II. PASSIVE FILTERS. H(jω) = Stop. Pass

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1 II. PASSIE FITES Frequencyselectve or flter crcuts pass to the output only those nput sgnals that are n a desred range of frequences (called pass band). The ampltude of sgnals outsde ths range of frequences (called stop band) s reduced (deally reduced to zero). Typcally n these crcuts, the nput and output currents are kept to a small value and as such, the current transfer functon s not an mportant parameter. The man parameter s the voltage transfer functon n the frequency doman, H v (j) = o /. As H v (j) s complex number, t has both a magntude and a phase, flters n general ntroduce a phase dfference between nput and output sgnals. To mnmze the number of subscrpts, hereafter, we wll drop subscrpt v of H v. Furthermore, we concentrate on the the openloop transfer functons, H vo, and denote ths smply by H(j). The mpact of loadng s sperately dscussed.. owpass Flters H(j ) An deal lowpass flter s transfer functon s shown. The frequency between the pass andstop bands s called the cutoff frequency ( c ). All of the sgnals wth frequences below c are transmtted and all other sgnals are stopped. Pass Band c Stop Band In practcal flters, pass and stop bands are not clearly defned, H(j) vares contnuously from ts maxmum toward zero. The cutoff frequency s, therefore, defned as the frequency at whch H(j) s reduced to / = 0.7 of ts maxmum value. Ths corresponds to sgnal power beng reduced by / as P. Κ 0.7Κ H(j ) c owpass flters A seres crcut as shown acts as a lowpass flter. For no load resstance ( openloop transfer functon), o can be found from the voltage dvder formula: o o = j o = j = j(/) We note H(j) = (/) ECE65 ecture Notes (F. Najmabad), Sprng 006

2 It s clear that H(j) s maxmum when denomnator s smallest,.e., 0 and H(j) decreases as s ncreased. Therefore, ths crcut allows lowfrequency sgnals to pass through whle blockng hghfrequency sgnals (.e., reduces the ampltude of the voltage of the hghfrequency sgnals). The reference to defne the low and hgh frequences s the cutoff frequency: low frequences mean frequences much lower than c. To fnd the cutoff frequency, we note that the H(j) Max = occurs at = 0 (alternatvely fnd d H(j) /d and set t equal to zero to fnd = 0 whch maxmzes H(j) ). Therefore, H(j) max = H(j) =c = H(j) max = ( c /) = ( ) c = c = Therefore, c = and j/ c Input Impedance: Usng the defnton of the nput mpedance, we have: Z = I = j The value of the nput mpedance depends on the frequency. For good voltage couplng, we need to ensure that the nput mpedance of ths flter s much larger than the output mpedance of the prevous stage. Snce we do not know the frequency of the nput sgnal, we need to ensure that good voltage couplng crtera s satsfed for all frequences (or all possble values of Z ). As such, the mnmum value of Z s an mportant number. Z s mnmum when the mpedance of the nductor s zero ( 0). Z mn = Output Impedance: The output mpdenace can be found by kllng the source and fndng the equvalent mpdenace between output termnals: Z o Z o = j ECE65 ecture Notes (F. Najmabad), Sprng 006

3 where the source resstance s gnored. Agan, the value of the output mpedance depends on the frequency. For good voltage couplng, we need to ensure that the output mpedance of ths flter s much smaller than the nput mpedance of the next stage for all frequences, the maxmum value of Z o s an mportant number. Z o s maxmum when the mpedance of the nductor s nfnty ( ). Z o max = Bode Plots and Decbel The voltage transfer functon of a twoport network (and/or the rato of output to nput powers) s usually expressed n Bel: ( ) Po Number of Bels = log 0 P or o Number of Bels = log 0 because P. Bel s a large unt and decbel (db) s usually used: o Number of decbels = 0 log 0 or o o = 0 log 0 db There are several reasons why decbel notaton s used: ) Hstorcally, the analog systems were developed frst for audo equpment. Human ear hears the sound n a logarthmc fashon. A sound whch appears to be twce as loud actually has 0 tmes power, etc. Decbel translates the output sgnal to what ear hears. ) If several twoport network are placed n a cascade (output of one s attached to the nput of the next), the overall transfer functon, H, s equal to the product of all transfer functons: H(j) = H (j) H (j)... 0 log 0 H(j) = 0 log 0 H (j) 0 log 0 H (j)... H(j) db = H (j) db H (j) db... makng t easer to fnd the overall response of the system. 3) Plot of H(j) db versus frequency has specal propertes that make analyss smpler. For example, the plot asymptotes to straght lnes at low and hgh frequences as s shown below. ECE65 ecture Notes (F. Najmabad), Sprng 006 3

4 Also, usng db defnton, we see that, there s a 3 db dfference between maxmum gan and gan at the cutoff frequency: 0 log H(j c ) 0 log H(j) max = 0 log [ H(jc ) H(j) max ] = 0 log ( ) 3 db Bode plots are plots of H(j) db (magntude) and H(j) (phase) versus frequency n a semlog format (.e., axs s a log axs). Bode plots of frstorder lowpass flters are shown below (W denotes c ). H(j) db H(j) At hgh frequences, / c, H(j) [ ] H(j) / db = 0 log = 0 log( c ) 0 log() c / c whch s a straght lne wth a slope of 0 db/decade n the Bode plot. It means that f s ncreased by a factor of 0 (a decade), H(j) db changes by 0 db. At low frequences, / c, H(j) whch s also a straght lne n the Bode plot. The ntersecton of these two asymptotc values s at = /(/ c ) or = c. Because of ths, the cutoff frequency s also called the corner frequency. The behavor of the phase of H(j) can be found by examnng tan (/ c ). At low frequences, / c, H(j) 0 and at hgh frequences, / c, H(j) 90. At cutoff frequency, H(j) 45. ECE65 ecture Notes (F. Najmabad), Sprng 006 4

5 General frstorder lowpass flters As we dscussed before, transfer functons characterze a twoport network. As such, t s useful to group twoport networks nto famles based on ther voltage transfer functons. To facltate ths groupng, the conventon s to smplfy the voltage transfer functon to a form such that the eal part of the denomnator of H(j) s unty (.e., the denomnator should be j or j ). As we wll see later n ths secton, ths groupng wll also help reduce the math that we do n analyzng varous crcuts. The lowpass flter dscussed before s part of the famly of frstorder lowpass flters (frst order means that appears n the denomnator wth an exponent of or. In general, the voltage transfer functon of a frstorder lowpass flter s n the form: K j/ c The maxmum value of H(j) = K s called the flter gan. Note that the exponent of n the denomnator s so that H(j) decreases wth frequency (thus,a lowpass flter): K H(j) = (/ c ) K ( ) K tan c For flter, K =, and c = /. Note that K can be negatve, and n that case, the mnus sgn adds 80 phase shft to the transfer functon as s denoted by K /K factor above. owpass C flters A seres C crcut as shown also acts as a lowpass flter. For no load resstance ( openloop transfer functon), o can be found from the voltage dvder formula: o = /(jc) /(jc) = jc j(c) C o We see that the voltage transfer functon of ths crcut s smlar to transfer functon of a general frstorder lowpass flter. So, ths s a lowpass flter wth K = and c = /C. (Note: we dentfed the crcut and found the cutoff frequency wthout dong any math!). ECE65 ecture Notes (F. Najmabad), Sprng 006 5

6 We could, of course, do the math followng the procedure n analyzng the lowpass flter to get the same answer. (Exercse: Show ths.). Followng the same procedure as for flters, we fnd nput and output Impedances Z = jc Z o = jc and and Z mn = Z o max = Termnated and C lowpass flters Now let us examn the effect of a load on the performance of our and C flters. For ths example, a resstve load s consdered but the analyss can be easly extended to an mpedance load. For example, consder the termnated C flter shown: C o From the crcut, o = /(jc) [/(jc) ] = / j( C) wth = Ths s smlar to the transfer functon for untermnated C flter but wth resstance beng replaced by. Therefore, c = C = ( )C and / j/ c We see that the mpact of the load s to reduce the flter gan (K = / < ) and to shft the cutoff frequency to a hgher frequency as = <. Input Impedance: Z = jc Output Impedance: Z o = jc Z mn = Z o max = We could have arrved at the same results usng the the relatonshp between openloop, H o (j), and termnated, H(j), transfer functons of a twoport network: Z Z Z o H o (j) = jc jc ECE65 ecture Notes (F. Najmabad), Sprng 006 6

7 (Exercse: show ths.) Also, note that the output mpdenace of the termnated crcut s exactly the same as the openloop verson. Furthermore, t can be seen that as long as Z o or Z o max = (our condton for good voltage couplng), and the termnated C flter wll look exactly lke an untermnated flter The flter gan s one, the shft n cutoff frequency dsappears, and nput and output resstances become the same as before. Termnated lowpass flters The parameters of the termnated flters can be found smlarly: oltage Transfer Functon: o = Input Impedance: Z = j, j/ c, c = ( )/. Z mn = Output Impedance: Z o = (j), Z o max = Here, the mpact of load s to shft the cutoff frequency to a lower value. Flter gan s not affected. Agan for Z o or Z o max = (our condton for good voltage couplng), the shft n cutoff frequency dsappears and the flter wll look exactly lke an untermnated flter. Exercse: Derve above equatons for the transfer functon and nput and output mpdenacess. ECE65 ecture Notes (F. Najmabad), Sprng 006 7

8 . Frstorder hgh pass flters In general, the voltage transfer functon of a frstorder hghpass flter s n the form: K j c / It s a frstorder flter because appears n the denomnator wth an exponent of. It s a hghpass flter because H = 0 for = 0 and H s constant for hghfreqneces. Paramter c s the cutoff freqnecy of the flter (Exercse: prove that H(j c ) s / = 0.7 of H(j) Max.) The maxmum value of H(j) = K s called the flter gan. H(j) = K K ( c /) K tan ( ) c Bode Plots of frstorder hghpass flters (K = ) are shown below. The asymptotc behavor of ths class of flters s: At low frequences, / c, H(j) (a 0dB/decade lne) and 90 At hgh frequences, / c, H(j) (a lne wth a slope of 0) and 0 H(j) H(j) ECE65 ecture Notes (F. Najmabad), Sprng 006 8

9 Hghpass C flters C A seres C crcut as shown acts as a hghpass flter. The openloop voltage transfer functon of ths flter s: o o = /(jc) = j(/c) Therefore, ths s a frstorder hghpass flter wth K = and C = /C. Input and output mpdenaces of ths flter can be found smlar to the procedure used for lowpass flters: Input Impedance: Z = jc Output Impedance: Z o = jc and and Z mn = Z o max = Hghpass flters A seres crcut as shown also acts as a hghpass flter. Agan, we fnd the openloop tranfuncton to be: o c = j c / Input Impedance: Z = j and Z mn = Output Impedance: Z o = j and Z o max = Exercse: Compute the voltage transfer functon and nput and output mpdenaces of termnated C and flters. ECE65 ecture Notes (F. Najmabad), Sprng 006 9

10 .3 Bandpass flters A band pass flter allows sgnals wth a range of frequences (pass band) to pass through and attenuates sgnals wth frequences outsde ths range. l : u : 0 l u : B u l : Q 0 B : ower cutoff frequency; Upper cutoff frequency; Center frequency; Band wdth; Qualty factor. H(j ) l Pass Band u As wth practcal low and hghpass flters, upper and lower cutoff frequences of practcal band pass flter are defned as the frequences at whch the magntude of the voltage transfer functon s reduced by / (or 3 db) from ts maxmum value. Secondorder bandpass flters: Secondorder band pass flters nclude two storage elements (two capactors, two nductors, or one of each). The transfer functon for a secondorder bandpass flter can be wrtten as H(j) = K ( jq 0 K ) 0 Q ( 0 0 ) K K tan [ ( Q 0 )] 0 The maxmum value of H(j) = K s called the flter gan. The lower and upper cutoff frequences can be calculated by notng that H(j) max = K, settng H(j c ) = K/ and solvng for c. Ths procedure wll gve two roots: l and u. H(j c ) = H(j) max = K = ( Q c ) ( 0 c = Q 0 0 c 0 c c 0 ± c 0 Q = 0 K Q ( c 0 0 c ) = ± ) ECE65 ecture Notes (F. Najmabad), Sprng

11 The above equaton s really two quadratc equatons (one wth sgn n front of fracton and one wth a sgn). Solvng these equaton we wll get 4 roots (two roots per equaton). Two of these four roots wll be negatve whch are not physcal as c > 0. The other two roots are the lower and upper cutoff frequences ( l and u, respectvely): l = 0 4Q 0 Q u = 0 4Q 0 Q Bode plots of a secondorder flter s shown below. Note that as Q ncreases, the bandwdth of the flter become smaller and the H(j) becomes more pcked around 0. H(j) db H(j) Asymptotc behavor: At low frequences, / 0, H(j) (a 0dB/decade lne), and H(j) 90 At hgh frequences, / 0, H(j) / (a 0dB/decade lne), and H(j) 90 At = 0, K (purely real) H(j) = K (maxmum flter gan), and 0. There are two ways to solve secondorder flter crcuts. ) One can try to wrte H(j) n the general form of a secondorder flters and fnd Q and 0. Then, use the formulas above to fnd the lower and upper cutoff frequences. ) Alternatvely, one can drectly fnd the upper and lower cutoff frequences and use 0 l u to fnd the center frequency and B u l to fnd the bandwdth, and Q = 0 /B to fnd the qualty factor. The two examples below show the two methods. Note that one can always fnd 0 and k rapdaly as H(j 0 ) s purely real and H(j 0 ) = k ECE65 ecture Notes (F. Najmabad), Sprng 006 3

12 Seres C Bandpass flters Usng voltage dvder formula, we have o = j /(jc) ( j ) C C o There are two approaches to fnd flter parameters, K, 0, u, and l. Method : We transform the transfer functon n a form smlar to general form of the transfer functon for second order bandpass flters: K ( jq 0 ) 0 Note that the denomnator of the general form s n the form j... Therefore, we dvde top and bottom of transfer functon of seres C bandpass flters by : ( j ) C Comparng the above wth the general form of the transfer functon, we fnd K =. To fnd Q and 0, we note that the magnary part of the denomnator has two terms, one postve and one negatve (or one that scales as and the other that scales as /) smlar to the general form of transfer functon of ndorder bandpass flters (whch ncludes Q/ 0 and Q 0 /). Equatng these smlar terms we get: Q 0 = Q 0 = C Q 0 = Q 0 = C We can solve these two equatons to fnd: 0 = Q = 0 C / = C ECE65 ecture Notes (F. Najmabad), Sprng 006 3

13 The lower and upper cutoff frequences can now be found from the formulas on page 3. Method : In ths method, we drectly calculate the flter parameters smlar to the procedure followed for general form of transfer functon n page 30. Some smplfcatons can be made by notng: ) At = 0, H(j) s purely real and ) K = H(j = j 0 ). Startng wth the transfer functon for the seres C flter: ( j ) C We note that the transfer functon s real f coeffcent of j n the denomnator s exactly zero (note that ths happens for = 0 ),.e., Also 0 0 C = 0 0 = C K = H(j = j 0 ) = = The cutoff frequences can then be found by settng: H(j c ) = K = ( c ) = c C whch can be solved to fnd u and l. Input and Output Impedance of bandpass C flters Z = j ( jc = j ) C Z mn = occurs at = 0 ( Z o = j ) Z o jc max = ECE65 ecture Notes (F. Najmabad), Sprng

14 WdeBand BandPass Flters Bandpass flters can be constructed by puttng a hghpass and a lowpass flter back to back as shown below. The hghpass flter sets the lower cutoff frequency and the lowpass flter sets the upper cutoff frequency of such a bandpass flter. H (j ) H (j ) H (j ) X H (j ) l = u= c c c c An example of such a bandpass flter s two C lowpass and hghpass flters put back to back. These flters are wdely used (when approprate, see below) nstead of an C flter as nductors are usually bulky and take too much space on a crcut board. C C o ow Pass Hgh Pass In order to have good voltage couplng n the above crcut, the nput mpedance of the hghpass flter (actually Z mn = ) should be much larger than the output mpedance of the lowpass flter (actually Z o max = ), or we should have. In that case we can use untermnated transfer functons: H (j) H (j) = c = /( C ) c = /( C ) j/ c j c / ( j/ c )( j c /) = ( c / c ) j(/ c c /) Agan, we can fnd the flter parameters by ether of two methods above. Transformng the transfer functon to a form smlar to the general form (left for students) gves: K = c / c Q = c / c c / c 0 = c c ECE65 ecture Notes (F. Najmabad), Sprng

15 One should note that the Bode plots of prevous page are asymptotc plots. The real H(j) dffers from these asymptotc plots, for example, H(j) s 3 db lower at the cutoff frequency. A comparson of asymptotc Bode plots for frstorder hghpass flters are gven n page 8. It can be seen that H (j) acheves ts maxmum value ( n ths case) only when / c < /3. Smlarly for the low pass flter, H (j) acheves ts maxmum value ( n ths case) only when / c > 3. In the bandpass flter above, f c c (.e., c 0 c ), the center frequency of the flter wll be at least a factor of three away from both cutoff frequences and H(j) = H H acheves ts maxmum value of. If c s not c (.e., c < 0 c ), H and H wll not reach ther maxmum of and the flter H(j) max = H H wll be less than one. Ths can be seen by examnng the equaton of K above whch s always less than and approaches when c c. More mportantly, we can never make a narrow band flter by puttng two frstorder hghpass and lowpass flters back to back. When c s not c, H(j) max becomes smaller than. Snce the cutoff frequences are located 3 db below the maxmum values, the cutoff frequences wll not be c and c (those frequences are 3 db lower than H(j) max = ). The lower cutoff frequency moves to a value lower than c and the upper cutoff frequency moves to a value hgher than c. Ths can be seen by examnng the qualty factor of ths flter at the lmt of c = c Q = c / c c / c = = 0.5 whle our asymptotc descrpton of prevous page ndcated that when c = c, bandwdth becomes vanshngly small and Q should become very large. Because these flters work only when c c, they are called wdeband flters. For these wdeband flters ( c c ), we fnd from above: K = Q = c / c j(/ c c /) 0 = c c We then substtute for Q and 0 n the expressons for cutoff frequences (page 3) to get: u = 0 4Q 0 Q = 0 Q ( ) 4Q l = 0 4Q 0 Q = ( ) 0 4Q Q ECE65 ecture Notes (F. Najmabad), Sprng

16 Ignorng 4Q term compared to (because Q s small),we get: u = 0 Q = c c c / c = c For l, f we gnore 4Q term compared to, we wll fnd l = 0. We should, therefore, expand the square root by Taylor seres expanson to get the frst order term: u ( 0 ) Q 4Q = 0 Q Q = 0 Q = c What are WdeBand and NarrowBand Flters? Typcally, a wdeband flter s defned as a flter wth c c (or c 0 c ). In ths case, Q 0.35 (prove ths!). A narrowband flter s usually defned as a flter wth B 0 (or B 0. 0 ). In ths case, Q 0. Example: Desgn a bandpass flter wth cutoff frequences of 60 Hz and 8 khz. The load for ths crcut s MΩ. As ths s wdeband, bandpass flter ( u / l = f u /f l = 50 ), we use two low and hghpass C flter stages smlar to crcut above. The prototype of the crcut s shown below: C The hghpass flter sets the lower cutoff frequency, and the MΩ load sets the output mpedance of ths stage. Thus: Z o max = MΩ 00 kω c (Hghpass) = l = C = π 60 C o ow Pass Hgh Pass C = 0 3 kω One should choose as close as possble to 00 kω (to make the C small) and C = 0 3 usng commercal values of resstors and capactors. A good set here are = 00 kω and C = 0 nf. The lowpass flter sets the upper cutoff frequency. The load for ths component s the nput resstance of the hghpass flter, Z mn = = 00 kω. Thus: Z o max = 00kΩ 0 kω c (owpass) = u = C = π C = 0 5 ECE65 ecture Notes (F. Najmabad), Sprng

17 As before, one should choose as close as possble to 0 kω and C = 0 5 usng commercal values of resstors and capactors. A good set here are = 0 kω and C = nf. In prncple, we can swtch the poston of lowpass and hghpass flter stages n a wdeband, bandpass flter. However, the lowpass flter s usually placed before the hghpass flter because the value of capactors n such an arrangement wll be smaller. (Try redesgnng the above crcut wth lowpass and hghpass flter stages swtched to see that one capactor become much smaller and one much larger.) Exercse: Desgn an C flter wth the specfcatons n the prevous example. (Hnt: Do not set = 00 kω as ths would make the value of the nductor very large.).4 Exercse Problems Problem. Desgn a C bandpass flter wth a lower cutoff frequency of khz and a bandwdth of 3 khz. What s the center frequency and Q of ths flter? Problem. We have an amplfer that amplfes a khz sgnal from a detector. The load for ths amplfer can be modeled as a 50 kω resstor. The amplfer output has a large amount of 60 Hz nose. We need to reduce the ampltude of nose by a factor of 0. Desgn a frstorder passve flter whch can be placed between the amplfer and the load and does the job. Would ths flter affect the khz sgnal that we are nterested n? If so, by how much? Problem 3. The tuner for an FM rado requres a bandpass flter wth a central frequency of 00 MHz (frequency of a FM staton) and a bandwdth of MHz. a) Desgn such a flter. b) What are ts cutoff frequences? Problem 4. A telephone lne carres both voce band (04 khz) and data band (5 khz to MHz). Desgn a flter that lets the voce band through and rejects the data band. The flter must meet the followng specfcatons: a) For the voce band, the change n transfer functon should be at most db; and b) The transfer functon should be as small as possble at 5 khz, the low end of the data band. ECE65 ecture Notes (F. Najmabad), Sprng

18 .5 Soluton to Exercse Problems Problem. Desgn a C bandpass flter wth a lower cutoff frequency of khz and a bandwdth of 3 khz. What s the center frequency and Q of ths flter? The crcut prototype s: For a nd order bandpass flter: B(Hz) = f u f l B(rad/s) = πb(hz) = f u = 3 = 4 khz C o u = πf u = l = πf l = = u l = B(rad/s) = 0 Q Q = = 0.67 For the seres C crcut: 0 = C C = 0 C = 0 Q = 0 / = = 0.63 µf ( ) = 0 Q = B(rad/s) = B = = 88 Ω Therefore, usng commercal values, the desgn values are = 0 mh, = 80 Ω, and C = 0.68 µf. ECE65 ecture Notes (F. Najmabad), Sprng

19 Problem. We have an amplfer that amplfes a khz sgnal from a detector. The load for ths amplfer can be modeled as a 50 kω resstor. The amplfer output has a large amount of 60 Hz nose. We need to reduce the ampltude of nose by a factor of 0. Desgn a frstorder passve flter whch can be placed between the amplfer and the load and does the job. Would ths flter affect the khz sgnal that we are nterested n? If so, by how much? We want to have khz sgnals to go through but reduce 60 Hz sgnals, so we need a hghpass flter. The prototype of the crcut s shown below. For ths crcut: o = c = C Z mn = j c / Invertng Amp. C o Z o max = As the output mpedance of the nvertng amplfer crcut s zero, we do not need to worry about the nput mpedance of our flter. The output mpedance of the flter s restrcted by Z o max = 50 kω 5 kω Ths flter should reduce the ampltude of 60 Hz ( 60 = π 60 = 0π rad/s) sgnal by a factor of 0,.e., H(j = j 60 ) = o 60 Hz = ( c / 60 ) = 0. ( c / 60 ) = 00 C = c 0 60 = 375 rad/s C = easonable choces are = 3.9 kω (to keep t below 5 kω) and C = 68 nf (f c 600 Hz). The mpact on khz sgnal ( 000 = 000π rad/s) can be found from: H(j = j 000 ) = ( c / 000 ) = (375/683) = 0.86 So the ampltude of khz sgnal s reduced by 4% (or by.3 db). ECE65 ecture Notes (F. Najmabad), Sprng

20 Problem 3. The tuner for an FM rado requres a bandpass flter wth a central frequency of 00 MHz (frequency of a FM staton) and a bandwdth of MHz. a) Desgn such a flter. b) What are ts cutoff frequences? Because ths s not a wdeband flter, the smplest flter wll be an C flter as s shown. For ths flter: 0 = C = π C o Q = 0 B = C = π00 06 π 0 6 = 50 Usng a = µh nductor: C = 4π 0 6 C = 4π C =.5 0 F Choose: C =. pf C =, 500 = Choose: = 3 Ω ( = µh and C =. pf). To fnd the cutoff freqneces, we not:, 500C = 0 6 = 8 = 3.5 Ω, B = f u f l = MHz f 0 = f u f = 00 MHz Soluton of the above two equatons n two unknowns wll gve f l 99 MHz and f u 0 MHz. ECE65 ecture Notes (F. Najmabad), Sprng

21 Problem 4. A telephone lne carres both voce band (04 khz) and data band (5 khz to MHz). Desgn a flter that lets the voce band through and rejects the data band. The flter must meet the followng specfcatons: a) For the voce band, the change n transfer functon should be at most db; and b) The transfer functon should be as small as possble at 5 khz, the low end of the data band. We need a lowpass flter as t should allow lowfrequency sgnals (voce band) to go through whle elmnatng hghfrequency sgnals (data band). The prototype of an C lowpass flter s shown and ts transfer functon s: j/ c = The cutoff frequency of the flter s not gven and t should be found from the specfcatons. Frst, we need the change n transfer functon to be at most db for the frequency range of 04 khz. The transfer functon of flters that satsfy ths constrant s the curve labeled n the fgure and any transfer functon located to the rght of ths curve (such as transfer functon labeled ). jf/f c C o 0 db H(j ) 4 5 f (khz) Second, the transfer functon should be as small as possble at 5 khz. Ths requres that we choose the cutoff frequency as small as possble. Therefore, the transfer functon of our flter should be curve labeled as t has the smallest possble value at 5 khz: 0 log ( H(jf = 4 khz) ) = db H(jf = 4 khz) = 0.89 Usng the expresson for H(j), we have: H(jf = 4 khz) = (f/f c ) = 0.89 f/f c = f c = f = 7.85 khz f c = πc = C = Choosng C = nf, we have = The commercal values then are C = nf and = 0 kω. ECE65 ecture Notes (F. Najmabad), Sprng 006 4