3 years ago

How can I use the flux density B formula if I don’t know magnetic flux

Engineering

1 answer:

BabaBlast [244]3 years ago

3 0

Answer:

use the dimensions shown in the figure

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

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

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

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

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

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

Whenever $G_c(s)$ has a double intergrator, the error $e(\infty)$ becomes zero

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3 years ago