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Unformatted text preview: mode term. For exam~~e, whe~ t~~
i nput is e (t, t he forced response (righthand column) has the form {3e . B ut If e
h appens t o b e a characteristic mode of t he s ystem, t he correct form .af .the forced
.
response IS {3t e ( t (see P air 2). I f t e(t also happens t o be at charactenstlc mode of
2(
d
t he s ystem, t he c orrect form o f t he forced response is {3t e , an so on.
Although this method can be used only for inputs with a finite number of
derivatives, this class of inputs includes a wide variety of t he most. commonly encountered signals in practice. Table 2.2 shows a variety o f such m puts a nd t~e
form of t he forced response corresponding t o each input. We shall demonstrate thiS
p rocedure with a n example.
• E xample 2.9
Solve t he differential equation (D2 + 3D + 2) y(t) = D f(t)
if the i nput f(t) = t 2 + 5 t + 3
a nd t he i nitial conditions are y(O+) = 2 a nd y(O+) = 3.
T he c haracteristic polynomial o f t he s ystem is
A2 + 3A+ 2 = ( A+ l )(A + 2) Therefore, t he c haracteristic modes are e  t a nd e  2t. T he n atural r esponse is t hen a l inear
combination of these modes, so t hat
Yn(t) = K le t + K 2e 2t
t 2: 0
Here t he a rbitrary c onstants K I a nd K2 must be determined from t he s ystem's initial
conditions. 142 2 TimeDomain Analysis of ContinuousTime Systems
The forced response to the input t 2 + 5t + 3, according to Table 2.2 (Pair 5 with ( = 0), is t:,. E xercise E 2.15
A n L TIC s ystem is specified b y t he e quation Y<t>(t) = f ht 2 + {3lt + (3o (D2 Moreover, Y<t>(t) satisfies the system equation [Eq. (2.52)]; t hat is,
(D2 N ow DY<t>(t) =
D 2 <t>(t)
Y and + 3 D + 2) Y<P(t) = D f(t) ft (2.53) 2
({32t + (3lt + (3o) = 2{32 t + {31 = ft 2
[t + 5t + 3] = 2 t+ 5 Substituting these results in Eq. (2.53) yields
or fi Answers: ( a) y<t>(t)=t2+~t_* (b)y(t)=5e2t3e3t+(t2+~t_*). Q(D) [{3e<t] = P (D)e<t =0
=2 =5 Y<t>(t) = t +1 = 1, {3I = 1, and (32 = O. Consequently Q (D)e(t = Q«()e(t t> 0 /3Q«()e(t = P«()e(t = K I e t +K2e 2t +t+1 and t >O P«()
/3 = Q«() +1 Setting t = 0 a nd substituting yeO) = 2 and yeO) = 3 in these equations, we have
2 a nd Therefore, Eq. (2.52) becomes + Y<P(t) yet) =  KIe t  2K2e 2t = ( 2e(t dt The total system response y( t) is the sum of the natural and forced solutions. Therefore so t hat ~ ( e(t)
2 D 2 e(t = Solving these three equations for their unknowns, we obtain {3o
Therefore yet) = Yn(t) (2.54) d
De<t = _ ( e(t) = ( e(t
dt Equating coefficients of similar powers on both sides of this expression yields 2{3o + 3{3I + 2{32 'V Now observe t hat 2{32 t2 + (2{31 + 6(32)t + (2{3o + 3{31 + 2(32) = 2t + 5 2{32 (D + l )J(t) T he i nput is J(t) = 6t 2. F ind ( a) t he forced response Y<P(t) ( b) t he t otal r esponse yet) if t he
i nitial conditions a re y(O+) =
a nd y(O+) = ~. 2{32 + 3(2{32t + (31) + 2({32t2 + (3lt + (3o) = 2t + 5 2{31 + 6{32 + 5D + 6) yet) = T he Exponential Input e (t
T he e xponential signal is...
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This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.
 Spring '13
 Bayliss
 Signal Processing, The Land

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