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Discrete Systems 39
every input sequence term on the right-hand side of the linear transformation.
To determine time-invariance, we will then compare it to the shifted
output sequence y(n − k), obtained after replacing every n by (n − k) on
the right-hand side of the linear transformation.
1. y(n) = L[x(n)] = 10 sin(0.1πn)x(n): The response due to shifted
input is
y k (n) =L[x(n − k)] = 10 sin(0.1πn)x(n − k)
while the shifted output is
y(n − k) =10 sin[0.1π(n − k)]x(n − k) ̸=y k (n).
Hence the given system is not time-invariant.
2. y(n) =L[x(n)] = x(n +1)− x(1 − n): The response due to shifted
input is
y k (n) =L[x(n − k)] = x(n − k) − x(1 − n − k)
while the shifted output is
y(n − k) =x(n − k) − x(1 − [n − k]) = x(n − k) − x(1 − n + k) ̸=y k (n).
Hence the given system is not time-invariant.
3. y(n) =L[x(n)] = 1 4 x(n)+ 1 2 x(n − 1) + 1 4x(n − 2): The response due
to shifted input is
y k (n) =L[x(n − k)] = 1 4 x(n − k)+ 1 2 x(n − 1 − k)+ 1 4x(n − 2 − k)
while the shifted output is
y(n − k) = 1 4 x(n − k)+ 1 2 x(n − k − 1) + 1 4 x(n − k − 2) = y k(n)
Hence the given system is time-invariant.
□
We will denote an LTI system by the operator LT I [·]. Let x(n) and
y(n) bethe input-output pair of an LTI system. Then the time-varying
function h(n, k) becomes a time-invariant function h(n − k), and the output
from (2.11) is given by
∞∑
y(n) =LT I [x(n)] = x(k)h(n − k) (2.13)
k=−∞
The impulse response of an LTI system is given by h(n). The mathematical
operation in (2.13) is called a linear convolution sum and is denoted by
y(n) △ = x(n) ∗ h(n) (2.14)
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