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A discrete system is represented by the difference equation

X1 (k + 1) 

a a - 1 

X1(k) 
X2 (k + 1 ) a + 1 a X2(k)
It has initial conditions X1 (0) = 1; X2 (0) = 0. The pole locations of the system for a = 1, are
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- 1 ± j0
- – 1 ± j0
- ± 1 + j0
- 0 ± j1
Correct Option: A
From the given difference equation,
| A = | ![]() | a | a - 1 | ![]() |
| a + 1 | a |
The pole locations of the system for a = 1.
| Then A = | ![]() | 1 | 0 | ![]() |
| 2 | 1 |
SI – A| ⇒ (s – 1)² = 0
S = 1 ± j0