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0066_Frame_C24 Page 19 Thursday, January 10, 2002 3:44 PM

in the natural modes. This last observation is consistent with the parameter choice in Example 1.6, which
made the mass-spring system to be overdamped.

Structure of the Forced Response
Consider the Eq. (24.102). Then, when the initial state is zero, the state will only exhibit the forced
component. However, the forced component will still include natural modes plus some additional forced
or particular modes, which depend on the nature of the system input u[t]. In general, the forcing modes
in the input will also appear in the state. However, special cases arise when a forcing mode in u[t] coincides
with a system natural mode.
System Stability
Stability in linear time-invariant systems can also be analyzed using the state matrix A d . As we said, all
systems variables can be expressed as linear functions of the state and the system input. When the system
input u[t] is a vector of bounded time functions, then the boundedness of the system variables depends
on the state to be bounded. We then have the following result:

Theorem 24.2 Consider a system with the state description (24.100) and (24.101) where B d , C d , and D d
have bounded elements. Then the system state is bounded for all bounded inputs if and only if the eigenvalues
of A d lies inside the unit disc, i.e., |h | < 1, ∀ .

Speed of Response and Resonances
We recall that the natural modes of discrete-time systems are the powers of the eigenvalues h . Since
those eigenvalues can always be described as complex quantities, we can then write the natural modes as

t
t jq t
jq
( h ) = ( |h |e ) = |h | e , where q = ∠ h (24.114)
t

Therefore, we have that
• 0 < |h | < ∞ determines the speed at which the mode decays to zero for stable systems (|h | < 1),
or grows to infinity for unstable systems (|h | > 1)
• −p < q ≤ p determines the frequency of the natural mode, measured in radians.
Although the natural modes of stable systems decay to zero, their nature determines the system transient
response.
To illustrate these issues the step response, with zero initial conditions, is frequently used.

Example 24.7
Consider the first order, single-input single-output discrete-time system
[
x t + 1] = h x t[] + u t[] (24.115)
y t[] = ( 1 h )x t[] (24.116)
–
∀
To obtain the step response, we can use the Eq. (24.103), where x o = 0, u[t] = 1, t ≥ . 0

t−1
y t[] = C d ∑ A d t−i−1   B d (24.117)

 i=0 
 t−1  – t –
= ( 1 h ) ∑ h t−i−1  = ( 1 h )h t−1 1 h (24.118)
–
----------------
–
 i=0  1 h – 1
–
= 1 h t (24.119)
–
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