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Signal and systems miscellaneous

Signals and Systems

  1. The DTFT of the given signal
    x[n] =
    1
    – nu[– n – 1]
    2
    1. ejΩ
      2 – e– jΩ
    2. 2ejΩ
      2 – e– jΩ
    3. ejΩ
      2 – ejΩ
    4. 2ejΩ
      2 – ejΩ
Correct Option: C

∞
X(ejΩ) =
x[n].e–jΩn
n = – ∞

∞
=
(1/2)-nu[- n - 1]e–jΩn
n = – ∞
2

∞
=
(1/2)-ne–jΩn
n = – ∞
2

∞
=
(1/2ejΩn)-n
n = – ∞
2

0
∞
∞
=∵
ak =
(1/a)k =
(a)-k
k = ∞
k = 0
k = 0

=
ejΩ
2
1 -
ejΩ
2

=
ejΩ
2 - ejΩ



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