Passive Filters. References: Barbow (pp ), Hayes & Horowitz (pp 32-60), Rizzoni (Chap. 6)

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Passve Flters eferences: Barbow (pp 6575), Hayes & Horowtz (pp 360), zzon (Chap. 6) 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 /. Subscrpt v of H v s frequently dropped. As H(j) s complex number, t has both a magntude and a phase, flters n general ntroduce a phase dfference between nput and output sgnals. LowPass Flters An deal lowpass flter s transfer functon s shown. The frequency between pass and stop bands s called the cutoff frequency ( c ). All of the sgnals wth frequences below c are transmtted and all other sgnals are stopped. H(j ) Pass Band Stop Band c 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 Lowpass L flters L A seres L crcut as shown acts as a lowpass flter. For no load resstance (output s open crcut, L ), o can be found from the voltage dvder formula: o = jl H(j) = o = jl = j(l/) o To fnd the cutoff frequency, we note H(j) = (L/) ECE65 Lecture Notes (F. Najmabad), Wnter 006 7

H(j) s maxmum when denomnator s smallest,.e., 0 (alternatvely fnd d H(j) /d and set t equal to zero to fnd = 0). In ths case, H(j) max = H(j c ) = 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. Thus, the mnmum value of Z s an mportant number. Z s mnmum when the mpedance of the nductor s zero ( 0). Output Impedance: The output mpdenace can be found by kllng the source and fndng the equvalent mpdenace between output termnals: L Z o Z o = jl where the source resstance s gnored. Agan, the value of the output mpedance also 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, the maxmum value of Z o s an mportant number. Z o s maxmum when the mpedance of the nductor s nfnty ( ). ECE65 Lecture Notes (F. Najmabad), Wnter 006 8

Lowpass C flters A seres C crcut as shown also acts as a lowpass flter. For no load resstance (output s open crcut, L ): o = H(j) = /(jc) /(jc) = jc j(c) C o To fnd c, we follow a procedure smlar to L flters above to fnd c = C and H(j) = j/ c smlar to the voltage transfer functon for lowpass L flters (only c s dfferent). Input and Output Impedances: Followng the same procedure as for L flters, we fnd: Z jc Z o jc and and Frstorder Lowpass flters L and C flters above are part of the famly of frstorder flters (they nclude only one capactor or nductor). In general, the voltage transfer functon of a frstorder lowpass flter s n the form: H(j) = K j/ c The maxmum value of H(j) = K s called the flter gan. For L and C flters, K =. K H(j) = (/ c ) H(j) = tan ( c ). 0.8 H 0.6 Phase 0.4 0. 0 0 4 6 8 0 f/f c 0 0 0 30 40 50 60 70 80 90 0 4 6 8 0 f/f c For lowpass L flters: c L For lowpass C flters: c = C ECE65 Lecture Notes (F. Najmabad), Wnter 006 9

Bode Plots and Decbel The rato of output to nput power n a twoport network s usually expressed n Bell: ( ) 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), t s easy to show that 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 agan make analyss smpler as s seen below. For example, usng db defnton, we see that, there s 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. Bode plots of frstorder lowpass flters (K = ) are shown below (W denotes c ). ECE65 Lecture Notes (F. Najmabad), Wnter 006 0

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 H(j) = tan (/ c ). At low frequences, / c, H(j) 0 and at hgh frequences, / c, H(j) 90. At cutoff frequency, H(j) 45. ECE65 Lecture Notes (F. Najmabad), Wnter 006

Termnated L and C flters Termnated twoport networks are referred to those wth a fnte load resstance. For example, consder ths termnated lowpass C flter: oltage Transfer Functon: 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 cutoff frequency to a hgher frequency as L <. Input Impedance: Z jc L Output Impedance: Z o jc, Z o max As long as L Z o or L (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 resstance become the same as before. Termnated L lowpass 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. L Output Impedance: Z o = (jl), Here, the mpact of load s to shft the cutoff frequency to a lower value. Flter gan s not affected. Agan for L Z o or L (our condton for good voltage couplng), the shft n cutoff frequency dsappears and the flter wll look exactly lke an untermnated flter. ECE65 Lecture Notes (F. Najmabad), Wnter 006

Hghpass C flters C A seres C crcut as shown acts as a hghpass flter. For no load resstance (output open crcut), we have: o H(j) = o = /(jc) = j(/c) The gan of ths flter, H(j), s maxmum when denomnator s smallest,.e., leadng to H(j) max =. Then, the cutoff frequency can be found from H(j c ) = H(j) max = whch leads to c = C H(j) = j 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 Hghpass L flters A seres L crcut as shown also acts as a hghpass flter. For no load resstance (output open crcut), we have: L o c L H(j) = j c / Input Impedance: Z jl and Output Impedance: Z o jl and ECE65 Lecture Notes (F. Najmabad), Wnter 006 3

Frstorder Hghpass Flters In general, the voltage transfer functon of a frstorder hghpass flter s n the form: H(j) = K j c / The maxmum value of H(j) = K s called the flter gan. For L and C hghpass flters, K =. H(j) = K H(j) = tan ( c /) ( ) c For hghpass L flters: c L For hghpass C flters: c = 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 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 006 4

Termnated C hghpass flters The parameters of the termnated C flters can be found smlarly: oltage Transfer Functon: From the crcut, C o L H(j) = o = L L /(jc) = 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 / Here, the mpact of the load s to shft the cutoff frequency to a hgher frequency (as L < ). Input Impedance: Z = jc L Output Impedance: Z o jc L As long as L Z o or L (our condton for good voltage couplng), and the termnated C flter wll look lke a untermnated flter The shft n cutoff frequency dsappears and nput and output resstance become the same as before. Termnated L hghpass flters The parameters of the termnated L flters can be found smlarly: oltage Transfer Functon: H(j) / j c / Input Impedance: Z (jl) L c L L Output Impedance: Z o = (jl) We see that the load lowers the gan, K / < and shfts the cutoff frequency to a lower value. As long as L Z o or L (our condton for good voltage couplng), and the termnated L flter wll look lke a untermnated flter. ECE65 Lecture Notes (F. Najmabad), Wnter 006 5

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 : Lower 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) = H(j) = K ( jq 0 K ) 0 Q ( 0 0 ( ) H(j) = 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 Lecture Notes (F. Najmabad), Wnter 006 6

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, H(j) = K (purely real) H(j) = K (maxmum flter gan), and H(j) = 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 Lecture Notes (F. Najmabad), Wnter 006 7

Seres LC Bandpass flters Usng voltage dvder formula, we have H(j) = o = H(j) = jl /(jc) ( j L ) C 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 ndorder bandpass flters (whch ncludes Q/ 0 and Q 0 /). Equatng these smlar terms we get: Q 0 = L Q 0 = C Q 0 = L Q 0 = C We can solve these two equatons to fnd: 0 = Q = 0 L LC /L C ECE65 Lecture Notes (F. Najmabad), Wnter 006 8

The lower and upper cutoff frequences can now be found from the formulas on page 4. 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 cutoff frequences can then be found by settng: H(j c ) = K = ( c L ) = c C whch can be solved to fnd u and l smlar to page 3. Input and Output Impedance of bandpass LC flters Z = jl ( jc = j L ) C occurs at = 0 ( Z o = jl ) Z o jc max ECE65 Lecture Notes (F. Najmabad), Wnter 006 9

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 LC flter as nductors are usually bulky and take too much space on a crcut board. C C o Low Pass Hgh Pass In order to have good voltage couplng n the above crcut, the nput mpedance of the hghpass flter (actually ) should be much larger than the output mpedance of the lowpass flter (actually ), or we should have. In that case we can use untermnated 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 006 30

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 and real ones for frstorder hghpass flters are gven n page. 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 = H(j) = c / c j(/ c c /) 0 = c c We then substtute for Q and 0 n the expressons for cutoff frequences (page 4) to get: u = 0 4Q 0 Q = 0 Q ( ) 4Q l = 0 4Q 0 Q = ( ) 0 4Q Q ECE65 Lecture Notes (F. Najmabad), Wnter 006 3

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. ECE65 Lecture Notes (F. Najmabad), Wnter 006 3

Example: Desgn a bandpass flter to pass sgnals between 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: C o MΩ 00 kω c (Hghpass) = l = C = π 60 Low Pass C = 0 3 kω Hgh Pass 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, = 00 kω. Thus: 00kΩ 0 kω c (Lowpass) = u = C = π 8 0 3 C = 0 5 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.) ECE65 Lecture Notes (F. Najmabad), Wnter 006 33