Example of a second order circuit responseese216/handouts/Chpt14LaplaceRLC...Example of a second...
1
Example of a second order circuit response Consider the following function transfer function of a 2 nd order (RLC) low-pass filter. ) 1 2 ( 1 () ( ) ( 2 + + = s s s V s V i o ζ in which the undamped natural frequency ω o =1 rad/s For a unit step function as the input (Vi(s)=1/s), the output can be written as, ) 1 2 ( 1 ) ( 2 + + = s s s s V o ζ The following plot shows the time response of the output for a unit step input, for the case of ζ=0.25 (under-damped), ζ=1 (critically damped) and ζ=2 (over-damped). 0 5 10 15 20 25 30 0 0.5 1 1.5 T=2pi=6.28sec Time, t (sec) vo (V) damping factor = 0.25 2 1 The expressions of the output are equal to: a) vo=1+1.03*exp(-0.25*t).*cos(0.97*t+165.54 o ) (case ζ=0.25) b) vo=1+0.077*exp(-3.732*t)-1.077*exp(-0.268*t) (case ζ=2) c) v0=1-t.*exp(-t)-exp(-t) (case ζ=1) ESE216 Chpt14LaplceRLCResponse2
Transcript of Example of a second order circuit responseese216/handouts/Chpt14LaplaceRLC...Example of a second...
Example of a second order circuit response Consider the following function transfer function of a 2nd order (RLC) low-pass filter.
)12(1
()()(
2 ++=
sssVsV
i
o
ζ
in which the undamped natural frequency ωo=1 rad/s For a unit step function as the input (Vi(s)=1/s), the output can be written as,
)12(1)( 2 ++
=sss
sVo ζ
The following plot shows the time response of the output for a unit step input, for the case of ζ=0.25 (under-damped), ζ=1 (critically damped) and ζ=2 (over-damped).
0 5 10 15 20 25 300
0.5
1
1.5T=2pi=6.28sec
Time, t (sec)
vo (V)
dampingfactor = 0.25
2
1
The expressions of the output are equal to: a) vo=1+1.03*exp(-0.25*t).*cos(0.97*t+165.54o) (case ζ=0.25) b) vo=1+0.077*exp(-3.732*t)-1.077*exp(-0.268*t) (case ζ=2) c) v0=1-t.*exp(-t)-exp(-t) (case ζ=1)
ESE216 Chpt14LaplceRLCResponse2
