LTI Continuous-Time Systems 131

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 that y zi ( 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 an Nth-order linear differential equation with constant coefficients, and with x(t) as the input and y(t) as the output:

dy(t)

d N y(t)

dx(t)

d M x(t)

t≥0 (2.12) The corresponding N initial conditions are y(0), d k y(t)/dt k | t=0 for k = 1, . . . , N − 1. Defining the

a 0 y(t) + a 1 +···+

= b 0 x(t) + b 1 +···+ b M

derivative operator as

d n y(t)

D [y(t)] =

n > 0, integer

dt n

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

we write the differential Equation (2.12) as

( a 0 + a 1 D+···+D N ) [y(t)] = (b 0 + b 1 D+···+b M D M ) [x(t)]

t≥0

D k [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 for t < 0—that is, the system is not energized for t < 0. However, many LTI systems represented by differential equations have nonzero initial conditions. Considering that the input signal x(t) is independent of the initial conditions, we can think of these as two different inputs. As such, using superposition we have that the complete solution of the differential equation is composed of a zero-input solution, due to the initial conditions when the input x(t) is zero, and the zero-state response due to the input x(t) with zero initial conditions.

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

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

( a 0 + a 1 D+···+D N ) [y(t)] = 0

(2.14) with zero initial conditions. If y zi ( t) is the response of the zero-input differential Equation (2.13),

( a 0 N + a 1 D+···+D ) [y(t)] = (b 0 + b 1 D+···+b M D M ) [x(t)]

and y zs ( t) the zero-state (or zero initial conditions) differential Equation (2.14), we have that the

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

complete response is their sum,

y(t) = y zi ( t) + y zs ( t)

Indeed, y zi ( t) and y zs ( t) satisfy

0 + a 1 D+···+D ) [y zi ( t)] = 0

D k [y zi ( t)] t=0 ,

k = 0, . . . , N − 1

0 + a 1 D+···+D ) [y zs ( t)] = (b 0 + b 1 D+···+b M D ) [x(t)] Adding these equations gives

0 a 1 D+···+D [y zi t) + y zs ( t)] = (b 0 + b 1 D+···+b M D ) [x(t)]

D k [y(t)] t=0 ,

k = 0, . . . , N − 1

indicating that y zi ( t) + y zs ( 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

operator a 0 + a 1 D+···+D N . We can do so by replacing D by a complex variable s, as the roots will

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

0 + a 1 s+···+s N =

( s−p k )

is then obtained. The roots of this polynomial are called the natural frequencies or eigenvalues and 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.

■ Example 2.6

Consider a circuit that is a series connection of a resistor R = 1  and an inductor L = 1 H, with

a voltage source v(t) = Bu(t), and I 0 amps is the initial current in the inductor. Find and solve the differential equation for B = 1 and B = 2 for initial conditions I 0 = 1 and I 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

di(t) v(t) = i(t) + dt

i(0) = I 0