Signals and Systems - Electrical Engineering

2.3 LTI Continuous-Time Systems 131

On the other hand, if the initial conditions are different from zero, when checking linearity and time invariance

we only change the input and do not change the initial conditions so thatyzi(t)remains the same, and thus

the system is nonlinear. The Laplace transform will provide the solution of these systems.

Most continuous-time dynamic systems with lumped parameters are represented by linear ordinary
differential equations with constant coefficients. By linear it is meant that there are no nonlinear
terms such as products of the input and the output, quadratic terms of the input and the output, etc.
If the coefficients change with time the system is time varying. The order of the differential equation
equals the number of independent elements capable of storing energy.

Consider a dynamic system represented by anNth-order linear differential equation with constant
coefficients, and withx(t)as the input andy(t)as the output:

a 0 y(t)+a 1

dy(t)

dt

+···+

dNy(t)

dtN

=b 0 x(t)+b 1

dx(t)

dt

+···+bM

dMx(t)

dtM

t≥ 0 (2.12)

The correspondingNinitial conditions arey( 0 ),dky(t)/dtk|t= 0 fork=1,...,N−1. Defining the
derivative operator as

Dn[y(t)]=

dny(t)

dtn

n>0, integer

D^0 [y(t)]=y(t)

we write the differential Equation (2.12) as

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

Dk[y(t)]t= 0 , k=0,...,N− 1

As indicated before, the system represented by this differential equation is LTI if the initial conditions
as well as the input are zero fort<0—that is, the system is not energized fort<0. However, many
LTI systems represented by differential equations have nonzero initial conditions. Considering that
the input signalx(t)is independent of the initial conditions, we can think of these as two different
inputs. As such, using superposition we have that thecomplete solutionof the differential equation is
composed of azero-input solution, due to the initial conditions when the inputx(t)is zero, and the
zero-state responsedue to the inputx(t)with zero initial conditions.

Thus, to find the complete solution we need to solve the following two related differential equations:

(a 0 +a 1 D+···+DN)[y(t)]= 0 (2.13)

with initial conditionsDk[y(t)]t= 0 ,k=0,...,N−1, and the differential equation

(a 0 +a 1 D+···+DN)[y(t)]=(b 0 +b 1 D+···+bMDM)[x(t)] (2.14)

with zero initial conditions. Ifyzi(t)is the response of the zero-input differential Equation (2.13),
andyzs(t)the zero-state (or zero initial conditions) differential Equation (2.14), we have that the