School of Electrical, Electronic, and Computer Engineering Signals and Systems 328

Tutorial-2

1. Calculate the transfer function Y/R of the system shown in the signal flow graph using the Mason’ gain formula.

2. (Dorf and Bishop P2.8) A bridged-T network is often used in ac control systems as a

filter network. The circuit of one bridged-T network is shown below. Show that the transfer function of the network is

2 21 1 2

2 21 2 1 2

( ) 1 2

( ) 1 (2 )o

in

V s R Cs R R C s

V s R R Cs R R C s

+ +=+ + +

.

3. (Dorf and Bishop P2.17) A mechanical system is shown below, which is subjected to

a known displacement 3( )x t with respect to the reference. (a) Determine the two independent equations of motion. (b) Obtain the equations of motions in terms of the Laplace transform, assuming that the initial conditions are zero. (c) Sketch a signal-flow graph representing the system of equations. (d) Obtain the relationship between

1( )X s and 3( )X s , 13( )T s , by using Mason’s signal-flow gain formula.

4. (Dorf and Bishop CDP2.1) We desire to accurately position a table for a machine as

shown below. A traction-drive motor with a capstan roller possesses several desirable characteristics compared to the more popular ball screw. The traction drive exhibits low friction and no backlash. However, they are susceptible to disturbances. Develop a model of the traction drive shown in the figure for the parameters given in the table. The drive uses a dc armature-controlled motor with a capstan roller attached to the shaft. The drive bar moves the linear slide-table. The slide uses an air bearing, so its friction is negligible. We are determining the open-loop model and its transfer function in this problem. Feedback will be introduced later.

5. (Ogata, 4th Ed. pp. 116) Simplify the block diagram shown in the figure and obtain

the transfer function (C(s)/R(s)) of the closed-loop system.

6. For MIMO system shown in the block diagram calculate the output C.

Solutions 1. Loops are: ABA, BYB, ABYA, and AYA and corresponding loop gains are:

3 1 4 2 3 4 3 2 1 2 2 3, , , ,and G H G H G G H G H H G H . There are no non-touching loops.

Forward paths are: RABY, and RAY and their corresponding gains are:

1 3 4 1 2, and G G G G G and their corresponding cofactors are: 1 2 1.D D= =

( )1 1 2 2

1 3 4 1 2

3 1 4 2 3 4 3 2 1 2 2 3

1

1

YP D P D

R DG G G G G

G H G H G G H G H H G H

= +

+=− − − − −

.

2. Define 1i , 2i and 1v as shown below.

Circuit equations are

in 1 11

v v isC

= + in 2 2ov v R i= + 1 21

ov v isC

= + 1 1 1 2( )v R i i= +

Dependent variables are 1i , 2i , 1v and ov . Rewriting the circuit equations as follows

1 in 11

v v isC

= − 2 in2 2

1 1oi v v

R R= − 1 2

1ov v i

sC= + 1 1 2

1

1i v i

R= −

yield the following signal-flow graph.

1v

1i

1−

1 sC

1 sC−

1

1

11 R

21 R−

21 R

1i 2ioV

1VinV

Signal-flow graph has 3 loops (1 1 1v i v , 2 2oi v i and 1 2 1 1ov v i i v ) and 3 open paths from

inv to ov ( in 2 ov i v , in 1 ov v v and in 2 1 1 ov i i v v ). The determinant and cofactors are

2 21 2 1 2

1 2 11

sCR sCR s C R R∆ = + + + , 1

1

11

sCR∆ = + , 2 3 1∆ = ∆ =

The stated transfer function follows from application of Mason’s Rule. Note above solution is not unique. One could rewrite the circuit equations in a different way to get a different signal-flow graph. The resulting transfer function must, however, be the same. (Obviously!)

3. (a) 1 1 1 1 2 1 1 3( ) ( ) 0M x K x x b x x+ − + − =ɺɺ ɺ ɺ

2 2 2 2 3 2 2 3 1 2 1( ) ( ) ( ) 0M x K x x b x x K x x+ − + − + − =ɺɺ ɺ ɺ

(b) 21 1 1 1 1 2 1 3( ) ( ) ( ) ( ) 0M s b s K X s K X s b sX s+ + − − =

22 2 1 2 2 2 2 3 1 1( ) ( ) ( ) ( ) ( ) 0M s b s K K X s b s K X s K X s+ + + − + − =

(c) Rewrite equations as follows

1 1

1 2 32 21 1 1 1 1 1

2 3

( ) ( ) ( )

( ) ( )

K b sX s X s X s

M s b s K M s b s K

AX s BX s

= ++ + + +

= +

1 2 2

2 1 32 22 2 1 2 2 2 1 2

1 3

( ) ( ) ( )

( ) ( )

K b s KX s X s X s

M s b s K K M s b s K K

CX s DX s

+= ++ + + + + +

= +

yields the following signal-flow graph.

B

D

C

AX2X3

X1

(d) It can be easily shown using Mason’s Rule that

13( )1

B ADT s

AC

+=−

Substituting in the expressions for A, B, C and D yields finally 3 2

2 1 1 2 1 1 1 2 2 1 1 213 4 3 2

1 2 1 2 2 1 1 1 1 2 2 1 1 2

1 1 1 2 2 1 1 2

( )( )

( ) ( )

( )

M b s b b s b K b K b K s K KT s

M M s M b M b s M K M K M K b b s

b K b K b K s K K

+ + + + +=+ + + + + +

+ + + +

4. Motor is an armature-controlled dc motor. Therefore, following Fig. 2.18 of Dorf

and Bishop (pp. 55), the block diagram is as shown below.

Kb

r X(s)θωVa(s)+_

back emf

1s

1

TJ s b+m

a a

KL s R+

In the block diagram, aR and aL are the armature resistance and inductance

respectively, mK and bK are motor constants that relate armature current to torque (torque constant) and rotor speed to back emf (back emf constant), respectively. The block labelled “1 s ” converts the rotor angular speed to actual angles while the block labelled “r” converts the rotor angle to displacement of the slide. r is the roller radius.

TJ is the total inertia on the rotor. It includes the rotor’s own inertia and that due to the mass of the slide and drive bar. Thus

2( )T m s bJ J M M r= + +

b is the total damping coefficient on the motor and equals only mb , the motor

damping coefficient, since it is assumed the slide uses an air bearing. The transfer function is given by (see eqn. 2.69, Dorf and Bishop)

( )

( ) ( )( )m

a a a T m b

K rX s

V s s L s R J s b K K s=

+ + +.

5.

6.