ $h(t)$ , you flip shift $h(t)$ $h(t)$ into $h(t-t_{0})$ $h(t-t_{0})$ on the $t_{0}$ $t_{0}$ axis, then you multiply it by $x(t_{0})$ $x(t_{0})$ to get $x(t_{0})h(t-t_{0})$ $x(t_{0})h(t-t_{0})$ , then you integrate with respect to $t_{0}$ $t_{0}$ , so that the convolution is: $x(t)*h(t)=\int _{-\infty }^{\infty }x(t_{0})h(t-t_{0})dt_{0}$ $x(t)*h(t)=\int _{-\infty }^{\infty }x(t_{0})h(t-t_{0})dt_{0}$ . The first animation shows this happening with sampled waveforms: $x_{s}(t)=x(t)\sum _{n=0}^{\infty }\delta (t-nT)=\sum _{n=0}^{\infty }x(nT)\delta (t-nT)$ $x_{s}(t)=x(t)\sum _{n=0}^{\infty }\delta (t-nT)=\sum _{n=0}^{\infty }x(nT)\delta (t-nT)$ and $h_{s}(t)=h(t)\sum _{n=0}^{\infty }\delta (t-nT)=\sum _{n=0}^{\infty }h(nT)\delta (t-nT)$ $h_{s}(t)=h(t)\sum _{n=0}^{\infty }\delta (t-nT)=\sum _{n=0}^{\infty }h(nT)\delta (t-nT)$ .
  - Discrete convolution animations
- Notice that this is really the same as Polynomial Multiplication.