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Syllabus Active Filters Introduction, Active versus Passive Filters, Types of Active Filters, First-Order Filters, The Biquadratic Function, Butterworth Filters, Transfer Function Realizations, Low pass Filters, High-Pass Filters, Band-Pass Filters, Band-Reject Filters, All-Pass Filters, Switched Capacitor Filters, Filter Design Guide Lines. Filter Basics • A filter is a frequency-selective circuit that passes a specified band of frequencies and blocks or attenuates signals of frequencies outside this band. • A filter is used to remove (or attenuate) unwanted frequencies in an audio signal • “Stop Band” – the part of the frequency spectrum that is attenuated by a filter. • “Pass Band” – part of the frequency spectrum that is unaffected by a filter. • Filters are usually described in terms of their “frequency responses,” e.g. low pass, high pass, band pass, band reject (or notch) Advantages of Active Filters over Passive Filters (i) The maximum value of the transfer function or gain may be greater than unity, (ii) The loading effect is minimal, which means that the output response of the filter is essentially independent of the load driven by the filter. (iii) The active filters do not exhibit insertion loss. Hence, the passband gain is equal to 0 dB. (iv) Complex filters can be realized without the use of inductors, (v) The passive filters using R, L and C components are realizable only for radio frequencies. Because, the inductors become very large, bulky and expensive at audio frequencies. Due to low Q at low frequency applications, high power dissipation is incurred. The active filters overcome these problems, (vi) Rapid, stable and economical design of filters for variety of applications is possible. (vii) The active filters are easily tunable due to flexibility in gain and frequency adjustments. (viii) The op-amp has high input impedance and low output impedance. Hence, the active filters using op-amp do not cause loading effect on the source and load. Therefore, cascading of networks does not need buffer amplifier. (ix) Active filters for fixed frequency and variable frequency can be designed easily. The adjustable frequency response is obtained by varying an external voltage signal. (x) There is no restriction in realizing rational function using active networks. (xi) Use of active elements eliminates the two fundamental restrictions of passivity and reciprocity of RLC networks. Limitations of Active Filters over Passive Filters (i) The high frequency response is limited by the gain-bandwidth product and slew rate of the practical op-amps, leading to comparatively lower bandwidth than the designed bandwidth. (ii) The design of active filters becomes costly for high frequencies. (iii) Active filters require dual polarity dc power supply whereas passive filters do not. (iv) The active element is prone to the process parameter variations and they are sensitive to ambient conditions like temperature. Hence, the performance of the active filter deviates from the ideal response. Ideal Filter Characteristics Filter Characteristics CLASSIFICATION OF FILTERS Types of Filters • Butterworth – Flat response in the pass band & stop band and called flat-flat filter. • Chebyshev – steeper roll-off but exhibits pass band ripple (making it unsuitable for audio systems) & flat stopband. • Cauer – It has equiripple both in pass & stop band. 9 passband passband stopband Butterworth filter magnitude response stopband Chebyshev filter magnitude response passband stopband Cauer filter magnitude response FILTERS BASED ON FREQUENCY Low pass filter (LPF) High pass filter (LPF) 20db/decade 20db/decade Understanding Poles and Zeros The transfer function provides a basis for determining important system response characteristics The transfer function is a rational function in the complex variable s = σ + jω, that is zi’s are the roots of the equation N(s) = 0, and are defined to be the system zeros, and the pi’s are the roots of the equation D(s) = 0, and are defined to be the system poles. Example A linear system is described by the differential equation Find the system poles and zeros. Solution: From the differential equation the transfer function is The system therefore has a single real zero at s = −1/2, and a pair of real poles at s = −3 and s = −2. Consider a Pole at Zero. Its response is constant. Consider Poles at +a and –a. The exponential responses are shown, for a function k/s+a, and k/s-a Consider conjugate poles +jω & -jω & their mirror image on the right side, along with their responces which is decaying sine wave and increasing sine wave. The equation shown has 3 poles & one Zero at -1. Zeros show how fast the amplitudes vary. Frequency Response of filters •Ideal •Practical • Filters are often described in terms of poles and zeros – A pole is a peak produced in the output spectrum – A zero is a valley (not really zero)a Order of the Filter Comparison of FIR & IIR Filter 1. 2. 3. 4. 5. 6. 7. 8. 9. FIR (Finite Impulse Response) (non-recursive) filters produce zeros. In signal processing, a finite impulse response (FIR) filter is a filter whose impulse response (or response to any finite length input) is of finite duration, because it settles to zero in finite time. Filters combining both past inputs and past outputs can produce both poles and zeros. FIR filters can be discrete- time or continuoustime, and digital or analog. FIR is always stable FIR has no limited cycles. FIR has no analog history. FIR filters are dependent upon linear-phase characteristics. FIR is dependent upon i/p only. 10. FIR’s delay characteristics is much better, but they require more memory. 11. FIR filters are used for tapping of a higherorder. 1. IIR (infinite Impulse Response) (recursive) filters produce poles. 2. This is in contrast to infinite impulse response (IIR) filters, which may have internal feedback and may continue to respond indefinitely (usually decaying). 3. IIR filters are difficult to control and have no particular phase. 4. IIR can be unstable 5. IIR is derived from analog. 6. IIR filters make polyphase implementation possible. 7. IIR filters are used for applications which are not linear. 8. IIR filters are dependent on both i/p and o/p. 9. IIR filters consist of zeros and poles, and require less memory than FIR filters. 10. IIR filters can become difficult to implement, and also delay and distortion adjustments can alter the poles & zeroes, which make the filters unstable. 11. IIR filters are better for tapping of lower-orders, since IIR filters may become unstable with tapping higher-orders. ACTIVE FILTERS USING OP-AMP: Filters are frequency selective circuits. They are required to pass a specific band of frequencies and attenuate frequencies outside the band. Filters using an active device like OPAMP are called active filters. Other way to design filters is using passive components like resistor, capacitor and inductor. ADVANTAGES OF ACTIVE FILTERS: Possible to incorporate variable gain Due to high Zi & Z0 of the OPAMP, active filters donot load the input source or load. Flexible design. FREQUENCY RESPONSE OF FILTERS: Gain of a filter is given as, G=Vo/Vin Ideal & practical frequency responses of different types of filters are shown below. First Order Low-Pass Butterworth Filter 1st-ORDER Butterworth LPF: Because of simplicity, Butterworth filters are considered. In 1st. order LPF which is also known as one pole LPF. Butterworth filter and it’s frequency response are shown above. • RC & values decide the cut-off frequency of the filter. • Resistors R1 & RF will decide it’s gain in pass band. As the OPAMP is used in the non-inverting configuration, the closed loop gain of the filter is given by RF AVF 1 R1 EXPRESSION FOR THE GAIN OF THE FILTER: jX C Voltage across the capacitor is, V1 Vin (1) R jX C Reactance of the capacitor is, X C Equation (1) becomes Output of the filter is, R Vin V0 AVF V1 1 F R1 1 j 2fRC V AVF 0 Vin f 1 j f H 1 2fC 1 j Vin 2fC jVin Vin V1 (2) 2 fRC 1 2fRC j 1 R j j 2 fC Vin= Vin 1 j 2fRC f = frequency of input signal. Where fH= 1/2πRC cut-off frequency. f = frequency of the input signal The gain magnitude and phase angle equations of the low-pass filter can be obtained by converting Equation (7-lb) into its equivalent polar form, as follows: The operation of the low-pass filter can be verified from the gain magnitude equation, (7-2a): 1. At very low frequencies, that is, f < fH, 2. At f = fH, 3. f < fH, DESIGN PROCEDURE: Step1: Choose the cut-off frequency fH Step2: Select a value of ‘C’ ≤ 1µF (Approximately between .001 & 0.1µF) Step3: Calculate the value of R using Step4: Select resistors R1 & R2 depending on the desired pass band gain. =2. So RF=R1 For a first order Butterworth LPF, calculate the cut –off frequency if R=10K & C=0.001µF.Also calculate the pass band voltage gain if R1=10K RF =100K 1 1 fH 15.915KHz 3 6 2RC 2 10 10 0.00110 1+100K/10K =11 Design a I order LPF for the following specification Pass band voltage gain = 2. Cut off frequency, fC = 10KHz. AVF = 2; Let RF = 10K RF/R1=1 Let C = 0.001µF 1 1 1 fH &R 2RC 2f H C 2 10 103 0.00110 6 R=15.9K 1st ORDER HPF: fL is shown for HPF Circuit diagram & frequency response are shown above. Again RC components decide the cut off frequency of the HPF where as RF & R1 decide the closed loop gain. EXPRESSION FOR THE GAIN: Voltage V1 R Vin R jX C 1 WhereX C 2fC f R R R j 2fC j V1 Vin Vin j 1 1 j 2fRC fL V R R in 2fC j 2fC f 1 j fl Output voltage = jf AVF V0 fL Gain = V f in 1 j fL jf AVF fL V V0 AVF .V1 f in 1 j fL Magnitude= SECOND-ORDER LOW-PASS BUTTERWORTH FILTER The gain of the second-order filter is set by R1, and RF, while the high cutoff frequency fH is determined by R2, C2, R3, and C3, as follows: High Cutoff frequency, Filter Design 1. Choose a value for the high cutoff frequency fH 2. To simplify the design calculations, set R2 = R3 = R and C2 = C3 = C. Then choose a value of C ≤ 1µF 3. Calculate the value of R using Equation for fH: 4. 4. Finally, because of the equal resistor (R2 = R3) and capacitor (C2 = C3) values, the pass band voltage gain AF = (1 + RF/R1) of the second-order low-pass filter has to be equal to 1.586. That is, RF = 0.586/R1 This gain is necessary to guarantee Butterworth response. Hence choose a value of R1 < 100 kΩ and calculate the value of RF . SECOND-ORDER HIGH-PASS BUTTERWORTH FILTER As in the case of the first-order filter, a second-order high-pass filter can be formed from a second-order low-pass filter simply by interchanging the frequency determining resistors and capacitors. Figure 7-8(a) shows the second-order high-pass filter. SECOND-ORDER HIGH-PASS BUTTERWORTH FILTER 7.8 (a) Second-order Hi pass-pass Filter Analysis i1 Replace VC i2 i3 4th Order Filter Q=fc/BW = fC/(fH-fL) Narrow Band-Pass Filter DESIGN EQUATIONS: Q Select C1 = C2 =C R1 2f C CAF Q Q R2 RB 2 f C C 2f C C 2Q AF R3 A is the gain at f =fC AF 2R1 Condition on gain AF<2Q2 Notch filterj 45 Shunted Twin T Filter with swapped R & C ALL PASS FILTER: It is a special type of filter which passes all the frequency components of the input signal to output without any attenuation. But it introduces a predictable phase shift for different frequency of the input signal. The all pass filters are also called as delay equalizers or phase correctors. Switched-Capacitor Filters • • • • Active RC filters are difficult to implement totally on an IC due to the requirements of large valued capacitors and accurate RC time constants The switched capacitor filter technique is based on the realization that a capacitor switched between two circuit nodes at a sufficiently high rate is equivalent to a resistor connecting these two nodes. Switched capacitor filter ICs offer a low cost high order filter on a single IC. Can be easily programmed by changing the clock frequency. Switched-Capacitor Filters S1 R + v1 i S2 + + v2 v1 - - + CR - v1 v 2 i R v1 v 2 R i q1 = CRv1 q2 = CRv2 Dq = q1-q2 = CR(v1-v2) v2 - the value of R is a function of CR and fC. For a fixed value of C, the value of R can be adjusted by adjusting fC Copyright © S.Witthayapradit.2009 q i qfC fC C R v1 v2 T T Requ 1 fC T v1 v2 1 i fC C R R equ + + v1 v2 - - jω X X Imaginary X -jω X -a X X X X +a Real X Zeros: roots of N(s) • Poles: roots of D(s) • Poles must be in the left half plane for the system to be stable • As the poles get closer to the boundary, the system becomes less stable • Pole-Zero Plot: plot of the zeros and poles on the complex s plane