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Aliun [14]
4 years ago
13

Don't break or crush mercury-containing lamps because mercury powder may be released.

Engineering
1 answer:
alexandr402 [8]4 years ago
5 0
A.

It would be released without a doubt so be careful!

Hope this helps :)
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If the feedforward path of a control system contains at least one integrating element, then the output continues to change as lo
Thepotemich [5.8K]

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The attached system shows that there’s an integrator between the point where disturbance enters the system and error measuring element. A any time when R(s)=0 then

\frac {C(s)}{D(s)}=\frac {G(s)}{1+G_c(s)G(s)} and considering that E(s)=D(s)-G_c(s)C(s) then

\frac {E(s)}{D(s)}=1-(\frac {C(s)}{D(s)})G_c(s)

\frac {E(s)}{D(s)}=1-(\frac {G(s)}{1+G_c(s)D(s)})G_c(s)

\frac {E(s)}{D(s)}=\frac {1}{1+G_c(s)G(s)}

E(s)=\frac {D(s)}{1+G_c(s)G(s)}

For ramp disturbance d(t)=at

D(s)=\frac {a}{s^{2}} therefore, the steady state error is given by

e(\infty)= \lim_{s \to 0} s E(s)

e(\infty)= \lim_{s \to 0} s [\frac {D(s)}{1+G_c(s)G(s)}]

e(\infty)= \lim_{s \to 0} s [\frac {a}{s^{2}+s^{2}G_c(s)G(s)}]

e(\infty)= \lim_{s \to 0} s [\frac {a}{s+sG_c(s)G(s)}]

e(\infty)= \lim_{s \to 0} s [\frac {a}{sG_c(s)G(s)}]

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