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 Derive the equations that relate the voltages across a
resistor, an inductor, and a capacitor in series as:
 the unit step function associated with voltage or current
source changes from 0 to 1 or
 a switch connects a voltage or current source into the
circuit.
 Describe the solution to the 2nd order equations when
the condition is:
 Overdamped
 Critically Damped
 Underdamped
 With a step function voltage source.
 You must determine the initial condition of the
inductor and capacitor at t < to and then find the final
conditions at t = ∞s.
 Since the voltage source has a magnitude of 0V at t < to
 i(to
-) = iL(to
-) = 0A and vC(to
-) = 0V
 vL(to
-) = 0V and iC(to
-) = 0A
 Once the steady state is reached after the voltage source
has a magnitude of Vs at t > to, replace the capacitor with
an open circuit and the inductor with a short circuit.
 i(∞s) = iL(∞s) = 0A and vC(∞s) = Vs
 vL(∞s) = 0V and iC(∞s) = 0A
 Initial Conditions
 i(to
-) = iL(to
-) = 0A and vC(to
-) = 0V
 vL(to
-) = 0V and iC(to
-) = 0A
 Final Conditions
 i(∞s) = iL(∞s) = 0A and vC(∞s) = Vs
 vL(∞s) = 0V and iC(∞s) = 0A
 Since the voltage across the capacitor is the only
parameter that has a non-zero boundary
condition, the first set of solutions will be for vC(t).
oossotoC
S
C
CC
SC
CC
CL
C
C
SL
L
C
tttvttvttv
LC
V
tv
LCdt
tdv
L
R
dt
tvd
Vtv
dt
tdv
RC
dt
tvd
LC
titi
dt
tdv
Cti
VRi
dt
tdi
Ltv
tv
when t)()()(
)(
1)()(
)(
)()(
)()(
)(
)(
0
)(
)(
0)(
2
2
2
2
 Similar to the solutions for the natural response, there
are three different solutions. To determine which one
to use, you need to calculate the natural angular
frequency of the series RLC network and the term .
L
R
LC
o
2
1
 Overdamped response ( > o)
where t-to = t
 Critically damped response ( = o)
 Underdamped response ( < o)
2
0
2
2
2
0
2
1
21
21
)(
s
s
eAeAtv tsts
C
t
C etAAtv )()( 21
22
21 )]sin()cos([)(
od
t
ddC etAtAtv
.The final condition of the voltages across the capacitor
is the steady state solution.
 vC(∞s) = Vs
 Overdamped response
 Critically damped response
 Underdamped response
VseAeAtv tsts
C
21
21)(
VsetAAtv t
C )()( 21
VsetAtAtv t
ddC )]sin()cos([)( 21
otttwhere
 Once the voltage across the capacitor is known, the
following equations for the case where t > to can be
used to find:
)()(
)(
)(
)()()()(
)(
)(
tRitv
dt
tdi
Ltv
titititi
dt
tdv
Cti
RR
L
L
RLC
C
C
 The set of solutions when t > to for the voltage across the capacitor in a
RLC network in series was obtained when power is applied to the
circuit at t = to.
 There are two components to the solution.
 The transient component, which has the same form as the transient solution for
the natural response of a series RLC circuit.
 The steady state component, which is the final condition for the voltage across the
capacitor is the steady state solution.
 Selection of equations is determine by comparing the natural
frequency o to
 Coefficients are found by evaluating the equation and its first
derivation at t = to
-.
 Voltage across the capacitor is equal to the initial condition when t < to
 Using the relationships between current and voltage, the current
through the capacitor and the voltages and currents for the inductor
and resistor can be calculated.
Step respponse of rlc circuit by Aditya Pratap Singh Delhi University

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Step respponse of rlc circuit by Aditya Pratap Singh Delhi University

  • 1.
  • 2.  Derive the equations that relate the voltages across a resistor, an inductor, and a capacitor in series as:  the unit step function associated with voltage or current source changes from 0 to 1 or  a switch connects a voltage or current source into the circuit.  Describe the solution to the 2nd order equations when the condition is:  Overdamped  Critically Damped  Underdamped
  • 3.  With a step function voltage source.
  • 4.  You must determine the initial condition of the inductor and capacitor at t < to and then find the final conditions at t = ∞s.  Since the voltage source has a magnitude of 0V at t < to  i(to -) = iL(to -) = 0A and vC(to -) = 0V  vL(to -) = 0V and iC(to -) = 0A  Once the steady state is reached after the voltage source has a magnitude of Vs at t > to, replace the capacitor with an open circuit and the inductor with a short circuit.  i(∞s) = iL(∞s) = 0A and vC(∞s) = Vs  vL(∞s) = 0V and iC(∞s) = 0A
  • 5.  Initial Conditions  i(to -) = iL(to -) = 0A and vC(to -) = 0V  vL(to -) = 0V and iC(to -) = 0A  Final Conditions  i(∞s) = iL(∞s) = 0A and vC(∞s) = Vs  vL(∞s) = 0V and iC(∞s) = 0A  Since the voltage across the capacitor is the only parameter that has a non-zero boundary condition, the first set of solutions will be for vC(t).
  • 7.  Similar to the solutions for the natural response, there are three different solutions. To determine which one to use, you need to calculate the natural angular frequency of the series RLC network and the term . L R LC o 2 1
  • 8.  Overdamped response ( > o) where t-to = t  Critically damped response ( = o)  Underdamped response ( < o) 2 0 2 2 2 0 2 1 21 21 )( s s eAeAtv tsts C t C etAAtv )()( 21 22 21 )]sin()cos([)( od t ddC etAtAtv
  • 9. .The final condition of the voltages across the capacitor is the steady state solution.  vC(∞s) = Vs
  • 10.  Overdamped response  Critically damped response  Underdamped response VseAeAtv tsts C 21 21)( VsetAAtv t C )()( 21 VsetAtAtv t ddC )]sin()cos([)( 21 otttwhere
  • 11.  Once the voltage across the capacitor is known, the following equations for the case where t > to can be used to find: )()( )( )( )()()()( )( )( tRitv dt tdi Ltv titititi dt tdv Cti RR L L RLC C C
  • 12.  The set of solutions when t > to for the voltage across the capacitor in a RLC network in series was obtained when power is applied to the circuit at t = to.  There are two components to the solution.  The transient component, which has the same form as the transient solution for the natural response of a series RLC circuit.  The steady state component, which is the final condition for the voltage across the capacitor is the steady state solution.  Selection of equations is determine by comparing the natural frequency o to  Coefficients are found by evaluating the equation and its first derivation at t = to -.  Voltage across the capacitor is equal to the initial condition when t < to  Using the relationships between current and voltage, the current through the capacitor and the voltages and currents for the inductor and resistor can be calculated.