Shift-invariance (time-invariance)
Definition
A discrete time system y(n) = T[x(n)] is shift-invariant if, for the all x(n) and all n0, we have: T[x(n-n0)] = y(n-n0).
That is applying a time delay to the input of a system is equivalent to applying it to the output.
Step 1. Determine the output y(n) corresponding to input x(n).
Step 2. Delay output y(n) by n0 units, resulting in y(n-n0).
Step 3. Determine output y(n, n0) corresponding to input x(n-n0).
Step 4. Determine if y(n, n0) = y(n-n0).
If equal, then the system is shift-invariant; else it is time-varying.
When we suppose that the system is time-varying a quite useful the alternative approach is to find a counter-example to disprove time-invariance, that is, use intuition to find an input signal for which the condition of shift-invariance is violated and which suffices to show that the system is not shift-invariant.
Example1 Determine whether y(n) = T[x(n)] = x(-n) is shift-invariant.
Answer Find the output for x(n), delay it by n0, and compare with output for x(n-n0). The output for x(n) can be given by
y(n) = T[x(n)] = x(-n) Delaying y(n) by n0 gives
y(n-n0) = x(-(n-n0)) = x(-n+n0) → (A)
As an aside this amounts to reflecting first and then shifting.
The output for x(n-n0) is denoted y(n, n0) and is given by y(n, n0) = T[x(n-n0)] = x(-n-n0) → (B)
As an aside this amounts to shifting first and then reflecting.
(A) and (B) are not equal. That is, y(n, n0) ¹ y(n-n0), so the system is time-varying.
Example2 Test the system y(n) = x(2n) for time-invariance.
Answer This system represents time scaling. That is, y(n) is a time-compressed version of x(n), compressed by a factor of 2. For example, the value of x that occurred at 2n is occurring at n in the case of y. Intuitively, then, any time shift in the input will also be compressed by a factor of 2, and it is for this reason that the system is not time-invariant.
This is demonstrated by counter-example. It can also be shown by following the formal procedure, which we shall do first below. For the input x(n) the output is y(n) = T[x(n)] = x(2n)
Delay this output by n0 to get
y(n-n0) = x(2(n-n0)) = x(2n-2n0) → (A)
Next, for the input x(n) = x(n-n0) the output is
y(n, n0) = x(2n) = x(2n-n0) → (B)
(A) and (B) are not equal. So the system is not time-variant.
By counter example To show that the system y(n) = x(2n) is not time-invariant by way of a counter example consider the x(n) below:
x(n) = 1, -2 ≤ n ≤ 2
0, otherwise
We shall show that y(n, 2) ≠ y(n-2). We express x(n) in terms of unit step functions as
x(n) = u(n+2) - u(n-3)
We determine y(n-2) by first obtaining y(n) and then delaying it by 2 units:
y(n) → y(n) = x(2n) = u(2n+2) - u(2n-3)
Delay → y(n-2) = x(2(n-2)) = u(2(n-2)+2) - u(2(n-2)-3) = u(2n-2) - u(2n-7)
Next we determine y(n, 2) by first obtaining x(n-2) and then the corresponding output y(n, 2):
x(n-2) → x(n-2) = u((n-2)+2) - u((n-2)-3) = u(n) - u(n-5) = x2(n), say
y(n, 2) → y(n, 2) = x2(2n) = u(2n) - u(2n-5)
We can easily sketch y(n, 2) and y(n-2) and see that y(n, 2) ≠ y(n-2), and therefore the system y(n) = x(2n) is not time-invariant.
Alternatively, this can be done entirely graphically.
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