ACTIVE FILTERS An important application of op-amp is the active filter. The word filter refers to the process of removing undesired portion of the frequency spectrum. The word active implies the use of one or more active devices, usually an operational amplifier, in the filter circuit. As an example of the application of op-amps in area of active filters, we will discuss the Butterworth filter. The discussion is only an introduction to the subject of the filter theory design. Two advantages of active filters over passive filters are: The maximum gain or the maximum value of the transfer function may be greater than unity. The loading effect is minimum, which means that the output response or the filter is essentially independent of the load driven by the filter.

Active Network Design From our discussion of frequency response, we know that RC-networks form filters. a is a simple example of a coupling capacitor circuit. The voltage transfer function for this circuit is T ( s ) = V 0 ( s ) V i ( s ) = R R + 1 sC = sRC 1 + sRC size 12{T \( s \) = { {V rSub { size 8{0} } \( s \) } over {V rSub { size 8{i} } \( s \) } } = { {R} over {R+ { {1} over { ital "sC"} } } } = { { ital "sRC"} over {1+ ital "sRC"} } } {}

a) Simple high-pass filter and b) Bode plot of transfer function amplitude The Bode plot of the voltage gain magnitude /T(jω)/ size 12{ lline T \( jω \) rline } {} is shown in a. The circuit is called a high-pass filter. (a) is another example of a simple RC network. Here, the voltage transfer function is T ( s ) = V 0 ( s ) V i ( s ) = 1 sC 1 sC + R = 1 1 + sRC size 12{T \( s \) = { {V rSub { size 8{0} } \( s \) } over {V rSub { size 8{i} } \( s \) } } = { { { {1} over { ital "sC"} } } over { { {1} over { ital "sC"} } +R} } = { {1} over {1+ ital "sRC"} } } {}

a) Simple low pass filter and b) Bode plot of transfer function The Bode plot of the voltage gain magnitude /T(jω)/ size 12{ lline T \( jω \) rline } {} for this circuit is shown in (b). This circuit is called a low-pass filter.

a) High-pass filter with voltage follower and b) low pass filter with voltage follower. Although these circuits both perform a basic filtering function, they may suffer from loading effects, substantially reducing the magnitude gain from the unity value shown in (b) and (b). Also, the cutoff frequency fL size 12{f rSub { size 8{L} } } {} and fH size 12{f rSub { size 8{H} } } {} may change when a load is connected to the output. The loading effect can essentially be eliminated by using a voltage follower as shown in . In addition, a non-inverting amplifier configuration can be incorporated to increase the gain, as well as eliminate the loading effects. These two filter circuits are called one-pole filters; the slope of the voltage gain magnitude curve outside the passband is 6 dB/octave or 20 dB/decade. This characteristic is called the rolloff. The rolloff becomes sharper or steeper with higher-order filters and is usually one of the specifications given for active filters. Two other categories of filters are bandpass and band-reject. The desired ideal frequency characteristics are shown in

Ideal frequency characteristics: a) band pass filter and b) band reject filter.

General Two-Pole Active Filter Consider with admittances Y1 size 12{Y rSub { size 8{1} } } {} through Y4 size 12{Y rSub { size 8{4} } } {} and an ideal voltage follower. We will derive the transfer function for the general network and will then apply specific admittances to obtain particular filter characteristics. A KCL equation at node Va size 12{V rSub { size 8{a} } } {} yields ( V i − V a ) Y 1 = ( V a − V b ) Y 2 + ( V a − V 0 ) Y 3 size 12{ \( V rSub { size 8{i} } - V rSub { size 8{a} } \) Y rSub { size 8{1} } = \( V rSub { size 8{a} } - V rSub { size 8{b} } \) Y rSub { size 8{2} } + \( V rSub { size 8{a} } - V rSub { size 8{0} } \) Y rSub { size 8{3} } } {} A KCL equation at node Vb size 12{V rSub { size 8{b} } } {} produces ( V a − V b ) Y 2 = V b Y 4 size 12{ \( V rSub { size 8{a} } - V rSub { size 8{b} } \) Y rSub { size 8{2} } =V rSub { size 8{b} } Y rSub { size 8{4} } } {} From the voltage follower characteristics, we have Vb=V0 size 12{V rSub { size 8{b} } =V rSub { size 8{0} } } {}. Therefore, becomes V a = V b ( Y 2 + Y 4 Y 2 ) = V 0 ( Y 2 + Y 4 Y 2 ) size 12{V rSub { size 8{a} } =V rSub { size 8{b} } \( { {Y rSub { size 8{2} } +Y rSub { size 8{4} } } over {Y rSub { size 8{2} } } } \) =V rSub { size 8{0} } \( { {Y rSub { size 8{2} } +Y rSub { size 8{4} } } over {Y rSub { size 8{2} } } } \) } {} Substituting into and again noting that Vb=V0 size 12{V rSub { size 8{b} } =V rSub { size 8{0} } } {}, we have V i Y 1 + V 0 ( Y 2 + Y 3 ) = V a ( Y 1 + Y 2 + Y 3 ) = V 0 ( Y 2 + Y 4 Y 2 ) ( Y 1 + Y 2 + Y 3 ) size 12{V rSub { size 8{i} } Y rSub { size 8{1} } +V rSub { size 8{0} } \( Y rSub { size 8{2} } +Y rSub { size 8{3} } \) =V rSub { size 8{a} } \( Y rSub { size 8{1} } +Y rSub { size 8{2} } +Y rSub { size 8{3} } \) =V rSub { size 8{0} } \( { {Y rSub { size 8{2} } +Y rSub { size 8{4} } } over {Y rSub { size 8{2} } } } \) \( Y rSub { size 8{1} } +Y rSub { size 8{2} } +Y rSub { size 8{3} } \) } {} Multiplying and rearranging terms, we get the following expression for the transfer function: T ( s ) = V 0 ( s ) V i ( s ) = Y 1 Y 2 Y 1 Y 2 + Y 4 ( Y 1 + Y 2 + Y 3 ) size 12{T \( s \) = { {V rSub { size 8{0} } \( s \) } over {V rSub { size 8{i} } \( s \) } } = { {Y rSub { size 8{1} } Y rSub { size 8{2} } } over {Y rSub { size 8{1} } Y rSub { size 8{2} } +Y rSub { size 8{4} } \( Y rSub { size 8{1} } +Y rSub { size 8{2} } +Y rSub { size 8{3} } \) } } } {}

General two pole active filter. To obtain a low-pass filter, both Y1 size 12{Y rSub { size 8{1} } } {} and Y2 size 12{Y rSub { size 8{2} } } {} must be conductances, allowing the signal to pass into the voltage follower at low frequencies. If element Y4 size 12{Y rSub { size 8{4} } } {} is a capacitor, then the output rolloff at high frequencies. To produce a two-pole function, element Y3 size 12{Y rSub { size 8{3} } } {} must also be a capacitor. On the other hand, if elements Y1 size 12{Y rSub { size 8{1} } } {} and Y2 size 12{Y rSub { size 8{2} } } {} are capacitors, then the signal will be blocked at low frequencies but will be passed into the voltage follower at high frequencies resulting in a high-pass filter. Therefore, admittances Y3 size 12{Y rSub { size 8{3} } } {} and Y4 size 12{Y rSub { size 8{4} } } {} must both be conductances to produce a two-pole high-pass transfer function.

Two-Pole Low-Pass Butterworth Filter To form a low-pass filter, we set Y1=G1=1/R1 size 12{Y rSub { size 8{1} } =G rSub { size 8{1} } = {1} slash {R rSub { size 8{1} } } } {}, Y2=G2=1/R2 size 12{Y rSub { size 8{2} } =G rSub { size 8{2} } = {1} slash {R rSub { size 8{2} } } } {}, Y3=sC3 size 12{Y rSub { size 8{3} } = ital "sC" rSub { size 8{3} } } {} and Y4=sC4 size 12{Y rSub { size 8{4} } = ital "sC" rSub { size 8{4} } } {}, as shown in . The transfer function, from , becomes T ( s ) = V 0 ( s ) V i ( s ) = G 1 G 2 G 1 G 2 + sC 4 ( G 1 + G 2 + sC 3 ) size 12{T \( s \) = { {V rSub { size 8{0} } \( s \) } over {V rSub { size 8{i} } \( s \) } } = { {G rSub { size 8{1} } G rSub { size 8{2} } } over {G rSub { size 8{1} } G rSub { size 8{2} } + ital "sC" rSub { size 8{4} } \( G rSub { size 8{1} } +G rSub { size 8{2} } + ital "sC" rSub { size 8{3} } \) } } } {} At zero frequency, s = j ω size 12{ω} {} = 0 and the transfer function is T ( s = 0 ) = G 1 G 2 G 1 G 2 = 1 size 12{T \( s=0 \) = { {G rSub { size 8{1} } G rSub { size 8{2} } } over {G rSub { size 8{1} } G rSub { size 8{2} } } } =1} {} In the high frequency limit, s=jω→∞ size 12{s=jω rightarrow infinity } {} and the transfer function approaches zero. This circuit therefore acts as a low-pass filter.

General two pass filter. A butterworth filter is a maximally flat magnitude filter. The transfer function is designed such that the magnitude of the transfer function is as flat as possible within the passband of the filter. This objective is achieved by taking the derivatives of the transfer function with respect to frequency and setting as many as possible equal to zero at the center of the passband, which is at zero frequency for the low-pass filter. Let G1=G2=G=1/R size 12{G rSub { size 8{1} } =G rSub { size 8{2} } =G= {1} slash {R} } {}. the transfer function is then T ( s ) = 1 R 2 1 R 2 + sC 4 ( 2 R + sC 3 ) = 1 1 + sRC 4 ( 2 + sRC 3 ) size 12{T \( s \) = { { { {1} over {R rSup { size 8{2} } } } } over { { {1} over {R rSup { size 8{2} } } } + ital "sC" rSub { size 8{4} } \( { {2} over {R} } + ital "sC" rSub { size 8{3} } \) } } = { {1} over {1+ ital "sRC" rSub { size 8{4} } \( 2+ ital "sRC" rSub { size 8{3} } \) } } } {} We define time constant as τ3=RC3 size 12{τ rSub { size 8{3} } = ital "RC" rSub { size 8{3} } } {} and τ4=RC4 size 12{τ rSub { size 8{4} } = ital "RC" rSub { size 8{4} } } {}. If we then set s = j ω size 12{ω} {}, we obtain T ( jω ) = 1 1 + j ωτ 4 ( 2 + j ωτ 3 ) = 1 ( 1 − ω 2 τ 3 τ 4 ) + j ( 2 ωτ 4 ) size 12{T \( jω \) = { {1} over {1+j ital "ωτ" rSub { size 8{4} } \( 2+j ital "ωτ" rSub { size 8{3} } \) } } = { {1} over { \( 1 - ω rSup { size 8{2} } τ rSub { size 8{3} } τ rSub { size 8{4} } \) +j \( 2 ital "ωτ" rSub { size 8{4} } \) } } } {} The magnitude of the transfer function is therefore / T ( jω ) / = [ ( 1 − ω 2 τ 3 τ 4 ) 2 ] − 1 / 2 size 12{ lline T \( jω \) rline = \[ \( 1 - ω rSup { size 8{2} } τ rSub { size 8{3} } τ rSub { size 8{4} } \) rSup { size 8{2} } \] rSup { size 8{ - 1/2} } } {} For the maximally flat filter (that is, a filter with a minimum rate of change), which defines a Butterworth filter, we set d / T / dω / ω = 0 = 0 size 12{ { {d lline T rline } over {dω} } \rline rSub { size 8{ω=0} } =0} {} Taking the derivative, we find d / T / dω = 1 2 [ ( 1 − ω 2 τ 3 τ 4 ) 2 + ( 2 ωτ 4 ) 2 ] − 3 / 2 [ − 4 ωτ 3 τ 4 ( 1 − ω 2 τ 3 τ 4 ) + 8 ωτ 4 2 ] size 12{ { {d lline T rline } over {dω} } = { {1} over {2} } \[ \( 1 - ω rSup { size 8{2} } τ rSub { size 8{3} } τ rSub { size 8{4} } \) rSup { size 8{2} } + \( 2 ital "ωτ" rSub { size 8{4} } \) rSup { size 8{2} } \] rSup { size 8{ - 3/2} } \[ - 4 ital "ωτ" rSub { size 8{3} } τ rSub { size 8{4} } \( 1 - ω rSup { size 8{2} } τ rSub { size 8{3} } τ rSub { size 8{4} } \) +8 ital "ωτ" rSub { size 8{4} } rSup { size 8{2} } \] } {} Setting the derivative equal to zero at ω=0 size 12{ω=0} {} yields d / T / dω / ω = 0 = [ − 4 ωτ 3 τ 4 ( 1 − ω 2 τ 3 τ 4 ) + 8 ωτ 4 2 ] = 4 ωτ 4 [ − τ 3 ( 1 − ω 2 τ 3 τ 4 ) + 2τ 4 ] size 12{ { {d lline T rline } over {dω} } \rline rSub { size 8{ω=0} } = \[ - 4 ital "ωτ" rSub { size 8{3} } τ rSub { size 8{4} } \( 1 - ω rSup { size 8{2} } τ rSub { size 8{3} } τ rSub { size 8{4} } \) +8 ital "ωτ" rSub { size 8{4} } rSup { size 8{2} } \] =4 ital "ωτ" rSub { size 8{4} } \[ - τ rSub { size 8{3} } \( 1 - ω rSup { size 8{2} } τ"" lSub { size 8{3} } τ rSub { size 8{4} } \) +2τ rSub { size 8{4} } \] } {} is satisfied when 2τ4=τ3 size 12{2τ rSub { size 8{4} } =τ rSub { size 8{3} } } {}, or C 3 = 2C 4 size 12{C rSub { size 8{3} } =2C rSub { size 8{4} } } {} Based on this condition, the transfer function magnitude is, from , / T / = 1 [ 1 + 4 ( ωτ 4 ) 4 ] 1 / 2 size 12{ lline T rline = { {1} over { \[ 1+4 \( ital "ωτ" rSub { size 8{4} } \) rSup { size 8{4} } \] rSup { size 8{1/2} } } } } {} The 3 dB, or cutoff, frequency occurs when /T/=1/2 size 12{ lline T rline =1/ sqrt {2} } {}, or when 4(ω3dBτ4)=1 size 12{4 \( ω rSub { size 8{3 ital "dB"} } τ rSub { size 8{4} } \) =1} {}. We then find that ω 3 dB = 2πf 3 dB = 1 τ 4 2 = 1 2 RC 4 size 12{ω"" lSub { size 8{3 ital "dB"} } =2πf rSub { size 8{3 ital "dB"} } = { {1} over {τ rSub { size 8{4} } sqrt {2} } } = { {1} over { sqrt {2} ital "RC" rSub { size 8{4} } } } } {} In general, we can write the cutoff frequency in the form ω 3 dB = 1 RC size 12{ω"" lSub { size 8{3 ital "dB"} } = { {1} over { ital "RC"} } } {} Finally, comparing , and yields C 4 = 0 . 707 C size 12{C rSub { size 8{4} } =0 "." "707"C} {} and C 3 = 1 . 414 C size 12{C rSub { size 8{3} } =1 "." "414"C} {} The two-pole low-pass Butterworth filter is shown in (a). The Bode plot of the transfer function magnitude is shown in (b). From , the magnitude of the voltage transfer function for the two pole low-pass Butterworth filter, can be written as / T / = 1 1 + ( f f 3 dB ) 4 size 12{ lline T rline = { {1} over { sqrt {1+ \( { {f} over {f rSub { size 8{3 ital "dB"} } } } \) rSup { size 8{4} } } } } } {}

a) Two-pole low-pass Butterworth filter and b)Bote plot transfer function magnitude. shows that the derivative of the voltage transfer function magnitude at ω=0 size 12{ω=0} {} is zero even without setting 2τ4=τ3 size 12{2τ rSub { size 8{4} } =τ rSub { size 8{3} } } {}. However, the added condition of 2τ4=τ3 size 12{2τ rSub { size 8{4} } =τ rSub { size 8{3} } } {} produces the maximally flat transfer characteristics of the Butterworth filter.

Two-Pole High-Pass Butterworth Filter To perform a high-pass filter, the resistors and capacitors are interchanging from those in the low-pass filter. A two-pole high-pass Butterworth filter is shown in (a). The analysis proceeds exactly the same as in the last section, except that the derivative is set equal to zero at s=jω=∞ size 12{s=jω= infinity } {}. Also, the capacitors are set equal to each other. The 3dB or cutoff frequency can be written in the general form ω 3 dB = 2πf 3 dB = 1 RC size 12{ω rSub { size 8{3 ital "dB"} } =2πf rSub { size 8{3 ital "dB"} } = { {1} over { ital "RC"} } } {}

a) Two-pole high-pass Butterworth filter and b) Bode plot transfer function magnitude. We find that R3=0.707R size 12{R rSub { size 8{3} } =0 "." "707"R} {} and R4=1.414R size 12{R rSub { size 8{4} } =1 "." "414"R} {}. The magnitude of the voltage transfer function for the two-pole high-pass Butterworth is / T / = 1 1 + ( f 3 dB f ) 4 size 12{ lline T rline = { {1} over { sqrt {1+ \( { {f rSub { size 8{3 ital "dB"} } } over {f} } \) rSup { size 8{4} } } } } } {} The Bode plot of the transfer function magnitude for the two-pole high-pass Butterworth filter is shown in (b).

High-Order Butterworth Filters The filter order is the number of poles and is usually dictated by the application requirements. An N-pole active low-pass filter has a high-frequency rolloff rate of Nx6 dB/octave. Similarly, the response of an N-pole high-pass filter increases at a rate of Nx6 dB/octave, up to the cutoff frequency. In each case, the 3 dB frequency is defined as f 3 dB = 1 2π RC size 12{f rSub { size 8{3 ital "dB"} } = { {1} over {2π ital "RC"} } } {} The magnitude of the voltage transfer function for a Butterworth Nth-order low-pass filter is / T / = 1 1 + ( f f 3 dB ) 2N size 12{ lline T rline = { {1} over { sqrt {1+ \( { {f} over {f rSub { size 8{3 ital "dB"} } } } \) rSup { size 8{2N} } } } } } {} For a Butterworth Nth-order high-pass filter, the voltage transfer function magnitude is / T / = 1 1 + ( f 3 dB f ) 2N size 12{ lline T rline = { {1} over { sqrt {1+ \( { {f rSub { size 8{3 ital "dB"} } } over {f} } \) rSup { size 8{2N} } } } } } {} (a) shows a three-pole low-pass Butterworth filter. The three resistors are equal, and the relationship between the capacitors is found by taking the first and second derivatives of the voltage gain magnitude with respect to frequency and setting those derivatives equal to zero at s=jω=0 size 12{s=jω=0} {}. (b) shows a three-pole high-pass Butterworth filter. In this case, the three capacitors are equal and the relationship between the resistors is also found through the derivatives. Higher-order filters can be created by adding additional RC networks. However, the loading effect on each additional RC circuit becomes more severe. The usefulness of active filters is realized when two or more op-amp filter circuits are cascaded to produce one large higher-order active filter. Because of the low output impedance of the op-amp, there is virtually no loading effect between cascaded stages.

a) Three-pole low-pass Butterworth fielter and b)Three-pole high-pass Butterworth fielter. (a) shows a four-pole low-pass Butterworth filter. The maximally flat response of this filter is not obtained by simply cascading two two-pole filters. The relationship between the capacitors is found through the first three derivatives of the transfer function. The four-pole high-pass Butterworth filter is shown in (b). Higher-order filters can be designed but are not considered here. Bandpass and band-reject filters use similar circuit configurations.

Four-pole low pass Butterworth filter and b)four-pole high-pass Butterworth filter.

OSCILLATORS In this section, we will look at the basic principles of sine-wave oscillators. In our study of feedback, we emphasized the need for negative feedback to provide a stable circuit. Oscillators, however, use positive feedback and, as such, are actually nonlinear circuits in some cases. The analysis and design of oscillator circuits are divided into two parts. In this first part, the condition and frequency for oscillation are determined; in the second part, means for amplitude control is addressed. We consider only the first step in this section to gain insight into the basic operation of oscillators.

Basic Principle of Oscillation The basic oscillator consists of an amplifier and a frequency-selective network connected in a feedback loop. shows a block diagram the fundamental feedback circuit, in which we are implicitly assuming that negative feedback is employed. Although actual oscillator circuits do not have an input signal, we initially include one here to help in the analysis. In previous feedback circuits, we assumed the feedback transfer function β size 12{β} {} was independent of frequency. In oscillator circuits, however, β size 12{β} {} is the principal portion of the loop gain that is dependent on frequency.

Block diagram of the fundamental feedback circuit. For the circuit shown, the ideal closed-loop transfer function is given by A f ( s ) = A ( s ) 1 + A ( s ) β ( s ) size 12{A rSub { size 8{f} } \( s \) = { {A \( s \) } over {1+A \( s \) β \( s \) } } } {} And the loop gain of the feedback circuit is T ( s ) = A ( s ) β ( s ) size 12{T \( s \) =A \( s \) β \( s \) } {} We know that the loop gain T(s) is positive for negative feedback, which means that the feedback signal vfb size 12{v rSub { size 8{ ital "fb"} } } {} subtracts from the input signal vs size 12{v rSub { size 8{s} } } {}. If T(s) = - 1, the closed-loop transfer function goes to infinity, which means that the circuit can have a finite output for a zero input signal. As T(s) approaches -1, an actual circuit becomes nonlinear, which means that the gain does not go to infinity. Assume that T(s) = -1 so that positive feedback exists over a particular frequency range. If a spontaneous signal (due to noise) is created at vs size 12{v rSub { size 8{s} } } {}, and the error signal vs size 12{v rSub { size 8{s} } } {} is reinforced and increased. This reinforcement process continues at only those frequencies for which the total phase shift around the feedback loop is zero. Therefore, the condition for oscillation is that, at a specific frequency, we have T ( jω 0 ) = A ( jω 0 ) β ( jω 0 ) = − 1 size 12{T \( jω rSub { size 8{0} } \) =A \( jω rSub { size 8{0} } \) β \( jω rSub { size 8{0} } \) = - 1} {} The condition that T(jω0)=−1 size 12{T \( jω rSub { size 8{0} } \) = - 1} {} is called the Barkhausen criterion. shows that two conditions must be satisfied to sustain oscillation: The total phase shift through the amplifier and feedback network must be Nx 3600 size 12{"360" rSup { size 8{0} } } {}, where N = 0, 1, 2, …, The magnitude of the loop gain must be unity. In the feedback circuit block diagram shown in , we implicitly assume negative feedback. For an oscillator, the feedback transfer function, or the frequency-selective network, must introduce an additional 180 degree phase shift such that the net phase around the entire loop is zero. For the circuit to oscillate at a single frequency ω0 size 12{ω rSub { size 8{0} } } {}, the condition for oscillation, from , should be satisfied at only that one frequency.

Phase-Shift Oscillator An example of an op-amp oscillator is the phase-shift oscillator. One configuration of this oscillator circuit is shown in . The basic amplifier of the circuit is the op-amp A3 size 12{A rSub { size 8{3} } } {}, which is connected as an inverting amplifier with its output connected to a three-stage RC filter. The voltage followers in the circuit eliminate loading effects between each RC filter stage.

Phase-shift oscillator circuit with voltage follower buffer stages. The inverting amplifier introduces a – 180 degree phase shift, which means that each RC network must provide 60 degrees of phase shift to produce 180 degrees required of the frequency-sensitive feedback network in order to produce positive feedback. Note that the inverting terminal of op-amp A3 is at virtual ground; therefore, the RC network between op-amps A2 size 12{A rSub { size 8{2} } } {} and A3 size 12{A rSub { size 8{3} } } {} functions exactly as the other two RC networks. We assume that the frequency effects of the op-amps themselves occur at much higher frequencies than the response due to the RC networks. Also, to aid in the analysis, we assume an input signal ( vi size 12{v rSub { size 8{i} } } {}) exists at one node as shown in the figure. The transfer function of the first RC network is v 1 = ( sRC 1 + sRC ) ( v i ) size 12{v rSub { size 8{1} } = \( { { ital "sRC"} over {1+ ital "sRC"} } \) \( v rSub { size 8{i} } \) } {} Since the RC networks are assumed to be identical, and since there is no loading effect of one RC stage on another, we have v 3 v i = ( sRC 1 + sRC ) 3 = β ( s ) size 12{ { {v rSub { size 8{3} } } over {v rSub { size 8{i} } } } = \( { { ital "sRC"} over {1+ ital "sRC"} } \) rSup { size 8{3} } =β \( s \) } {} where β(s) size 12{β \( s \) } {} is the feedback transfer function. The amplifier gain A(s) in and is actually the magnitude of the gain, or A ( s ) = / v 0 v i / = R 2 R size 12{A \( s \) = lline { {v rSub { size 8{0} } } over {v rSub { size 8{i} } } } rline = { {R rSub { size 8{2} } } over {R} } } {} The loop gain is then T ( s ) = A ( s ) β ( s ) = R 2 R ( sRC 1 + sRC ) 3 size 12{T \( s \) =A \( s \) β \( s \) = { {R rSub { size 8{2} } } over {R} } \( { { ital "sRC"} over {1+ ital "sRC"} } \) rSup { size 8{3} } } {} From , the condition for oscillation is that /T(jω0)/=1 size 12{ lline T \( jω rSub { size 8{0} } \) rline =1} {} and the phase of T(jω0) size 12{T \( jω rSub { size 8{0} } \) } {} must be 180 degrees. When these requirements are satisfied, then v0 will equal ( vi size 12{v rSub { size 8{i} } } {}) and a separate input signal will not be required. If we set s = j, becomes T ( jω ) = R 2 R ( jω RC 1 + jω RC ) 3 = − ( R 2 R ) ( jω RC ) ( ω RC ) 2 [ 1 − 3ω 2 R 2 C 2 ] + jω RC [ 3 − ω 2 R 2 C 2 ] size 12{T \( jω \) = { {R rSub { size 8{2} } } over {R} } \( { {jω ital "RC"} over {1+jω ital "RC"} } \) rSup { size 8{3} } = - \( { {R rSub { size 8{2} } } over {R} } \) { { \( jω ital "RC" \) \( ω ital "RC" \) rSup { size 8{2} } } over { \[ 1 - 3ω rSup { size 8{2} } R rSup { size 8{2} } C rSup { size 8{2} } \] +jω ital "RC" \[ 3 - ω rSup { size 8{2} } R rSup { size 8{2} } C rSup { size 8{2} } \] } } } {} To satisfy the condition T(jω0)=−1 size 12{T \( jω rSub { size 8{0} } \) = - 1} {}, the imaginary component of must equal zero. Since the numerator is purely imaginary, the denominator must become purely imaginary, or {1−3ω2R2C2}=0 size 12{ lbrace 1 - 3ω rSup { size 8{2} } R rSup { size 8{2} } C rSup { size 8{2} } rbrace =0} {} which yields ω = 1 3 RC size 12{ω= { {1} over { sqrt {3} ital "RC"} } } {} where ω size 12{ω} {} is the oscillation frequency. At this frequency, becomes T ( jω ) = − R 2 R ( j / 3 ) ( 1 / 3 ) 0 + ( j / 3 ) [ 3 − ( 1 / 3 ) ] = − R 2 R ( 1 8 ) size 12{T \( jω \) = - { {R rSub { size 8{2} } } over {R} } { { \( j/ sqrt {3} \) \( 1/3 \) } over {0+ \( j/ sqrt {3} \) \[ 3 - \( 1/3 \) \] } } = - { {R rSub { size 8{2} } } over {R} } \( { {1} over {8} } \) } {} Consequently, the condition T(jω)==1 size 12{T \( jω \) "=="1} {} is satisfied when R 2 R = 8 size 12{ { {R rSub { size 8{2} } } over {R} } =8} {} implies that if the magnitude of the inverting amplifier gain is greater than 8, the circuit will spontaneous begin oscillating and will sustain oscillation. Using , we can determine the effect of each RC network in the phase-shift oscillator. At the oscillation frequency ω0 size 12{ω rSub { size 8{0} } } {}, the transfer function of each RC network stage is jω 0 RC 1 + jω 0 RC = ( j / 3