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38 Chapter 2 DISCRETE-TIME SIGNALS AND SYSTEMS
which is not equal to
a 1 y 1 (n)+a 2 y 2 (n) =3a 2 1x 2 1(n)+3a 2 2x 2 2(n)
Hence the given system is nonlinear.
2. y(n) =2x(n − 2) + 5: Consider
T [ a 1 x 1 (n)+a 2 x 2 (n) ] =2[a 1 x 1 (n − 2) + a 2 x 2 (n − 2)]+5
= a 1 y 1 (n)+a 2 y 2 (n) − 5
Clearly, the given system is nonlinear even though the input-output
relation is a straight-line function.
3. y(n) =x(n +1)− x(1 − n): Consider
T [a 1 x 1 (n)+a 2 x 2 (n)] = a 1 x 1 (n +1)+a 2 x 2 (n +1)+a 1 x 1 (1 − n)
+ a 2 x 2 (1 − n)
= a 1 [x 1 (n +1)− x 1 (1 − n)]
+ a 2 [x 2 (n +1)− x 2 (1 − n)]
= a 1 y 1 (n)+a 2 y 2 (n)
Hence the given system is linear.
□
Linear time-invariant (LTI) system A linear system in which an
input-output pair, x(n) and y(n), is invariant to a shift k in time is called
a linear time-invariant system i.e.,
y(n) =L[x(n)] ⇒ L[x(n − k)] = y(n − k) (2.12)
For anLTI system the L[·] and the shifting operators are reversible as
shown here.
x(n) −→ L [·] −→ y(n) −→ Shift by k −→ y(n − k)
x(n) −→ Shift by k −→ x(n − k) −→ L [·] −→ y(n − k)
□ EXAMPLE 2.6 Determine whether the following linear systems are time-invariant.
1. y(n) =L[x(n)] = 10 sin(0.1πn)x(n)
2. y(n) =L[x(n)] = x(n +1)− x(1 − n)
3. y(n) =L[x(n)] = 1 x(n)+ 1 x(n − 1) + 1 x(n − 2)
4 2 4
Solution
First we will compute the response y k (n) △ = L[x(n − k)] to the shifted
input sequence. This is obtained by subtracting k from the arguments of
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