solution06

# solution06 - 1(a = k t i b k = J J Vin = iR eb = iR k t i =...

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1/ (a) ( ) in t t t in t t in t b in t V RJ k J k J b R k J k b R k V k R k V i k iR e iR V J k b i k J = + + + = = + = + = = Τ = θ θ θ θ θ θ θ θ θ θ θ θ & & & & & & & & & & & & / 2 (b) J k s J b R k s RJ k s H t t / ) / ( / ) ( 2 2 + + + = (c) i) When the derivates are set to zero rads 625 . 0 = = = in t in t V Rk k V RJ k J k θ θ ii) In the steady the shaft is stationary and the torque produced by the motor is balanced by the torque from the torsional spring. i k k t = = Τ θ As the shaft is stationary e b is zero and the current is R V i in / = Putting the two together gives the result obtained in (i).

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iii) rads 625 . 0 5 ) ( lim 5 ) ( ) ( ) ( ) ( 0 = = = = Rk k s sY s s H s X s H s Y t s (d) i) ( ) 4 3 5 . 2 5 ) ( ) ( ) ( ) ( 4 3 5 . 0 ) ( 2 2 + + = = = + + = s s s s s H s V s H s X s s s H c = -0.31250 + 0.35434i, -0.31250 – 0.35434i, 0.625 p = -1.5000 + 1.3229i, -1.5000 - 1.3229i, 0 rads 2935 . 2 deg 409 . 131 1 47245 . 0 1 = = = c c ) 2935 . 2 3229 . 1 cos( 5 . 1 9449 . 0 625 . 0 ) ( + + = t t e t θ at t = 5, output is 0.62454 ii) ( ) 4 4 5 . 2 5 ) ( ) ( ) ( ) ( 4 4 5 . 0 ) ( 2 2 + + = = = + + = s s s s s H s V s H s X s s s H 625 . 0 5 . 2 5 . 2 25 . 1 5 . 2 625 . 0 4 4 5 . 2 ) ( ) ( ) ( ) ( 0 , 2 , 2 2 2 2 1 2 2 0 2 3 3 3 2 1 2 1 1 3 2 1 = = = = = = + + = + + = = = = = = = = s s s s s s ds d c s c s s c p s c p s c p s c s X p p p
Matlab: c = -0.625, -1.25, 0.625 p = -2, -2, 0 t t te e t 2 2 25 . 1 625 . 0 625 . 0 ) ( = θ at t = 5, output is 0.62469 iii) ( ) 4 6 5 . 2 5 ) ( ) ( ) ( ) ( 4 6 5 . 0 ) ( 2 2 + + = = = + + = s s s s s H s V s H s X s s s H c = 0.10676, -0.73176,
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