3. Example 1
3
Determine System Properties
Y(n) = x(n2)
Solution:
1- Linear or non-Linear
Y1(n) = x1(n2)
Y2(n) = x2(n2)
Y1(n) + Y2(n) = x1(n2) + x2(n2) 1
x1(n2) + x2(n2) = y’(n) 2
From 1 and 2
y’(n) = Y1(n) + Y2(n) so the system is linear
2- Time-invariant or Time-variant
y’(n) = x (n2 -T) 1
Y(n –T) = x ((n-T)2) 2
From 1 and 2
y’(n) ≠ Y(n –T) so the system is time- variant
3-Static or dynamic and causal and non-causal
Y(1) = x(1)
Y(2) = x(4) Output depend on the future input only so the system is dynamic and non-
causal
Y(3) =x(9)
4- System is stable since it follow BIBO
If we substitute for x(n2) by any value we
will obtain a bounded value for Y.
4. Example 2
4
Determine System Properties
Y(n) = x(n) + n x(n+1)
Solution:
1- Linear or non-Linear
Y1(n) = x1(n) + n x1 (n+1)
Y2(n) = x2(n) + n x2 (n+1)
Y1(n) + Y2(n) = x1(n) + n x1 (n+1) + x2(n) + n x2 (n+1)
= x1(n) + x2(n) + n (x1 (n+1) + x2 (n+1)) 1
y’(n)= x1(n) + x2(n) + n (x1(n+1) + x2(n+1)) 2
From 1 and 2
y’(n) = Y1(n) + Y2(n) so the system is linear
2- Time-invariant or Time-variant
y’(n) = x (n -T) + n x(n-T+1) 1
Y(n –T) = x (n-T) + (n-T) x(n-T+1) 2
From 1 and 2
y’(n) ≠ Y(n –T) so the system is time- variant
3-Static or dynamic and causal and non-causal
Output depend on the present and on the future input so the system is dynamic and non-
causal
4- System is unstable since In the given
system equation, n∞, 𝐲∞ even if x(n)
is bounded, Hence the system is unstable
5. Example 3
5
Determine System Properties
Y(n) = ex(n)
Solution:
1- Linear or non-Linear
Y1(n) = ex1(n)
Y2(n) = ex2(n)
Y1(n) + Y2(n) = ex1(n) + ex2(n) 1
y’(n)= ex1(n)+ x2(n) = ex1(n) . ex2(n) 2
From 1 and 2
y’(n) ≠ Y1(n) + Y2(n) so the system is non-linear
2- Time-invariant or Time-variant
y’(n) = ex(n-T) 1
Y(n –T) = ex(n-T) 2
From 1 and 2
y’(n) = Y(n –T) so the system is time-invariant
3-Static or dynamic and causal and non-causal
Output depend on the present input only so the system is static and causal
4- System is stable since as long as x(n) is
bounded y(n) is also bounded
