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# Operational Amp Design ECE 4435

GPA 3.64

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This 0 page Class Notes was uploaded by Cassidy Effertz on Monday November 2, 2015. The Class Notes belongs to ECE 4435 at Georgia Institute of Technology - Main Campus taught by William Leach in Fall. Since its upload, it has received 10 views. For similar materials see /class/233859/ece-4435-georgia-institute-of-technology-main-campus in ELECTRICAL AND COMPUTER ENGINEERING at Georgia Institute of Technology - Main Campus.

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Date Created: 11/02/15

A Voltage Controlled Ampli er The object of this experiment is to assemble and test a voltage controlled ampli er VCA that might be used as a variable gain stage in a compressor a limiter or guitar effects box The variable gain element in the VGA is a JFET operated in its linear or triode region The basic circuit is shown in Figure 1 The drain source voltage at the JFET is R2 1 U U DS A I R1 R2 For op amp circuits powered by i15V power supplies a practical upper limit on the peak input voltage is 12 V For linear operation of the JFET its drain source voltage should not exceed VTOQ where VTO is its threshold voltage Thus a design speci cation for the circuit is R V 12 2 S 1 T01 R1 R2 2 The circuit is to be designed so that it has unity gain when the JFET is an open circuit ie when it is pinched off This requires RlRzRa 1n addition the circuit is to be designed so that the maximum attenuation when the JFET has its minimum resistance is 26 dB a linear attenuation of 120 This requires Ra 7 dsmjn 1132 i 1 F2 R1 7 dsmjn 1132 20 where Td5rmn is the minimum drain source resistance of the JFET given by 1 7 d5mjn 26 VTO and 6 is its transconductance parameter For your JFET use the design equations above to calculate Bl R2 and R3 Assemble the circuit using a TL071 TL081 or LF351 op amp Use RA RB 1015 Apply a sine wave to the input and a negative dc voltage to 110 Verify that that gain of the circuit can be varied between unity and 120 by varying 110 over the range VTO S 110 S 0 Do not apply a dc voltage outside this range or the JFET may be damaged Remember that VTO is negative Once the circuit is operational apply a triangle wave to 111 Connect the SC input to the oscilloscope to 111 and the y input to 110 You should observe an approximately straight line on the oscilloscope whose slope can be varied between 1 and 120 by varying 110 Note any curvature in this line This curvature adds undesired distortion to the signal The next step is to linearize the circuit by feeding the voltage 39UDs2 UA2 back into the JFET gate The circuit is shown in Figure 2 When the output voltage of A1 is negative the gate source voltage of the JFET is given by Ra Re Re 1105 UA UC Rg Rs R4 1 Figure 1 Basic VCA circuit The constraints on the elements are RgReil 323572 Ed R4 With R5 chosen to be some convenient value eg R5 1019 use these equations to calculate values for R4 and Re Use R7 10 k9 Note that there is no current through R7 when 1105 lt 0 so that this resistor has no effect on 1105 Figure 2 Circuit With the linearization voltage added to the JFET gate voltage Assemble the circuit in Figure 2 using the values calculated above Note that 110 must be positive in this circuit because A2 inverts the control voltage Perform the same tests as for the circuit in Figure 2 In the following weeks the VGA Will be used as the control element in a compressor circuit You should neaten up the circuit so that it occupies no more room 2 Dr Leach s Filter Potpourri Transfer Functions For sinusoidal time variations the input voltage to a lter can be written 111t Re Viejml where V is the phasor input voltage ie it has an amplitude and a phase and aim cos wt j sinwt A sinusoidal signal is the only signal in nature that is preserved by a linear system Therefore if the lter is linear its output voltage can be written 110t Re new where V is the phasor output voltage The ratio of V0 to V is called the voltage gain transfer function It is a function of frequency Let us denote V T 0 W V We can write T as follows T jw A w 61W where A w and p w are real functions of w A w is called the gain function and p w is called the phase function As an example consider the lter input voltage 111t V1 cos wt 0 Re Vlejgejml The corresponding phasor input and output voltages are W V16 V0 Vle A w ej w It follows that the lter voltage is 110t Re VlejeA w ejfwejm A w V1 cos wt 0 p This equation illustrates why A w is called the gain function and p w is called the phase function The complex frequency 8 is usually used in place of jw in writing transfer functions In general most transfer functions can be written in the form where K is a gain constant and N s and D s are polynomials in 8 containing no reciprocal powers of s The roots of D s are called the poles of the transfer function The roots of N s are called the zeros As an example consider the function TS42 s4l 4 s4l s65561 52ls3l The function has a zero at s 74 and poles at s 72 and s 73 Note that Too 0 Because of this some texts would say that T s has a zero at s 00 However this is not correct because Noo 0 Note that the constant terms in the numerator and denominator of T s are both unity This is one of two standard ways for writing transfer functions Another way is to make the coef cient of the highest powers of s unity In this case the above transfer function would be written 7 6 s 4 7 s 4 T 52556 T s2s3 Because it is usually easier to construct Bode plots with the rst form that form is used here Because the complex frequency 8 is the operator which represents d dt in the differential equation for a system the transfer function contains the differential equation Let the transfer function above represent the voltage gain of a circuit ie T s VaVi where V and V respectively are the phasor output and input voltages It follows that 51v4 1m T 8 When the operator 8 is replaced with d dt the following differential equation is obtained 1d2UO Edvo dUI 4 6dt2T6dtTUO dtTUI where 110 and 111 respectively are the time domain output and input voltages Note that the poles are related to the derivatives of the output and the zeros are related to the derivatives of the input How to Do Bode Plots A Bode plot is a plot of either the magnitude or the phase of a transfer function T as a function of w The magnitude plot is the more common plot because it represents the gain of the system Therefore the term Bode plot usually refers to the magnitude plot The rules for making Bode plots can be derived from the following transfer function T s K in where n is a positive integer For n as the exponent the function has n zeros at s 0 For n it has n poles at s 0 With 5 jw it follows that Tjw Kjiquot ww0iquot lT jw K ww0iquot and 1Tjw in X 90 lfw is increased by a factor of 10 lTjwl changes by a factor of 10iquot Thus a plot of lT versus w on log 7 log scales has a slope of log 10iquot in decadesdecade There are 20 st in a decade so the slope can also be expressed as i20n dBdecade As a rst example consider the low pass transfer function K T S 1 sw1 This function has a pole at s 7M1 and no zeros For 8 jw and ww1 ltlt lwe have Tjw 2 K lTjwl 2 K and 1Tjw 2 0 X 90 0 For ww1 gtgt 1 Tjw 2 Kjww1 1lTjwl2 Kww1 1 and T jw 2 71X 90 790 On log flog scales 2 the magnitude plot for the low frequency approximation has a slope of 0 while that for the high frequency approximation has a slope of 1 The low and high frequency approximations intersect when K Kw1w or when w wl For w w1Tjw K1j K and T arctan1 45 Note that this is the average value of the phase on the two adjoining asymptotes The Bode magnitude and phase plots are shown in Fig 1 Note that the slope of the asymptotic magnitude plot rotates by 1 at w 601 Because M1 is the magnitude of the pole frequency we say that the slope rotates by 1 at a pole A straight line segment that is tangent to the phase plot at w M1 would intersect the 0 level at Ml481 and the 90 level at 481w1 lTjwl Alma Km 0 km Cd 45 w 00 90 a1 b Figure 1 Bode plots a Magnitude b Phase As a second example consider the transfer function TsKlt1wil This function has a zero at s w1 For 8 jw and ww1 ltlt 1we have TOM 2 K Tjw 2 K and ZTjw 2 0 X 90 0 For ww1 gtgt 1 Tjw 2 Kjww11 Tjw 2 Kww1 and T 2 1 gtlt90 90 On log log scales the magnitude plot for the low frequency approximation has a slope of 0 while that for the high frequency approximation has a slope of 1 The low and high frequency approximations intersect when K Kww1 or when w 601 For M M1 K and ZTjw arctan1 45 Note that this is the average of the phase on the two adjoining asymptotes The Bode magnitude and phase plots are shown in Fig 2 Note that the slope of the asymptotic magnitude plot rotates by 1 at w 601 Because M1 is the magnitude of the zero frequency we say that the slope rotates by 1 at a zero A straight line segment that is tangent to the phase plot at w M1 would intersect the 0 level at Ml481 and the 90 level at 481w1 iTJ39w 4 I T0760 K 1 32 I 0 b Figure 2 Bode plots a Magnitude b Phase From the above examples we can summarize the basic rules for making Bode plots as follows 1 In any frequency band where a transfer function can be approximated by K jw w0iquot the slope of the Bode magnitude plot is in decdec The phase is in X 90 E0 Poles cause the asymptotic slope of the magnitude plot to rotate clockwise by one unit at the pole frequency 9quot Zeros cause the asymptotic slope of the magnitude plot to rotate counter clockwise by one unit at the zero frequency As a third example consider the transfer function T s S 1 sw1 1 This function has a pole at s 7M1 and a zero at s 0 For 8 jw and ww1 ltlt 1we have 1T jw 2 Kww1 and T jw 2 90 For ww1 gtgt 1 1T jw 2 K and T jw 2 0 On log 7 log scales the magnitude plot for the low frequency approximation has a slope of 1 while that for the high frequency approximation has a slope of 0 The low and high frequency approximations intersect when Kww1 K or when w M1 For M M1 1T and T 90 7 arctan1 45 The Bode magnitude and phase plots are shown in Fig 3 Note that the slope of the asymptotic magnitude plot rotates by 71 at the pole The transfer function is called a high pass function because its gain approaches zero at low frequencies lTjwl Alma K 90 1 45 A a w 1 1 a b Figure 3 Bode plots 3 Magnitude b Phase A shelving transfer function has the form 1 swg T S 1 sw1 The function has a pole at s 7M1 and a zero at s i602 We will consider the low pass shelving function for which M1 lt M2 For 8 jw and ww1 ltlt 1 we have 2 K and T 2 0 As w is increased the pole causes the asymptotic slope to rotate from 0 to 71 at M1 The zero causes the asymptotic slope to rotate from 71 back to 0 at U2 For UUg gtgt 1 1T 2 Kw1w2 The Bode magnitude plot is shown in Fig 4a 1f the transfer function did not have the zero the actual gain at M1 would be The zero causes the gain to be between and K Similarly the pole causes the actual gain at U2 to be between Kw1w2 and KM1w2 The actual plot intersects the asymptotic plot at the geometric mean frequency m The phase plot has a slope that approaches 0 at very low frequencies and at very high frequencies At the geometric mean frequency w the phase is approaching 790 1f the function only had a pole the phase at M1 would be 745 approaching 790 at higher frequencies However the zero causes the high frequency phase to approach 0 Thus the phase at M1 is more positive than 745 At the geometric mean frequency m the slope of the phase function is zero The Bode phase plot is shown in Fig 4b ITjwl Alma 0 CL 1 2 K o 0 Km 1 45 90 02 Cd 0 b Figure 4 Bode plots a Magnitude b Phase Classes of Filter Functions The lters considered in this experiment can be divided into four classes These are low pass high pass band pass and band reject Although it is impossible to realize an ideal lter the characteristics of the four classes of lters are simplest to describe for ideal lters An ideal low pass lter has a cutoff frequency below which the gain is independent of frequency and above which the gain is zero Fig 5a illustrates the magnitude response of an ideal low pass lter having a gain K The responses of two physically realizable lters are also shown An ideal high pass lter has a cutoff frequency above which the gain is constant and below which the gain is zero The magnitude responses of an ideal high pass lter and two physically realizable lters are illustrated in Fig 5b An ideal band pass lter has two cutoff frequencies between which the gain is constant and zero elsewhere The magnitude responses of an ideal band pass lter and two physically realizable lters are illustrated in Fig 5c An ideal band reject lter has two cutoff frequencies between which the gain is zero and constant elsewhere The magnitude responses of an ideal band reject lter and two physically realizable lters are illustrated in Fig 5d G Gain Tn deal 1 deal 0 co 0 W we we a b Gain Gain dew Ideal K K O w O W mcl c2 wc1 6062 C d Figure 5 a Low pass b High pass c Band pass d Band reject Low pass lters are used in applications where it is desired to remove the high frequency content of a signal For example aliasing distortion can occur if a signal is applied to the input of an analog to digital converter that has a frequency higher than one half the sampling frequency of the converter A low pass lter might be used to limit the bandwidth of the signal Similarly a high pass lter is used in applications where it is desired to remove the low frequency content of a signal For example the tweeter driver in a loudspeaker can be 5 damaged by low frequencies signals To prevent this a high pass lter called a crossover network must be connected to the tweeter A band pass lter is used in applications where it is desired to pass only the frequencies in a band For example to detect a low level tone that is buried in noise a band pass lter might be used to pass the tone and reject the noise A band reject lter is used in applications where it is desired to reject a particular frequency or band of frequencies For example a 60 HZ hum induced in the ampli er of a public address system might be ltered out with a band pass lter Frequency Transformations Filter transfer functions are normally derived as low pass functions Frequency transforma tions are then used to transform the low pass functions into either high pass band pass or band reject transfer functions For a low pass lter let the normalized frequency p be de ned by S p MC where wc is a normalization frequency For the case of low pass and high pass lters wc is the called the cutoff frequency of the lter Depending on the type of lter it is not necessarily the 73 dB cutoff frequency To distinguish between the two in the following M3 is used to denote the 73 dB cutoff frequency In the case of a band pass lter wc is the center frequency of the band pass response In the case of a band reject lter wc is the center frequency of the band reject response The frequency transformations are de ned as follows Low Pass to High Pass p gt m BIH 1 Low Pass to Band Pass p gt p p 1 71 Low Pass to Band Reject p gt B 1 p where the arrow is read is replaced by The parameter B determines the 73 dB bandwidth of the band pass and band reject functions Transformations of FirstOrder Functions As an example consider the rst order low pass lter function 1 TLP S 1 saw where a is a positive constant The function TLP p is given by 1 T LP P 1 pa The high pass band pass and band reject transfer functions are given by 7 1 7 up 711ap71ap 1 aBgtp 1 Bp 111 a 112 aBp1 High Pass THP p Band Pass TBP p 7 1 7 p2 1 71B 1p1p71a 122 1aBp1 Note that the order of the transfer function is doubled for the band pass and band reject transformations Band Reject TBR p Transformations of SecondOrder Functions Consider the second order low pass function 1 09ch 15 SWc 1 where a and b are positive constants The function TLP p is given by 1 17002 15 Iia 1 The high pass band pass and band reject transfer functions are given by High Pass THP p 2 1 2 m2 1047 15 104 1 up 15 up 1 TLP S TLP P 1 B p 111 M12 15 1131J 111 a1 1 Gig322 p4 Hat5133 mgB2 2p2 Ob53 1 Band Pass TBP p 1 341 111 10L2 11 341 111 1 a 1 p2 12 p4 1abBp3 10L2B2 2p2 1abBp 1 Butterworth Filter Transfer Functions The Butterworth Approximation The general form of a nth order low pass lter transfer function having no zeros can be Band Reject TBR p written 1 1 01swc 02 swc2 on swcquot where K is the dc gain constant wc is a normalization frequency and the c are positive constants The magnitude squared function is obtained by setting 8 jw and solving for TLP jw g This function contains only even powers of w and is of the form 1 1 01wwc2 02 tiwar 0 tiwagquot where the C are positive constants which are related to the 0 For the Butterworth lter the constants C are chosen so that TLP jw 2 approximates the magnitude squared function of an ideal low pass lter in the maximally at sense The magnitude squared function for the ideal lter is de ned by TLP S K 1TLPjw12 K2 1THwa Kgforwgwc Oforwgtwc To obtain the maximally at approximation the C are chosen to make as many derivatives as possible of lTLP jw Z equal to zero at w 0 1f the derivative of a function is zero the derivative of the reciprocal of the function is also zero It follows that the maximally at condition can be imposed by solving for the constants C which make as many derivatives as possible of HTLP jw glil equal to zero at w 0 Because the denominator polynomial of lTLP jw Z is an even function all odd order derivatives are already zero at w 0 For the second derivative to be zero we must have C1 0 For the fourth derivative to be zero we must have C2 0 This procedure is repeated to obtain C 0 for all i However we cannot set CT 0 because this would make the approximating function independent of frequency Therefore we set C 0 for all 1 S i S n 7 1 to obtain 1 T u 2 K2 1 LP 7 1 On JMC The rst 2n 7 1 derivatives of this function are zero at w 0 It is standard to choose CT to make lTLP jwc 2 K22 This forces the 73 dB frequency M3 to be equal to the normalization frequency wc This condition requires CT 1 Thus the magnitude squared function of the nth order Butterworth low pass lter becomes 2 1 WWW Figure 6 shows example plots of lTLp for 1 S n S 5 for the Butterworth low pass lter The plots assume that K 1 The horizontal axis is the normalized radian frequency 1 wwc Each function has the value 1 at U 0 the value 05 at U 1 and approaches 0 as U gt 00 As the order n increases the width of the at region in the passband is extended lTLP MHZ K and the lter exhibits a sharper cutoff The response characteristic is called maximally at because there are no ripples in the passband response Figure 6 Plots of the Butterworth magnitude response for 1 S n S 5 To illustrate how the maximally at condition is applied to a speci c lter transfer function consider the third order low pass function 1 1 01swc 02 swc2 03 swc3 The magnitude squared function is given by TLP S K 1 lTLP jwllg K2 2 2 2 4 2 6 1 01 i 202 WM 02 i 20103 wwc 03 WM For this to be maximally at with a 73 dB cutoff frequency of MC we must have of 7 202 0 0 7 20103 0 03 1 Solution for 01 and 02 yields 01 02 2 Thus the Butterworth third order low pass transfer function is K 1 2 3 1 2swc 2 swc swc 1 1 K X 2 1 Swc 1 Swc Swc The maximally at lters are called Butterworth lters after S Butterworth who de scribed the procedure for deriving the transfer functions in his 1930 paper On the Theory of Filter Ampli ers which was published in Wireless Engineer The resulting denomina tor polynomials for TLP s are called Butterworth polynomials The rst six Butterworth polynomials in factored form are TLP 8 b1 x x 1 11290 i 902 14142x 1 b3ac n1 21 11490 902 0765490 1 902 18478x 1 11590 x 1 902 0618090 1 902 16180x 1 11690 902 0517690 1 902 14142x 1 902 19319x 1 EvenOrder Butterworth Filters For an even order Butterworth low pass lter of order n the transfer function can be written in the product form n2 1 TL s 111 SWc2 150 Swc 1 The constants b are given by 1 2 sin 0 where the 0 are given by 23971 0 1 TL n X90 for1 i Example 1 Solve for the transfer functions of the secondiorder Butterworth lowipass and highipass lters Solution For n 2 there is only one second order transfer function The calculations are summarized as follows 1 1 45 1 The low pass transfer function is given by 1 Swc2 V5 Swc 1 The high pass transfer function is obtained by replacing swc with wcs to obtain Ts K 1 T quotKWJv wm1 K WM 09ch tswe 1 OddOrder Butterworth Filters For an odd order Butterworth low pass lter of order n the transfer function can be written in the product form 1 new2 1 TLPltSgtWX The constants bi are given by 1 bi 2sin0i where the 0139 are given by 23971 71 0i1 gtlt90 for1 i n n 2 Example 2 Solve for the transfer functions of the thirdiorder Butterwor th lowipass and highipass lters Solution For n 3 each transfer function contains one rst order polynomial and one second order polynomial The calculations for the second order polynomial are summarized as follows The low pass transfer function is given by 1 1 TLPSWXW The high pass transfer function is obtained by replacing swc with wcs to obtain 1 1 onwl w mm1 wm1emf mo THP S The Cutoff Frequency For the Butterworth low pass and high pass lter functions the cutoff frequency wc is the frequency at which the magnitude squared function is down by a factor of 12 This is the 73 dB frequency M3 For the band pass and band reject lter functions the cutoff frequency wc is the so called center frequency There are two frequencies one on each side of MC at which the magnitude squared function is down by a factor of 12 These are the two 73 dB frequencies Let these be denoted by Md and U62 1f these are speci ed the center frequency MC and the parameter B in the frequency transformations are given by MC w61w62 B w w 62 7 cl Chebyshev Filter Transfer Functions The Chebyshev Approximation In 1899 the Russian mathematician P L Chebyshev also written Tschebyscheff Tchebysh eff or Tchebicheff described a set of polynomials tn which have the feature that they ripple between the peak values of 1 and 71 for 71 S x S 1 His polynomials are widely used in lter approximations for frequencies that span the audio band to the microwave band The rst six Chebyshev polynomials are wc 16x5 7 20903 590 32906 i 48904 1890 71 w 5 t6 t1 ac t2 ac 2x2 7 1 t3 ac 4903 7 390 t 90 8904 7 8x2 1 4 Figure 7 shows the plots of the rst four of these polynomials over the range 72 S x S 2 Figure 7 Plots of Chebyshev polynomials for 1 S n S 4 The Chebyshev approximation to the magnitude squared function of a low pass lter is given by 1 5 0 1 622 wwc where K is the dc gain constant and E is a parameter which determines the amount of ripple in the approximation For M 0 it follows that lTLP jwMZ K2 For n odd ti 0 0 so that the numerator in lTLP jam2 has the value 1 For 0 S M S Ma the denominator ripples lTLPjw12 K2 11 between the values 1 and 1 62 This causes iTLp jam2 to ripple between the values K2 and K21 62 At w wc it has the value K21 62 For w gt wciTLpjw12 a 0 For n even ti 0 1 so that the numerator in iTLP jam2 has the value 1 62 For 0 S M S Ma the denominator ripples between the values 1 and 162 This causes iTLP 12 to ripple between the values K2 and K2 1 62 At w wc it has the value K2 For M gt MO iTLpjw12 gt 0 The major difference between the even and odd order approximations is that the odd order functions ripple down from the zero frequency value whereas the even order functions ripple up Figure 8 shows example plots of iTLP for the 05 dB ripple 4th and 5th order lters The plots assume that K 1 The horizontal axis is the normalized radian frequency 1 wwc The gure shows the 4th order approximation rippling up by 05 dB from its zero frequency value The 5th order approximation ripples down by 05 dB from its zero frequency Compared to the Butterworth lters the Chebyshev lters exhibit a sharper cutoff at the expense of ripple in the passband The more the ripple the sharper the cutoff 05 1 V 15 2 25 Figure 8 Plots of the magnitude responses of the 05 dB ripple 4th and 5th order Chebyshev filters The dB Ripple The dB ripple for a Chebyshev lter is the peak to peak passband ripple in the Bode mag nitude plot of the lter response The dB ripple determines the parameter E as follows dB ripple 10log1 62 This can be solved for E to obtain 6 The Cutoff Frequency Unlike the Butterworth lter the cutoff frequency for the Chebyshev lter is not the fre quency at which the response is down by 3dB The cutoff frequency is the frequency at which the Bode magnitude plot leaves the equal ripple box in the lter passband For the even order low pass lters the gain at the cutoff frequency is equal to the zero frequency gain For the odd order low pass lters the gain at the cutoff frequency is down from the zero frequency gain by an amount equal to the dB ripple For the nth order Chebyshev low pass lter the 73 dB frequency M3 can be related to the cutoff frequency ME by setting K22 The value of M which satis es this equation is the 73 dB frequency 603 1f we let x wwc this leads to the equation 1 tnxi g2tgoo The positive value of x which satis es this equation is the desired solution 1f the 73 dB frequency for the low pass and high pass lters is speci ed the required cutoff frequency wc is given by Low Pass of order n ma E 90 High Pass of order n ma xwg For the band pass and band reject lter functions the cutoff frequency wc is the so called center frequency There are two frequencies one on each side of MC at which the magnitude squared function is down by a factor of 12 These are the two 73 dB frequencies Let these be denoted by Mal and U62 1f these are speci ed the center frequency MC and the parameter B in the frequency transformations are given by 2 w 7 wcleZ 7 Mel 7 5062 7 5062 7 Mel wwc ME MG 90w Band Pass of order 2n wc xwdwcg B C 5062 7 Mel wcsc Band Reject of order 2n wc xwclwcg B 5062 7 Mel where M3 is the 73 dB radian frequency For any 90 there are two values of M3 which satisfy the band pass and band reject relations Denote these by Ma and Mb The geometric mean of these two frequencies must equal wc ie wc In a design from speci cations the 73 dB frequencies might be speci ed for the band pass and band reject lters In this case the required values for the center frequency and the parameter B in the frequency transformation can be solved for Example 3 An 4th7order 05 dB ripple low7pass lter is to have the 73 dB cuto frequency f3 10 kHz Calculate the required cuto frequency fc Solution The parameter E has the value 6 100510 7 1 034931 We must solve for the positive real root of the equation 1 84782 17 20 9 9 1 V0349312 The desired root is x 11063 1t follows that the required cutoff frequency is fa fgx 9039 kHz Example 4 An 4th7order 05 dB ripple high7pass lter is to have the 73 dB cuto frequency f3 10 kHz Calculate the required cuto frequency fc Solution The value of x from the low pass example gives fC xfg 11063 kHz Example 5 An 8th7order 05 dB ripple bandpass lter is to have a center frequency fc 10 kHz and a bandwidth fag 7 fcl QkHz Note that the transfer function is obtained from that for the 4th7order LPF so that the bandpass lter is 8th order Calculate the ualue ofB in the low7pass to band7pass transformation Solution The value of x from the low pass example gives B xfc fag 7 fcl 5532 Example 6 An 8th7order 05 dB ripple band7reject lter is to have a center frequency fc 10 kHz and a bandwidth fag 7 fcl QkHz Note that the transfer function is obtained from that for the 4th7order LPF so that the band7reject lter is 8th order Calculate the ualue of B in the low7pass to band7pass transformation Solution The value ofx from the low pass example gives B fcsfl fag 7 fcl 452 13 The Parameter h To write the Chebyshev transfer functions the parameter b is required Given the order n and the ripple parameter E h is de ned by h tanh sinh 1 n e EvenOrder Chebyshev Filters For an even order Chebyshev low pass lter of order n the transfer function can be written in the product form n2 1 ms i1 The constants ai and bi are given by 1 12 ailtmisin20i 1 1 12 bi lt1m for1 i n2 where the 0139 are given by 0i TL X90 for1 i n2 Example 7 Solve for the transfer function of the 05 dB ripple fourtheorder Chebyshev low pass and highepass lters Solution The parameters 6 and h are given by e 100510 7012 03493 b tanh lsinirl L 04166 4 03493 For n 4 there are two second order polynomials in each transfer function The calculations are summarized as follows 0i 1139 bi 1 225 0 10313 29406 2 675 0 059703 07051 The low pass transfer function is given by 1 s10313wc2 129406 510313wc 1 1 TLP S gtlt s059703wc2 107051 5059703wc 1 The high pass transfer function is obtained by replacing swc with wcs to obtain 1 THP S WC1031352 129406 we103135 1 1 X MC059703S2 107051 MC0597038 1 10313511162 10313swc2 129406 10313swc 1 059703511162 059703511162 107051 059703swc 1 OddOrder Chebyshev Filters For an odd order Chebyshev low pass lter of order n the transfer function can be written in the form X 1 new2 1 Tm saquot12 AJC1gtlt saiwc21bisaiwc1 The constants ai and bi are given by v172 awn2 for i n 1 2 1 12 ai m isingei 1011 gig n712 1 2 1 12 bi lt1m for1 i n712 where the 0139 are given by 221 0il gtlt90 for1 i ni12 n Example 8 Solve for the transfer function for the 1dB ripple thirdeorder Chebyshev low pass and highepass lters Solution The parameters 6 and h are given by e 10110 7012 05088 1 1 h 7 tanh smh For n 3 each transfer function contains one second order polynomial and one rst order polynomial The calculations are summarized as follows gt 04430 1 300 09971 20177 2 7 04942 The low pass transfer function is given by 1 1 X 50494211 1 50997111C2 120177 50997le 1 TLP S The high pass function is obtained by replacing swc with wcs to obtain 1 1 T s X H MC049428 1 uh0997152 120177 QUE099715 1 049425wc 0997151162 gtlt 049425wc 1 Damn11 120177 0997lswc 1 Elliptic Filter Transfer Functions The elliptic lter is also known as the Cauer Chebyshev lter lt is a lter which exhibits equal ripple in the passband and in the stop band lt can be designed to have a much steeper roll off than the Chebyshev lter The approximation to the magnitude squared function of the elliptic low pass lter is of the form 1 521 0 1 e232 cuw where K is the dc gain constant 6 determines the dB ripple and RT wwc is the rational Chebychev function This function has the form lTLP WW K2 712 qugl 9591 11 1 711 2 lt V 421 ass112 11 1 17011002 Ba for 71 even so for n odd where 0 lt q lt 1 From this de nition it follows that Ba satis es the following properties B 3101 RMquot 0 R 712 3 0 for 71 even 11 0 for 71 odd 521 lquot2 for 71 even lquot712 for 71 odd 712 RT H 2 for 71 even 00 for n odd It is beyond the scope of this treatment to describe how the q are speci ed For n even the transfer function can be put into the form TLP S nZ SCle2 T 1 H mm 1bsawc 1 For n odd the form is 1 ill2 sciwc2 1 T S samcf lt1bgtltsawcgt1 gtlt San12wc 1 These differ from the forms for the Chebyshev lter by the presence of zeros on the jw axis That is TLP jw 0 for w qwc Figure 9 shows the plot of lTLP for a 4th order elliptic lter where U wwc is the normalized radian frequency The lter has a dB ripple of 05 dB for U S 1 Like the Chebyshev lters the gain ripples up from its dc value for the even order lter There are two zeros in the response one at U 159 and the other at U 348 For 1 gt 15 the gain ripples between 0 and 00153 As 1 becomes large the gain approaches the value 00153 which is 363 dB down from the dc value Odd order elliptic lters have the property that the gain ripples down from the dc value and approaches 0 as 1 becomes large Figure 9 Magnitude response versus normalized frequency for the example elliptic filter The passband for the elliptic lter is the band de ned by M S Ma Like the Chebyshev lters the dB ripple is de ned in this band The even order lters ripple up from the dc gain whereas the odd order lters ripple down At w wc the gain of an even order lter is equal to the dc gain whereas the gain of an odd order lter is down by an amount equal to the dB ripple For M gt MO the gain decreases rapidly and is equal to zero at the zero frequencies of the transfer function Between adjacent zeros the gain peaks up in an equal ripple fashion ie with equal values at the peaks Let the gain at the peaks between adjacent zeros have a dB level that is down from the dc gain by 14min dB The stop band frequency ms is de ned as the frequency between the cutoff frequency MC and the rst zero frequency at which the gain is down by 14min dB For M gt MS the gain is down from the dc level by 14min dB or more The three parameters which de ne the alignment of an elliptic lter are the dB ripple in the passband the ratio uswC of the stop band frequency to the cutoff frequency and the minimum dB attenuation 14mm in the stop band The tables below give the values of the 1 1 and c in the elliptic transfer functions for 05 dB ripple lters having uswC values of 15 20 and 30 for orders n 2 through n 5 17 The Thompson Phase Approximation Let 11 t be the input voltage to a lter If the output voltage is given by 110 t 11 t 7 739 the lter is said to introduce a time delay of 739 seconds to the signal The transfer function of such a lter is given by For 8 jw the phase of the transfer function in radians is given by 775 w in 727rf739 Thus the phase lag through the lter is directly proportional to the frequency A lter having such a phase function is called a linear phase lter The Thompson approximation is a low pass lter which has a phase function that approximates a linear phase function It can only be realized as a low pass lter This is because a time delay cannot be realized with for example a high pass lter The Thompson lter is also called a Bessel lter This is because Bessel polynomials are used in obtaining the lter coe icients Given the phase function p w for any lter there are two delays which can be de ned These are the phase delay 70 and the group delay 7399 These are given by w ltW1 77 27rf T de 7 WW 9 dw 27r df These are equal for a linear phase lter Let n be the order of the lter For n even the transfer function can be put into the form n2 1 ms i1 For n odd the form is 1 new2 1 X San12wc 1 11 saiwc2 115 saiwc 1 TLP S where the time delay of the lter is given by 739 lwc lt is beyond the scope of the treatment here to describe how the a and b are obtained The tables below give the values for n l to n 6 Also given is the ratio of the 73 dB frequency f3 to the cutoff frequency fa forl n 4 for5 n 6 Figure 10 shows the plot of lT as a function of frequency for the 4th order Thomp son lter for the case wc l rads This choice gives a delay of 1 second The 73 dB frequency of the lter is f3 21139w027r 03364 Hz lf the 73 dB frequency is multiplied by a constant the delay time is divided by that constant Fig ll shows the phase response of the lter The phase delay is shown plotted in Fig 12 lt can be seen from this gure that the delay is approximately equal to 1 second up to the 73 dB frequency of the lter Figure 10 Plot of lTj27rfl versus f SecondOrder Filter Topologies Sallen Key LowPass Filter Figure 13 shows the circuit diagram of the second order Sallen Key low pass lter The voltage gain transfer function is given by V l K27 W swo 1Q Swo 1 where K is the dc gain constant we is the radian resonance frequency and Q is the quality factor These are given by R4 K1 33 Figure 11 Plot of the phase in degrees of T j27rf versus f Figure 12 Plot of the phase delay in seconds of T j27rf versus f 1 V RleCng V RleCng 17 K 13101 131 132 02 do Q Figure 13 Second order Sallen Key low pass filter Special Case 1 Let K 1 R4 a short and R3 an open If Bl R2 we and Q are speci ed C1 and C2 are given by Q 1 1 C 1 U0 R1 R2 1 7 Qwo R1 32 The lter is often realized with El R2 02 Special Case 2 Capacitor values are often di icult to obtain It is possible to specify C1 and C2 and calculate the resistors Let K 1 R4 a short and R3 an open If Cl 02 we and Q are speci ed R1 and R2 are given by 02 1i 1 7 4 2 i Q 01 Any value for the ratio 0201 can be chosen provided 4Q2C392C1 S 1 Note that the values for El and R2 are interchangeable 1 R R 17 2 QQMOCZ Example 9 Design a unityigain secondiordei SalleniKey lowipass lter with f0 1kHz and Q Solution Let 01 01 MF and 02 0022 HF R1 and R2 are given by 1 li 14gtlt gtlt022 Either value may be assigned to El The other value is then assigned to R2 2 X 27r1000 gtlt 0022 X 10 6 129 k9 and 894 k9 R17 R2 Special Case 3 Let R1 R2 R and C1 C2 C 1f C we and Q are speci ed R and K are given by 1 Rw00 1 K37 Q In niteGain MultiFeedback LowPass Filter Figure 14 shows the circuit diagram of the second order in nite gain multi feedback low pass lter The voltage gain transfer function is given by V 1 7K7 W swof 1Q swogt 1 where K is the dc gain constant we is the radian resonance frequency and Q is the quality factor These are given by 7 R3 7 R1 1 V RgRgCng Q R2RaC1C2 R2 Ra 323331 K do Figure 14 Second order infinite gain multi feedback low pass filter Special Case 1 Let R1 R2 R3 so that K 71 1f R we and Q are speci ed C1 and C2 are given by Q 2 1 C 1 we R1R2 1 Qwo R1 232 The lter is often designed with R1 R2 02 Special Case 2 Capacitor values are often di cult to obtain It is possible to specify C1 and C2 and calculate the resistors Let K 71 R3 Bl 1fC1 02 we and Q are speci ed 312 and R2 are given by 02 1i 178 2 1 Q 01 Any value for the ratio 0201 can be chosen provided SQZOgCl S 1 Note that the values for 312 and R2 are interchangeable R1 P 2 4Qw002 Example 10 Design a unityegam secondeorder in niteegam multiefeedback lowepass lter with f0 1kHz and Q 1 Solution Let 01 01 MF and 02 001 HF R12 and R2 are given by 1 1 i 1 7 8 X E X 01 The lter can be designed either with R1 25415 and R2 999 k9 or with R1 2015 and R2 12715 R1 xi 7R2 i 2 4 X 27r1000 X 001 gtlt10 6 127 kg and 999 kg SecondOrder LowPass Filter 3dB Cutoff Frequency The 73 dB radian cutoff frequency 913 of the lters in Figures 13 and 14 is given by 12 1 1 2 11131110 li QQZ 1 li 2Q2 Sallen Key HighPass Filter Figure 15 shows the circuit diagram for the second order Sallen Key high pass lter The voltage gain transfer function is given by Va 89102 W swof 19 swogt 1 where K is the asymptotic high frequency gain we is the resonance frequency and Q is the quality factor These are given by R4 K 1 R3 1 V R1 B20102 V 31320102 R1 C1 Cg1 K R265 do Q Figure 15 Second order Sallen Key high pass filter Special Case 1 Let C1 C2 C and K 1 RF a short and R3 an open If C we and Q are speci ed R1 and R2 are given by 1 R 1 QQwOC 2Q R 2 U00 Special Case 2 Let R1 R2 R and C1 C2 C If R we and Q are speci ed C and K are given by 1 0 CODE 1 K37 Q In niteGain MultiFeeclback HighPass Filter The second order in nite gain multi feedback high pass lter is not a stable circuit in prac tice This is because its input node connects through two series capacitors to the V node of the op amp At high frequencies the input node becomes shorted to a virtual ground This can cause oscillation problems in the source that drives the lter Therefore this topology is not recommended SecondOrder HighPass Filter 3dB Cutoff Frequency The 73 dB lower radian cutoff frequency M3 for the circuit in Figure 15 is given by 1 1 2 We 1729 1722 Sallen Key BandPass Filter Figure 16 shows the circuit diagram of the second order Sallen Key band pass lter The voltage gain transfer function is given by v 1Q swogt W swof 1Q swogt 1 712 where K is the gain at resonance we is the resonance frequency and Q is the quality factor These are given by K X K03302 R1 R2 R1HRz C1 C2 3302 1 KORl R1 32 1 R1HR2 330102 R1HR2 330102 R1HR2 C1 C2 3302 1 KORl R1 32 where K0 is the gain from the non inverting input of the op amp to the output This is given by do Rs K1 0 R4 Figure 16 Second order Sallen Key band pass filter Special Case Let Bl R2 R3 R and C1 C2 C If Q we and C are speci ed R K0 and K are given by 1 R K 47 K4 271 MOO 0 Q2 Q In niteGain MultiFeedback BandPass Filter Figure 17 shows the circuit diagram of the second order in nite gain multi feedback band pass lter The voltage gain transfer function is given by m K 1Q cm W swof 1Q swogt 1 where K is the gain at resonance we is the resonance frequency and Q is the quality factor These are given by R301 R1 01 C2 1 R1HR2 330102 330102 RlHR2 C1C2 K do Q Figure 17 Second order infinite gain multi feedback band pass filter Special Case Let C1 C2 C For a speci ed K we and Q Bl R2 and R3 are given by Q R1 2Q R R KwOO 2 2Q2K 71 3 we SecondOrder BandPass Filter Bandwidth The 73dB bandwidth Awg of a second order band pass lter is related to the resonance frequency we and the quality factor Q by R1 we Awe Q Let the lower and upper 73dB cutoff frequencies respectively be denoted by Ma and uh These frequencies are related to the bandwidth and the quality factor by Awg wb iwa U0 iwawb A Biquad Filter A biquadratic transfer function or biquad for short is a transfer function that is the ratio of two second order polynomials It has the form 09an 1Qn SWn 1 SWd2 1Qd SWd 1 where K w Q Md and Qd are constants The function can be thought of as the sum of three transfer functions a low pass a band pass and a high pass For the case QT gt 00 the zeros are on the jw axis and the transfer function exhibits a notch at w wc ie TBQ 0 Band reject lters and elliptic lters have biquad terms of this form In general the simplest biquad circuits to realize are the ones which require more than TBQ S K one op amp Compared to the single op amp biquads these circuits can be realized with only one capacitor value and they are usually easier to tweak Fig 18 shows the circuit diagram of an example three op amp biquad For this circuit we can write 1 W V V 7 R lt 4HCS R3 R V V V0 R a 1 mg 71 V V V quot b CsltRR1 Figure 18 Biquad filter Although these equations look deceptively simple a lot of algebra is required to solve for The solution is of the form of TBQ 8 given above For a speci ed K wd Qd on Q and C the design equations for the circuit are 1 R de R R1 E w 2 R 131 Md Ung R 3 wdQdiwnQn R4 QdR Example 11 Design a thirdeorder elliptic lowepass lter hauing 05 dB ripple a notch in its transfer function at f 15734 Hz a stopeband attenuation of 219 dB and a stop band to cuto frequency ratio of fSfc 15 The gain at dc is to be unity Solution From the fCfS 15 Elliptic lter table the transfer function is given by 7 1 X s16751wc2 1 sO76695wc 1 s10720wc2 123672 510720wc 1 F S where wc is the cutoff frequency The transfer function has a null at w 16751wc For the null to be at 157 734 Hz we have wc 271391573416751 597 017 Thus the transfer function is 1 98859 2 1 F S X 2 8 545263 1 563266 12367 563266 1 The biquadratic term can be realized with the three op amp biquad Let C 1000 pF The element values are given by R 158115 1 632660 R R1 T 158115 R m 98859 2 63266132 632662367 i 9885900 R4 2367R374215 R 2367R2 913715 co The rst order term can be realized as a passive lter at either the input or the output of the biquad The preferred realization does not require a buffer stage for isolation One way of accomplishing this is to divide Bl R2 and R3 each into two series resistors where the ratio of the resistors in each pair is the same By voltage division the voltage at the node where the resistors in each pair connect is the same for the three pairs Thus these three nodes can be connected together A capacitor to ground from this node then realizes the rst order term Let the ratio of the resistors in each pair be 033067 Divide R1 into series resistors of values 03331 521715 and 06731 105915 Similarly divide R2 into resistors of values 03332 127415 and 06732 258615 and divide R3 into resistors of values 03333 301515 and 06733 612115 Let the smaller of the resistors in each pair connect to the V node These three resistors are in parallel in the circuit and can be replaced with a single resistor of value 5217111274113015 329715 The time constant for the capacitor which sets the single pole term in the transfer function is 377C where I77 3297111059112586116121 220915 Thus C 1 45263 X 2209 001 HF The completed circuit is given in Fig 19 The magnitude response is shown in Fig 20 Figure 19 Completed elliptic filter Figure 20 Magnitude response versus frequency in kHz A Second Biquad Filter A second biquad lter that has its zeros on the jw axis is shown in Fig 21 The following equations can be written for the circuit Va VC 7 W 7 Va B1018 B3018 V l a VG R V0 32 b 28lt 2H025 These equations can be solved for VaW to obtain where w 7 1R7R 09an 1 SWd2 1Qd SWd 1 K Ra Rs w R1 R1R2R40102 R R2 02 R3 01 d 7 7 Qd lt1R6 7 R2 R3 Cl 02 Figure 21 A second biquad circuit Because there are more element values than equations that relate them values must be assigned to some of the elements before the others can be calculated Let values for Cl 02 K R5 and l 3736 be speci ed The other element values are given by 1 R7R6 2 Qd KQdClwd deZ Rs R 1 1 3736 2 R3 KR1 R4 K E w ThirdOrder Sallen Key Filter Circuits The lter circuits given in this section consist of a rst order stage in cascade with a second order stage with no buffer ampli er isolating the two stages The transfer functions are di cult to derive Low Pass Filters Figure 22 Third order Sallen Key low pass filter Figure 22 shows the circuit diagram of a third order unity gain Sallen Key low pass lter For R1 R2 R3 B it can be shown that the transfer function for this lter is given by V 1 W 3301020353 2R20102 03 52 R 301 03 5 1 For a cutoff frequency wc the element values for Butterworth and Chebyshev lters are given in the following table The Butterworth lters are 3 dB down at 2wC whereas the Chebyshev lters are down by an amount equal to the dB ripple HighPass Filters Figure 23 shows the circuit diagram of a third order unity gain Sallen Key high pass lter For C1 C2 C3 C it can be shown that the transfer function for this lter is given by W i R1R2R30383 W 31132330353 P2 131 31330252 2 132 R3 Cs 1 For a cutoff frequency wc the element values for Butterworth and Chebyshev lters are given in the following table The Butterworth lters are 3 dB down at 2wC whereas the Chebyshev lters are down by an amount equal to the dB ripple 30 R2 C1 02 03 Vi R3 R1 Figure 23 Third order Sallen Key high pass filter Impedance Transfer Functions RC Network The impedance transfer function for a two terminal RC network which contains only one capacitor and is not an open circuit at dc can be written 1739zs Z c Rd17ps where R16 is the dc resistance of the network 739 is the pole time constant and 72 is the zero time constant The pole time constant is the time constant of the network with the terminals open circuited The zero time constant is the time constant of the network with the terminals short circuited Figure 24a shows the circuit diagram of an example two terminal RC network The impedance transfer function can be written by inspection to obtain 1R208 11R1R2Cs CR2 L R2 R R 1 1 a b ZR Figure 24 Example RC and BL impedance networks RL Network The impedance transfer function for a two terminal BL network which contains only one inductor and is not a short circuit at dc can be written 1 728 1 7175 ZRdc where R16 is the dc resistance of the network 739 is the pole time constant and 72 is the zero time constant The pole time constant is the time constant of the network with the terminals open circuited The zero time constant is the time constant of the network with the terminals short circuited Figure 24b shows the circuit diagram of an example two terminal BL network The impedance transfer function can be written by inspection to obtain 1 Z RlllR21 LR1 325 Voltage Divider Transfer Functions RC Network The voltage gain transfer function of a BC voltage divider network containing only one capacitor and having a non zero gain at dc can be written V0 k1 728 V T 1 75 where k is the dc gain C an open circuit 739 is the pole time constant and Tz is the zero time constant The pole time constant is the time constant of the network with V 0 and V0 open circuited The zero time constant is the time constant of the network with V 0 and V open circuited Figure 25a shows the circuit diagram of an example RC network The voltage gain transfer function can be written by inspection to obtain Va R2 Ra 1 R2iiRa CS X W R1 R2 R3 1R1 R2 iiRal Cs Figure 25b shows the circuit diagram of a second example RC network The voltage gain transfer function can be written by inspection to obtain W7 R3 X Vi R1 Ra 1 R1iiRa Rzl CS C R2 C R1 v V0 V V0 R1 R5 R2 b C Figure 25 Example RC voltage divider networks HighPass RC Network The voltage gain transfer function of a high pass RC voltage divider network containing only one capacitor can be written V0 7 775 V 1 75 where k is the in nite frequency gain C a short circuit and 77 is the pole time constant The pole time constant is calculated with V 0 and V0 open circuited Figure 25c shows 32

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