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Unformatted text preview: Moreover, C onsequently, t he l oop e quation (1.48) c an b e e xpressed as (D (1.58) f (t) Multiplying b oth s ides o f t he a bove e quation b y D ( that is, differentiating t he a bove
e quation), we o btain
(1.59a)
(15D + 5)x(t) = D f(t) y (t) J<t) 91 S ystem M odel: I nputOutput D escription + l )y(t) (1.60) = f (t) ( 1.55)
or w hich i s i dentical t o E q. ( 1.52).
R ecall t hat E q. ( 1.55) is n ot a n a lgebraic e quation, a nd D2 + 3 D + 2 i s n ot an
a lgebraic t erm t hat m ultiplies y (t)j i t i s a n o perator t hat o perates o n y (t). I t m eans
t hat w e m ust p erform t he f ollowing o perations o n y (t): t ake t he s econd d erivative
o f y (t) a nd a dd t o i t 3 t imes t he f irst d erivative o f y (t) a nd 2 t imes y (t). C learly,
a p olynomial i n D m ultiplied b y y (t) r epresents a c ertain d ifferential o peration o n 3 dy
dt • E xample 1 .11
F ind t he e quation r elating t he i nput t o o utput for t he series R C c ircuit o f Fig. 1.32
if t he i nput is t he v oltage f (t) a nd o utput is ( a) t he loop c urrent x (t) ( b) t he c apacitor
v oltage y (t).
T he loop e quation for t he c ircuit is R x(t) + 11' c 00 x(T)dT = f(t) 5['00 X(T) dT = f(t) • E xercise E 1.lS For the R LC circuit in Fig. 1.31, find the inputoutput relationship if the output is t he
inductor voltage VL(t).
Answer: (D2 + 3 D + 2) VL(t) = D2 f(t) 'V
Hint: VL(t) = L Dy(t) = D y(t). For the R LC circuit in Fig. 1.31, find the inputoutput relationship if the o utput is the
capacitor voltage v c(t).
Hint: v c{t) = d riy(t) = t>y(t).
Answer: (D2 + 3 D + 2) v c{t) = 2 f(t) 'V ' T(t) (1.56) ) ~i1=Jr or 15x(t) + (1.61) E xercise E 1.l7 !:!. y (t). !:!. + y(t) = f (t) B 8(t) (1.57)
F ig. 1 .33 B (b) A rmature c ontrolled dc m otor. W ith o perational n otation, t his e quation c an b e e xpressed as
numerator as well as in the denominator. This happens, for instance, in circuits with allinductor
loops or allcapacitor cutsets. To eliminate this problem, avoid the integral operation in system
equations so t hat t he resulting equations are differential rather than integradifferential. In electrical circuits, t his can be done by using charge (instead of current) variables in loops containing
capacitors and using current variables for loops without capacitors. In the literature this problem
of commutativity of D and 1 / D is largely ignored. As mentioned earlier, such procedure gives erroneous results only in special systems, such as the circuits with allinductor loops or allcapacitor
cutsets. Fortunately such systems constitute a very small fraction of the systems we deal with.
For further discussion of this topic and a correct method of handling problems involving integrals,
see Ref. 4 • E xample 1 .12
I n r otational s ystems, t he e quations o f m otion a re s imilar t o t hose in t ranslational
s ystems. I n p lace o f force F , we have t orque T . I n p lace o f m ass M , we have m oment
o f i nertia J ( the r otational m ass), a nd in place o f l inear accel...
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 Spring '13
 Bayliss
 Signal Processing, The Land

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