NO LATE HOMEWORKDue: Wednesday, 6/16/04

ME 3015 -Homework #5Assigned: Wednesday, 6/9/04

O'~--~

Problem 1 (25 points total)L

-~OO~ -I

;Rl

~ , ~__~=:~A L.. S I-;)- I. 3

#~2 R, ~Pr

.£ ~o\

:>--

~;R2JEt

\a

Iei

I

c\ R2~3eo

I;R3

11)1-

6ef+ IC»f'J I<~= l<2.-'r~3

(rl~Y\t-I"f)

(N()~e/tJ

-:.0

.3

_1-T :::0~S.1l.~

._-=1_-.>< '2X). -= (, (t ,'tf.t.J --LX1t-1-~

0

Jr=C>=b

-\-I-l-c..

"Zl~ti~~

)

a

oj

~) ZsIJ. -RI13 -ReI3

3) "I,=JJ.-'rJ[~

4) rEo:: ReI3-e\'Mjt-./"'i~ I, "12 I

j )

d~..r;jfole.~ ~ X, =)(2XI"-£:'.)("). ':, -e eo R~

5) -7 :2)

-L [E' -LsJ3J { \1r~ s .L~--';-~~~ J -~ \ -t'R e )JL 3

B-6- 7. Fran the circui t~ shown tothe right, we obtain

ei -eAil =Rl0

eA.--]2 -R2

0

is-R. '.(1)

/.' ..A\ ~If--o.~!"--~~~~-'~'-"-~~I~~=~,,- ,',(1) I.' t J R3

~eA -eoi3 = R3

eo.--15 -R4

Sincei]. = i2 + is

we have

ei -eA

Rl

eA eo=-+-R2 R4

Also I sincei4 + i3 = is

we obtain

eo-R4

eA -eo

R3

Equation (1) can be written as

eoR4

e.1 ---Rl

-L+ 1 )Rl ~ eA +

,.

Laplcice transforming this; equation and simplifying, we get

Ei(S)Rl

Eo(s)R4

= (3)

Laplalce transforming Equation (2), we obtain

iran 'which we obtain

(4)

Ei(S)

Rl

Rl + R2

R1R2

R3R4Cs + RJ + R4R4(RJCS + 1) Eo(s)

]~o(s)---,,- =

R4

fran ,mich the transfer f\mction Eo(s)/Ei(s) can be obtained as

EOI(S)--EiltS)

= RZR4{R3Cs + 1

[RIRZR3 + {Rt + R2)R3~]CS + RIR2 + (Rl + R2){R3 + R4)

-.

hJ S XoR,

~7;~

'_0

~ ~ + ~ J(il -i2) dt + R(il -i2) + * 5 ildt = 0

di J2 1. ..1L2 -+ -c 12 dt + R(12 -11 + -cdt 3 2

which can be rewri tten as

..1Llql + -(ql

C2

.n

Using the force-voltage analogy, we can convert thelast two equations as follows:

Fran these equations an analogous lIEChanical systemcan be obtained as shown to the right.

B-6-26.";n}-en---

Define the vol ta~Je at point A as eA'

~~- 1Ei(S) -RICs + 1

Define the voltage at point B as ~.Then

R3EB(s) = u'-'R2 + R3

Noting that this circuit involves~ti ve feedback, the differentialinput voltage ~s zero. Thus

E",( s)

-

EA(S) = EB(s)Hence

1 -EA(s) = RICS + 1 Ei{S) = EB(S) = Eo(s)r

RJ

fran wbi,ch we obtain

Eo(s)--Ei(s)

~

~ equation.s for the liquid-level system are

~

to = hl/ql and R2 = h2/Q2' the system equations can be rewri tten asSince Rl

dh hI

1 q - q --C -=q- 1 -R1 dt 1

=.!:1-_~Rl R2

dh2C2 ~ :: ql -q2

From Equations (1) and (2) we obtain

(3)dhl dh2 h2

C -+c -=q--1 dt 2 dt R2

By differentiating Equation (2) with respect to t, we get

d~2 1 dh2 -1 dhl-+ dt2 R2 dt Rl dt (4)

C2

.an

dZqzRl Cl R2C2 -dt:I

dq2+ R2C2) at + q = q2+ (RICl

1Q2(S)~-

Q(s)=