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Unformatted text preview: y~k) (0) is t he value of t he k th d erivative o f Yn(t) a t t = O. We c an e xpress
t his c ondition f or various values of n ( the s ystem order) as follows:
n = 1:
= 4: =1
Yn(O) = 0 a nd Yn(O) = 1
Yn(O) = Yn(O) = 0 a nd iin(O) = 1
Yn(O) = Yn(O) = iin(O) = 0 a nd iin(O) =
Yn(O) 1 (2.21) a nd so on.
I f t he o rder o f P (D) is less t han t he o rder of Q (D), bn = 0, a nd t he i mpulse
t erm bn 8(t) i n h ( t ) is zero.
• I n t he above discussion, we have assumed m ::::: n , a s specified by Eq. (2.17b).
A ppendix 2.1 shows t hat t he e xpression for h (t) a pplicable t o all possible values of
m a nd n is given b y
h (t) = P (D)[Yn(t)u(t)] where Y n(t) is a linear combination of t he c haracteristic modes of t he s ystem s ubject
t o i nitial conditions (2.20). T his e xpression reduces t o Eq. (2.19) when m ::::: n .
D etermination o f t he i mpulse response h (t) u sing t he p rocedure in t his s ection
is relatively simple. However, in C hapter 6 we shall discuss a nother, even simpler
m ethod using t he L aplace transform.
f::,. Determine the unit impulse response of LTIC systems described by the equations:
( e) E xample 2 .3 Determine t he unit impulse response h( t} for a system specified by the equation
(D2 + 3D + 2) yet} This is a second-order system (n = 2) = D f(t} 2C2 e- 2t (2.23a) (2.23b) T he initial conditions are [see Eq. (2.21) for n = 2]
and Yn(O} = 0 Setting t = 0 in E qs. (2.23a) and (2.23b), and substituting the above initial conditions, we
Solution of these two simultaneous equations yields
- Cl - This is a second-order system with bn = b2 = O. First we find the zero-input component
for initial conditions y(O-} = 0, and y(O-} = 1
Y zi=dsoive('D2y+3*Dy+2*y=0', 'y(O} = 0', ' Dy(O} = 1', ' t')
Yzi • - exp(-2*t)+exp(-t)
Since P (D) = D , we differentiate the zero-input response:
PYzi • 2 *exp(-2*t)-exp(-t) T herefore h(t} 0=C!+C2
1= C omputer E xample C 2.3 Determine the impulse response h(t} for an LTIC system specified by the differential
(D2 + 3D + 2)y(t} = D f(t} Differentiation of t his equation yields = - cle- t - Answers: (a) 38(t) - e- 2t u(t) ( b) (2 - e- 2t )u(t) (e) (1 - t)e-tu(t) \ l 8 (A2 + 3A + 2) = (A + 1}(A + 2) Yn(t} (D + 2)y(t) = (3D + 5 )f(t)
D(D + 2)y(t) = (D + 4 )f(t)
(D2 + 2D + l)y(t) = D f(t) (2.22) having the characteristic polynomial T he characteristic roots of this system are A = - 1 and A = - 2. Therefore
Yn(t} = c le- t + c2e- 2t E xercise E 2.4 (2.24) = b28(t) + [Dyo(t)]u(t} = (2e- 2t - e-t)u(t} 0 118 2 Time-Domain Analysis of Continuous-Time Systems S ystem Response t o Delayed Impulse
I f h(t) is t he r esponse of a n LTIC system t o t he i nput 8(t), t hen h(t - T) is t he
response of this s ame s ystem t o t he i nput 8(t - T). T his conclusion follows from
t he t ime-invariance p roperty of LTIC systems. Thus, by knowing t he u nit impulse
response h(t), we c an d etermine t he system response t o a delayed impulse 8(t - T). 2.4 System Response t o E xternal Input: T he Z ero-State Response 119 t [ ( I) [ (nil,;) 2 .4 System Response t o External Input: Z ero-state Response
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