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Chapter 14 Analog Filters

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Chapter 14 Analog Filters 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. z11=+σ jω 1z21= σ − jω 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= ω1 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 ) dω0 −1 = 2 dR1 R1 C1 dω 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 < > RC()() C+ C R R RC C+ C R R 11 1 2 1 2 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 =− ±ωn 1 − 2Q 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 2 2 γ Hj()ω = 2 2 α = β = 0 22 ⎛⎞ωn ()ωn −+ωω⎜⎟ ⎝⎠Q H(ω p ) 2 Q> 2 ∂ Hj()ω 2 = 0 H(ω p ) 2 Q> ∂ω 2 γ 2 ωn 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. 2 2 γ Hj()ω = 2 2 2 22 ⎛⎞ωn ()ωωnn−+⎜⎟ ⎝⎠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 ss22++n ω β = γ = 0 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 βω22 β 22Q frequency = ω 2 ()()ωω222−+n ω 22ωn n Q ⎡ 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 ωn RC11 LC 11 Q 1 C1 ωn ==, QR1 LC11 L1 ωn 1 pj1,2 =− ±ωn 1 − 2Q 4Q 2 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) occurs when a solution exists for 2 dD 22 =−2(R111CL )( R 1 − RCL 111ω ) + L 1 d()ω2 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 ω ss22++n ω Q n 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.
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