352_32115_EC434_2012_4__3_1_Chapter_03

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Chapter 3: Passive Filters and Transfer Functions


In this chapter we will look at the behavior of certain circuits by examining their transferfunctions !ne important class of circuits is filters " good example is trying to tune in a radiostation If you want to listen to an F# station broadcasting at $%3#h&' you want to process thesignal coming from your antenna and allow any signal very close to $%3#(& to pass on while

blocking' or attenuating' all other signals Circuits that allow voltages at some fre)uencies topass while attenuating those at other fre)uencies are called filters

A: FIRST ORDER LOW PASS FILTERS

The simplest filters' and crudest' are first order high pass filters and first order low passfilters These can be made of a resistor and acapacitor or made of a resistor and an inductor*+esistors and capacitors are usually used at lowfre)uencies,-e looked at one in the previouschapter and it is shown in fig .3 which I/veduplicated at the right This is often called a low

pass +C filter -e are primarily interested in thecomplex transfer function' (*0, and particularly

E*t, C

+

Input !utput

in its polar representation' 1(*0,1 exp*0, The magnitude of (*0,' or its amplitude' is *frome)n .2,

( )( ) ( ) .. 456

6

+C56

605(

+=

+= 36

where I/ve used 7 +C The phase of (*0,' ' is 0ust7 8 tan96*, 3.

-e usually plot these as a function of for a given ,but first let/s look at the general behavior

of these functions *ote that all circuits with the same product of + and C yield the same curveor behavior,

Case 6: If ;; 6' the magnitude of (*0, < 6 and < In this case the output isapproximately the same as the input' ie the same amplitude and little phase shift

Case .: If == 6' the magnitude of (*0, < 6>;; 6 and the phase shift < 8 >. or8 %o

Case 3: If 7 6' the magnitude of (*0, 7 6>?. 7 @@ and the phase shift = 8/4or8 2Ao

This suggests that the circuit attenuates and shifts the phase of signals whose fre)uency = 6/6/is usually called the cutoff fre)uency' cfor the circuit' since that is where the behavior

changes' even though the change is gradual for this type of circuit *The cutoff fre)uency isusually definedas the fre)uency where the output amplitude is B 3d from the input amplitude,!ne could guess this result by noting that the impedance of the capacitor decreases as thefre)uency increases "t low fre)uency' the capacitors impedance is large and almost all thevoltage drop is across the capacitor "t 7 6 the magnitude of the capacitors impedance e)ualsthe resistors' and the voltage is split between them *ote that each can have @@D of thevoltage drop instead of AD because the maximum voltages drops are out of phase with eachother by >2 at that point' ie they do not occur at the SAMETIME "t the same time' the sum of

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the two drops must e)ual the input voltage, "t high fre)uencies' the impedance of the capacitoris small and almost all the voltage drop is across the resistor

Eince we are often interested in the behavior over a wide range of fre)uencies andattenuations' we use a logarithmic scale -e plot .og6*1(*0,1, versus the og6*>c, otethat ( is dimensionless' so it is ok to take the log of ( *I/ve

dropped the G0H "lso' if I use og or log that will be base 6 andln will be the natural logarithm' or base e, -hen we use .og6*1(*0,1,' this is 1(*0,1 in d or decibels The table at the rightshows several useful relations for d ote that if 1(1 ; 6 it/s log isnegative The ( at the right is not one for any particular circuitI/ve 0ust made up some values for an example

1(*0,1 1(*0,1 in d

6 .d6 d

@@ 83d

A 8Jd

6 8 .d

6 8 2d

ote that 1(1 is the ratio of the magnitudes' or amplitudes' of the output voltage to the inputvoltage Thus the d scale here uses the input amplitude as a reference and measures the outputamplitude as a fraction of the input/s amplitude *-e also use a d scale for other )uantities likesound intensity' where the reference sound intensity is 696.->m., For the low pass filterexample' if 7 6' the magnitude of ( 7 8 3d' or the output is 3d belowthe input level or@@ times the input level The two plots below show 1(1 as a function of f 7 >.with the 1(1in d and f as og6*f>fc, *ote that f>fc7 >c, The second plot shows the phase vsog6*f>fc, -e usually plot these vs og6*f>fc, because they scale as *f>fc, Every first orderlow pass filter will ave te sa!e sape wen plotted tis way"

|H| vs Log(f/fc)

-80.0

-60.0

-40.0

-20.0

0.0

-3.00 -2.00 -1.00 0.00 1.00 2.00 3.00 4.00

Log(f/fc)

|H

|dB

Fig 36

Phase vs Log(f/fc)

-2.00

-1.50

-1.00

-0.50

0.00

-3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0

Log(f/fc)

Phase

(radians)

Fig 3.

The filter is called a first order filter because in the region where there is attenuation the outputamplitude is approximately proportional to *6>)to the first power " second order low passfilter would be approximately proportional to *6>)2in that region *(ow would a 3rdorder lowpass filter depend on *6>) when ==cK,

Try to answer the following 3 )uestions from the first grapha, If f c7 6(&' where will the output amplitude be 6>6 the input amplitudeKb, If fc7 6(&' where will the output amplitude be 6>6 of the input amplitudeKc, If I want the output amplitude at J (& to be 6>6 of the input amplitude' what should

the cutoff fre)uency fcbeK

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If ==6' or == c' (*0, < 6>*0,' or the complex Loute)uals the integral of Eintimes6> *Mou should verify this statement, Tis #ir#$it appro%i!ately inte&rates te inp$tvolta&e if ''("

): FIRST ORDER *I+* PASS FILTERS

" first order high pass filter will be similar to the low passfilter' but the capacitor and resistor will be interchanged' ie theoutput voltage will be the voltage across the resistor The circuit isshown at the right "gain the input is a sinusoidal voltage and wewill use its complex representation This circuit is 0ust a dividercircuit' but with the impedances NC and N+reversed in positionfrom the low pass example The complex output voltage will begiven by

L in7 Loe0 t

C

+ Lout

fig 33 (igh Pass Filter

+C056

+C05L

C056+

+L

NN

NLL inin

C+

+inout

+=

+=

+=

33

The transfer function (*0, is 0ust the coefficient of Linor' using +C 7 '

( )( )

( )

+=

+= 456tan

.

O0exp

.456

45

4056

40505( 32

The last term has been converted into polar coordinates *Mou should verify that this is thecorrect form, The magnitude and phase of ( are given by

( )( )

( )45tan.

Ophaseand

456

4505( 6

.

==+

= 3A

"gain we will consider three cases:Case 6: If ;; 6' 1(*0,1 < and < >. or %o The output is smaller than the input'

and the phase shift approaches %oCase .: If == 6' 1(*0,1 < 6 and the phase shift < 0Case 3: If 7 6' 1(*0,1 7 6>?. 7 @@ and the phase shift = /4or 2Ao

This is called a high pass filter because fre)uencies above c7 6>tend to be passed with littleattenuation or phase shift while those below ctend to be attenuated otice the phase shift hereis positive while for the low pass it was negative The magnitude of ( *in d, and the phase of( *in radians, are plotted below versus og*f>fc, It is called a 6storder filter because 1(1 goes

..

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|H(jw )| vs log(f/fc)

-80.0

-60.0

-40.0

-20.0

0.0

-4.00 -2.00 0.00 2.00 4.00

Log(f/fc)

|H

|in

dB

Fig 32

Phase of H vs Log(f/fc)

0.00

0.50

1.00

1.50

2.00

-4.00 -2.00 0.00 2.00 4.00

Log(f/fc)

Phase

(radians)

Fig 3A

like *f>fc, to the first power when f ;; fc *Can you guess how a .ndorder high pass filter/stransfer function would behave when f ;; fcK, All (st order i& pass filters ave te sa!esape wen plotted tis way" The transition from the region of little attenuation' f == fc' to theregion of strong attenuation is not very sharp with this type of filter' the transition region being

roughly from *f>fc, 7 6>3 to *f>fc, 7 3 That is the 1((P1 7 %A at *f>fc, 7 3' within AD of the highfre)uency limit and 1((P1 7 36J at *f>fc, 7 6>3' where the approximate expression yields 333'again a deviation of about AD For many applications we can approximate 1((P1 7 6 if *f>fc, = 3and 1((P1 7 *f>fc, if *f>fc, ; 6>3

The first order high pass filter blocks the QC or constant part of a signal' and only passesthe part that depends on time For example' if the input is AL R "cos*t, and ==c' the outputwill be 0ust "cos*t, The AL will be blocked and disappear The inputs to some devices' egoscilloscopes' have a choice or A, or D, #o$plin& QC coupling passes all parts of a signal"C coupling puts the input through a high pass filter' which blocks the lower fre)uencies The+C time constant for an oscilloscope is usually around 6s' producing a cutoff fre)uency ofabout 6J (& -hen a signal goes through a high pass filter' it is shifted so that for times == '

the average of the output voltage is volts"t fre)uencies below the cutoff fre)uency' this circuit approximately differentiates the

input and multiplies it by or 6>c' ie (*0, < or /c *Mou should verify this,Finally' it is often helpful to write the transfer functions of these filters in terms of the

cutoff fre)uency c7 .fc76> Then the first order low pass filter has a transfer function givenby

=

+

= c

6

.

c

FPf

ftanand'

f

f6

6(

LP

3J

and the first order high pass filter has a transfer function given by

=

+

= c

6

.

c

c(P

f

ftan

.

Oand'

f

f6

ff

( HP3@

.3

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If one defines % - f.f#- .#/ tey ta0e on a so!ewat neater for!"

( )xtanand'x6

6( 6

.FP

=+

= LP 3$

( )xtan.

Oand'

x6

x( 6

.

(P=

+= HP

3%A $sef$l appro%i!ation is tat for % '' (/

6(andx

6( (PFP

Si!ilarly/ for % 11(

xHandHHPLP

6

,: SE,O2D ORDER R,3LI4E FILTERS

The simplest second order filteris two first order filters that arecascaded' ie one follows the other inthe circuit (owever' for it to behavenicely' the stages should be non9interacting This will beapproximately the case when theThevenin impedance of the first stage*its output impedance, is much largerthan the series impedance of the two

E*t, C.7C6>63

+.7 63+

6

Input

+6

C6 !utput

Fig 3J

elements of the second stage ote that in the circuit in fig 3J' a second order low pass filter'

I/ve chosen +.7 6+6and C.7 C6>6 which ensures that the series impedance of the secondstage is at least 6 times the output impedance of the first stage That is usually sufficient Ifthat is the case' the transfer function for the entire circuit is' to a good approximation' the productof the two separate transfer functions' or (total*0, < (6*0,S(.*0, This makes it much easier toanaly&e the circuit In this example both circuits accidentallyU have the same +C time constant *They don/t have to have the same ' but I/ll leave some of those cases for you to figure out,"s a result'

( )( )

( )( )45.0tanexp456

6

4056

605( 6

.

.

total

+=

+

= 36

where the last expression is in polar form "gain we will look at three cases

Case 6: If ;; 6' the magnitude of (*0, < 6 and < In this case the output isapproximately the same as the input' ie the same amplitude and no phase shift

Case .: If == 6' 1(*0,1 < *6>)2;; 6 and the phase shift < 8 or 8 6$oCase 3: If 7 6' the magnitude of (*0, 7 6>. and the phase shift = 8/2or 8 %o

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The reason it is called a second orderfilter is that the amplitude falls off like*6>)to the second power if == 6Mou should also note that the amplitude of(totalis the product of the two individual

amplitudes and the phase is the sum of thetwo individual phases "t highfre)uencies' the transfer function )2. In this region a doubling of thefre)uency results in a reduction of 2 in theamplitude' or the amplitude changes by86.d for every doubling of thefre)uency *In audio systems they usually

2nd Order Low Pass Filter

-100.00

-80.00

-60.00

-40.00

-20.00

0.00

-4.00 -3.00 -2.00 -1.00 0.00 1.00 2.00 3.00

Log(f/fc)

|H|dB

Fig 3@

say 86.d per octave,

Cascading two non9interacting filters means that the net transfer function is the

product of the two individual transfer functions -hen I use the d or logarithmic scale'. og*1(net1, 7 . og* 1(61S1(.1, 7 . og*1(61, R . og*1(.1,'

or in d' 1(net1 7 1(61 R 1(.1 In te d) s#ale tey add This is a useful shortcut whenplotting the 1(net1 on a d scale It is interesting to note that the phase shifts also add

This is not a very sharp filter' ie the transition from 1(1 < 6 to the region above 7 6 where itis decreasing rapidly occurs over a range of fre)uencies For instance if you want to pass asignal of fre)uency fsand not attenuate it by more than 6D' ie 1(1 7 %' you need to set fcofthe individual filters such that

( ).

c>fsf6

63%3

+

= 366

or *fs>fc, 7 6>3 This means that fc7 3fs' the cutoff fre)uency of each filter stage has to be 3 timesthe signal fre)uency " noise signal around fcwould only be reduced in amplitude to V itsoriginal amplitude The noise signal would have to be at 3fc7 %fsto be reduced in amplitude bya factor of 6 *8.d,

There are three different versions of cascading two first order +C type filters to get asecond order filter The first is cascading two low pass filters The second is cascading two highpass filters and the third is the cascade a low pass and a high pass filter to produce a band passfilter The band pass filter would like the one below' where I/ve again made the two cutofffre)uencies the same I/ve also show a plot of 1(1 ote that 1(1 has a maximum of B Jd or 6>.*!ften you want the low pass cutoff fre)uency higher than the high pass cutoff fre)uency, I willleave these for you to investigate

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E*t,

C.7C6>63

+.7 63+6Input

+6

C6 !utput

Fig 3$

Band Pass Filter

-80.0

-60.0

-40.0

-20.0

0.0

-4.00 -2.00 0.00 2.00 4.00

Log(f/fc)

Fig 3%

D: SE,O2D ORDER L,R FILTERS

" betterU second order filter can be made from an inductor' capacitor and a resistor inseries Mou can get all three behaviors by looking at thevoltage drop across different elements If your output isthe voltage across the capacitor' you will have a second

order low pass filter If the output is across the inductor'you will have a second order high pass filter If theoutput is across the resistor' you will have a band passfilter Consider the low pass filter version It is shownat the right This looks 0ust like another divider circuitIf the input voltage is Loexp*0t,' the output voltage

E*t, C

+

!utput

F

Fig 36

Loutwill be

( )t05expoLNNN

NoutL

+FC

C

++= 36.

and the transfer function (P*0, will be given by Lout7 ( Lin Therefore

( )+C055FC6

6

+F05C05

6

C056

NNN

N05(

.+FC

CFP

+=

++=

++= 363

ormally one sets C 7 *6>o,.since C has units of 6 over fre)uency s)uared and this ois theangular fre)uency of free oscillations for an C circuit' sometimes called the natural fre)uencyor undamped resonant fre)uency for the circuit !ne can write

( )+C50555

5

+C055

56

605(

.o

..o

.o

.

o

FP +=

+

=362

The magnitude of ( is 0ust

( )( ) ( ).....o

.

o

...

.

.

o

.

FP

F+555

5

C+55

56

605(

+=

+

=36A

The phase of (' ' is

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=

..o

6FP

55

F+5

tan 36J

"gain we will consider three cases:Case 6: If ;; ' the magnitude of (P*0, < 1and P< 0 In this case the output is

e)ual to the inputCase .: If == o' the magnitude of (P*0, < *o>,.and the phase shift P< 8or

86$o The output/s magnitude is much less than the input/sCase 3: If 7 o' the phase shift P= 8/2or 8%oand the magnitude of (*0, is

( )C

F

+

6

+C5

605(

ooFP == 36@

where I/ve put 7 o7 6>*C,6>.in e)n 366 The interesting thing is that the output can belarger than the input for certain choices of +' and C For instance if 7 6m(' C7 6nF' then*>C,6>.7 6 If + ; 6k' the output is larger than the input at this fre)uency There are twoterms associated with this circuit' the W of the circuit and the damping factor' d #ost peopledefine d 7 6>*.W,' but some define d as d 7 6>W *It is d that changes from author to author' notthe W, I5ll $se d - (.6789

.d

6

C

F

+

6W == 36$

W and d are dimensionless "t 7 o' 1(1 7 W If d 7 6>?. 7 @@' the output amplitude isnever larger than the input/s If d ; @@' then for some fre)uency' the output amplitude will belarger than the input/s If d 76' the circuit is said to be critically damped' and if d =6' the circuitis said to be overdamped These terms are also used to describe mechanical oscillations wherethe symbol for the damping factor is usually If one lets x 7 *>o,' one can rewrite 1(1 as

( )

( ) .... 26

6

dxx

jHLP

+=

36%

Xsually ; d ; 6' so for x ;;6 and x==6 the magnitude of ( is not sensitive to d It is in theregion around x76 that the value of d is important' typically for *6>3, ; x ; 3 Mou should alsonote that the maximum value of 1(1 is not necessarily at x 7 6' although it goes to x 7 6 as d getssmaller The maximum of ( occurs when the denominator in 36A is a minimum If you take thedenominators derivative and set it 7 ' you/ll find that the maximum for 1(1 occurs atx.7 *6 8 .d., if d ; 6>?. If d = 6>?.' the maximum is at x 7' or 7 Yote that I cansubstitute f>fo7 for >oin 36A above This is because the ./s cancel It is often convenient touse x7f>foinstead of x7>owhen doing calculations, I have plotted 1(1 in the region 6 ; x ;A for several d/s below Te d-( #ase is te sa!e as two #as#aded R, filters wit te sa!e

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|H| for various d's vs log (f/fo)

-30

-25

-20

-15

-10

-5

0

5

-1 -0.5 0 0.5 1

log(f/fo)

|H|dB

d=0.4

d = 0.6

d=0.7

d = 1

Fig 366The d/s range from 2 to 6 !nce you get away from the x 7 6 region' or log*f>fo, 7 ' they areall similar For d less than @' the fall off is sharper but you get some overshoot' ie 1(1 =6 nearx 7 6 I have not shown the phase shift' or phase of (' but you can get a rough idea from e)n36J

The i& passversion looks at the voltage across the inductor I recommend you try towrite the magnitude of the transfer function as a function of x 7 >o7 f>foand d Mou simplyreplace NCin the n$!eratorof first expression in e)n 3% by Nand do the algebra The

magnitude of that transfer function should look like the mirror image of the one above' iereflected left to right through log*f>fo, 7 For this case 1((P1 and (Pare

( )( )

=+

=.(P

....

.

(Px6

.xd6tanOand

d2xx6

x05( 3.

The band passoutput would be the voltage drop across the resistor (ere you replace NCinthe n$!eratorof first expression in e)n 3% by N+and do the algebra Mou should get 1(P1 is

( )( )

=

+=

.

6P

....P

x6

.xdtan

.

Oand

d2xx6

.xd05( 3.6

The maximum of 1(P1 occurs at x 76' or f 7 foand it doesn/t depend on d 1(P1max7 6 "t that

fre)uency' the phase shift is &ero' and the output is the same as the input (owever' thesharpness or narrowness of the band pass does depend on the damping factor d The width of theband pass is defined in terms of the two fre)uencies where 1(P1 7 6>?. If the lower fre)uencyis f6and the upper is f.' the width is f.8 f67f The narrowness of the band pass filter is givenby f>foand the W of the circuit is related to the narrowness by

of

Zf

W

6= 3..

.$

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I/ve show a plot of 1(P1 for a couple of different d/s ote that as d gets smaller' the range offre)uencies that pass with little attenuation becomes smaller and smaller -e say the filter issharper or narrower "ll of the ones shown are narrower than the two9stage +C band pass filter'whose d 7 6

|H| for !and "ass L#$ filter

-50

-40

-30

-20

-10

0

-1 -0.5 0 0.5 1

Log(f/fo)

|H|dB

d=0.707

d = 0.3

d = 0.1

d = 0.03

Fig 36.!n old radios' the way you selected a station was by turning a dial that changed the

capacitance of a variable capacitor that was part of a band pass circuit This would change fotomatch the fre)uency of the radio station you wanted to listen to Then the signal of that stationwas passed and the signals from other stations were attenuated In that case you needed a verynarrow band pass filter' ie a large W or a small d

!ne last note about C+ filters The voltage across the capacitor and inductor are 6$ooutof phase' ie they have the opposite phase y arranging the circuit to have the inductor and

capacitor together and measure the voltage across both for the output' one gets a not#filterThis attenuates the input close to fo' but does not attenuate it far away from fo' 0ust the opposite ofa band pass filter Mou use notch filters to get rid of interference or noise that occurs at a specificfre)uency " common example is a J(& notch filter to remove noise picked up from our J(&power mains Yote that the series impedance of an ideal inductor and capacitor' NR NC' goesto &ero at o7 6>*C,6>.[

I should warn you that these analyses assume ideal behavior for the resistor' the capacitorand the inductor arger inductors' = 6( are seldom ideal They often have A9.ofresistance and some capacitance between the windings If you take a 6m( inductor and look atthe magnitude of its impedance' you will probably find that at low fre)uencies the < 69. ofresistance it is likely to have will mean that N does not \ Eimilarly at high fre)uencies theinterwinding capacitance means that its impedance will not continue to increase as the fre)uencyincreases' instead at some point it will decrease Capacitors also exhibit some inductance' butyou can usually minimi&e its effect at fre)uencies below 6#(& *This is done by choosing theright type of capacitor ]lectrolytic capacitors often have more inductance, ecause it is harderto get good inductors for filters below 6k(&' people often use active filters in the lowerfre)uency ranges "ctive filters use operational amplifiers' or op amps for short' resistors andcapacitors to simulateU C+ circuits -e will look at them when we discuss op amps

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I have not discussed parallel C+ circuits there 0ust isn/t time to discuss everything Moumight want to try them on your own

E: FRE8E2,; DOMAI2 A2D TIME DOMAI2

-hen we describe the response of a circuit to a sinusoidal input' we refer to the description

as the fre

352_32115_EC434_2012_4__3_1_Chapter_03

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