![A2(ω) provides a single zero at z1A3(ω) provides a single pole at p1
Analog Electronics Circuits
8
A2(ω) [dB]
ω<<z1 ω=z1 ω=10z1 ω>>z1
0 3 20 20log(ω/z1)
A3(ω) [dB]
ω<<p1 ω=p1 ω=10p1 ω>>p1
0 -3 -20 -20log(ω/p1)
The overall plot of A(ω) is shown for two cases: z1<p1 and
z1>p1, respectively and for simplicity: K= p1/z1. The break
frequencies occur at z1 and p1. Note that when |z1|=|p1| and
z1>p1, A(ω)=A1(ω) (all pass filter).](https://image.slidesharecdn.com/part2basedonmanuscript1-200317115057/85/Active-RC-Filters-Design-Techniques-8-320.jpg)
Active RC Filters Design Techniques
- 1. Add Symbol Analog Electronic Circuits 1 0402xxx: Analog Electronic Circuits Chapter 3: continued Active RC Filters Design Techniques
- 2. 5. First Order Active RC Sections ▪ First order active RC section has transfer function in the s-plane of the following form: T(S) is the transfer function of the first order filter (Bilinear Transfer Function) where N(S) and D(S) are linear function with real constants (K, z1=ωz1, and p1=ωp1). ▪ Depending on the pole and the zero locations (p1,z1) of T(S), the magnitude frequency response (|T(jω)|) can be a low pass response(|z1|>|p1| both located at LHP(special case z1 at ∞ ideal integrator)), a high pass response (|z1|<|p1| both located at LHP(special case z1 at → ideal differentiator)), or an all pass response (|z1|=|p1| and z1>p1 → zero at RHP). 𝑇 𝑆 = 𝑉𝑜𝑢𝑡 𝑉𝑖𝑛 = 𝑁(𝑆) 𝐷(𝑆) = 𝐾 𝑆 + 𝑧1 𝑆 + 𝑝1 = 𝐾 𝑧1 𝑝1 ∙ 1 + Τ𝑆 𝑧1 1 + Τ𝑆 𝑝1 Analog Electronic Circuits 2 Add your footnotes
- 3. Notes: • In general, minimum phase transfer function has all zeros located at the LHP. • Non-minimum phase transfer function has zeros in the RHP. If all zeros of transfer function are all reflected about the jω-axis, there is no change in the magnitude transfer function and only difference is in the phase-shift characteristics (All pass network). • If the phase responses of two different transfer functions (two different systems) are compared, the net phase shift over the frequency range from zero to infinity is less for the transfer function with all its zeros in the LHP (minimum phase transfer function). • The range of phase shift of a minimum phase curve is least possible or minimum corresponding to a given magnitude curve, while the range of the non-minimum phase curve is the largest possible for the given magnitude curve. • All pass network is an example of non-minimum phase network which passes all frequencies with equal gain while provides a phase (angle) delay for all frequencies. • The phase lead network is a network that provides a positive phase angle over the frequency range of interest, yielding a system to have adequate phase margin. • The phase lag network is a network that provides a negative phase angle and a significant attenuation over the frequency range of interest. Analog Electronic Circuits 3
- 4. What is the transfer function? Analog Electronics Circuits 4 ▪ The transfer function (network function) of a linear, stationary (time-invariant constant parameter) system is defined as the ratio of the Laplace transform of the output variable signal to the Laplace transform of the input variable signal, with all initial condition assumed to be zero. It gives input-output description of the behaviour of a system without including any information about the internal structure of the system and its behaviour. It is analytical tool for finding the frequency response of the circuit. ▪ Transfer function representation ▪ Magnitude and phase transfer functions 𝑇 𝑗𝜔 = 𝑇(𝑗𝜔) 𝑒 𝑗𝜃 𝜔 𝑇 𝑗𝜔 = ሽ𝑅𝑒 𝑁 𝑗𝜔 + 𝑗𝐼𝑚{𝑁(𝑗𝜔) ሽ𝑅𝑒 𝐷 𝑗𝜔 + 𝑗𝐼𝑚{𝐷(𝑗𝜔) 𝑇 𝑗𝜔 = 𝑅𝑒 𝑁 𝑗𝜔 2 + ሽ𝐼𝑚{𝑁(𝑗𝜔) 2 𝑅𝑒 𝐷 𝑗𝜔 2 + ሽ𝐼𝑚{𝐷(𝑗𝜔) 2 𝜃 𝜔 = tan−1 𝐼𝑚 𝑁 𝑗𝜔 𝑅𝑒 𝑁 𝑗𝜔 − tan−1 𝐼𝑚 𝐷 𝑗𝜔 𝑅𝑒 𝐷 𝑗𝜔
- 5. Bilinear Transfer Function 𝑇 𝑗𝜔 = 𝐾 𝑗𝜔 + 𝑧1 𝑗𝜔 + 𝑝1 = 𝐾 𝑧1 𝑝1 ∙ 1 + Τ𝑗𝜔 𝑧1 1 + Τ𝑗𝜔 𝑝1 𝑇 𝑗𝜔 = 𝐾 𝑧1 𝑝1 ∙ 1 + 𝜔 𝑧1 2 1 + 𝜔 𝑝1 2 𝜃 𝜔 = (0° 𝑓𝑜𝑟 𝐾 > 0 𝑜𝑟 180° 𝑓𝑜𝑟 𝐾 < 0) + tan−1 𝜔 𝑧1 − tan−1 𝜔 𝑝1 Analog Electronics Circuits 5
- 6. Bode Plots for Bilinear Transfer Function It is a simple technique exists for obtaining an approximate plot of the magnitude and phase of the transfer function given its poles and zeros. The technique is useful particularly in the case of real poles and zeros. The method was developed by H. Bode, and the resulting diagram are called Bode plots. The plots of the magnitude transfer function |T(jω)| (in dB) and the phase transfer function θ(ω) (in degree) versus ω in logarithmic representation is based on the asymptotic behaviour of |T(jω)| and θ(ω). In case bilinear transfer function, define A(jω)=20log(|T(jω)|) and θ(ω) 𝐴 𝑗𝜔 = 20 log 𝐾 𝑧1 𝑝1 ∙ 1 + 𝜔 𝑧1 2 1 + 𝜔 𝑝1 2 𝜃 𝜔 = (0° 𝑜𝑟 180°) + tan−1 𝜔 𝑧1 − tan−1 𝜔 𝑝1 & Analog Electronics Circuits 6
- 7. 𝐴 𝜔 = 𝐴1 𝜔 + 𝐴2 𝜔 + 𝐴3 𝜔 𝐴 𝜔 = 20 log 𝐾 𝑧1 𝑝1 + 20 log 1 + 𝜔 𝑧1 2 − 20 log 1 + 𝜔 𝑝1 2 A1(ω) is a constant term, A2(ω) provides a single zero at z1, A3(ω) provides a single pole at p1 Analog Electronics Circuits 7
- 8. A2(ω) provides a single zero at z1A3(ω) provides a single pole at p1 Analog Electronics Circuits 8 A2(ω) [dB] ω<<z1 ω=z1 ω=10z1 ω>>z1 0 3 20 20log(ω/z1) A3(ω) [dB] ω<<p1 ω=p1 ω=10p1 ω>>p1 0 -3 -20 -20log(ω/p1) The overall plot of A(ω) is shown for two cases: z1<p1 and z1>p1, respectively and for simplicity: K= p1/z1. The break frequencies occur at z1 and p1. Note that when |z1|=|p1| and z1>p1, A(ω)=A1(ω) (all pass filter).
- 9. P1(ω) is a straight line (0ᵒ (case I) or 180ᵒ (case II) P2(ω) provides a single zero at z1 (provides 45ᵒ/decade) P3(ω) provides a single pole at p1 (provides -45ᵒ/decade) Analog Electronics Circuits 9 𝜃 𝜔 = (0° 𝑜𝑟 180° ) + tan−1 𝜔 𝑧1 − tan−1 𝜔 𝑝1 𝑃 𝜔 = 𝑃1 𝜔 + 𝑃2 𝜔 + 𝑃3 𝜔 P2(ω) Case I | Case II ω=0.1z1 ω=z1 ω=10z1 0ᵒ -180ᵒ 45ᵒ -135ᵒ 90ᵒ -90ᵒ P3(ω) ω=0.1p1 ω=p1 ω=10p1 0ᵒ -45ᵒ - 90ᵒ P2(ω) Case I P3(ω) P2(ω) Case II
- 10. Analog Electronics Circuits 10 Case a. The overall θ(ω) for |z1|<|p1| under case I, and for |z1|<|p1| under case II (both z1 and p1 located at LHP and provide minimum phase transfer function). For case I, it is obvious that the phase shift ranges over less than 80ᵒ). The network representing this feature is called phase-lead network. It will exhibits like a differentiator if |z1|<<|p1|. P(ω) Case I P(ω) Case II
- 11. Analog Electronics Circuits 11 Case b. In the case |z1|>|p1| under K>0 (case I), and for |z1|>|p1| under K<0 (case II) both z1 and p1 located at LHP, the network representing this feature is called phase-lag network. It will exhibits like a integrator if |z1|>>|p1|. P(ω) Case I P(ω) Case II
- 12. Analog Electronics Circuits 12 Case c. In the case |z1|<|p1| and z1 located at RHP, the complete θ(ω) will be P(ω) Case I P(ω) Case II
- 13. Analog Electronics Circuits 13 Case d. When |z1|=|p1| and z1 located at RHP (all pass filter), it provides non-minimum phase transfer function. This is due to z1 at RHP. The phase shift ranges over 180ᵒ. P(ω) Case I P(ω) Case II
- 14. Analog Electronics Circuits 14 Notes: • In case a. and case c., p1 has same location but z1 location is changed with same magnitude. The only difference is in the phase-shift characteristics. If the phase characteristics of the two system functions are compared, it can be readily shown that the net phase shift over the frequency range from zero to infinity is less for the system with all its zero in the LHP (the phase shift ranges over less than 80ᵒ). Hence, the transfer function with all its zeros in the LHP is called minimum phase transfer function. Case a. P(ω) Case I Case c. P(ω) Case II Minimum Phase Transfer Function Non-minimum Phase Transfer Function
- 15. Analog Electronics Circuits 15 Notes: • The range of phase shift of a minimum phase transfer function is the least possible or minimum corresponding to a given amplitude curve, whereas the range of the non- minimum phase curve is the greatest possible for the given amplitude curve, as shown below. The zero is reflected about jω-axis and there is no change in the magnitude transfer function. The only difference is in the phase-shift characteristics.
- 16. Analog Electronics Circuits 16 Exercises: 1. Obtain the Bode plot for the transfer function 𝑇 𝑆 = 6 𝑆+0.5 𝑆+3 2. Obtain the Bode plot for the 𝑇 𝑆 = 𝐾 𝑑 𝑆 (ideal differentiator) 3. Obtain the Bode plot for the 𝑇 𝑆 = 𝐾 𝑖 𝑆 (ideal integrator)
- 17. Analog Electronics Circuits 17 Example: Given the below Bode plot (magnitude frequency response), find the magnitude transfer function. Solution: It is obvious that the plot represents high order magnitude transfer function. The three break frequencies (poles and zeros) are a. Double zero at the origin b. Double pole at ωp1 = ω2 c. A single pole at ωp2 = ω3 Therefore, A1(ω) is given by 𝑇 𝑗𝜔 = 𝑗 𝜔 𝜔1 2 1 + 𝑗 𝜔 𝜔2 2 1 + 𝑗 𝜔 𝜔3 𝑇 𝑆 = 𝜔3 𝜔2 𝜔1 2 𝑆2 𝑆 + 𝜔2 2 𝑆 + 𝜔3 𝐴1 𝜔 = 20 log 𝑇 𝑗𝜔
- 18. Analog Electronics Circuits 18 𝑇 𝑗𝜔 = 𝑗 𝜔 𝜔1 2 1 + 𝑗 𝜔 𝜔2 2 1 + 𝑗 𝜔 𝜔3 𝑇 𝑆 = 𝜔3 𝜔2 𝜔1 2 𝑆2 𝑆 + 𝜔2 2 𝑆 + 𝜔3 𝐴1 𝜔 = 20 log 𝜔 𝜔1 2 1 − 𝜔 𝜔2 2 2 + 2 𝜔 𝜔2 2 ∙ 1 + 𝜔 𝜔3 2 𝜃 𝜔 = 2 tan−1 𝜔 𝜔1 − 2 tan−1 𝜔 𝜔2 − tan−1 𝜔 𝜔3
- 19. Analog Electronics Circuits 19 Exercise: For the Bode plot (magnitude frequency response), find out the magnitude transfer function.
- 20. 5. First Order Active RC Sections ▪ First order sections based on inverting op-amp configuration (Realization of the bilinear transfer function using the inverting op-amp configuration) Analog Electronic Circuits 20 Add your footnotes The aim is to realize the bilinear transfer function using the active RC sections based on inverting operational amplifier. 𝑇 𝑆 = 𝑉𝑜𝑢𝑡 𝑉𝑖𝑛 = − 𝑍2 𝑍1 𝑍2 𝑍1 = 𝐾 𝑆 + 𝑧1 𝑆 + 𝑝1 𝑇 𝑆 = 𝑉𝑜𝑢𝑡 𝑉𝑖𝑛 = −𝐾 𝑆 + 𝑧1 𝑆 + 𝑝1 If Z1 and Z2 are both series RC network, then Z1 = R1+(1/SC1), Z2 = R2+R2/(SC2), and K= R2.
- 21. 5. First Order Active RC Sections ▪ First order sections based on inverting op-amp configuration (Realization of the bilinear transfer function using the inverting op-amp configuration) Analog Electronic Circuits 21 Add your footnotes 𝑇 𝑆 = 𝑉𝑜𝑢𝑡 𝑉𝑖𝑛 = − 𝑍2 𝑍1 = − 𝑅2 + 1 𝑆𝐶2 𝑅1 + 1 𝑆𝐶1 = − 𝑅1 𝑅2 ∙ 1 + 1 𝑆𝑅2 𝐶2 1 + 1 𝑆𝑅1 𝐶1 where Kz1=1/C2 and p1=1/C1. Alternative form, in term of the admittances − 𝑍2 𝑍1 = 𝑌1 𝑌2
- 22. Analog Electronics Circuits 22 Design Example: Based on the following Bode plot (magnitude frequency response), design a practical first order active RC op-amp section that meets the given characteristics of Bode plot. Solution: The two break frequencies (pols and zero) are a. A single zero at ω=0.5rad/sec b. A single pole at ω =6rad/sec c. The constant term is -6dB (20 log 𝐾 𝑧1 𝑝1 =-6dB → K=6) 𝑇 𝑆 = −6 0.5 6 ∙ 1 + 𝑆 0.5 1 + 𝑆 6 = −6 ∙ 𝑆 + 0.5 𝑆 + 6 = − 𝑅2 + 1 𝑆𝐶2 𝑅1 + 1 𝑆𝐶1 Z1=R1+(1/SC1)=1+(1/6S), Z2=R2+(1/SC2)=6+(1/3S)
- 23. Analog Electronics Circuits 23 R1=1Ω, C1=(1/6)F, R2=6 Ω, C2=(1/3)F To get the practical design, we need to do the frequency scaling and the magnitude scaling. Frequency Scaling: Assume the circuit is designed for a certain frequency band and it is desired that the frequency band be changed, keeping the same magnitude characteristics (frequency scaling kf = 1000rad/sec (frequency shift by 1000)). Thus, z1 = 500rad/sec and p1 = 6000rad/sec. How this will affect the original circuit? New C = C/kf while all R do not change by frequency scaling since they impedances do not depend on the frequency. New C1=(1/6)mF and new C2=(1/3)mF. Magnitude Scaling: The above transfer function contains RC as ratio of R and as product of RC. Then multiplying all R by scaling factor (km) and dividing all C by the same quantity will not change the transfer function.
- 24. Analog Electronics Circuits 24 The magnitude of the elements (R and C) are scaled yielding to a practical design. Let km=10,000; thus practical values of R and C are: R1=10kΩ, R2=60kΩ, C1≈16nF, C2≈33nF Element values before frequency scaling and magnitude scaling Element values after frequency scaling and magnitude scaling R kmR C C/(kmkf)
- 25. 5. First Order Active RC Sections ▪ First order sections based on non-inverting op-amp configuration (Realization of the bilinear transfer function using the non-inverting op-amp configuration) Analog Electronic Circuits 25 Add your footnotes The aim is to realize the bilinear transfer function using the active RC sections based on non-inverting operational amplifier. 𝑇 𝑆 = 𝑉𝑜𝑢𝑡 𝑉𝑖𝑛 = 1 + 𝑍2 𝑍1 𝑇 𝑆 = 𝑉𝑜𝑢𝑡 𝑉𝑖𝑛 = 𝐾 𝑆 + 𝑧1 𝑆 + 𝑝1 1 + 𝑍2 𝑍1 = 𝐾 𝑆 + 𝑧1 𝑆 + 𝑝1 𝑍2 𝑍1 = 𝐾 − 1 𝑆 + 𝐾𝑧1 − 𝑝1 𝑆 + 𝑝1 Since Z1 and Z2 are impedances, there are positive real values. K, z1, and p1 must satisfy the following conditions: 𝐾 ≥ 1 and 𝐾𝑧1 > 𝑝1
- 26. Analog Electronics Circuits 26 Two cases: K=1 and K=p1/z1>1 Case I: K=1 This implies z1>p1, that is, if K is unity, the non-inverting configuration can only realize a low pass transfer function, and 𝑍2 𝑍1 = 𝑧1 − 𝑝1 𝑆 + 𝑝1 = 𝑧1 − 𝑝1 𝑆 1 + 𝑝1 𝑆 Therefore, the following assignments for Z1 and Z2 can be made: Z1 = 1+(p1/S) = R1+(1/SC1), which yields, R1 = 1 and C1 = 1/p1. And Z2 = (z1-p1)/S = 1/SC2, which yields, C2 = 1/( z1-p1).
- 27. Analog Electronics Circuits 27 Two cases: K=1 and K=p1/z1>1 Case II: K=p1/z1>1 This implies z1<p1, that is the non-inverting configuration realizes a high pass transfer function. One possible realization is Z1 = 1+(1/(S/p1)) = R1+(1/SC1), and Z2 = (p1-z1)/z1 = R2. 𝑍2 𝑍1 = 𝑝1 𝑧1 − 1 𝑆 𝑆 + 𝑝1 = 𝑝1 − 𝑧1 𝑧1 1 + 1 𝑆 𝑝1
- 28. Analog Electronics Circuits 28 ▪ Cascading Connection: 5. First Order Active RC Sections It is required to design high order filters to achieve new filters (Band Pass filters and Band Reject filters) which cannot be obtained in case the first order filter. Also, high order filters have better bandwidth and selectivity. The figure below demonstrate a cascade of N circuits with bilinear transfer functions of T1, T2, and TN.
- 29. Analog Electronics Circuits 29 Loading Effect: Having minimum loading effect is considered in the cascaded connections based on the Thevenin’s equivalent circuit. The Thevenin’s impedance taken from the output section needs to be very smaller than the input impedance of the next cascaded section. Thus, the Thevenin’s voltage will be the input voltage to the next cascaded section. Using op-amp in designing RC filter is a good choice since it has very high input impedance and having very low output impedance. Non-inverting op-amp is preferable since input voltage is applied directly to non-inverting input. In case using inverting op-amp, then voltage buffer can be a good choice to connect at the output of each section. Zin of non-inverting op-amp Zout of non-inverting op-amp
- 30. 5. First Order Active RC Sections ▪ First order all pass filter (lattice network=phase correction circuit=phase shaping) Analog Electronic Circuits 30 Add your footnotes Circuit with all pass magnitude response is called phase correcting circuit, since the magnitude of the output is constant for all frequency and only the phase of the output is function of frequency. With this property, these circuits can be connected in cascade with other circuits to correct for any desired phase (without changing the magnitude response). Also, it can be used to design high order filter (such as notch filter or comb filter), by using cascaded all pass filters and adder connected in feedforward connection. First order all pass filter is characterized by the following transfer function: 1-LH pole and 1-RH zero at |a|.𝑇 𝑆 = 𝑉out 𝑉𝑖𝑛 = 𝐾 𝑆 − 𝑎 𝑆 + 𝑎 = 𝐾 𝑆 − 𝑧1 𝑆 + 𝑝1
- 31. Analog Electronics Circuits 31 To have zero at RHP, it can be realized by a voltage difference (difference op-amp with one input), shown below, that generates negative numerator coefficient. By direct analysis, the transfer function of voltage difference is given by: where k = RF/R, z1 = kGG, and p1 = GG. One possibility is by replacing Ga with SC (a capacitor). where k = RF/R=1, z1 = k/(RGC), and p1 = -(1/RGC). 𝑇 𝑆 = 𝑉𝑜𝑢𝑡 𝑉𝑖𝑛 = 𝐺 𝑎 − 𝑘𝐺 𝐺 𝐺 𝑎 + 𝐺 𝐺 𝑇 𝑆 = 𝑉𝑜𝑢𝑡 𝑉𝑖𝑛 = 𝑆𝐶 − 𝑘𝐺 𝐺 𝑆𝐶 + 𝐺 𝐺 = 𝑆 − 𝑘 𝑅 𝐺 𝐶 𝑆 + 1 𝑅 𝐺 𝐶 = − 1 − 𝑆𝐶𝑅 𝐺 1 + 𝑆𝐶𝑅 𝐺
- 32. Analog Electronics Circuits 32 The phase transfer function is given by And the following design equation holds for desired θd at a given frequency ωd Design the previous first order all pass filter to provide phase shift of 135ᵒ at 100rad/sec frequency. Solution: 𝜃 𝜔 = 180° − 2 tan−1 𝜔𝑅 𝐺 𝐶 ; 𝑤ℎ𝑒𝑟𝑒 0° ≤ 𝜃 ≤ 180° 𝑅 𝐺 𝐶 = cot 𝜃 𝑑 2 𝜔 𝑑 Example: 𝑅 𝐺 𝐶 = cot 𝜃 𝑑 2 𝜔 𝑑 = cot 135° 2 100 = 0.4142 100 = 4.142𝑚𝑠𝑒𝑐 C=0.1µF RG=R=RF=41.42kΩ
- 33. Analog Electronics Circuits 33 LTspice Simulation Magnitude and phase responses of the all pass filter: θd = 135ᵒ at ωd = 100rad/sec= 2π*(15.9Hz) This APF is considered as lead filter providing a positive phase angle over the frequency range of interest. Phase response --- dashed line Magnitude response ꟷ solid line
- 34. Analog Electronics Circuits 34 If we need a lag filter; i.e. negative phase, it can be achieved by replacing GG with SC instead of Ga. One possibility is by replacing GG with SC (a capacitor). where k = RF/R=1, z1 = 1/(kRaC), and p1 = -(1/RaC). The phase transfer function is given by 𝑇 𝑆 = 𝐺 𝑎 − 𝑘𝑆𝐶 𝐺 𝑎 + 𝑆𝐶 = −𝑘 𝑆 − 1 𝑘𝑅 𝑎 𝐶 𝑆 + 1 𝑅 𝑎 𝐶 = − 𝑆 − 1 𝑅 𝑎 𝐶 𝑆 + 1 𝑅 𝑎 𝐶 = 1 − 𝑆𝐶𝑅 𝑎 1 + 𝑆𝐶𝑅 𝑎 𝜃 𝜔 = −2 tan−1 𝜔𝑅 𝑎 𝐶 ; 𝑤ℎ𝑒𝑟𝑒 − 180° ≤ 𝜃 ≤ 0°
- 35. Analog Electronics Circuits 35 The following design equation holds for desired θd at a given frequency ωd: Design the previous first order all pass filter that has constant magnitude response and provides phase shift of -30ᵒ at 100rad/sec frequency. NB: Both the phase lead network and the phase lag network can be used as a phase correction network depending on the system needs. 𝑅 𝑎 𝐶 = − tan 𝜃 𝑑 2 𝜔 𝑑 Example: 𝑅 𝑎 𝐶 = tan −𝜃 𝑑 2 𝜔 𝑑 = tan −30° 2 100 = 0.2679 100 = 2.679𝑚𝑠𝑒𝑐 C=0.1µF Ra=R=RF=26.8kΩ
- 36. Analog Electronics Circuits 36 ▪ (1) R. Schaumann, H. Xiao, and M. E. Valkenburg, Design of Analog Filters, 2nd ed., Oxford University Press, 2010. ▪ (2) C. K. Alexander and M. N. O. Sadiku, Fundamentals of Electric Circuits, 3rd ed., McGraw-Hill Companies, Inc., 2007. References