Electric Circuit Theory Nam Ki Min nkmin@korea.ac.kr 010-9419-2320
Chapter 15 Active Filter Circuits Nam Ki Min nkmin@korea.ac.kr 010-9419-2320
Contents and Objectives 3 Chapter Contents 15.1 First-Order Low-Pass and High-Pass Filters 15.2 Scaling 15.3 Op Amp Bandpass and Bandreject Filters 15.4 Higher Order Op Amp Filters 15.5 Narrowband Bandpass and Bandreject Filters Chapter Objectives 1. Know the op amp circuits that behave as first-order low-pass and high-pass filters and be able to calculate component values for these circuits to meet specifications of cutoff frequency and passband gain. 2. Be able to design filter circuits starting with a prototype circuits and use scaling to achieve desired frequency response characteristics and component values. 3. Understand how to use cascaded first- and second-order Butterworth filters to implement lowpass, high-pass, bandpass, and bandreject filters of any order. 4. Be able to use the design equations to calculate component values for prototype narrowband, bandpass, and bandreject filters to meet desired filter specifications.
Some Preliminaries 4 Three Major Limits to The Passive Filters First, they cannot generate gain greater than 1; passive elements cannot add energy to the network. Second, they may require bulky and expensive inductors. Third, they perform poorly at frequencies below the audio frequency range (300 Hz < f <3000 Hz). Nevertheless, passive filters are useful at high frequencies. Advantages and Disadvantages of Active Filters over Passive RLC Filters Active filters consist of combinations of resistors, capacitors, and op amps. They offer some advantages over passive RLC filters. First, they are often smaller and less expensive, because they do not require inductors. This makes feasible the integrated circuit realizations of filters. Second, they can provide amplifier gain in addition to providing the same frequency response as RLC filters. Third, active filters can be combined with buffer amplifiers (voltage followers) to isolate each stage of the filter from source and load impedance effects. This isolation allows designing the stages independently and then cascading them to realize the desired transfer function. However, active filters are less reliable and less stable. The practical limit of most active filters is about 100 khz most active filters operate well below that frequency.,
Some Preliminaries 5 Types of Active Filters Filters are often classified according to their order (or number of poles) or their specific design type. - First-order lowpass filter - First-order highpass filter - Bandpass filter - Bandreject (or Notch) filter
Bode Plots Some Preliminaries Bode Plots A more systematic way of obtaining the frequency response is to use Bode plots. Before we begin to construct Bode plots, we should take care of two important issues: the use of logarithms and decibels in expressing gain. Some Properties of Logarithms 6 log P 1 P 2 = log P 1 + log P 2 log P 1 /P 2 = log P 1 log P 2 log P n = n log P log 1 = 0 Decibel Scale In communications systems, gain is measured in bels. Historically, the bel is used to measure the ratio of two levels of power or power gain G; that is, G = Numver of bels = log 10 P 2 P 1 The decibel (db) provides us with a unit of less magnitude. It is 1/10 th of a bel and is given by G db = 10log 10 P 2 P 1 (1)
Decibel Scale G db = 10log 10 P 2 P 1 Some Preliminaries Bode Plots (1) 7 When P 1 = P 2, there is no change in power and the gain is 0 db. If P 2 = 2P 1, the gain is G db = 10log 10 2P 1 P 1 = 10log 10 2 = 3 db (2) If P 2 = 0.5P 1, the gain is G db = 10log 10 0.5P 1 P 1 = 10log 10 0.5 = 3 db (3) Equations (1) and (2) show another reason why logarithms are greatly used: The logarithm of the reciprocal of a quantity is simply negative the logarithm of that quantity.
Decibel Scale Some Preliminaries Alternatively, the gain G can be expressed in terms of voltage or current ratio. P 2 V 2 2 G db = 10 log 2 /R 2 10 = 10 log P 10 1 V 2 2 1 /R 1 = 10 log 10 V 2 V 1 2 + 10 log 10 R 1 R 2 Bode Plots (4) G db = 10log 10 P 2 P 1 (1) 8 G db = 20log 10 V 2 V 1 10 log 10 R 2 R 1 (5) For the case when R 2 = R 1, a condition that is often assumed when comparing voltage levels, Eq.(5) becomes G db = 20 log 10 V 2 V 1 (6)
Decibel Scale Some Preliminaries Bode Plots If P 1 = I 1 2 R 1 and P 2 = I 2 2 R 2, for R 2 = R 1, we obtain 9 G db = 20 log 10 I 2 I 1 (7) The db value is a logarithmic measurement of the ratio of one variable to another of the same type. Therefore, it applies in expressing the transfer function H in Eqs.(8) and (9), which are dimensionless quantities, H ω H ω = voltage gain = V o(ω) V i (ω) = current gain = I o(ω) I i (ω) (8) (9) but not in expressing H in Eqs.(10) and (11). H ω H ω = Transfer Impedance = V o(ω) I i (ω) = Transfer admittance = I o(ω) V i (ω) (10) (11)
Bode Plots Some Preliminaries Bode Plots The frequency range required in frequency response is often so wide that it is inconvenient to use a linear scale for the frequency axis(fig.a). Also, there is a more systematic way of locating the important features of the magnitude and phase plots of the transfer function. For these reasons, it has become standard practice to use a logarithmic scale for the frequency axis and a linear scale in each of the separate plots of magnitude and phase. Such semilogarithmic plots of the transfer function known as Bode plots have become the industry standard(fig.b). 10 Fig.a: A linear scale for the frequency axis
Bode Plots Some Preliminaries Bode Plots Bode plots differ from the frequency response plots in Chapter 14 in two important ways. Instead of using a linear axis for the frequency values, a Bode plot uses a logarithmic axis. This permits us to plot a wider range of frequencies of interest. Normally we plot three or four decades of frequencies, say from 10 2 khz to 10 6 khz, or 1 khz to 1 MHz, choosing the frequency range where the transfer function characteristics are changing. If we plot both the magnitude and phase angle plots, they again share the frequency axis. 11
Bode Plots Some Preliminaries Bode Plots The Bode magnitude is plotted in decibels (db) versus the log of the frequency. Briefly, if the magnitude of the transfer function is H(jω), its value in db is given by 12 A db = 20 log 10 H(jω) - When A db = 0, 20 log 10 H(jω) = 0 H(jω) = 1 - When A db < 0, 20 log 10 H(jω) < 0 0 < H(jω) < 1 - When A db > 0, 20 log 10 H(jω) > 0 1 < H(jω) - When H(jω) = 1/ 2, 20 log 10 12 = 3 db
Bode Plots Some Preliminaries Bode Plots 13 At cutoff frequency ω c = 2πf c, H(jω) = 1 2 20 log 10 12 = 20 log 10 2 1 = 20 log10 2 1 2 = 20 1 2 log 102 = 20 1 2 0.3 = 3 db Actual response of a RC filter The lowpass filter reduces the overall voltage gain of an amplifier by 3 db when the frequency is reduced to the cutoff frequency ω c = 2πf c. Bode plot
Bode Plots Some Preliminaries Bode Plots As the frequency continues to increase beyond ω c = 2πf c, the overall voltage gain also continues to decrease. The rate of decrease in voltage gain with frequency is called roll-off. For each ten times increase in frequency beyond f c, there is a 20 db reduction in voltage gain. 14 20 log 10 0.1 = 20 db A ten-times change in frequency is called a decade. The attenuation is reduced by 20 db for each decade that the frequency increases beyond the cutoff frequency. This causes the overall voltage gain to drop 20 db per decade. V o (s) = Z c 1 V R + Z i (s) H jω = c 1 + jωrc
Bode Plots Some Preliminaries Bode Plots The -20dB/decade roll-off rate for the gain of a basic RC filter means that at a frequency of 10f c, the output will be -20dB(10%) of the input. This roll-off rate is not a particularly good filter characteristic because too much of the unwanted frequencies (beyond the passband) are allowed through the filter. In order to produce a filter that has a steeper transition region (and hence form a more effective filter), it is necessary to add additional circuitry to the basic filter. Responses that are steeper than -20dB/decade in the transition region cannot be obtained by simply cascading identical RC stages (due to loading effects). However, by combining an op-amp with frequency - selective feedback circuits, filters can be designed with roll-off rates of -40, -60, or more db/decade. Filters that include one or more op-amps in the design are called active filters. These filters can optimize the roll-off rate or other attribute (such as phase response) with a particular filter design. In general, the more poles the filter uses, the steeper its transition region will be. The exact response depends on the type of filter and the number of poles. 15
Summary Some Preliminaries Bode Plots Bode plots describe the behavior of a filter by relating the magnitude of the filter's response (gain) to its frequency. The key feature of this graph is that both axes have logarithmic scales. An example of this type of plot is shown in Figure Pass band stop band 16 Gain Pass-band ripple stop band Pass band Attenuation rate A filter's Bode plot can show key features of a filter, such as cutoff frequency, attenuation rate, and passband ripple. Cutoff frequency (ω c = 2πωf c ) Frequency (Hz)
Active Low-Pass Filter Circuits First-Order Low-Pass and High-Pass Filters First-Order Low-Pass Filter 17
Qualitative Analysis First-Order Low-Pass and High-Pass Filters First-Order Low-Pass Filter At very low frequencies (ω o 0), the capacitor acts like an open circuit, and the op amp circuit acts like an amplifier with a gain of R 2 /R 1. 18 At very high frequencies(ω o ), the capacitor acts like a short circuit, thereby connecting the output of the op amp circuit to ground. The op amp circuit in Fig. 15.1 thus functions as a low-pass filter with a passband gain of R 2 /R 1.
Quantitative Analysis Output voltage V o = Z f Z i V i Z f = R 2 Transfer function: First-Order Low-Pass and High-Pass Filters First-Order Low-Pass Filter 1 sc R 2 + 1 sc = R 2 1 + sr 2 C 19 H s = V o = Z R 2 f 1 + sr = 2 C = R 2 1 V i Z i R 1 R 1 1 + sr 2 C = R 1 2 R 2 C R 1 s + 1 R 2 C H jω = R 2 R 1 1 1 + jωr 2 C ω c = 1 R 2 C K = R 2 R 1 = K ω c s + ω c (15.3) (15.2) (15.1) At ω = ω c 1 + jωr 2 C = 1 + j 1 + j = 2
Active High-Pass Filter Circuits First-Order Low-Pass and High-Pass Filters First-Order High-Pass Filter 20 A first-order(single pole) high-pass filter
Qualitative Analysis First-Order Low-Pass and High-Pass Filters First-Order High-Pass Filter At very low frequencies (ω o 0), the capacitor acts like an open circuit, thereby connecting the input of the op amp circuit to ground(fig.b). v o 0 At very high frequencies(ω o ), the capacitor acts like a short circuit, and the op amp circuit acts like an amplifier with a gain of R 2 /R 1.(Fig.c) The op amp circuit in Fig. 15.4 thus functions as a high-pass filter with a passband gain of R 2 /R 1. 21
Quantitative Analysis Output voltage First-Order Low-Pass and High-Pass Filters First-Order High-Pass Filter 22 V o = Z f Z i V i Z i = R 1 + 1 sc Transfer function: H s = V o V i = Z f Z i = R 2 R 1 + 1 sc = R 2 R 1 s s + 1 R 1 C = K s s + ω c (15.4) H jω = R 2 R 1 1 1 + 1 jωr 1 C ω c = 1 R 1 C K = R 2 R 1 (15.6) (15.5) At ω = ω c 1 + 1 jωr 1 C = 1 j 1 j = 2
Scaling 23 Scaling In the design and analysis of both passive and active filter circuits, working with element values such as 1Ω, 1 H, and 1 F is convenient. After making computations using convenient values of R, L, and C, the designer can transform the convenient values into realistic values using the process known as scaling. There are two types of scaling: magnitude and frequency. Magnitude scaling We scale a circuit in magnitude by multiplying the impedance at a given frequency by the scale factor k m. R = k m R, L = k m R, C = C k m (15.7) k m = scale factor: a positive real number that can be either less than or greater than 1 - Unprimed variables represent the initial values of the parameters. - Primed variables represent the scaled values of the variables
Scaling 24 Scaling Frequency scaling We change the circuit parameters so that at the new frequency, the impedance of each element is the same as it was at the original frequency. Because resistance values are assumed to be independent of frequency, resistors are unaffected by frequency scaling. Both inductors and capacitors are multiplied by the frequency scaling factor,k f. R = R, L = L/k f, and C = C/k f (15.8) k f = frequency scale factor: a positive real number that can be either less than or greater than 1 Circuit scaling A circuit can be scaled simultaneously in both magnitude and frequency. The scaled values (primed) in terms of the original values (unprimed) are R = k m R L = k m k f L C = 1 k m k f C (15.9)
Scaling 25 The Use of Scaling in the Design of Op Amp Filters To use the concept of scaling in the design of op amp filters, (a) First select the cutoff frequency, ω c, to be 1 rad/s (if you are designing low- or highpass filters), or select the center frequency, ω o, to be 1 rad/s (if you are designing bandpass or bandreject filters). (b) Then select a 1 F capacitor and calculate the values of the resistors needed to give the desired passband gain and the 1 rad/s cutoff or center frequency. (c) Finally, use scaling to compute more realistic component values that give the desired cutoff or center frequency.
Op Amp Bandpass and Bandreject Filters Op Amp Bandpass Filters The Easiest Way of Creating a Bandpass Filter; Initial Approach While there is a wide variety of such op amp circuits, our initial approach is motivated by the Bode plot construction shown in Figure. The bandpass filter consists of three separate components: A unity-gain low-pass filter whose cutoff frequency is ω c2, the larger of the two cutoff frequencies. A unity-gain high-pass filter whose cutoff frequency is ω c1, the smaller of the two cutoff frequencies. A gain component to provide the desired level of gain in the passband. These three components are cascaded in series. 26 ω c1 ω c2
Op Amp Bandpass and Bandreject Filters Op Amp Bandpass Filters The Easiest Way of Creating a Bandpass Filter; Initial Approach These three components are cascaded in series. 27 ω c2 ω c1 K ω c2 ω c1 ω c1 ω c2
Analysis of the Cascaded Bandpass Filter Op Amp Bandpass and Bandreject Filters Op Amp Bandpass Filters The transfer function of the cascaded bandpass filter is the product of the transfer functions of the three cascaded components: 28 H s = V o V i = ω c2 s + ω c2 = s s + ω c1 R f R i = Kω c2 s s 2 + ω c1 + ω c2 s + ω c1 ω c2 (15.10) Kω c2 s s + ω c1 s + ω c2
Analysis of the Cascaded Bandpass Filter Op Amp Bandpass and Bandreject Filters Op Amp Bandpass Filters Eq.(15.10) is not in the standard form for the transfer function of a bandpass filter discussed in Chapter 14. H BP = βs s 2 + sβ + ω o 2 H s = Kω c2 s s 2 + ω c1 + ω c2 s + ω c1 ω c2 (15.10) In order to convert Eq.(15.10) into the form of the standard transfer function for a bandpass filter, we require that ω c2 ω c1 (15.11) When Eq.(15.11) holds, 29 ω c1 + ω c2 ω c2 and the transfer function for the cascaded bandpass filter in Eq.(15.10) becomes H s = Kω c2 s s 2 + ω c2 s + ω c1 ω c2 (1)
Analysis of the Cascaded Bandpass Filter Op Amp Bandpass and Bandreject Filters Op Amp Bandpass Filters Once we confirm that Eq.(15.11) (ω c2 ω c1 ) holds for the cutoff frequencies specified for the desired bandpass filter, we can design each stage of the cascaded circuit independently and meet the filter specifications. Now we compute the values of R L and C L in the low-pass filter and the values R H and C H in the high-pass filter to provide the desired cutoff frequencies. - Low-pass filter: ω c2 = 1 R L C L - High-pass filter: ω c1 = 1 R H C H (15.12) (15.13) 30 ω c1 ω c2 ω c2 ω c1
Analysis of the Cascaded Bandpass Filter Op Amp Bandpass and Bandreject Filters Op Amp Bandpass Filters We compute the values of R i and R f in the inverting amplifier to provide the desired passband gain. To do this, we consider the magnitude of the bandpass filter s transfer function(eq.1), evaluated at the center frequency, ω o : 31 H jω 0 = Kω c2 jω 0 jω 0 2 + ω c2 jω 0 + ω c1 ω c2 = Kω c2 ω c2 = K (15.14) K = R f R i : the gain of the inverting amp Therefore, H jω 0 = R f R i (15.15)
Op Amp Bandpass and Bandreject Filters Op Amp Bandreject Filters A component approach to the design of op amp bandreject Like the bandpass filter, the bandreject filter consists of three separate components. There are important differences, however: A unity-gain low-pass filter has a cutoff frequency of ω c1, which is the smaller of the two cutoff frequencies. A unity-gain high-pass filter has a cutoff frequency of ω c2, which is the larger of the two cutoff frequencies. The gain component provides the desired level of gain in the passbands. The most important difference is that these three components cannot be cascaded in series, because they do not combine additively on the Bode plot. Instead, we use a parallel connection and a summing amplifier, as a circuit in Fig. 15.13(a). 32
Op Amp Bandpass and Bandreject Filters Op Amp Bandreject Filters A component approach to the design of op amp bandreject Again, it is assumed that the two cutoff frequencies are widely separated, so that the resulting design is a broadband bandreject filter, and ω c2 ω c1. Then each component of the parallel design can be created independently, and the cutoff frequency specifications will be satisfied. The transfer function of the resulting circuit is the sum of the low-pass and high-pass filter transfer functions. 33 H s = R f R i ω c1 s + ω c1 + s s + ω c2 = R f ω c1 s + ω c2 + s s + ω c1 R i s + ω c1 s + ω c2 = R f R i s 2 + 2ω c1 s + ω c1 ω c2 s + ω c1 s + ω c2 (15.16)
Op Amp Bandpass and Bandreject Filters Op Amp Bandreject Filters A component approach to the design of op amp bandreject The cutoff frequencies are given by the equations ω c1 = 1 R L C L (15.17) ω c2 = 1 R H C H (15.18) 34 In the two passbands (as s 0 and s ), the gain of the transfer function is K = R f R i (15.19)
Op Amp Bandpass and Bandreject Filters Op Amp Bandreject Filters A component approach to the design of op amp bandreject The magnitude of the transfer function in Eq.(15.16) at the center frequency, ω o = ω c1 ω c2 ; K = R f R i 35 H jω 0 = R f jω 2 0 + 2ω c1 jω 0 + ω c1 ω c2 R i jω 2 0 + ω c1 + ω c2 jω 0 + ω c1 ω c2 = R f R i If ω c2 ω c1, then 2ω c1 ω c1 + ω c2 R f R i 2ω c1 ω c2 (15.19) H jω 0 2R f R f 2ω c1 R i ω c2 R i (15.20) H(s) = R f R i s 2 + 2ω c1 s + ω c1 ω c2 s + ω c1 s + ω c2 (15.16) So, the magnitude at the center frequency is much smaller than the passband magnitude. Thus the bandreject filter successfully rejects frequencies near the center frequency, again confirming our assumption that the parallel implementation is meant for broadband bandreject designs.
Higher Order Op Amp Filters Cascading Identical Filters How can we realize a filter with a sharper transition at the cutoff frequency? 36 All of the filter circuits we have examined so far, both passive and active, are nonideal. An ideal filter has a discontinuity at the point of cutoff, which sharply divides the passband and the stopband. Although we cannot hope to construct a circuit with a discontinuous frequency response, we can construct circuits with a sharper, yet still continuous, transition at the cutoff frequency. Idea filter In order to produce a filter that has a steeper transition region (and hence form a more effective filter), it is necessary to add additional circuitry to the basic filter. Responses that are steeper than -20dB/decade in the transition region cannot be obtained by simply cascading identical RC stages (due to loading effects). However, by combining an op-amp with frequency-selective feedback circuits, filters can be designed with roll-off rates of -40, -60, or more db/decade.
A Cascade of Identical Low-pass Filters Higher Order Op Amp Filters Cascading Identical Filters The Bode magnitude plots of a cascade of identical lowpass filters includes plots of just one filter, two in cascade, three in cascade, and four in cascade. It is obvious that as more filters are added to the cascade, the transition from the passband to the stopband becomes sharper. 37 - One filter : 20 db/dec - Two filters : 20+20=40 db/dec - Three filters : 60 db/dec - Four filters : 80 db/dec - n-element cascade of identical low-pass filters: n20db/dec
A Cascade of Identical Low-pass Filters Higher Order Op Amp Filters Cascading Identical Filters The transfer function for a cascade of n prototype low-pass filters. 38 H s = 1 s + 1 1 s + 1 1 s + 1 = 1 n s + 1 n (15.21) The order of a filter is determined by the number of poles in its transfer function. From Eq.(15.21), we see that a cascade of first-order low-pass filters yields a higher order. In fact, a cascade of n first-order filters produces an nth -order filter, having n poles in its transfer function and a final slope of 20n db/dec in the transition band.
Higher Order Op Amp Filters Cascading Identical Filters Cutoff frequency of nth-order low-pass filter As the order of the low-pass filter is increased by adding prototype low-pass filters to the cascade, the cutoff frequency also changes. 39 If we start with a cascade of n low-pass filters, we can compute the cutoff frequency for the resulting nth-order low-pass filter. H s = n 1 s + 1 n (15.21) H jω cn = 1 jω cn + 1 n = 1 2 jω cn + 1 = ω 2 cn + 1 1 ω 2 cn + 1 n = 1 2 1 2/n 1 ω 2 cn + 1 = 1 2 1/n ω 2 cn + 1 = 1 2 ω 2 n cn + 1 = 2 n ω cn = 2 1 1 2 + 1 = 1 2 2/n = 1 2 1/n = 1 n 2 ω cn (15.22)
Higher Order Op Amp Filters Cascading Identical Filters Cutoff frequency of nth-order low-pass filter 40 n ω cn = 2 1 (15.22) To demonstrate the use of Eq.(15.22), let s compute the cutoff frequency of a fourth-order unity-gain low-pass filter constructed from a cascade of four prototype lowpass filters: 4 ω c4 = 2 1 = 0.435 rad/s (15.23) Thus, we can design a fourth-order low-pass filter with any arbitrary cutoff frequency by starting with a fourth-order cascade consisting of low-pass filters and then scaling the components by k f = ω c /0.435 to place the cutoff frequency at any value of desired.
Higher Order Op Amp Filters Nonideal passband behavior of low-pass nth-order cascade The transfer function for a unity-gain low-pass nth-order cascade. H s = n ω cn s + ω n cn Cascading Identical Filters 41 H jω = = n ω cn ω 2 2 + ω cn 1 n ω/ω cn 2 + 1 n (15.24) - When ω ω cn, H jω 1 - When ω ω cn, H jω < 1 Because the cascade of low-pass filters results in this nonideal behavior in the passband, other approaches are taken in the design of higher order filters. One such approach is examined next.
Filter Response Characteristics. Higher Order Op Amp Filters Butterworth Filters The filter frequency responses are typically characterized by the shape of the response curve. A general comparison of the three response characteristics for a low-pass filter response curve is shown in Figure. The Butterworth( 버터워스 ) characteristic provides a very flat amplitude response in the passband and a roll-off rate of -20dB/decade/pole. Filters with the Butterworth response are normally used when all frequencies in the passband must have the same gain. The Butterworth response is often referred to as a maximally flat response. Filters with the Chebyshev( 체비셰프 ) response characteristic are useful when a rapid roll-off is required because it provides a roll-off rate greater than -20dB/decade/pole. This type of filter response is characterized by overshoot or ripples in the passband (depending on the number of poles) and an even less linear phase response than the Butterworth. The Bessel response exhibits a linear phase characteristic, meaning that the phase shift increases linearly with frequency. The result is almost no overshoot on the output with a pulse input. For this reason, filters with the Bessel response are used for filtering pulse waveforms without distorting the shape of the waveform. 42 Comparative plots of three types of filter response characteristics.
Butterworth low-pass filter Higher Order Op Amp Filters Butterworth Filters Butterworth filters are characterized by a maximally flat pass-band frequency response characteristic. A unity-gain Butterworth low-pass filter has a transfer function whose magnitude is given by H jω = 1 1 + ω/ω c 2n n: the order of filter 43 - The cutoff frequency is ω c for all values of n. - If n is large enough, the denominator is always close to unity when ω < ω c. - In the expression for H jω the exponent of ω/ω c is always even. Butterworth low-pass filter frequency response
Higher Order Op Amp Filters Butterworth Filters Butterworth low-pass filter Given an equation for the magnitude of the transfer function, how do we find H jω? 44 or H jω 2 = H(jω) H( jω) = H s H( s) 1 = 1 + ω 2n = 1 = 1 + (ω 2 ) n 1 1 + ( s 2 ) n = 1 1 + ( 1) n s 2n s = jω s 2 = ω 2 H s H( s) = 1 1 + ( 1) n s 2n 1 + ( 1) n s 2n : Butterworth polynomials, given in Table 15.3 in factored form.
Butterworth low-pass filter Higher Order Op Amp Filters Butterworth Filters They are the product of first- and second-order factors; therefore, we can construct a circuit whose transfer function has a Butterworth polynomial in its denominator by cascading op amp circuits, each of which provides one of the needed factors. A block diagram of such a cascade is shown in Fig. 15.20, using a fifth-order Butterworth polynomial as an example. 45
Butterworth low-pass filter Higher Order Op Amp Filters Butterworth Filters All odd-order Butterworth polynomials include the factor (s + 1), so all odd-order Butterworth filter circuits must have a subcircuit that provides the transfer function. This is the transfer function of the prototype low-pass op amp filter from Fig. 15.1. 46 ω c = 1 R 1 C H s = ω c s + ω c = 1 s + 1 So what remains is to find a circuit that provides a transfer function of the form H s = 1 s 2 + b 1 + 1 Such a circuit is shown in Fig. 15.21.
Analysis of a Butterworth Filter Circuit Output voltage Higher Order Op Amp Filters Butterworth Filters 47 V a V i R + V a V o sc 1 + V a V o R = 0 2 + RC 1 s V a 1 + RC 1 s V o = V i (15.31) V a V o R = V o sc 2 V a + 1 + RC 2 s V o = 0 (15.32) Using Cramer s rule with Eqs. 15.31 and 15.32, we V o = 2 + RC 1 s V i 1 0 2 + RC 1 s 1 + RC 1 s 1 1 + RC 2 s = V i R 2 C 1 C 2 s 2 + 2RC 2 s + 1 (15.33)
Analysis of a Butterworth Filter Circuit Transfer function H s = V o V i = If we set R = 1Ω H s = 1 R 2 C 1 C 2 s 2 + 2 RC 1 s + 1 C 1 C 2 s 2 + 2 C 1 s + 1 C 1 C 2 Higher Order Op Amp Filters 1 R 2 C 1 C 2 Butterworth Filters (15.34) (15.35) V o = = 2 + RC 1 s V i 1 0 2 + RC 1 s 1 + RC 1 s 1 1 + RC 2 s V i R 2 C 1 C 2 s 2 + 2RC 2 s + 1 48 (15.33) In order to get a transfer function of the form H s = 1 s 2 + b 1 + 1 we choose capacitor values so that b 1 = 2 C 1 1 = 1 C 1 C 2 (15.36)
Butterworth High-Pass Filter 15.4 Higher Order Op Amp Filters Butterworth Filters 49 The second-order Butterworth low-pass filter has a transfer function of the form H s = 1 s 2 + b 1 + 1 An nth-order Butterworth high-pass filter has a transfer function with the nth-order Butterworth polynomial in the denominator, just like the nth-order Butterworth low-pass filter. But in the high-pass filter, the numerator of the transfer function is s n whereas in the low-pass filter, the numerator is 1. To produce the second-order factors in the Butterworth polynomial, we need a circuit with a transfer function of the form H s = s 2 s 2 + b 1 s + 1
Butterworth High-Pass Filter Transfer function 15.4 Higher Order Op Amp Filters Butterworth Filters 50 V a V i 1 sc V a V o 1 sc + V a V o R 1 = V o 0 R 2 + V a V o 1 sc = 0 1 + 2R 1 Cs V a 1 + R 1 Cs V o = scv i R 2 Cs V a 1 + R 2 Cs V o = 0 V o = s 2 s 2 + 2 R 2 C s + 1 R 1 R 2 C 2 V i H s = V o V i = s 2 s 2 + 2 R 2 C s + 1 R 1 R 2 C 2 (15.47) Setting C = 1 F yields H s = s 2 s 2 + 2 R 2 s + 1 R 1 R 2 (15.48)
Butterworth High-Pass Filters H s = s 2 s 2 + 2 R 2 s + 1 R 1 R 2 Higher Order Op Amp Filters Butterworth Filters (15.48) H s = s 2 s 2 + b 1 s + 1 51 b 1 = 2 R 2 1 = 1 R 1 R 2 (15.49) First, the high-pass circuit in Fig.15.25 was obtained from the low-pass circuit in Fig. 15.21 by interchanging resistors and capacitors. Second, the prototype transfer function of a highpass filter can be obtained from that of a low-pass filter by replacing s in the low-pass expression with 1/s (see Problem 15.46).
Active High-Q Bandpass Filter The narrowband filter requires a high-q. Narrowband Bandpass and Bandreject Filters Narrowband Bandpass Filters The transfer functions for cascaded bandpass and parallel bandreject filters have discrete real poles. With discrete real poles, the highest quality bandpass filter (or bandreject filter) we can achieve has 52 Q = ω o β = 1 2 To build active filters with high quality factor values, we need an op amp circuit that can produce a transfer function with complex conjugate poles. Figure 15.26 depicts one such circuit for us to analyze
Active High-Q Bandpass Filter Transfer function Node b V a 0 1/sC = 0 V a R 3 Narrowband Bandpass and Bandreject Filters Narrowband Bandpass Filters V a = V o sr 3 C (15.54) 53 Node a V i V a R 1 = V a V o 1/sC + V a 0 1/sC + V a 0 R 2 V i = 1 + 2sR 1 C + R 1 /R 2 V a sr 1 CV o (15.55) Substituting Eq.(15.54) into Eq.(15.55) and then rearranging, we get an expression for the transfer function H s = s R 1 C s 2 + 2 R 3 C s + 1 R eq R 3 C 2 (15.56) R eq = R 1R 2 R 1 + R 2
Active High-Q Bandpass Filter Narrowband Bandpass and Bandreject Filters Narrowband Bandpass Filters The standard form of the transfer function for a bandpass filter is 54 H s = H s = Kβs s 2 2 + βs + ω o s R 1 C s 2 + 2 R 3 C s + 1 R eq = R 1R 2 R 1 + R 2 R eq R 3 C 2 (15.56) Equating terms and solving for the values of the resistors give β = 2 R 3 C Kβ = 1 R 1 C ω o 2 = 1 R eq R 3 C 2 (15.57) (15.58) (15.59) ω o : a specified center frequency Q: quality factor K: passband gain
Active High-Q Bandpass Filter Narrowband Bandpass and Bandreject Filters Narrowband Bandpass Filters The prototype version of the circuit in Fig. 15.25 55 ω o = 1 rad/s C = 1 F Q = ω o β = 1 β From Eq.(15.58), ω o = ω o = 1 LC ω o 1 LC Kβ = 1 R 1 C R 1 = 1 Kβ = Q K From Eq.(15.57), β = 2 R 3 C R 3 = 2 β = 2Q ω o : a specified center frequency Q: quality factor K: passband gain From Eq.(15.59), ω o 2 = 1 R eq R 3 C 2 R eq = 1 R 3 = 1 2Q = R 1R 2 R 1 + R 2 R 2 = Q/(2Q 2 K)
Active High-Q Bandreject Filter 15.5 Narrowband Bandpass and Bandreject Filters Narrowband Bandreject Filters The parallel implementation of a bandreject filter that combines low-pass and high-pass filter components with a summing amplifier has the same low-q restriction as the cascaded bandpass filter. An active high-q bandreject filter known as the twin-t notch filter because of the two T-shaped parts of the circuit at the nodes labeled a and b. 56
Active High-Q Bandreject Filter Analysis of twin-t notch filter Node a : Narrowband Bandpass and Bandreject Filters V a V i sc + V a V o sc + 2(V a σv o ) R Narrowband Bandreject Filters = 0 57 V a 2sCR + 2 V o scr + 2σ = scrv i (15.60) Node b : V b V i R + V b V o R + V b σv o 2sC = 0 V b 2 + 2RCs V o 1 + 2σRCs = V i (15.61) Noninverting input terminal of the top op amp : V a V o sc + V b V 0 R = 0 srcv a V b + src + 1 V o = 0 (15.62)
Active High-Q Bandreject Filter Analysis of twin-t notch filter Narrowband Bandpass and Bandreject Filters Narrowband Bandreject Filters 58 V a 2sCR + 2 V o scr + 2σ = scrv i V b 2 + 2RCs V o 1 + 2σRCs = V i (15.60) (15.61) srcv a V b + src + 1 V o = 0 (15.61) Using Cramer s rule to solve for V o gives V o = 2 RCs + 1 0 scrv i 0 2(RCs + 1) V i RCs 1 0 2(RCs + 1) 0 (RCs + 2σ) 0 2(RCs + 1) (2σRCs + 1) RCs 1 RCs + 1 = R 2 C 2 s 2 + 1 V i R 2 C 2 s 2 + 4RC 1 σ s + 1 (15.63)
Active High-Q Bandreject Filter Analysis of twin-t notch filter Narrowband Bandpass and Bandreject Filters Narrowband Bandreject Filters Rearranging Eq.(15.63), we can solve for the transfer function: H s = V o V i = s 2 + 1 R 2 C 2 s 2 + 4 1 σ RC s + 1 R 2 C 2 (15.64) 59 The standard form for the transfer function of a bandreject filter: H s = s2 + ω 0 2 s 2 + βs + ω 0 2 (15.65) Equating Eqs.(15.64) and (15.65) gives ω o 2 = 1 R 2 C 2 β = 4 1 σ RC (15.66) (15.67)
Active High-Q Bandreject Filter Analysis of twin-t notch filter Narrowband Bandpass and Bandreject Filters Narrowband Bandreject Filters In this circuit, we have three parameters (R, C, and σ) and two design constraints (ω o and β). Thus one parameter is chosen arbitrarily; it is usually the capacitor value because this value typically provides the fewest commercially available options. Once C is chosen, R = 1 ω o C (15.68) 60 And σ = 1 β = 1 1 4ω o 4Q (15.69)