Electronic – What should be DC gain of the system

controlcontrol system

Step response of system is

\$c(t) = 5 – 0.5e^{-2t} + 8e^{-t}\$

then find steady state gain. Answer is -7.5

Now in book solution is given as :

Since this is step response we will find impulse response by differentiating with time, then "new c(t)" = y(t)

\$y(t) = e^{-2t} + -8^{-t}\$

convert into s-domain

\$Y(s) = \frac{1}{s + 2} – \frac{8}{s + 1}\$,

by final value theorem taking s->0 DC gain is -7.5

I understood this much. But my doubt is that why can't I just use
\$\frac{C(s)}{R(s)}\$ where \$R(s) = \frac{1}{s}\$ as that is step response. Then by taking s->0 I get DC gain as 5 which is not the correct answer. Please illuminate!

Best Answer

Because when taking C(s)/R(s) you are not calculating the gain, but the final DC operating point, which is trivial to find. Just take \$t\rightarrow\infty\$ and see that \$c(\infty) = 5\$. The DC gain is not the same as the response at DC conditions!

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