Given H (s) = X=s + 1. x (t) = cos t 0º. The phasor Y

Given H (s) = X =
s + 1 . x (t) = cos t 0º. The phasor Y is given by—
Y s2 + s + 1

Given H (S) =	X	=	s + 1	.....(1)
	Y		s² + s + 1

X (t) = cost = 1∠ 0º.
From input x (t) = cos t it is clear that ω = 1. By putting S = jω in equation (1), with ω = 1, we get

	X	=	jω + 1	=	j + 1
	Y		j ² ω² + jω + 1		–1 + jω + 1

	j + 1	= √2	∠ 45°	= j₂ ∠ – 45°
	j		∠ 90°

⇒	1 ∠ 0°	= √2∠ – 45°
	y

or y =	1	– 45°
	√2

Given H (s) =		X	=	s + 1	. x (t) = cos t 0º. The phasor Y is given by—
		Y		s2 + s + 1

2	tan^{– 1} 1– tan^{– 1} 2
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