1) Design a second order low pass filter with a cutoff frequency of 1 khz.

Transcription

1 ECE 44 Filter Leture and Homework 4 Solution FILTE LECTUE ) Deign a eond order low pa ilter with a uto requeny o khz eond order ilter ha the pole at angle o + and 45 degree rom the negative real axi, whih are:, 77* j77* Combining the pole to determine the ilter traner untion: H () ( )( ) Setting thi equal to the eond order ative ilter traner untion give: H () ( )( ) ( k) C ( C) where k = the gain o the ampliier tage It i obviou that ω =/C and -k = or k = 586 We an hooe any value or and C but it i bet to hooe the C baed on available tandard value or C =uf Thi give an o 595K or a tandard value o 6K The ampliier i a non-inverting op-amp ampliier and k = + / Thereore the ratio o the eedbak reitor to the reitor between the + terminal and ground i 586 Chooing tandard value o 68K and K, give a ratio 57 ) Deign a third order ilter with a uto requeny o khz by aading a irt order with a eond order ilter NOTE: The pole are at dierent loation or eah etion third order ha pole paed at 6 degree and a pole on the negative real axi The ilter will be built rom a irt order etion ollowed by a eond order etion, but not the deign value rom above but with the pole at:, 5* j866* Combining the pole a above give a denominator o H() o: Equating thi to the repone o the eond order ative ilter give: ω =/C and -k = or k = The irt order etion and the eond order etion ue the ame reitor and apaitor ombination a above, uf and 6K The non-inverting op-amp ampliier will have a gain o two, thereore the eedbak reitor and the reitor to ground are any two equal reitor K would be pratial

2 ) Deign a ourth order ilter with a uto requeny o khz by aading two eond order ilter NOTE: The pole are in dierent loation or eah etion The pole on the ourth order ilter would be paed at 45 degree tarting at 5 degree above and below the negative real axi: -675, -5, 5, and 675 degree The ilter would be made o two eond order etion, one ater the other Their order i not important Their pole would be:, 8* j94* and, 94* j8* Combining thee give: 764 and 848 gain the reitor and apaitor pair or the ilter are all the ame: uf and 6K The only dierene between the two etion i the gain o the op-amp ampliier In the irt etion -k=764 or k=6 Thi ould be realized with a eed bak reitor o k and a reitor to ground o k In the eond etion, -k=848 or k =5 Thi ould be realized with a eedbak reitor o 5K and a reitor to ground o K HOMEWOK 4 ) Deign a low pa ilter that ha a pa band to Hz and the band begin at KHz with a minimum attenuation o -6 db Uing the ormula or band attenuation in a low pa Butterworth ilter: N where N i the ilter order, we an olve or N uing 6 N whih yield n=78 We mut round UP to to enure our ilter meet peiiation The deign o the ilter proeed exatly a problem in the ilter leture homework ) Deign a low pa ilter that ha a pa band to Hz and the band begin at KHz with a minimum attenuation o -8dB

3 Uing the proedure outlined above, the order o the ilter an be determined a 4, rounded up to 4 The ilter an be realized in two eond order etion a outlined in problem in the ilter leture homework The gain or the inverting op-amp ampliier will be exatly the ame: 6 and 5 The and C value to determine the uto requeny will be dierent and determined by: ω =/C Chooing uf again or C, i796k, the nearet tandard value i 8K ) Deign a high pa ilter that ha a pa band beginning at 5 Hz and the band end at Hz with a minimum attenuation o -4 db In thi ae the deign proeed lightly dierently The attenuation ormula or a HIGH pa Butterworth ilter i: N giving an order o 8, rounded up to The deign an proeed or a third order ilter an proeed a in the third order low pa ilter outlined above The dierene i that the and C reitor and apaitor with plae The gain o the eond order tage i the ame a beore and ω =/C Uing uf again, =8K or K or a tandard value 4) Deign a bandpa ilter that ha a enter requeny o Hz and a bandwidth o Hz Thi ilter i deined by it enter requeny, Hz, and it Q whih i the enter requeny divided by the bandwidth or C Vi C - + Vo We an hooe a onvenient apaitor o uf and, auming C= C, alulate uing:

4 Q whih give 8K, or the nearet tandard value o K To ind, the parallel C o ombination o and, we ue: ' whih give 7958 ohm The gain o thi ilter would be: or 7 To adjut the gain, 4Q ' we would ue and a a voltage divider and adjut their value o that the deign gain i: while keeping ' ) Deign a bandpa ilter that ha a enter requeny o 9KHz and a bandwidth o KHz Thi ilter i deined by it enter requeny, 9Hz, and it Q whih i the enter requeny divided by the bandwidth or 9 We an hooe a onvenient apaitor o uf and, auming C= C, alulate uing: Q whih give 59K, or the nearet tandard value o 6K To ind, the parallel C o ombination o and, we ue: ' 4Q whih give ohm The gain o thi ilter would be: or 76 To adjut the gain, ' we would ue and a a voltage divider and adjut their value o that the deign gain i: while keeping ' 6) Deign a Colpitt rytal ontrolled oillator at MHz uing a BJT The rytal ha a erie reitane o ohm and the BJT ha a beta o 5 and i biaed to have a g m o 85 What value would you hooe or the two apaitor in the eedbak iruit?

5 Q C C Large L I we ue the rule o thumb to make the real part o the impedane looking into the Colpitt iruit rom the emitter o the BJT to be 5 gm or 99 ohm in thi ae, we an ue that value in the ormula: C e( Z i ) where i the erie reitane o the rytal, to alulate C In thi ae it give 988 pf, or whih the nearet tandard value i pf To ind the eond apaitor, we need to look at the tranormer ation perormed by C and C, uh that the tep up ratio i: C C N C olletor with a gain o Thi orm the beta o our eedbak loop bak to the bae The ampliier i a ommon e e g m Sine e =e(z i )=5/g m, the gain i 5/5 or98 Thereore to atiy the oillation riterion o, N would be Thi would mean that C would be muh larger than C In reality, the better value would be to make to enure the oillator tart Thi would make C =C