I have a transfer function as follows –
$$
G(s) = \frac{320 (1-4s)e^{-3s}}{24s^2+28s+4}
$$
I calculated its time delay as 3 sec, gain as 80, but confused in finding the time constant.
I know the time constant of standard first and second order transfer function but not for non-standard. So I found its inverse laplace and got
$$
G(s) = 80(\frac{e^{0.5-t/6}}{3}-e^{3-t})
$$
and on putting
$$
\frac{e^{0.5-t/6}}{3}-e^{3-t} = \frac{1}{e}(\frac{e^{0.5}}{3}-e^3)
$$
got 2 roots, around 1 and 18. Please shed some light on how to find answer from laplace domain. TIA
Best Answer
I would first rewrite the equation in a low-entropy format, with a leading term corresponding to the gain:
\$G(s) = \frac{320 (1-4s)e^{-3s}}{24s^2+28s+4}=G_0\frac{1-\frac{s}{\omega_z}}{1+s(\tau_1+\tau_2)+s^2(\tau_1\tau_2)}e^{-s\tau}\$
When you factor 4 in the denominator \$D(s)\$, you have \$G_0=\frac{320}{4}=80\$ and \$D(s)=1+7s+6s^2\$.
The coefficient for \$s\$ has the dimension of time so it is a sum of time constants: \$7=\tau_1+\tau_2\$ while the coefficient for \$s^2\$ has a dimension of time squared: \$6=\tau_1\tau_2\$. If you solve this simple two-unknown two-equation system, you find that \$\tau_1=6\;\mathrm s\$ and \$\tau_2=1\;\mathrm s\$ and the right-half-plane zero is located at \$\omega_z=0.25\;\mathrm{rd/s}\$. The exponential term adds a pure delay \$\tau\$ of 3 s.
You can now rewrite the equation considering two cascaded poles:
\$G(s)=G_0\frac{1-\frac{s}{\omega_z}}{(1+\frac{s}{\omega_{p1}})(1+\frac{s}{\omega_{p2}})}e^{-s\tau}\$ with \$\omega_{p1}=1\;\mathrm{rd/s}\$ and \$\omega_{p2}=0.166\;\mathrm{rd/s}\$
If you want to further simplify this expression, you could also replace the delay by its first-order Padé approximant: \$e^{-st}\approx\frac{1-\frac{s}{\omega_{\tau}}}{1+\frac{s}{\omega_{\tau}}}\$ with \$\omega_{\tau}=\frac{2}{\tau}\$. You would add another RHPZ and a third time constant.