88324 control system mid term2007

8/9/2019 88324 Control System Mid Term2007

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88-324Mid-Term Test Summer 2007

Name: Student Number:

1.(30) Given a differential equation

y + 3y + 2y = r(t), y(0) = a, y(0) = b,

where a and b are constants,

a)(10). Assuming that the input signal is a unit step signal: r(t) = 1(t), find the expression of the

output signal y(t).

Note that

s2Y(s) as b + 3sY(s) 3a + 2Y(s) = R(s), Y(s) =as + b + 3a

s2 + 3s + 2

1

s2 + 3s + 2R(s)

If r(t) = 1(t), then R(s) = 1/s, so

Y(s) =as + b + 3a

s2 + 3s + 2+

1

s(s2 + 3s + 2)=

2a + b

s + 1+

a + b

s + 2+

1

2s

1

s + 1+

1

2(s + 2)

So

y(t) = (2a + b 1)et + (a + b +1

2)e2t +

1

2.

b)(10). Find the final value of limt

y(t) when r(t) = (t), i.e., a unit impulse signal.

If r(t) = (t), then we have R(s) = 1. So

Y(s) =

as + b + 3a

s2 + 3s + 2 +

1

s2 + 3s + 2 , limt y(t) = lims0 sY(s) = 0

c)(10). Assuming that a = b = 0, show that the unit-impulse response in (b) is the derivative of the

unit-step response in (a).

Note that in this case, for r(t) = (t), we have

Y(s) =1

s2 + 3s + 2=

1

s + 1

1

s + 2y(t) = et e2t

and for r(t) = 1(t), we can obtain from (a) that

y(t) = et +1

2e2t +

1

2

Obviously,

d

dt(et +

1

2e2t +

1

2) = et e2t.


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2.(30) Given an electric network shown in the following Figure:

The equations describing this network are stated as follows:

R1i1 + Ld(i1 i2)

dt= v,

R2i2 + vc + Ld(i2 i1)

dt= 0

i2 = Cdvcdt

,

where v(t) is a voltage source.

a)(15). Derive transfer functions from v(s) to i1(s).

Applying Laplace transformation to the equations above, we obtain

R1i1(s) + sL(i1(s) i2(s)) = v(s), R2i2(s) + vc(s) + sL(i2(s) i1(s)) = 0, i2(s) = sCvc

So i1(s)v(s) can be solved as

i1(s)

v(s)=

LCs2 + R2Cs + 1

(R1 + R2)LCs2 + (R1R2C+ L)s + R1

b)(15). Let v(s) = 1/s. Find the steady state current limt i1(t) and limt i2(t).

Now if v(s) = 1/s, i.e., v(t) = 1(t), then

i1(s) =LCs2 + R2Cs + 1

s [(R1 + R2)LCs2 + (R1R2C+ L)s + R1]

Solimt

i1(t) = lims0

si1(s) =1

R1

Note that, if i1() =1R1

, then vR1() = 1 = v(t). Therefore vL() = 0 and hence

limt i2(t) = 0.


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3.(40) A armature-controlled dc servomotor system is shown in the following figure

where Ra = 1, La 0, ke = kt = 0.5, n =N1N2

= 20, JL = 100kg m2, and DL = 100Nm

s/rad. This system can be described by following equations:

ea = Raia + kem, Tm = ktia, L = nm,1

nTm = JLL + DLL

where Tm is the torque generated by the motor.

a)(20). Find transfer function model for this electro-mechanical system from ea to L and L.

Note that, after implementing the values given, the Laplace transformed equations are

ea(s) = ia(s) + 0.5sm(s), Tm = 0.5ia(s), L(s) = 20m(s),1

20Tm = 100s

2L(s) + 100sL

Therefore, L(s)ea(s)

and L(s)ea(s)

can be solved as

L(s)

ea(s)

=1

s(4000s + 4000.025)

,L(s)

ea(s)

=1

4000s + 4000.025b)(20). Let ea(t) = 1(t)v be a unit-step testing signal. Find the motor shaft angle L(t).

If ea(t) = 1(t)v, then ea(s) = 1/s. Note that 4000.025/4000 1. So

L(s) =1

s2(4000s + 4000.025)=

1

4000s2

1

4000s+

1

4000(s + 1)

and

L(t) =1

4000t

1

4000+

1

4000et.

88324 control system mid term2007

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