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Linear Waveshaping
Presented by APN Rao, Dept ECE, GRIET, Hyderabad. Jan 2012 1
Text Book:
Pulse, Digital and Switching Waveforms
Jacob Millman, Herbert Taub
McGraw-Hill Kogakusha Ltd (1965)
Introduction to Linear Waveshaping
In a linear circuit, all components are assumed to be operating within their
respective linear regions.
Linear circuits preserve shape of a sinusoidal signal, and alter shapes of non-
sinusoidal signals.
Linear Waveshaping is the process by which a non-sinusoidal signal is altered
by transmission through a linear network.
Presented by APN Rao, Dept ECE, GRIET, Hyderabad. Jan 2012 2
by transmission through a linear network.
Responses of simple RC, RL and RLC circuits to some standard input
waveforms will be studied in this unit.
DC Value of a Waveform
For energy signals, the total integral (or the net area under the waveform) is
termed as its dc value. Zero dc value implies positive area equals negative
area. For a periodic waveform, dc or average value is the integral over one
period (net area under one cycle) divided by the period.
Modelling Resistor, Capacitor and Inductor
In time domain
At any instant, v across and i passing through an element are related as:
( ) ( )
( )
R R
cc
v t i t R
dvi t C
dt
di
=
=
Presented by APN Rao, Dept ECE, GRIET, Hyderabad. Jan 2012 3
In frequency domain
Capacitive reactance (current leads voltage)
Inductive reactance (voltage leads current)
Resistance is independent of frequency
( ) LL
div t L
dt=
1cX
j Cω=
LX j Lω=
( 2 )fω π=
Modelling R, L, C: Inferences
In a resistor
Voltage and current are always in time phase.
Voltage decreases in the direction of current.
In a capacitor
Voltage varies as time-integral of current. Constant charging current produces linearly varying
voltage.
Voltage across a capacitor can not change discontinuously for finite-valued currents.
Presented by APN Rao, Dept ECE, GRIET, Hyderabad. Jan 2012 4
Voltage across a capacitor can not change discontinuously for finite-valued currents.
Open circuit (infinite impedance, zero current) to dc voltage and short for high frequencies
In an inductor
Current is time-integral of voltage. Constant voltage produces linearly varying current.
Current through an inductor can not change discontinuously for finite-valued voltages.
Short circuit (zero impedance, zero voltage drop) to dc current and open circuit to high
frequencies.
Modelling High Pass RC Circuits
C
Rvi(t) vo(t)
(Input) (Output)
gain( )
( )( )
o
i
V fA f
V f=
1
( )
( ) 1
Zero gain is offered to dc component of
( ) ; 2
( ) 1
Gain increases with frequency,
input.
1Lower 3-dB frequency
max gain =
1.
i
o
i
o
RC
s
s
V j RCf
V
V s sRC
V
j
s sRC
f
RC
ω ωω π
ω ω
=+
=
=+
= =+
(Hz)
High Pass RC Circuits
Presented by APN Rao, Dept ECE, GRIET, Hyderabad. Jan 2012 5
f0 1
1Lower 3-dB frequency
2f
RCπ=
1
1 1
2
1
( ) 1
( )1
1( ) ; ( ) tan
(H
1
High Pass RC circuit is a
z)
phase lead network.
o
i
V f
V f fjf
fA f f
ff
f
θ −
=
−
= =
+
1 i o o
i o o
v v dt vRC
dv v dv
dt RC dt
= +
= +
∫
Step ResponseC
vo(t) V e-t/RC
V
vi(t)
V
RC = time constant
t0
Rvi(t)= V u(t) vo(t) = V e-t/RC
+ _
i(t)
High Pass RC Circuits
0V ( ) 11
V sRC Vs
s sRCs
RC
= × =+ +
∴ (Assumed v (0) = 0)
( )t
RCv t Ve−
=
Presented by APN Rao, Dept ECE, GRIET, Hyderabad. Jan 2012 6
t= __xRC v0 = __xV
0.5 0.607
1 0.368
2 0.135
3 0.0498
4 0.018
5 0.007
When step is applied, positive discontinuity
of V volts is passed to output (as capacitor can not change its
voltage instantaneously, that is v (0 )= v (0 ). T
Physical
hereafte
Reasoning
r,
capacitor c
:
C C
− +
0
harges increasing v with polarity shown, decreasing
the loop current i(t) exponentially with time-constant = RC.
At steady state, v ( ) = V, i( ) = 0 and v ( ) = 0
C
C ∞ ∞ ∴ ∞
Smaller the RC, faster the decay: Find and sketch the impulse response of the high pass RC ckt.
Justify the sketch with physical reasoning.
Exercise
∴ (Assumed vc(0) = 0)
( ) RCov t Ve=
Note the special feature of exponential decay- equal attenuation over equal intervals
Response to Square Pulse
vi(t)
V
0
T t
v0(t)
V1
0T
t
+
_
V1′
V ( )
T RC
1 1 2 1
t RC
0
V = V; V Ve ; V V V
for 0 t < T, v (t) Ve
−
−
′ ′= = −
< =
iv (t) Vu(t) Vu(t-T)= −
High Pass RC Circuits
Presented by APN Rao, Dept ECE, GRIET, Hyderabad. Jan 2012 7
0T _
V2
V ( )t TT RC RC
0for t > T, v (t) Ve V e
−− −
= −
DC component is absent in the output, indicated by its integral being zero. Positive area
equals the negative area making the net area under the output waveform zero. (Verify by
integration.) As the transfer function has a zero at origin, dc component gets zero gain.
The ability of high pass RC circuit to block dc is used by a series capacitor inserted
between two stages for ac coupling, while preventing dc coupling.
Think it over: Step response of HPRC does not integrate to zero! Why is DC component
not blocked in this case?
Response to Periodic Square Wave
vi(t)
V
0
T1tT2
V = peak-to-peak voltage of input ; Period T = T1 + T2
vo(t)
V
High Pass RC Circuits
Presented by APN Rao, Dept ECE, GRIET, Hyderabad. Jan 2012 8
0
T1tT2
Note that the transient portion of the response where dc level is shifting. At steady state,
the response reaches zero dc level, with positive and negative areas equalling over each
period. Therefore, steady state output is independent of input dc level. The waveform
becomes periodic at steady state.
The transient portion appears due to any dc bias in the input, which causes corresponding
step response to be present in the output. Steady state is reached when the step response
dies down.
Steady State Response to Periodic Square Wave
T1 T2
Period T = T1 + T2vo(t)
V1
t
V2
V1′
V2′V2′
V1
V2
V1′
V2′
V1
V2
V1′
For general case
( )
Relations for symmetric Steady state values of
symmetric square wave
High Pass RC Circuits
Presented by APN Rao, Dept ECE, GRIET, Hyderabad. Jan 2012 9
1
2
1 1
2 1
2 2
1 2
T
RC
T
RC
V V e
V V V
V V e
V V V
−
−
′ =
′= −
′ =
′= +
( )1 2
2
2
1 1
2 1
2 2
1 2
T T T/2square wave
T
RC
T
RCV V e
V V V
V V e
V V V
−
−
= =
′ =
′= −
′ =
′= +
1 2
1
2
2 1
2 1
symmetric square wave
response
VV
1
VV
1
V V
V V
T
RC
T
RC
e
e
−=
+
′ =
+= −
′ ′= −
Steady State Response to Periodic Square Wave: Effect of RC
VariationSmall time-constant RC<<T/2: Alternating positive and negative spikes each
with amplitude V and equal areas, coincide with upward and downward
discontinuities of the input. Steady state is reached within the first half-period.
Large time-constant RC>>T/2: Steady state is reached after many cycles. Shape
of response at steady state is nearly identical to the input waveform, except for
the absence of dc level tops.
High Pass RC Circuits
Presented by APN Rao, Dept ECE, GRIET, Hyderabad. Jan 2012 10
Shape distortion: The flat tops of input square wave become tilted in the
response. Tilt is small for large RC, and increases with decrease in RC. The
measure of tilt for symmetric square wave response at steady state is as below:
Peak-to-peak value of response: is nearly V for large RC, increases with
decrease in RC, approaching 2V for small RC.
2
1 1 1
2
1%Tilt 100 200; for large RC, 100 100
2 21
T
RC
T
RC
V V fe T
V RC fe
π−
−
′− −× = × ≈ × = ×
+
≜
Response to Exponential Input
High Pass RC Circuits
0
0
1
11
( ); ( ) ;
1 ( )
( ) 11
/ / ,
( )
[1 ] i
i
t
i
RC
RCRC
V V s sV s
V s ss s
V A BV s
s ss s
V VRC VnA B where
v t V e
RCn
τ τ
ττ
τττ τ
−= =
++
= = + + ++ +
= − = = =
−=
≜
Presented by APN Rao, Dept ECE, GRIET, Hyderabad. Jan 2012 11
0
1
/ / ,
1 11
( ) ; 1
t
R
t
C
RC
V VRC VnA B where
RC n
Vnv t e e further
n
RCn
τ
ττ τ
τ τ− −
= − = = =−− −
= − −
≜
definingt
xτ
≜
0( )1
x
xnVn
v x e en
− − = − −
Response to Exponential Input
High Pass RC Circuits
0v
v
input
n=100
1.0
0.9
0.8
0.7
0.6
0.5
Presented by APN Rao, Dept ECE, GRIET, Hyderabad. Jan 2012 12
v
tx
τ=
RCn
τ=
n increasing
n=0.1
n=1
n=10
0 5 10 15 20 25 30 35 40 45 50 55 60
0.5
0.4
0.3
0.2
0.1
0
Response to Ramp Input
High Pass RC Circuits
02
0
1; V ( ) ; V ( )
At steady st
( ) (
ate,
)
( )
ii
RC
v t t u s ss
s s
t
t
v RC
α α
α
α = = +
= ∞
=
=
αRC
vi(t) = αt
t0
v0(t)
0( ) 1t
RCv t RC eα−
= −
∴
Presented by APN Rao, Dept ECE, GRIET, Hyderabad. Jan 2012 13
t0
vi(t)
t0 v0(t)T
Response to sawtooth
(sweep) pulsevi(t)
t0
v0(t)
T
Response to
Limited Ramp
αT
( ) ( )Deviation from linearity for response to a sweep pulse
( )
( sweep duration) for RC T2
i o
i
v T v T
v T
TT
RC
−
= ≈
≜
≫ (Verify this result)
HP RC as a Differentiator
0
For low values of RC, HPRC circuit functions as a differentiator. This feature is explained
as below:
1( )
approximates to ) Transfer function for small RC values. In time
domain, thi
( ) 1i
V s sRCsRC
V s sRC=
+
0v ( ) , is very small in co
s corresponds to differentiation of
mparison
input by the circuit.
2) For t RC, at that instant. The to n v ,v i c ivt RCα≫ ≃≃
High Pass RC Circuits
Presented by APN Rao, Dept ECE, GRIET, Hyderabad. Jan 2012 14
0
0
0
( ) , ( )
3) 1
i i
t
RC
dv dvi t C v t RC
dt dt
vdv
dtRC eα α
−
∴
= − → =
≃ ≃
0 0 00
(constant) for t RC.Thus the input-
output relation + approximates to , resultin n
g i ( )
t
i
iR
i
C
idv v dv dv v dvv t RC
dt
dve
RC dt dt RC dt
dtα
−=
= =
≫≪
≃
How small should be RC for a good differentiator?
High Pass RC Circuits
Think of input waveform approximated satisfactorily by straight-line segments,
making it a step-ramp construction using suitably sized time-intervals. For good
differentiation, RC should be much smaller than the smallest of those time-
intervals.
Sinusoidal response of an ideal differentiator is also a sinusoidal wave with 900
phase lead. A practical circuit provides a phase lead = tan-1(1/ωRC), which is less
than 900. Better the differentiation, closer would be the phase-lead to 900. A
criterion for good differentiation is to specify the minimum phse-lead as 89.4o.
This implies ωRC < 0.01, at the frequency corresponding to the smallest time
Presented by APN Rao, Dept ECE, GRIET, Hyderabad. Jan 2012 15
This implies ωRC < 0.01, at the frequency corresponding to the smallest time
interval of appreciable variation (ω=2π/T).
Need for Amplifier after Differentiator
Good differentiator produces very low voltage responses, therefore needs to be
followed by a high gain amplifier.
Low Pass RC Circuit
R
vi(t) vo(t)
(Input) (Output)
C
1
1
( ) 1
( ) 1
Unity gain is given to dc component
( ) 1 ; 2
( ) 1
gain
decreasing w ith frequency (low pas
,
U pper 3 -dB f
s).
requency
o
i
o
i
RC
RCs
Vf
V s
V s sRC
CV j R
ωω π
ω ω
=+
=+
= =+
gain ( )( )
( )
o
i
V fA f
V f=
1
Presented by APN Rao, Dept ECE, GRIET, Hyderabad. Jan 2012 16
2
1 = (H z)
2
fCRπ
2
1
2
2
2
( ) 1
( )1
1( ) ; ( ) tan
1
phase laLow Pass RC circuit is a networkg
.
o
i
V f
fV fj
f
fA f f
ff
f
θ −
=
+
= = −
+
oi o
dvv v RC
dt= +
0 f
1
Note: f2 is the 3 dB bandwidth
of the Low Pass circuit.
Low Pass RC Circuits
Step Response
C
R
vi(t)= V u(t) vo(t) = V[1- e-t/RC ]+
_ i(t)
0
1V ( )
11
VV RCss sRC
s sRC
= × =+ +
∴ (Assumed vc(0) = 0)
( ) 1t
RCov t V e
− = −
vo(t) V[1- e-t/RC ]
V
vi(t)
V
RC = time constant
t
0
Presented by APN Rao, Dept ECE, GRIET, Hyderabad. Jan 2012 17
At t = 0 , output is zero (as capacitor can
not change its voltage instantaneously, that is v (0 )= v (0 ).
Thereafter, capacitor charges increasing v with
Physi
pola
cal Reas
rity shown,
decreas
oning:
C C
C
+
− +
0
ing the loop current exponentially with time-constant = RC.
At steady state, capacitor is fully charged and v ( ) v ( ) = V. C∞ = ∞
1
: F in d a n d sk e tc h th e im p u ls e r e s p o n s e o f th e low p a s s R C c k t.
J u s tify th e s k e tc h w ith p h ys ic a l r e a so n in g .
: F o r th e n o n -z e ro in it ia l c o n d it io n , v
1
( 0 ) , sh ow t 2 h ao
E x e rc is e
E x e r i Vc s e − =
1
t
; w h e re , V is th e s te a d y s ta te v a luv ) ( ) e( .t
R Co t V V V e
−= + −
∴ (Assumed vc(0) = 0)o
Rise Time tr of a low pass circuit is
the time for step response to rise
from 10% to 90% of steady state
value.
2
0.352.2 rt RC
f= =
Response to Square Pulse
Low Pass RC Circuits
vi(t)
V
0
T t
v0(t)
V
0T t
V2
( )
t RC
0
T RC
2 0
t TT RC RC
0
for 0 t < T, v (t) V 1 e
V v (T) V 1 e
for t > T, v (t) V 1 e e
−
−
−− −
< = −
= = −
= −
iv (t) Vu(t) Vu(t-T)= −
(Assumed v0(0) = 0)
Presented by APN Rao, Dept ECE, GRIET, Hyderabad. Jan 2012 18
0T t
0
DC component is passed to output with unity gain, indicated by integral (net area) of response
being equal to integral of input . Verify. The two hatched areas shown are equal.
v0(t)
V
0
T t
Observe that as RC decreases, the response approaches the
input square shape. However the leading upper edge is
rounded off and a tail is added. For most practical
purposes, the square pulse is adequately passed if
2
1T
f= Then T = 6.3 RC or RC=0.16 T
RC decreasing
Steady State Response to Periodic Square Wave
Low Pass RC Circuits
vi(t)
V′
0
T1tT2
Period T = T1 + T2
vo(t)
0T tT
V1 V1 V1
V2 V2 V2
V′ V′
V′′V′′
V′′
V′′
Presented by APN Rao, Dept ECE, GRIET, Hyderabad. Jan 2012 19
0T1
tT2
Steady state response is also periodic with T, and with the same dc value of the input.
[ ] [ ]
[ ]( )
[ ]
1
1 2
01 1 2 1
02 2 1 2
v V + V V ; V V + V V
v V + V V ; V V + V V
Tt
RC RC
t T T
RC RC
e e
e e
− −
−− −
′ ′ ′ ′= − = −
′′ ′′ ′′ ′′= − = −
2
2 12
: Show that for a symmetrical square wave input with zero average, the steady
stV 1 V
V tanh2
ate values are 2
41
TRC
TRC
e TV
RCe
Exercise
−
−
− = = = − +
Response to Exponential Input
Low Pass RC Circuits
( )o
From the equations derived earlier for exponential response of high pass RC,
exponential response for low pass RC circuit can be obtained as:
v
1 , 1V 1 1
xx
nx e n
e nn n
− −= + − ≠
− −
( ) 1 1 1
, xx e n−= − + =
Response to Ramp Input
Presented by APN Rao, Dept ECE, GRIET, Hyderabad. Jan 2012 20
Response to Ramp Input
( )
Similarly, from ramp response of HP RC, we obtain that for LP RC circuit
as : v ( )t
RCo t t RC RC eα α
−= − +
T>>RC
RC
αRC
0
( )i o
i
for ramp time TTransmission error
v ( ) v ( ) RC
e v (
R
) T
C
t
T T
T
−≈
≫
≜
Low Pass RC Circuit as Integrator
Low Pass RC Circuits
( ) 1 11) for large enough RC, (integrator)
( ) 1
2) large RC (small output condition)
Example
1 v ,
For RC , using power series expans
s
ion and approximatin
o
i
o i
o oi o o i
V s
V s RCs RCs
v v
dv dvv RC RC v v dt
dt dt RC
t
= ≈+
⇒
∴ = + ≈ ∴ ≈ ∫
≪
≫2
g,
Vt tα
Presented by APN Rao, Dept ECE, GRIET, Hyderabad. Jan 2012 21
( )
2
0
1
step response and ramp response 2
Ideal integrator provides 90 phase delay to sinusoidal response.
Practical RC integr
Criterion for good integra
ator gives phase delay of tan . A
t
ion
Vt t
RC RC
RC
α
ω−
≈ ≈
( )
0
good
differentiator may be specified by phase lag > 89.4 or 95.5.
satisfies this RC co>15 nditi nT o
RCω >
Good integrator produces very low voltage responses, therefore needs to be followed by a high gain
amplifier.
Signal Attenuators
R
R2
R1
vi(t) vo(t)
2o i
1 2
1 2
Rv (t) = v (t), where = attenuation factor
R +R
R ,R chosen large (in M ) to prevent source loading.
a a =
Ω
1 2R R
Attenuation ignoring stray capacitance
Effect of stray capacitance on attenuation
Presented by APN Rao, Dept ECE, GRIET, Hyderabad. Jan 2012 22
R2
R1
vi(t) vo(t)C2vo(t)C2
1 2R + R
2i
1 2
RV (t)
R +R≡
Stray capacitance shunts high frequency components of signal (low pass).
Step response has a transient with time constant RC = (R1||R2)C2
considered large for many applications.
Signal Attenuators: Compensation
C1
C2
R1
R2
vi(t)
vo(t)
X YR2
R1
vi(t) vo(t)C2
C1
≡
Compensating capacitor C1 is added (shunting R1) with C1 = R2C2/R1. Then the
Presented by APN Rao, Dept ECE, GRIET, Hyderabad. Jan 2012 23
1 1 1 2 2 1
equivalent bridge is balanced with zero current in arm X-Y.
Step response: At t=0+, the voltage jump at input appears across C1 - C2 in
series. Then
At steady state, C1 - C2 are open and
As the output initial value = final value (for the C1 chosen), we get ideal step
response with no distortion. (Note: Above analysis ignores source resistance)
2o
1 2
Rv ( ) = V
R +R∞
2
1 2
C+ 1o
C C 1 2
X Cv (0 ) = V = V
X +X C +C
Compensated Attenuator: Step Response
2o
1 2
Rv ( ) = V
R +R∞2 2
1
1
R Cv (0 ) for C =
Ro
+
t0
2 21
1
R Cv (0 ) for C <
Ro
+
2 21
1
R Cv (0 ) for C >
Ro
+
Over-compensation
Under-compensation
Correct compensation+ 1o
1 2
Cv (0 ) = V
C +C
Time Constant of Transient Decay
Presented by APN Rao, Dept ECE, GRIET, Hyderabad. Jan 2012 24
1 2 eff
1 21 2
1 2
1 2
Effective RC for exponential R R = (C +C )
decay to steady
Seen inw
stat
ard from output terminals,
R R R ; C
e
= C
R
C
+R
eff
=
For Larger Attenuation Factor: Reff reduces, leads to smaller effective RC therefore causing lesser
waveform distortion.
(Note: Above analysis ignores source resistance)
Time Constant of Transient Decay
Effect of Introducing Compensated Attenuator on Waveform
≡ ≡
Assumptions: 1) Compensation is exact, and arm X-Y maintains zero current. 2) RS << R1, R2
1 2C CEffective RC with Compensated Attenuator = R
Presented by APN Rao, Dept ECE, GRIET, Hyderabad. Jan 2012 25
1 2
1 2
1 2S 2
1 2 1 2
Effective RC with Compensated Attenuator = RC +C
C RRC without attenuator = R C , and Attenuation Factor
C +C R +R
Compensated attenuator reduces RC by the value of attenuation factor, a
S
= =
∴ nd
therefore improves waveform. Circuit for
Direct Coupling
An oscilloscope, used for monitoring waveforms at points in active circuits has to deal with RS which may
not be small. To improve waveform transmission, the probe includes a fixed attenuator (eg. 1/100)
R-L Circuits: Equivalence with RC Circuits
Low Pass RC Circuits
vi(t) vo(t)
(Input) (Output)C
C
Rvi(t) vo(t)
(Input) (Output)
R
( )( )
( )
( ) 1
o
i
RC sV s
V s RC s=
+ ( )( ) 1
( ) 1
o
i
V s
V s RC s=
+
HP RC LP RC
Presented by APN Rao, Dept ECE, GRIET, Hyderabad. Jan 2012 26
L
R΄vi(t) vo(t)
(Input) (Output)vi(t) vo(t)
(Input) (Output)L
R΄
( )( )/( )
( ) 1 /
o
i
L R sV s
V s L R s
′=
′+ ( )( ) 1
( ) 1 /
o
i
V s
V s L R s=
′+
HP RL LP RL
Parallel Circuits for Processing Current Waveforms
Low Pass RC Circuits
CR io(t)
( )( )
( )
( ) 1
o
i
RC sI s
I s RC s=
+ ( )( ) 1
( ) 1
o
i
I s
I s RC s=
+
HP RC LP RC
ii(t)
C R io(t)ii(t)
Presented by APN Rao, Dept ECE, GRIET, Hyderabad. Jan 2012 27
L R΄
( )( )/( )
( ) 1 /
o
i
L R sI s
I s L R s
′=
′+ ( )( ) 1
( ) 1 /
o
i
I s
I s L R s=
′+
HP RL LP RL
io(t)ii(t)
LR΄
ii(t)
io(t)
RLC Circuits
vi(t) vo(t)C
R
L0
2 2
0 0
2
0
0
V ( ) /; second order system
V ( ) / 1/
(standard form for system analysis)
1where natural (resonant) frequency
1 = = damping const
V ( )
ant; natural per
2=
V ( )
o
2
i d T2
o
i
o
i
s s RC
s s s RC LC
LC
L
s k s
k
s
R C
s k s
ωω ω
ω
=+ +
=
+
=
+
2 LCπ= 2
1 2 0 0
Roots of characteristic equat o
,
i n
1s s k kω ω= − ± −
Case 1: ( = 1 ); s ,s =Critically damped k ω− Case 2: ( > 1)over-damped k
Presented by APN Rao, Dept ECE, GRIET, Hyderabad. Jan 2012 28
( )0
1 2 0
00 2
0
0
0
0
0
Case 1: ( = 1 ); s ,s =
2 VStep response V ( ) ;
1 1 2 2Other substitutions:
2
Critically damp
ed
v
( )2
V
to
k
ss
tt e
R
RC L TLC
tx
T
ω
ω
ω
ω
π
ω
ω
−
−
=+
= = = =
=
=
( ) ( )0
0
1 2 0 0 2
0 0 2
0 0
00
1
0 0
22
Case 2: ( > 1)
1s ,s = 1
1for 1, 1
2
/
over-dampe
2 , 2
2 VStep response V
d
( )/ 2 2
(for k>>1)v ( )
V
Rt tt
k to k RCL
k
k kk
k k k
te e e
k
k k
ks
s
e
k s k
ωω
ω ω
ω ω
ω ω
ωω ω
− −−−
− ± −
≈ − ± − ≈ − −
=+
= − = −
+
≫
Q-factor
; indicates how close the RLC circuit is to resonant cond1
i onQ 2
tik
≜
: If C 0, with R & L fixed, step response (same as high pass RL circuit). If R 0, with L & C fix )ed, ( Rt
o oLv
Notv
e u tV
eV
−→ →→ →
RLC Circuits
( )
( )0
2
1 2 0 0
00 2
2 2
0 0
2
02
Case 3: ( < 1 )
s ,s = 1
2 VStep response V ( )
1
under-damped
v ( ) 2 sin 1
V 1
ko t
k
k j k
ks
s k
t ke k t
k
k
ω
ω ω
ω
ω ω
ω−
− ± −
=
= − −
+ + −
Presented by APN Rao, Dept ECE, GRIET, Hyderabad. Jan 2012 29
( )( )0
2
2
0damped
V 1
freque 1ncy d
k
k ωω ω
−
= − <
Ringing Circuit
A highly underdamped RLC circuit is also called a Ringing Circuit. The number N of
oscillations in its step response before the amplitude falls to 1/e of initial value, is a
measure of the damping constant.
and 1 1
2 2k Q N
N kπ
π= = =