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A Note on Derivation of Peak Time, Tp (Slide 27 in System Response)
The output response is
)sin(1
1)(2
φωζ
ζω
+−
−=−
tetc d
tn
(1)
with 21 ζωω −= nd and ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ −= −
ζζ
φ2
1 1tan .
Differentiating Eqn (1) by parts, we have
( )dd
t
d
tn tete
dttcd nn
ωφωζ
φωζ
ζω ζωζω
)cos(1
)sin(1
)()(22
+−
−+−
−−=
−−
[ ])cos()sin(1 2
φωωφωζωζ
ζω
+−+−
=−
ttedddn
tn
( ) ( )[ ]φωφωωφωφωζωζ
ζω
sin)sin(cos)cos(sin)cos(cos)sin(1 2
ttttedddddn
tn
−−+−
=−
Noting that ζφ =cos and 21sin ζφ −= , we have
[ ])cos()cossin()sin()sincos(1
)(2
ttetc ddnddn
tn
ωφωφζωωφωφζωζ
ζω
−++−
=−
&
( )[ ])cos()11()sin()1(1
2222
2tte
dnndnn
tn
ωζωζζζωωζωωζζ
ζω
−−−+−+−
=−
[ ])sin(1 2
tedn
tn
ωωζ
ζω
−=
−
Letting 0)( =tc& at ptt = , we thus have
[ ] 0)sin(1 2
=−
−
pdn
t
te pn
ωωζ
ζω
.
Since 0≠− pnte ζω for ∞<pt ,
we have 0)sin( =pd tω giving K,3,2,,0)sin( πππω =pd t
Choosing the first peak gives
d
ptωπ
=
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