written 8.0 years ago by | • modified 8.0 years ago |
i) Static or Dynamic
In y(n) = x(2n) – x(n-1) ,
2n indicates past(for negative n) or future(for positive n) signal;
Also, n-1 indicates signal in past
Hence, the given system is dynamic.
ii) Linear or non-linear
$y_1(n) = x_1(2n) – x_1(n-1) \\ y_2(n) = x_2(2n) – x_2(n-1) \\ \therefore y_1(n) + y_2(n) = x_1(2n) – x_1(n-1) + x_2(2n) – x_2(n-1) ….. (1) \\ Replacing x(n) by x_1(n) + x_2(n); \\ y(n) = x_1(2n) + x_2(2n) – x_1(n-1) – x_2(n-1)………...……(2)$
from (1) & (2), $y(n) = y_1(n) + y_2(n)$
$\therefore$ Linear System
iii) Shift invariant or variant
$y(n) = x(2n) – x(n-1) \\ \therefore y(n,k) = x(2n-k) – x(n-1-k) …. . . . .. (1) \\ y(n-k) = x(2(n-k)) – x(n-k-1) \\ i.e., y(n-k) = x(2n-2k)–x(n-k-1)… . . ... (2) \\ from (1) (2), y(n,k) \not= y(n-k)$
$\therefore$ System is Shift Variant.
iv) Causal or non-causal
2n indicates future signal for positive n,
Hence, the given system is Non-Causal.
v) Stable or unstable
y(n) = x(2n) – x(n-1) provides Bounded output for bounded input,
Hence, the system is Stable.