Chapter 14 Analog Filters
¾ 14.1 General Considerations ¾ 14.2 First-Order Filters ¾ 14.3 Second-Order Filters ¾ 14.4 Active Filters ¾ 14.5 Approximation of Filter Response
1 Outline of the Chapter
CH 14 Analog Filters 2 Why We Need Filters
¾ In order to eliminate the unwanted interference that accompanies a signal, a filter is needed.
CH 14 Analog Filters 3 Filter Characteristics
sufficiently narrow
bad stopband attenuation attenuated amplified
¾ Ideally, a filter needs to have a flat pass band and a sharp roll- off in its transition band. ¾ Realistically, it has a rippling pass/stop band and a transition band.
CH 14 Analog Filters 4 Example 14.1: Filter I
Problem : Adjacent channel interference is 25 dB above the signal. Determine the required stopband attenuation if Signal to Interference ratio must exceed 15 dB.
Solution: A filter with stopband attenuation of 40 dB
CH 14 Analog Filters 5 Example 14.2: Filter II
Problem: Adjacent 60-Hz channel interference is 40 dB above the signal. Determine the required stopband attenuation to ensure that the signal level remains 20dB above the interferer level.
Solution: A high-pass filter with stopband attenuation of 60 dB at 60Hz.
CH 14 Analog Filters 6 Example 14.3: Filter III
¾ A bandpass filter around 1.5 GHz is required to reject the adjacent Cellular and PCS signals.
CH 14 Analog Filters 7 Classification of Filters I
CH 14 Analog Filters 8 Classification of Filters II
QCV121=
QCV222=
Continuous-time Discrete-time
if VVC122> , absorbs charge from V 1
and delivers it to V2 ⇒≈ a resistor
CH 14 Analog Filters 9 Classification of Filters III
Passive Active
CH 14 Analog Filters 10 Summary of Filter Classifications
CH 14 Analog Filters 11 Filter Transfer Function
(a) (b)
¾ Filter (a) has a transfer function with -20dB/dec roll-off. ¾ Filter (b) has a transfer function with -40dB/dec roll-off and provides a higher selectivity.
CH 14 Analog Filters 12 General Transfer Function
(sz−−)( sz) ( sz −) Hs()= α 12 m ()()sp−−12 sp () sp −n
z = zero frequencies k = σ + jω pk = pole frequencies
CH 14 Analog Filters 13 Example 14.4 : Pole-Zero Diagram
1 RC12 s+1 Cs1 Hsa ()= Hsb ()= Hsc ()= 2 R11Cs+1 RC11()1+ C 2 s+ RLCs11 1+ Ls 1+ R 1
CH 14 Analog Filters 14 Example 14.5: Position of the poles
Impulse response contains
exp(pkkkttjt )= exp(σ )exp(ω )
Poles on the RHP Poles on the jω axis Poles on the LHP Unstable Oscillatory Decaying (no good) (no good) (good)
CH 14 Analog Filters 15 Transfer Function
()s −−zsz( ) ( sz −) Hs()= α 12 m ()()s −−psp12 () sp −n
¾ The order of the numerator m ≤ The order of the denominator n Otherwise, H(s)→∞ as s→∞. ¾ For a physically-realizable transfer function, complex zeros or poles occur in conjugate pairs. z =+σ jω z = σ − jω ω 11 121 1 ¾ If a zero is located on the jω axis, z1,2=± jω1 , H(s) drops to zero at ω1. 22 The numerator contains a product such as (sj−+=+ω11)( sjωω) s 1 ,
which vanishes at sj= 1
CH 14 Analog Filters 16 Imaginary Zeros
¾ Imaginary zero is used to create a null at certain frequency. For this reason, imaginary zeros are placed only in the stop band.
CH 14 Analog Filters 17 Sensitivity
P=Filter Parameter P dP dC SC = P C C=Component Value Example:
In simple RC filter, the -3dB corner frequency is given by 1/(R1C1) Errors in the cut-off frequency: (a) the value of components varies with process and temperature in ICs (b) The available values of components deviate from those required by the design
¾ Sensitivity indicates the variation of a filter parameter due to variation of a component value.
CH 14 Analog Filters 18 Example 14.6: Sensitivity
Problem: Determine the sensitivity of ω0 with respect to R1.
ω0 =1/(R1C1 ) ω d 0 −1 = 2 dR1 R1 C1 ωd dR 0 = − 1 ω0 R1 Sω0 = −1 R1
¾ For example, a +5% change in R1 translates to a -5% error in ω0.
CH 14 Analog Filters 19 First-Order Filters
sz+ Hs()= α 1 sp+ 1
¾ First-order filters are represented by the transfer function shown above. ¾ Low/high pass filters can be realized by changing the relative positions of poles and zeros. CH 14 Analog Filters 20 Example 14.8: First-Order. Filter I
V RRCs(1)+ out ()s = 211 VRRCCsRRin 12() 1+++ 2 1 2
pCCRR=−[( + ) || ]−1 z111= −/1(RC ), 11212
R2 C ω →⇒0 ω →∞⇒ 1 R + R 12 CC12+
11 11 < > RC11()() C 1+ C 2 R 1 R 2 RC()() C+ C R R 11 1 2 1 2 R C > R C CR R2C2 < R1C1 CR 2 2 1 1 11+<+21 11+>+21 CR 12 CR12
CH 14 Analog Filters 21 Example 14.9: First-Order Filter II
1 −||()R2 Vout Cs2 Iin ()s = Iin 1 Vin R1 || Cs1 R RCs+1 virtual short =−211 ⋅ R122RCs+1
R2 C ω →⇒−0 ω →∞⇒− 1 R 1 C2
R2C2 < R1C1 R2C2 > R1C1 CH 14 Analog Filters 22 Second-Order Filters
General transfer function αss2 + βγ+ Hs()= ω ss22++n ω Q n
ωn 1 pj1,2 =− ±n 1 − 2Q ω 4Q2
¾ Second-order filters are characterized by the “biquadratic” equation with two complex poles shown above. ¾ When Q increases, the real part decreases while the imaginary part
approaches ± ωn. => the poles look very imaginary thereby bringing the circuit closer to instability.
CH 14 Analog Filters 23 . Second-Order Low-Pass Filter ω 2 2 γ Hj()= 2 2 α = β = 0 22 ⎛⎞n ()ωn −+ωω⎜⎟ω ⎝⎠Q
H(ω p ) 2 Q> 2
∂ Hj()ω 2 = 0 H(ω p ) 2 Q> ∂ ω 2
γ 2 n ω 2 for Q > 2 Peak magnitude normalized to the passband magnitude: QQ/−1(4)21−
CH 14 Analog Filters 24 Example 14.10: Second-Order LPF
Problem: Q of a second-order LPF = 3. Estimate the magnitude and frequency of the peak in the frequency response. ω 2 2 γ Hj()= 2 ωω2 2 22 ⎛⎞n ()nn−+⎜⎟ω ⎝⎠3 Q = 3 QQ/11/(4)− 2 ≈ 3
2 ωnn11/(2)− Q ≈ ω
CH 14 Analog Filters 25 Second-Order High-Pass Filter
The zero(s) must fall below the poles α s 2 Hs()= ω β = γ = 0 ss22++n ω Q n
2 Frequency of the peak: 2 for Q > ωn /−/11(2)Q 2 Peak magnitude normalized to the passband magnitude: QQ/−1(4)21−
CH 14 Analog Filters 26 Second-Order Band-Pass Filter
β s Hs()= α = γ = 0 22ωn ss++ωn The magnitude approaches zero Q for both sÆ0 and sÆ ∞, reaching a maximum in between
CH 14 Analog Filters 27 Example 14.2: -3-dB Bandwidth
Problem: Determine the -3dB bandwidth of a band-pass response. β s Hs()= ω ()ss22++n ω Q n
The response reaches 1/ 2 times its peak value at -3dB βω β22 22Q frequency = 2 ()()ωω222−+ ωn 22ωn n Qω
⎡ 11⎤ ωω1,2=+± 0 ⎢ 1 2 ⎥ ⎣⎢ 4Q 2Q ⎦⎥ ω BW = 0 Q
CH 14 Analog Filters 28 LC Realization of Second-Order Filters
ω0
Ls 1 1 Z01 = at ω =∞ 0 and ZLs11==()|| 2 Z10=∞ at ωω = Cs1 LC11 s + 1
1/2 ¾ An LC tank realizes two imaginary poles at ±j/(L1C1) , which 1/2 implies infinite impedance at ω=1/(L1C1) .
CH 14 Analog Filters 29 Example 14.13: LC Tank
1 Ls1 ZLs11==()|| 2 Cs1 LC11 s + 1
¾ At ω=0, the inductor acts as a short. Z0= at ω =∞ 0 and ¾ At ω=∞, the capacitor acts as a short. 1
CH 14 Analog Filters 30 RLC Realization of Second-Order Filters
Ls111 RLs ZR21==|| 22 LCs11+++1 RLCs 111 Ls 1 R 1 RLs RLs ==11 11 2 11 22ωn RLC111() s++ s RLC111() s++ s n RC11 LC 11 Q ω
1 C1 ωn ==, QR1 LC11 L1
ωn 1 pj1,2 =− ±n 1 − 2 2Q ω 4Q
11 L1 =− ±j 1 − 2 2RC11 LC11 4R11C
¾ With a resistor, the poles are no longer pure imaginary which implies there will be no infinite impedance at any ω.
CH 14 Analog Filters 31 Voltage Divider Using General Impedances
V Z out ()s = P VZZin S+ P
Low-pass High-pass Band-pass
CH 14 Analog Filters 32 Low-pass Filter Implementation with Voltage Divider
ZLsS = 1 →∞ as s →∞ 1 ZRsP = ||1 →0 as → ∞ Cs1
Vout R1 ()s = 2 Vin RCLs111++ Ls 1 R 1
CH 14 Analog Filters 33 Example 14.14: Frequency Peaking
V out R1 ()s = 2 Vin RC111 Ls+ Ls 1+ R 1
2 2 222 Let DRRCLL=−( 1111ω ) + 1ω
Voltage gain greater than unity (peaking) occursω when a solution exists for
2 dD 22 =−2(R111CL )( R 1 − RCL 111ω ) + L 1 d()2 11 = 0 ⇒=ω −22 >0 LC11 2RC 1 1
C 1 Thus, when QR=⋅1 > , C C 1 2R2 1 > 1 ∵ QR= 1 L1 2 1 L 1 L 1 1 peaking occurs. CH 14 Analog Filters 34 Example 14.15: Low-pass Circuit Comparison
(a) (b)
Good Bad
¾ The circuit (a) has a -40dB/dec roll-off at high frequency. ¾ However, the circuit (b) exhibits only a -20dB/dec roll-off since the parallel combination of L1 and R1 is dominated by R1
because L1ω→∞, thereby reduces the circuit to R1 and C1.
CH 14 Analog Filters 35 High-pass Filter Implementation with Voltage Divider
α s 2 ∵ Hs( )= for high pass filter 22ωn ss++n Q ω
V ()||Ls R LCRs2 out ()s ==11 111 1 2 V in RC111 Ls+ Ls 1+ R 1 ()||Ls11 R + Cs1
CH 14 Analog Filters 36 Band-pass Filter Implementation with Voltage Divider
Zp must contain both a capacitor and an inductor so that it approaches zero as s Æ 0 or s Æ ∞
Z β s p ∵ Hs()= ω ss22++n ω Q n
1 ()||Ls V 1 Cs Ls out ()s ==11 1 2 Vin R111CLs+ Ls 1+ R 1 ()||Ls11+ R Cs1
CH 14 Analog Filters 37 Summary
¾ Analog filters prove essential in removing unwanted frequency components that may accompany a desired signal. ¾ The frequency response of a filter consists of a passband, stopband, and a transition band between the two. The passband and stopband may exhibit some ripple. ¾ Filters can be classified as LP, HP, BP, or BR topologies. They can be realized as continuous-time or discrete-time configurations, and as passive or active circuits. ¾ The frequency response of filters has dependences on various component values and, therefore, suffers from sensitivity to component variations. ¾ First-order passive or active filters can readily provide a LP or HP response, but their transition band is quite wide and stopband attenuation only moderate. ¾ Second-order filters have a greater stopband attenuation and are widely used. For a well-behaved frequency and time response, the Q of these filters is typically maintained below 2/2 Why Active Filter?
¾ Passive filters constrain the type of transfer function.
¾ They may require bulky inductors.
CH 14 Analog Filters 39 Sallen and Key (SK) Filter: Low-Pass
VVout≈ X
By applying KCL
VVCsRV= + Y out 2 2 out VCsout 2 V 1 out ()s = 2 2 αss++βγ Vin R1212RCCs+ R 1++() R 2 Cs 2 1 Hs()= ω ss22++n ω 1 n 1 C1 ω = Q QRR= 12 n RR12+ C 2 RRCC1212
¾ Sallen and Key filters are examples of active filters. This particular filter implements a low-pass, second-order transfer function.
CH 14 Analog Filters 40 Example 14.16: SK Filter with Voltage Gain
V1RRVout=+( 3 4) X
R 1+ 3 V R out ()s = 4 Vin 2 ⎛⎞R3 R1212RCCs+ ⎜⎟ RC 12+− RC 22 R 1 C 1 s +1 ⎝⎠R4
CH 14 Analog Filters 41 Example 14.17: SK Filter Poles
Problem: Assuming R1=R2, C1=C2, Does such a filter contain complex poles?
R 1+ 3 V R Q ↑ out ()s = 4 R Vin 2 3 R1212RCCs+ ()1 RC 12+− RC 22 R 1 C 1 s + R4 1 RC RC RC R =+−12 22 11 3 QRCRCRCR21 11 22 4
1 Q = R 2 − 3 R4
¾ The poles begin with real, equal values for RR 34 /= 0 and become complex for RR 34 /> 0 .
CH 14 Analog Filters 42 Sensitivity in Low-Pass SK Filter
the gain of filter V out = VX KRR= 1+ 34
1 dω ∵ω = n = n 1 RC12 RC 22 RC 11R3 dR1 ∵ =+− RRCC1212 QRCRCRCR21 11 22 4
1 ⎛⎞ ωωωωnnnn QQ1 R22CRC 12 SSSSRRCC= ===−SS=− =− + Q⎜⎟ + 1212 CC12 ⎜⎟ 2 2 ⎝⎠R11CRC 21
1 RC R C SSQQ=− =− + Q22 SQKQ = 11 RR12 K 2 RC11 R22C
CH 14 Analog Filters 43 Example 14.18: SK Filter Sensitivity I
Problem: Determine the Q sensitivities of the SK filter for the common choice R1=R2=R, C1=C2=C.
QQ11 SSRR=− =− + 1223− K QQ12 SSCC=− =− + 1223− K K S Q = K 3 − K
With K =1, SSQQ==0 RR12
QQQ1 SSSCC===K 12 2
CH 14 Analog Filters 44 Integrator-Based Biquads
2 ωn ωn 1 − Vsout () − . Vs() s2 Qsout
2 Vout αs α 2 ()s = ωωnn1 V 22ω Vsout()=− Vs in (). Vs out() − Vs out () in ss++n ω Qs s2 Q n
¾ It is possible to use integrators to implement biquadratic transfer functions.
CH 14 Analog Filters 45 KHN (Kerwin, Huelsman, and Newcomb) Biquads
Simplified diagram
111 VR+ VR⎛⎞ R R VVVV=−, =− = VVV=+−in546 X 1 6 XRCs out Y RC s X RRCC s2 out out⎜⎟ Y 11 2 2 1212 RR45+ ⎝⎠ R 3 R 3
α 2 ωωnn1 Comparing with Vsout()=− Vs in () . Vs out() − Vs out () Qs s2
R56⎛⎞R ωn R4 1 ⎛⎞R6 2 R6 1 α =+⎜⎟1 =+..1⎜⎟ωn = . R45+ RR⎝⎠ 3QRRRCR4511+ ⎝⎠ 3 R31212RRCC
CH 14 Analog Filters 46 Calculation of Vout with Simplified Circuit
To obtain V,A
R4 R5 VVA = X VVA = in V0in = V0X = RR45+ RR45+ VA R4XVRV+ 5in VVAA=+= V A V0in== V0 X RR45+
To obtain VV,Vout for given A we ground Y
VA ⎛⎞R6 VRRV1out=+=+() 3 6 A ⎜⎟ R33⎝⎠R
To obtain VV,Vout for given Y we ground A
VY ⎛⎞R6 VRVout=− 6 = Y ⎜⎟ R33⎝⎠R Versatility of KHN Biquads
V αs2 High-pass: out ()s = ω Vin ss22++n ω Q n
2 VX αs −1 V 1 Band-pass: ()s = . ∵ V =− out VRCsin 22ωn 11 X ss++n R11sC Q ω 2 VY αs 1 Low-pass: ()s = . VX 1 2 ∵ VY =− Vin 22ωn R1212RCCs ss++n R22sC Q ω CH 14 Analog Filters 48 Sensitivity in KHN Biquads
ω S n = 0.5 RRCC12124536,,,,,,, RRRR
R SSQQ= 0.5, =<5 1, RRCC1212,,, RR 45 , RR45+ Q RR− RC if RR= , S Q = 36 22 36 RR36, Q 2 R5 RRRC3611 then S vanishes 1+ RR36, R4 CH 14 Analog Filters 49 Tow-Thomas Biquad
V 1 out ⋅ RCs22 R 4
VVX = − Y
Vin R 1 VVRCsYout= − ()22
⎛⎞⎛⎞VVout11 in ⎜⎟⎜⎟⋅+R3 =−Vout ⎝⎠⎝⎠RCs22 R 4 R 1 sC 1
CH 14 Analog Filters 50 Tow-Thomas Biquad
BFP β s ∵ Hs()= ω ss22++n ω Q n LFP γ ∵ Hs()= ω ss22++n ω Q n
VRRRout 234 Cs2 Band-pass: =− . 2 VRin 1 RRRCCs23412+ RRCs 242+ R 3
VY RR34 1 Low-pass: = . 2 VRin 1 R23412RRCCs+ RRCs 242+ R 3
CH 14 Analog Filters 51 Differential Tow-Thomas Biquads
¾ An important advantage of this topology over the KHN biquad is accrued in integrated circuit design, where differential integrators obviate the need for the inverting stage in the loop.
CH 14 Analog Filters 52 Example 14.20: Tow-Thomas Biquad
α s2 + β s + γ Hs()= ω ss22++n ω Q n
VY RR34 1 Low-pass: = . 2 VRin 1 RRRCCs23412+ RRCs 242+ R 3
Note that ωn and Q of the Tow-Thomas filter can be adjusted (tuned) independently.
1 1 R RC ω n = Q − 1 = 242 R 2412RCC RC31
Adjusted by R2 or R4 Adjusted by R3
CH 14 Analog Filters 53 Antoniou General Impedance Converter
VVVV135= ==X
VX VZV44= + X Z5 I X IZ 3 VV43− VZX 4 IZ 3 = = ⋅ Z3 Z53Z
VX IZ 3 VVZI2323= − Z Z5 VZX 4 = VZX −⋅2 ⋅ Z53Z
VVX − 2 I X = ZZ13 Z1 Zin = Z5 Z Z = V 24 ZZ24 X Z135ZZ ¾ It is possible to simulate the behavior of an inductor by using active circuits in feedback with properly chosen passive elements.
CH 14 Analog Filters 54 Simulated Inductor
Z1 Z2
Z5 Z3 Z4
ZZ13 ∵ZZin = 5 ZRRCsin= X Y ZZ24 Thus, LRRCeq= X Y
¾ By proper choices of Z1-Z4, Zin has become an impedance that increases with frequency, simulating inductive effect.
CH 14 Analog Filters 55 High-Pass Filter with SI
2 Vout Ls1 ()s = 2 Vin RCLs111+ Ls 1+ R 1
¾ With the inductor simulated at the output, the transfer function resembles a second-order high-pass filter.
CH 14 Analog Filters 56 Example 14.22: High-Pass Filter with SI
VVVVout = 135==
Node 4 can also serve as an output.
RRXY⎛⎞ ⎛⎞RY VVVV= ⇒=+1 ⇒ VV4 =+out ⎜⎟1 5445⎜⎟ R RRYX+ ⎝⎠ R X ⎝⎠X
¾ V4 is better than Vout since the output impedance is lower.
CH 14 Analog Filters 57 Low-Pass Filter with Super Capacitor
¾ How to build a floating inductor to derive a low-pass filter? Not possible. So use a super capacitor.
Z1 Z2 1 Zin = Cs() RX Cs +1 Z Z4 5 γ Hs()= Z ω 3 ss22++n ω Q n VZ out= in VZRin in + 1 1 = 22 RR11X C s+ RCs+ 1
CH 14 Analog Filters 58 Example 14.23: Poor Low-Pass Filter
VVVVout = 135==
⎛⎞1 VCsout ⎜⎟+ ⎝⎠RX
⎡⎤⎛⎞1 VV4 =++=+⎢⎥out⎜⎟ CsRVV X out out()2 RCs X ⎣⎦⎝⎠RX
¾ Node 4 is no longer a scaled version of the Vout. Therefore the output can only be sensed at node 1, suffering from a high impedance.
CH 14 Analog Filters 59 Summary
¾ Continuous-time passive second-order filters employ RSC sections, but they become impractical at very low frequencies (because of large physical size of inductors and capacitors). ¾ Active filters employ op amps, resistors, and capacitors to create the desired frequency response. The Sallen and Key topology is an example. ¾ Second-order active (biquad) sections can be based on integrators. Examples include the KHN biquad and the Tow- Thomas biquad. ¾ Biquads can also incorporate simulated inductors, which are derived from the “general impedance converter” (GIC). The GIC can yield large inductor or capacitor values through the use of two op amps. Frequency Response Template
¾ With all the specifications on pass/stop band ripples and transition band slope, one can create a filter template that will lend itself to transfer function approximation.
CH 14 Analog Filters 61 Butterworth Response
1 Hj()ω = 2n ⎛⎞ 1+ ⎜⎟ ⎝⎠ω0 ω
ω-3dB=ω0, for all n.
¾ The Butterworth response completely avoids ripples in the pass/stop bands at the expense of the transition band slope.
To obtain the poles, we make a reverse substitution, ω = sj / , and set the denominarot to zero:
CH 14 Analog Filters 62 Butterworth Response
2n 22nn22nn2n ⎛⎞s sj+ ()ωω=⇒00 sj +()() = 10+ = 00 ⎜⎟ n 2n jω0 2n ⎝⎠ sj+ ()−=()0 0 This polynomial has 2n roots given by ω jkπ ⎛⎞21− ω pjknk ==0 exp exp⎜⎟ , 1,2,...,2 22⎝⎠n ωωωω− jπ jπ exp exp For 2nd order filter, 4 4 jjjj3579π πππ p = exp , exp , exp , exp k =1,2,3,4 04444 0 0 0 negative real positive real But only the roots having a negative real part are acceptable ωπ jkπ ⎛⎞21− pk ==0 exp exp⎜⎟jkn , 1,2, , 22⎝⎠n CH 14 Analog Filters 63 Poles of the Butterworth Response
2nd‐Order nth‐Order ()()()− pp− ⋅⋅⋅ − p Hs()= 12 n ()()()spsp−−⋅⋅⋅−12 spn where the factor in the numerator is included to yield H(s=0)=1 CH 14 Analog Filters 64 Example 14.24: Order of Butterworth Filter
Specification: passband flatness of
0.45 dB for f < f1=1 MHz, stopband attenuation of 9 dB at f2=2 MHz.
|=Hf(2MHz)0355 |=. ≈ −9 dB | Hf(1MHz)0951 =|=.≈ −0.45 dB 2
1 2 1 2 2n = 095. 2n =.0355 ⎛⎞2π f1 ⎛⎞2π f 1+ ⎜⎟ 1+ ⎜⎟2 ⎝⎠ω0 ⎝⎠ω0
2n ⎛⎞ f2 ff= 2 ⎜⎟= 64.2 21 n ==×.3,ωπ0 2 (1 45MHz) ⎝⎠f1 ¾ The minimum order of the Butterworth filter is three.
CH 14 Analog Filters 65 Example 14.25: Butterworth Response
Using a Sallen and Key topology, design a Butterworth filter for the response derived in Example 14.24.ωωω jjj234π ππ p = exp , exp , exp k=1,2,3π 0333 0 0 = −ω0
π ⎛⎞22π π pMHzj1 =+2*(1.45 )*cos⎜⎟ sin 2 ⎝⎠33 π 3 2nd-order SK ⎛⎞22π π pMHzj3 =−2*(1.45 )*cos⎜⎟ sin 4 ⎝⎠33 π 3 p2 = 2π *(1.45MHz ) RC section
CH 14 Analog Filters 66 Example 14.25: Butterworth Response (cont’d)
C1
αss2 + βγ+ R Hs()= ω R1 2 22n R3 ss++ω C Q n 2 C2 ωπ πππ2 ()()−−pp13 [2π ×. (1 45MHz)] HsSK ()==2 2 (spsp−−13 )( ) s −×.[] 4 (1 45MHz)cos(2 /+×. 3) s [2 (1 45MHz)] 2π 1 ⎛⎞ ∵ωn = n = 2×. (1 45MHz) and Q = 1/⎜⎟ 2cos =→ 1 ⎝⎠3 R1212RCC 1 C ∵QRR= 1 12 RR12= =Ω1k , C 2 =54.9pF, and C 1 =4 C 2 RR12+ C 2 1 ωn = 1 4R22C =×.2π (1 45MHz) →=ΩRC33 1k and = 109.8pF 11 RC33
CH 14 Analog Filters 67 Chebyshev Response
1 Hj()ω = 22⎛⎞ω 1+ ε Cn ⎜⎟ ⎝⎠0 ω Chebyshev Polynomial
ε : the amount of ripple 2 Cnn (/ωω0 ):the "Chebyshev polynomial" of th order
¾ The Chebyshev response provides an “equiripple” pass/stop band response.
CH 14 Analog Filters 68 Chebyshev Polynomial
1 Hj()ω = ⎛⎞ω 22⎛⎞ω H ⎜⎟ 1+ ε Cn ⎜⎟ ω ⎝⎠0 ⎝⎠0 ω 1 1 1+ ε 2
Chebyshev polynomial for Resulting transfer function for n=1,2,3 n=2,3
⎛⎞ωω ⎛−1 ⎞ Cnn ⎜⎟= ⎛⎞−1 ω = cosh⎜⎟n coshωω , ω > ω0 ⎝⎠0 CH 14 Analog Filters 69 ω Chebyshev Response 2 Peak-to-peak Ripple|dB =+ 20log 1 ε 1 1 |=HjPB ()|ω |=HjSB ()|ω 22⎛⎞− 1ω ⎛⎞ω 1coscos+ ε ⎜⎟n 1+ ε 22 coshn cosh− 1 ω ⎜⎟ ⎝⎠0 ⎝⎠ω0 CH 14 Analog Filters 70 Example 14.26: Chebyshev Response Suppose the filter required in Example 14.24 is realized with third- order Chebyshev response. Determine the attenuation at 2MHz. 1 1 ε Hj()ω = =.0 95 →ε =. 0 329 3 2 2 ⎡ ⎤ 1+ 2 ⎛⎞ 14+−⎢ ⎜⎟ 3⎥ ⎣⎢ ⎝⎠00⎦⎥ ε ωω ω =2π (1MHz) 0 ωω Hj( 2π (2MHz)) ==− 0.116 18.7dB ¾ A third-order Chebyshev response provides an attenuation of -18.7 dB a 2MHz. CH 14 Analog Filters 71 Example 14.27: Order of Chebyshev Filter Specification: Passband ripple: 1 dB Bandwidth: 5 MHz Attenuation at 10 MHz: 30 dB What’s the order? 1 dB= 20log 1+→=.εε2 0 509 Attenuation at ω = 2ω0 = 10 MHz: 30 dB 1 =.0 0316 1+. 0 50922 cosh (n cosh− 1 2) cosh2 (1. 317nnn )=→>.→= 3862 3 66 4 CH 14 Analog Filters 72 Example 14.28: Chebyshev Filter Design Using two SK stages, design a filter that satisfies the requirements in Example 14.27. ωω (21kk−−)ππ⎛⎞11−−11( 21) ⎛⎞ 11 pjk =−00sin sinh⎜⎟ sinh + cos cosh ⎜⎟ sinh 22nn⎝⎠ε n ⎝⎠ nε p1,4=−0.140ω 0 ± 0.983 jω 0 pj2,3= −±0.337ω 0 0.407 ω 0 SK1 SK2 CH 14 Analog Filters 73 Example 14.28: Chebyshev Filter Design (cont’d) ()()−−pp14 ()()− pp23− HsSK1()= HsSK 2 ()= ()()spsp−−14 ()()spsp−−23 0986. ω 2 0. 279ω 2 = 0 = 0 22ss22+.0 674ω +. 0 279ω ss+.028ω00 +. 0986ω 00 ωn10=.0 993ωπ = 2 × (4 . 965MHz) ωn20= 0.=×. 528ωπ 2 (2 64MHz) Q =.355 1 Q2 =.0783 . R12==ΩRCC1 k , 1 =. 50 4 2 R12==ΩRCC1 k , 1 =. 2 45 2 1 1 =×.2π (4 965MHz) =×.2(264π MHz) 245. RC 50. 4RC12 12 →=.C 38 5 pF, C = 94.3 pF →=CC214.52 pF, = 227.8 pF 21 CH 14 Analog Filters 74 Summary ¾ The desired filter response must in practice be approximated by a realizable transfer function. Possible transfer functions include Butterworth and Chebyshev responses. ¾ The Butterworth response contains n complex poles on a circle and exhibits a maximally-flat behavior. It is suited to applications that are intolerant of any ripple in the passband ¾ The Chebyshev response provides a sharper transition than Butterworth at the cost of some ripple in the passband and stopbands. It contains n complex poles on an ellipse.