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3 years ago

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A controller for a satellite attitude control with transfer function G = 1 s 2 has been designed with a unity feedback structure

and has the transfer function D(s) = 8(s+2) s+4 . Find the system type and the corresponding error constant for this system.

1 answer:

S_A_V [24]3 years ago

5 0

Answer:

type 2, k = 4

Explanation:

(a) The transfer function of the controller for a satellite attitude control is

$G = \frac{1}{s^2}$

The transfer function of unity feedback structure is

$D(s) = \frac{10(s+2)}{s+5}$

To determine system type for reference tracking, identify the number of poles at origin in the open-loop transfer function.

For unity feedback system, the open-bop transfer function

$G(s)D_c(s)=\frac{1}{s^2}\frac{10(s+2)}{s+5}$

$=\frac{10(s+2)}{s^2(s+5)}$

Determine the poles in G(s)4(s).

s = 0,0,-5

Type of he system is decided by the number of poles at origin in the open loop transfer function.

Since, there are two poles at origin, the type of the system will be 2.

Therefore, the system type is

Type 2

check the attached file for the concluding part of the solution

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A 11.5 nC charge is at x = 0cm and a -1.2 nC charge is at x = 3 cm ..At what position or positions on the x-axis is the electric

diamong [38]

Answer:

Explanation:

Given

$q_1=11.5\ nC$ charge is placed at $x=0\ cm$

another charge of $q_2=-1.2\ nC$ is at $x=3\ cm$

We know that Electric field due to positive charge is away from it and Electric field due to negative charge is towards it.

so net electric field is zero somewhere beyond negatively charged particle

Electric Field due to $q_2$ at some distance r from it

$E_2=\frac{kq_2}{r^2}$

Now Electric Field due to $q_1$ is

$E_1=\frac{kq_1}{(3+r)^2}$

Now $E_1+E_2=0$

$\frac{k\times 11.5}{(r+3)^2}\frac{k\times (-1.2)}{r^2}=0$

$\frac{3+r}{r}=(\frac{11.5}{1.2})^{0.5}$

$\frac{3+r}{r}=3.095$

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A flywheel made of Grade 30 cast iron (UTS = 217 MPa, UCS = 763 MPa, E = 100 GPa, density = 7100 Kg/m, Poisson's ratio = 0.26) h

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Answer:

N = 38546.82 rpm

Explanation:

$D_{1}$ = 150 mm

$A_{1}= \frac{\pi }{4}\times 150^{2}$

= 17671.45 $mm^{2}$

$D_{2}$ = 250 mm

$A_{2}= \frac{\pi }{4}\times 250^{2}$

= 49087.78 $mm^{2}$

The centrifugal force acting on the flywheel is fiven by

F = M ( $R_{2}$ - $R_{1}$ ) x $w^{2}$ ------------(1)

Here F = ( -UTS x $A_{1}$ + UCS x $A_{2}$ )

Since density, $\rho = \frac{M}{V}$

$\rho = \frac{M}{A\times t}$

$M = \rho \times A\times t$ $M = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t$

$M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37$

$M = 8252963901$

∴ $R_{2}$ - $R_{1}$ = 50 mm

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F = 33618968.38 N --------(2)

Now comparing (1) and (2)

$33618968.38 = 8252963901\times 50\times \omega ^{2}$

∴ ω = 4036.61

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$\omega = \frac{2\pi N}{60}$

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