Derivation of Peak Time
A Note on Derivation of Peak Time, Tp (Slide 27 in System Response) The output response is c(t ) = 1 − e −ζω nt 1−ζ 2
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A Note on Derivation of Peak Time, Tp (Slide 27 in System Response) The output response is c(t ) = 1 −
e −ζω nt 1−ζ 2
sin(ω d t + φ ) ⎛ 1−ζ 2 ⎜ ζ ⎝
ω d = ω n 1 − ζ 2 and φ = tan −1 ⎜
with
(1) ⎞ ⎟. ⎟ ⎠
Differentiating Eqn (1) by parts, we have (−ζω n e −ζω nt ) d c(t ) e −ζω nt (cos(ω d t + φ )ω d ) =− sin(ω d t + φ ) − dt 1−ζ 2 1−ζ 2 e −ζω nt
=
1−ζ 2 e −ζω nt
=
1−ζ 2
[ζω n sin(ω d t + φ ) − ω d cos(ω d t + φ )] [ζω n (sin(ω d t ) cos φ + cos(ω d t ) sin φ ) − ω d (cos(ω d t ) cos φ − sin(ω d t ) sin φ )]
Noting that cos φ = ζ and sin φ = 1 − ζ 2 , we have c&(t ) =
=
=
e −ζω nt
1−ζ 2
e −ζω nt
1−ζ
[(ζω n cos φ + ω d sin φ ) sin(ω d t ) + (ζω n sin φ − ω d cos φ ) cos(ω d t )]
[(ζ ω 2
2
e −ζω nt
1−ζ 2
n
+ ω n (1 − ζ 2 ) )sin(ω d t ) + (ζω n 1 − ζ 2 − 1 − ζ 2 ω nζ ) cos(ω d t )
[ω n sin(ω d t )]
Letting c&(t ) = 0 at t = t p , we thus have e
−ζω n t p
1−ζ 2 Since e
−ζω n t p
we have
[ω
n
]
sin(ω d t p ) = 0 .
≠ 0 for t p < ∞ ,
sin(ω d t p ) = 0 giving sin(ω d t p ) = 0, π ,2π ,3π ,K
Choosing the first peak gives tp =
π ωd
]