Signals and Systems - Electrical Engineering

132 C H A P T E R 2: Continuous-Time Systems

complete response is their sum,

y(t)=yzi(t)+yzs(t)

Indeed,yzi(t)andyzs(t)satisfy

(a 0 +a 1 D+···+DN)[yzi(t)]= 0

Dk[yzi(t)]t= 0 , k=0,...,N− 1

(a 0 +a 1 D+···+DN)[yzs(t)]=(b 0 +b 1 D+···+bMDM)[x(t)]

Adding these equations gives

(a 0 +a 1 D+···+DN)[yzi(t)+yzs(t)]=(b 0 +b 1 D+···+bMDM)[x(t)]

Dk[y(t)]t= 0 , k=0,...,N− 1

indicating thatyzi(t)+yzs(t)is the complete solution.

To find the solution of the zero-input and the zero-state equations we need to factor out the derivative

operatora 0 +a 1 D+···+DN. We can do so by replacingDby a complex variables, as the roots will

be either real or in complex-conjugate pairs, simple or multiple. Thecharacteristic polynomial

a 0 +a 1 s+···+sN=

∏

k

(s−pk)

is then obtained. The roots of this polynomial are called thenatural frequenciesoreigenvaluesand

characterize the dynamics of the system as it is being represented by the differential equation. The

solution of the zero-state can be obtained from a modified characteristic polynomial.

The solution of differential equations will be efficiently done using the Laplace transform in the next

chapter.

nExample 2.6

Consider a circuit that is a series connection of a resistorR= 1 and an inductorL=1 H, with

a voltage sourcev(t)=Bu(t), andI 0 amps is the initial current in the inductor. Find and solve

the differential equation forB=1 andB=2 for initial conditionsI 0 =1 andI 0 =0, respectively.

Determine the zero-input and the zero-output responses. Under what conditions is the system

linear and time invariant?

Solution

The first-order differential equation representing this circuit is given by

v(t)=i(t)+

di(t)

dt

i( 0 )=I 0