Solution Found!
In Chapter 5, we used various approximations for the time delay, one of which is the
Chapter 6, Problem 6.65(choose chapter or problem)
In Chapter 5, we used various approximations for the time delay, one of which is the first order Padé:
\(e^{-T_{d} s} \cong H_{1}(s)=\frac{1-T_{d} s / 2}{1+T_{d} s / 2} .\)
Using frequency response methods, the exact time delay
\(\mathrm{H}_{2}(\mathrm{~s})=\mathrm{e}^{-\mathrm{T}_{\mathrm{d}^{s}}}\)
can be obtained. Plot the phase of \(\mathrm{H}_{1}\) (s) and \(\mathrm{H}_{1}\) (s), and discuss the implications.
Questions & Answers
QUESTION:
In Chapter 5, we used various approximations for the time delay, one of which is the first order Padé:
\(e^{-T_{d} s} \cong H_{1}(s)=\frac{1-T_{d} s / 2}{1+T_{d} s / 2} .\)
Using frequency response methods, the exact time delay
\(\mathrm{H}_{2}(\mathrm{~s})=\mathrm{e}^{-\mathrm{T}_{\mathrm{d}^{s}}}\)
can be obtained. Plot the phase of \(\mathrm{H}_{1}\) (s) and \(\mathrm{H}_{1}\) (s), and discuss the implications.
ANSWER:Step 1 of 4
The expression for the \(H_{1}(s)\) is given by,
\(H_{1}(s)=\frac{1-T_{d}\left(\frac{s}{2}\right)}{1-T_{d}\left(\frac{s}{2}\right)}\)