Z transform using Convolution Integral

Question

Let our transfer function be $G(s) = \frac{10e^{-s}}{5s+1}$. We know that for sampling period of $T = 1$, we have $G(z) = \frac{2}{z-0.8187}$ (You can verify this in MATLAB using c2d function). What I want to do now is to use convolution integral method to verify this result. In Ogata's book it is said that $G(z) = \sum{}[\text{residue of } \frac{G(s)z}{z - e^{Ts}} \text{ at pole of G(s)}] = \frac{10e^{0.2}}{1}\frac{z}{z - e^{-0.2}} $. When this formula is applied, I get different result which is wrong. What's missing in here?

Answer 1

I am not familiar with the formula you gave from O'Hara's book, but it could be that it doesn't work for non-rational transfer functions (transfer functions which aren't a quotient of polynomials in $s$), as is the case with the delay term here.

In the case of $T=1$ it can be noted that $e^{-s}=z^{-1}$. If you would first substitute in that expression and then applying the formula, then it does yield the same result.

Also note that there are different ways of performing a continues to discrete conversion of transfer functions. The default method for c2d() is zero order hold. But there are also other methods, such as first order hold and Tustin method (c2d() also has others but I haven't really used any of them, but you can find them in the documentation of the function).