Digital Control Systems: Difference between revisions

Revision as of 22:24, 29 April 2014

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.

This is how the c2d zero order hold works in MATLAB/octave. It uses the "exact" solution to the discretization problem.
Also check Chapter 7.3 of Greg Starr's book.

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 *Lectures
 **[http://www.stanford.edu/class/archive/ee/ee392m/ee392m.1056/Lecture5_Digital.pdf Sampled Time Control]
+**[http://www.ece.rutgers.edu/~orfanidi/intro2sp/orfanidis-i2sp.pdf Introduction to Signal Processing]
 ===MATLAB/Octave===