Lecture06 New Signal Conditioning - Filterx

8/10/2019 Lecture06 New Signal Conditioning - Filterx

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0


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A A (( )) ,,

.. ::

0

bb

2211

1122

RRRR

VVRRVVRR

outoutaaVV ++++++++

++++++++========


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A A ,, ((00))

2211

22

RRRR VVininRR

outoutVV ++++++++========

0

1

2


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0

. .

.


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Three types of filterThree types of filter low, high and bandlow, high and band--pass filterspass filters

0


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CLASSIFICATION OF FILTERSCLASSIFICATION OF FILTERSSignal Filter

Analog Filter Digital Filter

0

emen ype requency an

Active Passive Low-Pass

High-Pass

Band-Pass

Band-Reject

All-Pass


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: C : C

A plot of H(f) transfer function is a powerful way to analyse

Bode plot magnitude and phase

Magnitude Bode plot of a LPF.

0

y axis: absolute value ofH(f) = |H(f)|

fC is the frequency on one

side of which H(f)~0 orthe circuit blocks thosefrequencies. It is thecutoff frequency.

fC

~70.7%

Blocked


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A A


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A A , ,

// = = CC/(/(++CC)) = =

== 1 C = 1 C = 11 C = 1 C = 1 CC == 11

Vout

Vin

)|( j|H

0 0

As fincreases from 0 onwards, VOUT/VIN (the TRANSFERFUNCTION) goes from 1 to 0. Thus high frequencies are

attenuated. Thus this is a LOW PASS FILTER.

++++++++========

++++++++

========fRCfRCjj

VV

fCfCjjRR

fCfCjjVVVV ININININOUTOUT

2211

11

22

11

22

11p s


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FIGURE 2.13 Circuit for the low-pass RCfilter.

0 Curtis Johnson

Process Control Instrumentation Technology, 8e]

Copyright 2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458

All rights reserved.


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ResponseResponse of the lowof the low--passpass RCRCfilter as a function of the frequency ratio.filter as a function of the frequency ratio.

0 Curtis Johnson

Process Control Instrumentation Technology, 8e]



X-axis log of ratio of input signal freq to critical frequency


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B B

C C / / 0.7070.707

RCfc

2

1=

0

outout

22//1122

11

11

++++++++

========

cc

inin

ff

ffVV

VV


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C C 11 1 1

0


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A A

0


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A A , ,

VVOUTOUT/V/VININ = Z= ZRR/(Z/(ZRR+Z+ZCC))

ZZRR = R= R

ZZCC = 1/CD = 1/Cs = 1/= 1/CD = 1/Cs = 1/jjCC == 1/1/j2fCj2fCVout

Vin

)|( j|H

0

As fincreases from 0 onwards, VOUT/VIN (the TRANSFERFUNTION) goes from 0 to 1. Thus low frequencies are

attenuated. Thus this is a HIGH PASS FILTER.

s p

++++++++========

++++++++

======== VV

fCfCjjRR

VVVV ININININOUTOUT

fRCfRCjj 2211

22

11

RR fRCfRCjj 22


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FIGURE 2.16 Response of the high-pass RCfilter as a function of frequency ratio.

0 Curtis JohnsonProcess Control Instrumentation Technology, 8e]




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C C A A B B C C / / 0.7070.707

ffcc

11========

0 0

outout

22//1122

11

//

++++++++

========

cc

cc

inin

ff

ff

ffff

VV

VV


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C C

0


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Low pass filter using op ampLow pass filter using op amp

Cf

RfI1 I2

I3

I2

I3

0

Summation of current at summing point

I1 + I2 + I3 = 0


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Therefore

+= outf

f

outin

sVCR

V

R

V

1

outf

ffin VR

sRC

R

V

+=

1

1

0

Hence, he output voltage is given as

in

ff

f

out V

sRCR

RV

+

=

1

1

1


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At high frequency, the capacitive reactance isgetting smaller.

Gain is getting smaller

The cutoff frequency is given as

or 11==

0

s = j, the filter gain is

ffff

( )

+

==2

1 1

1

ff

f

in

out

RCR

R

V

VG


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A AC A AC

R1

RF

C2( ) ( )

22 2

11CR

fCR F

cF

c ==

The -3 dB cut-off frequency:

1RR

G Fo =

The overall filter Gain:

0

Design a low pass filter withcut-off frequency of 5kHz,

and DC gain of 10:

Two equations, three unknowns

-

+

Vin

Vout

A

B

Transfer Function:Transfer Function:

))))))))RR ++++++++outout

22(((((((( )))))))) ((((((((

++++++++========

================

1122

11 //11

11

11

11

cc

ff

ffff

ff

inin RR

RR

RRCC

RR

VV

VVGG


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: :

A A

Rf

I1 I2

0


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= = 11,, = =

11 11 22 11,,

== ++

11

11

1111111111

11

1111II

ssCC

ssRRCCIIRRII

ssCCVVinin

++++++++========++++++++========

0

22 , ,

11 ++ 22 = 0= 0


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::

out

f

in VRVsRC

sC 1

1 11

1

=+

inf

out VsRC

sCRV11

1

1+=

0

CC

//

112

1

CRfc

=

11

1CR

c =


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jRC

jCR

V

V f

in

out

11

1

1+=

CR

0

,

( )2111 RC+=

1R

RG

f

o =


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A AC A AC

-R1

RFC1

The -3 dB cut-off frequency:

The filter Gain:

( ) ( )1111 2

11CR

fCR cc

==

1RR

G F=

0 0

Vout+

Vin

A

B

wo equa ons, ree un nowns

Select one component based on

other conditions, and

determine the values of the

other two components.Transfer Function:Transfer Function:

(((((((( )))))))) (((((((( ))))))))22221111

11

//11

//

11 cc

ccff

inin

outout

RRCC

CCRR

VV

VV

++++++++========

++++++++========


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: :

BA A BA A

CfI2

0

f

1


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Current I1:

Current I2:

in

VsRC

sCI

11

1

1

1+=

out

f

ff

VR

sCR

I

+

=

12

0

I1 = -I2 outf

ff

in VR

sCRVsRC

sC +=

+

1

111

1

( )( ) inff

f

out VsRCsCR

sCRV

11

1

11 ++=


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The upper and lower cut-off frequency for

the bandpass filter:

and Gain:

112

1

CR

fcl

=

ffch CR

f2

1=

V

0

For cl


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0


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Lecture06 New Signal Conditioning - Filterx