Solution Final Exam Bee 3143

QUATION: From the system below, determine: (a). Block diagram (b). Closed transfer function with vi as input variable and x as output variable. (c). Function x(t )

v

k

o

R R

R2 R

vf vi

b x

3

M

4

f (t )

1 _ +

R

vs

L

R

C

1

N

S R

2

Solution:

vs

L

R3

C

eb

i KVL in RLC circuit di 1 v s = iR + L + ∫ idt + eb , eb is a back emf dt C Back emf effect of x displacement dx eb = K 1 , K1 is a constant and x is the mass (M) position dt Electromagnetic force generate by solenoid t f (t ) = K 2 i , K 2 is a constant Laplace form equation (3) is F (s) = K 2 I (s) Substitution equation (2) into equation (1): di 1 dx v s = iR + L + ∫ idt + K 1 dt C dt Laplace form of equation (5)

Markah 1

Markah 1 Markah 1

Vs ( s) = LsI ( s) + RI ( s ) +

1 I ( s ) + K 1 sX ( s) Cs

 CLs 2 + RCs + 1   I ( s) + K 1 sX ( s ) Vs ( s) =  Cs  

Markah 1

Force related with mass M kX (s )

b s X (s )

M

M s 2 X (s) F (s) f (t ) = Mx + bx + kx Laplace form of equation (7) F ( s ) = Ms 2 X ( s) + bsX ( s) + kX ( s ) F ( s ) = ( Ms 2 + bs + k ) X ( s ) Equation (4) = equation (8), therefore K 2 I ( s) = ( Ms 2 + bs + k ) X ( s ) Then ( Ms 2 + bs + k ) I (s) = X ( s) K2 Substitution equation (9) into equation (6):   CLs 2 + RCs + 1  Ms 2 + bs + k  Vs ( s) =  X ( s)  + K 1 sX ( s) Cs K2     (CLs 2 + RCs + 1)( Ms 2 + bs + k )   X ( s ) + K 1 sX ( s ) Vs ( s) =  K 2 Cs     (CLs 2 + RCs + 1)( Ms 2 + bs + k )    + K 1 s  X ( s ) Vs ( s) =    K 2 Cs    Open loop transfer function K 2 Cs X (s) = 2 Vs ( s) (CLs + RCs + 1)( Ms 2 + bs + k ) + K 1 K 2 Cs 2

Feedback voltage

Markah 1

Markah 1

Markah 1

Markah 2

R3 vo R3 + R4 Laplace form R3 V f (s) = Vo ( s ) R3 + R4 Op-amp has function as comparator and gain R v s = 2 (vi − v f ) R1 Laplace form equation (13) R Vs ( s) = 2 (Vi ( s) − V f ( s )) R1 (a) Block diagram of system is vf =

Markah 1

Markah 1

Markah 1

Markah 4

V s(s) V i(s )

+

R R

_

2 1

K 2C s (C L s + R C s + 1)(M s 2 + b s + k ) + K 1K 2C s 2

X (s) 2

V (s) f

R3 R3 + R

4

(b) Then the closed loop transfer function is R2 K 2 Cs 2 R1 (CLs + RCs + 1)( Ms 2 + bs + k ) + K 1 K 2 Cs 2 X (s) = R3 R K 2 Cs Vi ( s ) 1+ 2 2 2 2 R1 (CLs + RCs + 1)( Ms + bs + k ) + K 1 K 2 Cs R3 + R4

X (s) = Vi ( s )

R2 K 2 Cs R1 R3 R (CLs + RCs + 1)( Ms + bs + k ) + K 1 K 2 Cs + 2 K 2 Cs R1 R3 + R4 2

2

Markah 4

2

(c) Function x in time with vi as input   R  R3 R X ( s ) (CLs 2 + RCs + 1)( Ms 2 + bs + k ) + K 1 K 2 Cs 2 + 2 K 2 Cs  = Vi ( s ) 2 K 2 Cs  R1 R3 + R4  R1    4 3 2 (CLM ) s X ( s) + (bCL + CMR ) s X ( s ) + (ckL + bCR + M + K 1 K 2 C ) s X ( s ) + (CkR + b +

R  R2 R3 K 2 C ) sX ( S ) + kX ( s ) = Vi ( s ) 2 K 2 Cs  R1 ( R3 + R4 )  R1 

Markah 2

d 4x d 3x d 2x − ( bCL + CMR ) − ( ckL + bCR + M + K K C ) + 1 2 dt 4 dt 3 dt 2  R R K C dx  R (CkR + b + 2 3 2 ) +  2 K 2 C vi (t ) Markah 1 R1 ( R3 + R4 ) dt  R1  kx(t ) = −(CLM )

(CLM ) d 4 x (bCL + CMR ) d 3 x (ckL + bCR + M + K 1 K 2 C ) d 2 x − − − k k k dt 4 dt 3 dt 2  R R K C dx  R (CkR + b + 2 3 2 ) +  2 K 2 C vi (t ) Markah 2 R1 ( R3 + R4 ) dt  R1 

x(t ) = −