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If x = Re G(jθ), and y = lm G(jω) then for ω → 0+, the Nyquist plot for
G(s) = 1 s(s + 1)(s + 2)
becomes asymptotic to the line
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- x = 0
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x = - 3 4 - x = y – 1 / 6
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x = y √3
Correct Option: B
| Given , G(s) = | ||
| s(s + 1)(s + 2) |
| G(jω) = | = | |||
| jω(1 - jω)(2 + jω) | ω(1 - ω²)(4 + ω²) |
| = | = | |||
| ω(1 + ω²)(4 + ω²) | ω(1 + ω²)(4 + ω²) |
| = | - j | |||
| (1 + ω²)(4 + ω²) | ω(1 + ω²)(4 + ω²) |
| ⇒ x = | ||
| (1 + ω²)(4 + ω²) |
| y = - | ||
| ω(1 + ω²)(4 + ω²) |
| At ω → 0 , x → | , y → - ∞ | |
| 4 |
