Unformatted text preview: fficients. Because d ky/dt k <=> s ky(s), t he Laplace transform of a differential
e quation is a n a lgebraic equation which can be readily solved for Y (s). N ext we
t ake t he inverse Laplace transform o f Y (s) t o find t he desired solution y (t). T he
following e xamples d emonstrate t he L aplace transform procedure for solving linear
differential e quations w ith c onstant coefficients.
• (6.42a) if the initial conditions are y(O-) = 2, y(O-) = 1, and the input f (t) = e - 4t u(t).
The equation is
d t; + 5 d~ + 6y(t) = dt + f (t)
y (t) < =} Y (s) Then from Eq. (6.34)
dt < =} 8Y(8) _ y (O-) = 8Y(8) - 2 = 13/2 _ __ _ 3 /2
s +4 The inverse Laplace transform of the above equation yields
(6.44) ca~ E xample 6 .9
Solve t he second-order linear differential equation (D2 + 5D + 6) y(t) = (D + l )f(t) Y (8) ' (6.43b) Therefore 1 (8 - = s +4' Taking the Laplace transform of Eq. (6.42b), we obtain T his e xact r esult a ppeared earlier in Eq. (6.11). C (8)= 2
8 Y(8) - 8Y(0-) - y(O-) = 82Y(8) - 28 - 1 = e - 4t u(t),
1 (6.40) < =} 391 This example demonstrates t he ease with which t he Laplace transform
solve linear differ.ential ~quatio~s w ith c onstant coefficients. T he m ethod is g eneral
and can solve a lmear differentIal equation w ith c onstant coefficients of any order.
Zero-Input and Z ero-State Components o f Response .
T he L aplace transform m ethod gives t he t otal response, which includes zeromput an~ zero-sta.te. ~ompon~n.ts. I t is possible t o s eparate t he two components if
we so deSire. ~he mltlal. c ondition t erms in t he r esponse give rise t o t he z ero-input
res~onse. For mstance, m Example 6.9, t he t erms a rising due t o i nitial conditions
~(.O. ) = 2 ~~d y(O-) = 1 in Eq. (6.43a) generate t he z ero-input response. These
1~ltlal condl~lOn t erms a re - (28 + 11), as seen in Eq. (6.43b). T he t erms on t he
nght-ha~d Side a re exclusively due t o t he i nput. E quation (6.43b) is r eproduced
below With t he p roper labeling of t he t erms. 392 6 Continuous-Time System Analysis Using t he Laplace Transform 6.3 Solution of Differential a nd Integro-Differential Equations. always determine t he n atural a nd t he forced components from t he z ero-input and
t he z ero-state components (see Eq. (2.51b), b ut t he converse is n ot true. Because
of these and some other problems, electrical engineers (wisely) s tarted d iscarding.
t he L:+ version in t he e arly sixties.
I t is i nteresting t o n ote t he t ime domain duals of these two Laplace versions.
he classical m ethod is t he d ual of t he L:+ m ethod, a nd t he convolution (zeromput/zero-state) m ethod is t he d ual of t he L:_ m ethod. T he first pair uses t he
i nitial conditions a t 0+, a nd t he second pair uses those a t t = 0 -. T he first pair
(the classical method a nd t he L:+ version) is useless in t he t heoretical s tudy of linear
system analysis. I t was no coincidence t hat t he L:_ version was adopted immediately
after t he i ntroduction o f t he state-space analysis (...
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