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

=