linear wave shaping

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.


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

=

=


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.


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


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)


0V ( ) 11

V sRC Vs

s sRCs

RC

= × =+ +

∴ (Assumed v (0) = 0)

( )t

RCv t Ve−

=


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)= −



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



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



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.



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


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

−= =

++

= = + + ++ +

= − = = =

−=

≜


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

− − = − −



0v

v

input

n=100

1.0

0.9

0.8

0.7

0.6

0.5


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


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−

= −

∴


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≫ ≃≃



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?


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


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


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


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


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)


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


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′′


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

−

−

− = = = − +



( )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−= − + =




( )

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


( ) 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


( )

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


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


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


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


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


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


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


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)


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


( )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−

− ± −

=

= − −

+ + −


( )( )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= = =

linear wave shaping

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