II. PASSIVE FILTERS. H(j ω) Pass. Stop

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1 II. PASSIE FILTES Frequency-selectve 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 open-loop transfer functons, H vo, and denote ths smply by H(j). The mpact of loadng s sperately dscussed.. Low-Pass Flters H(j ) An deal low-pass flter s transfer functon s shown. The frequency between the pass- and-stop bands s called the cut-off 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 cut-off 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 Low-pass L flters A seres L crcut as shown acts as a low-pass flter. For no load resstance ( open-loop transfer functon), o can be found from the voltage dvder formula: L o o = jl H(j) = o = jl = j(l/) We note H(j) = (L/) ECE65 Lecture Notes (F. Najmabad), Wnter 009 3

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 low-frequency sgnals to pass through whle blockng hgh-frequency sgnals (.e., reduces the ampltude of the voltage of the hgh-frequency sgnals). The reference to defne the low and hgh -frequences s the cut-off frequency: low -frequences mean frequences much lower than c. To fnd the cut-off 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 L/) = ( ) c L = c L = Therefore, c = L and H(j) = j/ c Input Impedance: Usng the defnton of the nput mpedance, we have: Z = I = jl 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 = jl L Z o ECE65 Lecture Notes (F. Najmabad), Wnter 009 4

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 two-port 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 two-port 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 Lecture Notes (F. Najmabad), Wnter 009 5

4 Also, usng db defnton, we see that, there s a 3 db dfference between maxmum gan and gan at the cut-off 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 sem-log format (.e., axs s a log axs). Bode plots of frst-order low-pass L 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 cut-off frequency s also called the corner frequency. The behavor of the phase of H(j) can be found by examnng H(j) = tan (/ c ). At low frequences, / c, H(j) 0 and at hgh frequences, / c, H(j) 90. At cut-off frequency, H(j) 45. ECE65 Lecture Notes (F. Najmabad), Wnter 009 6

5 General frst-order low-pass flters As we dscussed before, transfer functons characterze a two-port network. As such, t s useful to group two-port 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 low-pass L flter dscussed before s part of the famly of frst-order low-pass flters (frst order means that appears n the denomnator wth an exponent of or. In general, the voltage transfer functon of a frst-order low-pass flter s n the form: H(j) = 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 low-pass flter): K H(j) = (/ c ) H(j) = K ( ) K tan c For L flter, K =, and c = /L. 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. Low-pass C flters A seres C crcut as shown also acts as a low-pass flter. For no load resstance ( open-loop transfer functon), o can be found from the voltage dvder formula: o = H(j) = /(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 frst-order low-pass flter. So, ths s a low-pass flter wth K = and c = /C. (Note: we dentfed the crcut and found the cut-off frequency wthout dong any math!). ECE65 Lecture Notes (F. Najmabad), Wnter 009 7

6 We could, of course, do the math followng the procedure n analyzng the low-pass L flter to get the same answer. (Exercse: Show ths.). Followng the same procedure as for L flters, we fnd nput and output Impedances Z = jc Z o = jc and and Z mn = Z o max = Termnated L and C low-pass flters Now let us examn the effect of a load on the performance of our L 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: From the crcut, C o L H(j) = o = /(jc) L [/(jc) L ] = / j( C) wth = L Ths s smlar to the transfer functon for untermnated C flter but wth resstance beng replaced by. Therefore, c = C = ( L )C and H(j) = / j/ c We see that the mpact of the load s to reduce the flter gan (K = / < ) and to shft the cut-off frequency to a hgher frequency as = L <. Input Impedance: Z = jc L Output Impedance: Z o = jc Z mn = Z o max = We could have arrved at the same results usng the the relatonshp between open-loop, H o (j), and termnated, H(j), transfer functons of a two-port network: H(j) = Z L Z L Z o H o (j) = L L jc jc ECE65 Lecture Notes (F. Najmabad), Wnter 009 8

7 (Exercse: show ths.) Also, note that the output mpdenace of the termnated crcut s exactly the same as the open-loop verson. Furthermore, t can be seen that as long as L Z o or L 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 cut-off frequency dsappears, and nput and output resstances become the same as before. Termnated L low-pass flters The parameters of the termnated L flters can be found smlarly: oltage Transfer Functon: H(j) = o = Input Impedance: Z = jl L, j/ c, c = ( L )/L. Z mn = L Output Impedance: Z o = (jl), Z o max = Here, the mpact of load s to shft the cut-off frequency to a lower value. Flter gan s not affected. Agan for L Z o or L Z o max = (our condton for good voltage couplng), the shft n cut-off 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 Lecture Notes (F. Najmabad), Wnter 009 9

8 . Frst-order hgh pass flters In general, the voltage transfer functon of a frst-order hgh-pass flter s n the form: H(j) = K j c / It s a frst-order flter because appears n the denomnator wth an exponent of. It s a hgh-pass flter because H = 0 for = 0 and H s constant for hgh-freqneces. Paramter c s the cut-off 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 H(j) = K ( c /) K tan ( ) c Bode Plots of frst-order hgh-pass 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 H(j) = 90 At hgh frequences, / c, H(j) (a lne wth a slope of 0) and H(j) = 0 H(j) H(j) ECE65 Lecture Notes (F. Najmabad), Wnter

9 Hgh-pass C flters A seres C crcut as shown acts as a hgh-pass flter. The open-loop voltage transfer functon of ths flter s: H(j) = o = /(jc) = j(/c) C o Therefore, ths s a frst-order hgh-pass flter wth K = and C = /C. Input and output mpdenaces of ths flter can be found smlar to the procedure used for low-pass flters: Input Impedance: Z = jc Output Impedance: Z o = jc and and Z mn = Z o max = Hgh-pass L flters A seres L crcut as shown also acts as a hgh-pass flter. Agan, we fnd the open-loop tranfuncton to be: L o c = L H(j) = j c / Input Impedance: Z = jl and Z mn = Output Impedance: Z o = jl and Z o max = Exercse: Compute the voltage transfer functon and nput and output mpdenaces of termnated C and L flters. ECE65 Lecture Notes (F. Najmabad), Wnter 009 3

10 .3 Band-pass 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 : Lower cut-off frequency; Upper cut-off frequency; Center frequency; Band wdth; Qualty factor. H(j ) l Pass Band u As wth practcal low- and hgh-pass flters, upper and lower cut-off 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. Second-order band-pass flters: Second-order band pass flters nclude two storage elements (two capactors, two nductors, or one of each). The transfer functon for a second-order band-pass flter can be wrtten as H(j) = H(j) = K ( jq 0 K ) 0 Q ( 0 0 ) H(j) = K [ ( K tan Q 0 )] 0 The maxmum value of H(j) = K s called the flter gan. The lower and upper cut-off 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 Lecture Notes (F. Najmabad), Wnter 009 3

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 cut-off frequences ( l and u, respectvely): l = 0 4Q 0 Q u = 0 4Q 0 Q Bode plots of a second-order 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, H(j) = K (purely real) H(j) = K (maxmum flter gan), and H(j) = 0. There are two ways to solve second-order flter crcuts. ) One can try to wrte H(j) n the general form of a second-order flters and fnd Q and 0. Then, use the formulas above to fnd the lower and upper cut-off frequences. ) Alternatvely, one can drectly fnd the upper and lower cut-off 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 Lecture Notes (F. Najmabad), Wnter

12 Seres LC Band-pass flters Usng voltage dvder formula, we have L C H(j) = o = H(j) = jl /(jc) ( j L ) 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: H(j) = 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 LC bandpass flters by : H(j) = ( L 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 nd-order band-pass flters (whch ncludes Q/ 0 and Q 0 /). Equatng these smlar terms we get: Q = L 0 Q 0 = C Q 0 = L Q 0 = C We can solve these two equatons to fnd: 0 = Q = 0 L LC /L = C The lower and upper cut-off frequences can now be found from the formulas on page 33. ECE65 Lecture Notes (F. Najmabad), Wnter

13 Method : In ths method, we drectly calculate the flter parameters smlar to the procedure followed for general form of transfer functon n page 3. 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 LC flter: H(j) = ( j L ) 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 L 0 C = 0 0 = LC K = H(j = j 0 ) = = The cut-off frequences can then be found by settng: H(j c ) = K = ( c L ) = c C whch can be solved to fnd u and l. Input and Output Impedance of band-pass LC flters Z = jl ( jc = j L ) C Z mn = occurs at = 0 ( Z o = jl ) Z o jc max = ECE65 Lecture Notes (F. Najmabad), Wnter

14 Wde-Band Band-Pass Flters Band-pass flters can be constructed by puttng a hgh-pass and a low-pass flter back to back as shown below. The hgh-pass flter sets the lower cut-off frequency and the low-pass flter sets the upper cut-off frequency of such a band-pass flter. H (j ) H (j ) H (j ) X H (j ) l = u= c c c c An example of such a band-pass flter s two C low-pass and hgh-pass flters put back to back. These flters are wdely used (when approprate, see below) nstead of an LC flter as nductors are usually bulky and take too much space on a crcut board. C o LowPass HghPass C In order to have good voltage couplng n the above crcut, the nput mpedance of the hgh-pass flter (actually Z mn = ) should be much larger than the output mpedance of the low-pass flter (actually Z o max = ), or we should have. In that case we can use un-termnated transfer functons: H(j) = H (j) H (j) = c = /( C ) c = /( C ) j/ c j c / H(j) = ( 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 Lecture Notes (F. Najmabad), Wnter

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 frst-order hgh-pass flters are gven n page 30. 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 band-pass 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 cut-off 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 frst-order hghpass and low-pass flters back to back. When c s not c, H(j) max becomes smaller than. Snce the cut-off frequences are located 3 db below the maxmum values, the cut-off frequences wll not be c and c (those frequences are 3 db lower than H(j) max = ). The lower cut-off frequency moves to a value lower than c and the upper cut-off 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, band-wdth becomes vanshngly small and Q should become very large. Because these flters work only when c c, they are called wde-band flters. For these wde-band flters ( c c ), we fnd from above: K = Q = c / c 0 = c c H(j) = j(/ c c /) We then substtute for Q and 0 n the expressons for cut-off frequences (page 33) to get: u = 0 4Q 0 Q = 0 Q ( ) 4Q l = 0 4Q 0 Q = ( ) 0 4Q Q ECE65 Lecture Notes (F. Najmabad), Wnter

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: l ( 0 ) Q 4Q = 0 Q Q = 0 Q = c What are Wde-Band and Narrow-Band Flters? Typcally, a wde-band flter s defned as a flter wth c c (or c 0 c ). In ths case, Q 0.35 (prove ths!). A narrow-band flter s usually defned as a flter wth B 0 (or B 0. 0 ). In ths case, Q 0. Example: Desgn a band-pass flter wth cut-off frequences of 60 Hz and 8 khz. The load for ths crcut s MΩ. As ths s wde-band, band-pass flter ( u / l = f u /f l = 50 ), we use two low- and hgh-pass C flter stages smlar to crcut above. The prototype of the crcut s shown below: The hgh-pass flter sets the lower cut-off frequency, and the MΩ load sets the output mpedance of ths stage. Thus: Z o max = MΩ 00 kω c (Hgh-pass) = l = C = π 60 C C o LowPass C = 0 3 kω HghPass 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 low-pass flter sets the upper cut-off frequency. The load for ths component s the nput resstance of the hgh-pass flter, Z mn = = 00 kω. Thus: Z o max = 00kΩ 0 kω c (Low-pass) = u = C = π C = 0 5 ECE65 Lecture Notes (F. Najmabad), Wnter

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 low-pass and hgh-pass flter stages n a wdeband, band-pass flter. However, the low-pass flter s usually placed before the hgh-pass flter because the value of capactors n such an arrangement wll be smaller. (Try redesgnng the above crcut wth low-pass and hgh-pass flter stages swtched to see that one capactor become much smaller and one much larger.) Exercse: Desgn an LC 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.) ECE65 Lecture Notes (F. Najmabad), Wnter

18 .4 Exercse Problems In crcut desgn, use 5% commercal resstor and capactor values (,.,.,.3,.5,.6,.8,,.,.4,.7, 3., 3.3, 3.6, 3.9, 4.3, 4.7, 5., 5.6, 6., 6.8, 7.5, 8., 9. 0 n where n s an nteger). Youcan also use 0 mh nductors Problem. Desgn a LC bandpass flter wth a lower cut-off 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 frst-order 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 band-pass flter wth a central frequency of 00 MHz (frequency of a FM staton) and a bandwdth of MHz. a) Desgn such a flter (Use a µh nductor). b) What are ts cut-off frequences? Problem 4. A telephone lne carres both voce band (0-4 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 Lecture Notes (F. Najmabad), Wnter

19 .5 Soluton to Exercse Problems Problem. Desgn a LC bandpass flter wth a lower cut-off 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 band-pass flter: B(Hz) = f u f l f u = 3 = 4 khz L C o B(rad/s) = πb(hz) = u = πf u = l = πf l = = u l = B(rad/s) = 0 Q Q = = 0.67 For the seres LC crcut: 0 = LC LC = 0 C = L 0 Q = 0 /L = = 0.63 µf ( ) L = 0 Q = B(rad/s) = LB = = 88 Ω Therefore, usng commercal values, the desgn values are L = 0 mh, = 80 Ω, and C = 0.68 µf. ECE65 Lecture Notes (F. Najmabad), Wnter 009 4

20 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 frst-order 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 hgh-pass flter. The prototype of the crcut s shown below. For ths crcut: H(j) = o = c = C Z mn = j c / Invertng Amp. C o L 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 Lecture Notes (F. Najmabad), Wnter 009 4

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

22 Problem 4. A telephone lne carres both voce band (0-4 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 low-pass flter as t should allow low-frequency sgnals (voce band) to go through whle elmnatng hgh-frequency sgnals (data band). The prototype of an C low-pass flter s shown and ts transfer functon s: H(j) = j/ c = jf/f c The cut-off 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 0-4 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 ). 0 db H(j ) C o 4 5 f (khz) Second, the transfer functon should be as small as possble at 5 khz. Ths requres that we choose the cut-off 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 Lecture Notes (F. Najmabad), Wnter