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Vi w R Va SC₁ w Va R Vo + + + 1 店 SC₂ Vo HE by with such a transfer function. directly or cuit whose transfer function has a Butterworth polynomial in its denominator product of first- and second-order factors; therefore, we can construct a cir- Notice the form of the Butterworth polynomials in Table 15.1. They are the y cascading active filter circuits, each of which provides one of the needed factors. Figure 15.20 presents a block diagram of a cascade whose transfer function has a fifth-order Butterworth polynomial in its denominator. All odd-order Butterworth polynomials include the factor (s+ 1), so all odd-order Butterworth filter circuits must have a subcircuit with the transfer function H(s) = 1/(s + 1). This is the transfer function of the prototype low-pass active filter from Fig. 15.1. So what remains is to find a circuit whose transfer function has the form H(s) = 1/(s² + b₁s + 1). Such a circuit is shown in Fig. 15.21. To analyze this circuit, write the -domain KCL equations at the noninverting terminal of the op amp and at the node labeled V₁: Va - Vi R 0, Va - Vo +(Va - Vo)SC₁ + = R Vo - Va VosC2 + = 0. R ear Simplifying the two KCL equations yields - (2 + RC1s) Va (1 + RC₁s)V = Vi, simony-Va + (1 + RC25)V = 0. SC1 MAX3 w + R Va R + Vo SC2 1 s+1 1 s²+0.618s +1 1 Vo s²+1.618s +1 Figure 15.20 ▲ A cascade of first- and second-order circuits with the indicated trans- fer functions yielding a fifth-order low-pass Butterworth filter with we = 1 rad/s. Figure 15.21 ▲ A circuit that provides the second- order transfer function for the Butterworth filter cascade.

Vi w R Va SC₁ w Va R Vo + + + 1 店 SC₂ Vo HE by with such a transfer function. directly or cuit whose transfer function has a Butterworth polynomial in its denominator product of first- and second-order factors; therefore, we can construct a cir- Notice the form of the Butterworth polynomials in Table 15.1. They are the y cascading active filter circuits, each of which provides one of the needed factors. Figure 15.20 presents a block diagram of a cascade whose transfer function has a fifth-order Butterworth polynomial in its denominator. All odd-order Butterworth polynomials include the factor (s+ 1), so all odd-order Butterworth filter circuits must have a subcircuit with the transfer function H(s) = 1/(s + 1). This is the transfer function of the prototype low-pass active filter from Fig. 15.1. So what remains is to find a circuit whose transfer function has the form H(s) = 1/(s² + b₁s + 1). Such a circuit is shown in Fig. 15.21. To analyze this circuit, write the -domain KCL equations at the noninverting terminal of the op amp and at the node labeled V₁: Va - Vi R 0, Va - Vo +(Va - Vo)SC₁ + = R Vo - Va VosC2 + = 0. R ear Simplifying the two KCL equations yields - (2 + RC1s) Va (1 + RC₁s)V = Vi, simony-Va + (1 + RC25)V = 0. SC1 MAX3 w + R Va R + Vo SC2 1 s+1 1 s²+0.618s +1 1 Vo s²+1.618s +1 Figure 15.20 ▲ A cascade of first- and second-order circuits with the indicated trans- fer functions yielding a fifth-order low-pass Butterworth filter with we = 1 rad/s. Figure 15.21 ▲ A circuit that provides the second- order transfer function for the Butterworth filter cascade.

Electricity for Refrigeration, Heating, and Air Conditioning (MindTap Course List)
Electricity for Refrigeration, Heating, and Air Conditioning (MindTap Course List)
10th Edition
ISBN:9781337399128
Author:Russell E. Smith
Publisher:Russell E. Smith
Chapter7: Alternating Current, Power Distribution, And Voltage Systems
Section: Chapter Questions
Problem 5RQ
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Why is the voltage at the red-marked node labeled as VO?

Vi w R Va SC₁ w Va R Vo + + + 1 店 SC₂ Vo HE
Transcribed Image Text:Vi w R Va SC₁ w Va R Vo + + + 1 店 SC₂ Vo HE
by with such a transfer function. directly or cuit whose transfer function has a Butterworth polynomial in its denominator product of first- and second-order factors; therefore, we can construct a cir- Notice the form of the Butterworth polynomials in Table 15.1. They are the y cascading active filter circuits, each of which provides one of the needed factors. Figure 15.20 presents a block diagram of a cascade whose transfer function has a fifth-order Butterworth polynomial in its denominator. All odd-order Butterworth polynomials include the factor (s+ 1), so all odd-order Butterworth filter circuits must have a subcircuit with the transfer function H(s) = 1/(s + 1). This is the transfer function of the prototype low-pass active filter from Fig. 15.1. So what remains is to find a circuit whose transfer function has the form H(s) = 1/(s² + b₁s + 1). Such a circuit is shown in Fig. 15.21. To analyze this circuit, write the -domain KCL equations at the noninverting terminal of the op amp and at the node labeled V₁: Va - Vi R 0, Va - Vo +(Va - Vo)SC₁ + = R Vo - Va VosC2 + = 0. R ear Simplifying the two KCL equations yields - (2 + RC1s) Va (1 + RC₁s)V = Vi, simony-Va + (1 + RC25)V = 0. SC1 MAX3 w + R Va R + Vo SC2 1 s+1 1 s²+0.618s +1 1 Vo s²+1.618s +1 Figure 15.20 ▲ A cascade of first- and second-order circuits with the indicated trans- fer functions yielding a fifth-order low-pass Butterworth filter with we = 1 rad/s. Figure 15.21 ▲ A circuit that provides the second- order transfer function for the Butterworth filter cascade.
Transcribed Image Text:by with such a transfer function. directly or cuit whose transfer function has a Butterworth polynomial in its denominator product of first- and second-order factors; therefore, we can construct a cir- Notice the form of the Butterworth polynomials in Table 15.1. They are the y cascading active filter circuits, each of which provides one of the needed factors. Figure 15.20 presents a block diagram of a cascade whose transfer function has a fifth-order Butterworth polynomial in its denominator. All odd-order Butterworth polynomials include the factor (s+ 1), so all odd-order Butterworth filter circuits must have a subcircuit with the transfer function H(s) = 1/(s + 1). This is the transfer function of the prototype low-pass active filter from Fig. 15.1. So what remains is to find a circuit whose transfer function has the form H(s) = 1/(s² + b₁s + 1). Such a circuit is shown in Fig. 15.21. To analyze this circuit, write the -domain KCL equations at the noninverting terminal of the op amp and at the node labeled V₁: Va - Vi R 0, Va - Vo +(Va - Vo)SC₁ + = R Vo - Va VosC2 + = 0. R ear Simplifying the two KCL equations yields - (2 + RC1s) Va (1 + RC₁s)V = Vi, simony-Va + (1 + RC25)V = 0. SC1 MAX3 w + R Va R + Vo SC2 1 s+1 1 s²+0.618s +1 1 Vo s²+1.618s +1 Figure 15.20 ▲ A cascade of first- and second-order circuits with the indicated trans- fer functions yielding a fifth-order low-pass Butterworth filter with we = 1 rad/s. Figure 15.21 ▲ A circuit that provides the second- order transfer function for the Butterworth filter cascade.
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