The expression for the frequency of the oscillation in the terms of C, R and R v .

Question

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Answer 1

Question

Chapter 15, Problem 15.31P

a.

To determine

The expression for the frequency of the oscillation in the terms of C, R and R_v.

a.

Expert Solution

Answer to Problem 15.31P

The frequency of oscillation:

f0=12πRC2+4RRRV

Explanation of Solution

Given:

The circuit is given for the phase shift oscillator:

MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL), Chapter 15, Problem 15.31P , additional homework tip 1

Redrawing the given circuit

MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL), Chapter 15, Problem 15.31P , additional homework tip 2

Nodal analysis at the node v1 :

v1−v0RV+v1−v2R+v11 sC=0v1(1 R V +1R+sC)−v2R=v0RV...........(1)

Nodal analysis at the node v2 :

v2−v1RV+v2−v3R+v21 sC=0v2+v2−v3+v2sRC=v1v1=v2(2+sRC)−v3..........(2)

Nodal analysis at the node v3 :

v3−v2RV+v3−0R+v31 sC=0−v2+v2−v3+v3(sRC)=0v3(2+sRC)=v2v3=v22+sRC........(5)

Now, solving the equation 2 and 3:

v1=v2(2+sRC)−v22+sRC=v2(2+sRC−1 2+sRC)v1=v2( ( 2+sRC ) 2 −1 2+sRC)v2=v1( 2+sRC ( 2+sRC ) 2 −1).........(4)

Nodal analysis at the inverting terminal:

0−v0RF+0−v3R=0v0RF=−v3Rv0=−RFRv3

From equation (3):

v0=−( R F R)(1 2+sRC)v2−( R F R)(1 2+sRC)v2

From equation 4:

v0=−( R F R)(1 2+sRC)( 2+sRC ( 2+sRC ) 2 −1)v1−( R F R)(1 ( 2+sRC ) 2 −1)v1

Substituting values of v0 and v2 in equation (1):

{ v 1( 1 R V + 1 R +sC)− 1 R( 2+sRC ( 2+sRC ) 2 −1 ) v 1}=−(1 R V )( R F R)(1 ( 2+sRC ) 2 −1)v11RV+1R+sC−2+sRCR[ ( 2+sRC ) 2−1]=−(1 R V )( R F R)(1 ( 2+sRC ) 2 −1)

Now simplifying more:

1[ R VR ( 2+sRC ) 2−R R V]{R( 2+sRC)2−RRV( 2+sRC)2−RV+sRRVC( 2+sRC)2−sRRVC−2RV−sRRVC}={( 1 R V )( R F R)( 1 ( 2+sRC ) 2 −1)}

{R(4+4sRC+ s 2 R 2 C 2)−R+RV(4+4sRC+ s 2 R 2 C 2)−3RV+sRVRC(4+4sRC+ s 2 R 2 C 2)−2sRRVC}=−RF4R+4sR2C+s2R2C2−R+4RV+4sRRVC+s2R2RVC2−3RV+4sRRVC+4s2R2RVC2+s3R3RVC3−2sRRVC=−RF{3R+4sR2C+s2R3C2+RV+5s2R2RVC2+s3R3RVC3+6sRRVC}=−RF Evaluating the frequency of oscillation, considering s=jω :

{3R+j4ωR2C+ω2R3C2+RV+5ω2R2RVC2+jω3R3RVC3+j6ωRRVC}=−RF{3R+ω2R3C2+RV−5ω2R2RVC2+j(4ω R 2C− ω 3 R 3 R V C 3+6ωR R VC)}=−RF

Now equating imaginary part to 0:

4ωR2C−ω3R3RVC3+6ωRRVC=0ωRC(4R−ω2R2RVC2+6RV)=04R−ω2R2RVC2+6RV=06RV+4R=ω2R2RVC2

Now simplifying:

ω2=1R2C2( 2 R V +4 R V +4R R V )ω2=1R2C2(2+4R[ R V +R R R V ])ω2=1R2C2(2+ 4R R R V )

Squaring on the both sides:

ω=1 R 2 C 2 ( 2+ 4R R R V )ω=1RC2+ 4R R R V

Since, the frequency is given as:

2πf=1RC2+4RRRV

Hence, the required frequency of oscillation is

f0=12πRC2+4RRRV

b.

To determine

The range in the frequency of the oscillation.

b.

Expert Solution

Answer to Problem 15.31P

The range of frequency of oscillation:

19.45≤f0≤22.66kHz

Explanation of Solution

Given:

The circuit is given for the phase shift oscillator:

MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL), Chapter 15, Problem 15.31P , additional homework tip 3

R=25kΩ,C=0.001μF and 15≤RV≤30kΩ

The expression for the frequency of the oscillation is given as:

f0=12πRC2+4RRRV

Substitute the known values in the above expression:

f0=12π( 25× 10 3 )( 0.001× 10 −6 )2+4( ( 25× 10 3 25× 10 3 ×15× 10 3 25× 10 3 +15× 10 3 ))=12π( 25× 10 −6 )2+4( 25× 10 3 9.375× 10 3 )=1157.08× 10 −62+4( 2.667)

More simplifying:

f0=1157.08× 10 −612.667=3.599157.08× 10 −6=22657.464=22.66kHz

If, RV=30kΩ , the frequency is given as:

f0=12πRC2+4RR□RV

Substitute the known values:

f0=12π( 25× 10 3 )( 0.001× 10 −6 )2+4( ( 25× 10 3 25× 10 3 ×30× 10 3 25× 10 3 +30× 10 3 ))=12π( 25× 10 −6 )2+4( 25× 10 3 13.636× 10 3 )=1157.08× 10 −62+4( 1.833)

More simplifying:

f0=1157.08× 10 −69.333=3.055157.08× 10 −6=0.01945×10−6=19.45kHz

Hence, the required range of frequency of oscillation:

19.45≤f0≤22.66kHz

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Answer 2

Question

Chapter 15, Problem 15.31P

a.

To determine

The expression for the frequency of the oscillation in the terms of C, R and R_v.

a.

Expert Solution

Answer to Problem 15.31P

The frequency of oscillation:

f0=12πRC2+4RRRV

Explanation of Solution

Given:

The circuit is given for the phase shift oscillator:

MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL), Chapter 15, Problem 15.31P , additional homework tip 1

Redrawing the given circuit

MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL), Chapter 15, Problem 15.31P , additional homework tip 2

Nodal analysis at the node v1 :

v1−v0RV+v1−v2R+v11 sC=0v1(1 R V +1R+sC)−v2R=v0RV...........(1)

Nodal analysis at the node v2 :

v2−v1RV+v2−v3R+v21 sC=0v2+v2−v3+v2sRC=v1v1=v2(2+sRC)−v3..........(2)

Nodal analysis at the node v3 :

v3−v2RV+v3−0R+v31 sC=0−v2+v2−v3+v3(sRC)=0v3(2+sRC)=v2v3=v22+sRC........(5)

Now, solving the equation 2 and 3:

v1=v2(2+sRC)−v22+sRC=v2(2+sRC−1 2+sRC)v1=v2( ( 2+sRC ) 2 −1 2+sRC)v2=v1( 2+sRC ( 2+sRC ) 2 −1).........(4)

Nodal analysis at the inverting terminal:

0−v0RF+0−v3R=0v0RF=−v3Rv0=−RFRv3

From equation (3):

v0=−( R F R)(1 2+sRC)v2−( R F R)(1 2+sRC)v2

From equation 4:

v0=−( R F R)(1 2+sRC)( 2+sRC ( 2+sRC ) 2 −1)v1−( R F R)(1 ( 2+sRC ) 2 −1)v1

Substituting values of v0 and v2 in equation (1):

{ v 1( 1 R V + 1 R +sC)− 1 R( 2+sRC ( 2+sRC ) 2 −1 ) v 1}=−(1 R V )( R F R)(1 ( 2+sRC ) 2 −1)v11RV+1R+sC−2+sRCR[ ( 2+sRC ) 2−1]=−(1 R V )( R F R)(1 ( 2+sRC ) 2 −1)

Now simplifying more:

1[ R VR ( 2+sRC ) 2−R R V]{R( 2+sRC)2−RRV( 2+sRC)2−RV+sRRVC( 2+sRC)2−sRRVC−2RV−sRRVC}={( 1 R V )( R F R)( 1 ( 2+sRC ) 2 −1)}

{R(4+4sRC+ s 2 R 2 C 2)−R+RV(4+4sRC+ s 2 R 2 C 2)−3RV+sRVRC(4+4sRC+ s 2 R 2 C 2)−2sRRVC}=−RF4R+4sR2C+s2R2C2−R+4RV+4sRRVC+s2R2RVC2−3RV+4sRRVC+4s2R2RVC2+s3R3RVC3−2sRRVC=−RF{3R+4sR2C+s2R3C2+RV+5s2R2RVC2+s3R3RVC3+6sRRVC}=−RF Evaluating the frequency of oscillation, considering s=jω :

{3R+j4ωR2C+ω2R3C2+RV+5ω2R2RVC2+jω3R3RVC3+j6ωRRVC}=−RF{3R+ω2R3C2+RV−5ω2R2RVC2+j(4ω R 2C− ω 3 R 3 R V C 3+6ωR R VC)}=−RF

Now equating imaginary part to 0:

4ωR2C−ω3R3RVC3+6ωRRVC=0ωRC(4R−ω2R2RVC2+6RV)=04R−ω2R2RVC2+6RV=06RV+4R=ω2R2RVC2

Now simplifying:

ω2=1R2C2( 2 R V +4 R V +4R R V )ω2=1R2C2(2+4R[ R V +R R R V ])ω2=1R2C2(2+ 4R R R V )

Squaring on the both sides:

ω=1 R 2 C 2 ( 2+ 4R R R V )ω=1RC2+ 4R R R V

Since, the frequency is given as:

2πf=1RC2+4RRRV

Hence, the required frequency of oscillation is

f0=12πRC2+4RRRV

b.

To determine

The range in the frequency of the oscillation.

b.

Expert Solution

Answer to Problem 15.31P

The range of frequency of oscillation:

19.45≤f0≤22.66kHz

Explanation of Solution

Given:

The circuit is given for the phase shift oscillator:

MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL), Chapter 15, Problem 15.31P , additional homework tip 3

R=25kΩ,C=0.001μF and 15≤RV≤30kΩ

The expression for the frequency of the oscillation is given as:

f0=12πRC2+4RRRV

Substitute the known values in the above expression:

f0=12π( 25× 10 3 )( 0.001× 10 −6 )2+4( ( 25× 10 3 25× 10 3 ×15× 10 3 25× 10 3 +15× 10 3 ))=12π( 25× 10 −6 )2+4( 25× 10 3 9.375× 10 3 )=1157.08× 10 −62+4( 2.667)

More simplifying:

f0=1157.08× 10 −612.667=3.599157.08× 10 −6=22657.464=22.66kHz

If, RV=30kΩ , the frequency is given as:

f0=12πRC2+4RR□RV

Substitute the known values:

f0=12π( 25× 10 3 )( 0.001× 10 −6 )2+4( ( 25× 10 3 25× 10 3 ×30× 10 3 25× 10 3 +30× 10 3 ))=12π( 25× 10 −6 )2+4( 25× 10 3 13.636× 10 3 )=1157.08× 10 −62+4( 1.833)

More simplifying:

f0=1157.08× 10 −69.333=3.055157.08× 10 −6=0.01945×10−6=19.45kHz

Hence, the required range of frequency of oscillation:

19.45≤f0≤22.66kHz

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Answer 3

Question

Chapter 15, Problem 15.31P

a.

To determine

The expression for the frequency of the oscillation in the terms of C, R and R_v.

a.

Expert Solution

Answer to Problem 15.31P

The frequency of oscillation:

f0=12πRC2+4RRRV

Explanation of Solution

Given:

The circuit is given for the phase shift oscillator:

MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL), Chapter 15, Problem 15.31P , additional homework tip 1

Redrawing the given circuit

MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL), Chapter 15, Problem 15.31P , additional homework tip 2

Nodal analysis at the node v1 :

v1−v0RV+v1−v2R+v11 sC=0v1(1 R V +1R+sC)−v2R=v0RV...........(1)

Nodal analysis at the node v2 :

v2−v1RV+v2−v3R+v21 sC=0v2+v2−v3+v2sRC=v1v1=v2(2+sRC)−v3..........(2)

Nodal analysis at the node v3 :

v3−v2RV+v3−0R+v31 sC=0−v2+v2−v3+v3(sRC)=0v3(2+sRC)=v2v3=v22+sRC........(5)

Now, solving the equation 2 and 3:

v1=v2(2+sRC)−v22+sRC=v2(2+sRC−1 2+sRC)v1=v2( ( 2+sRC ) 2 −1 2+sRC)v2=v1( 2+sRC ( 2+sRC ) 2 −1).........(4)

Nodal analysis at the inverting terminal:

0−v0RF+0−v3R=0v0RF=−v3Rv0=−RFRv3

From equation (3):

v0=−( R F R)(1 2+sRC)v2−( R F R)(1 2+sRC)v2

From equation 4:

v0=−( R F R)(1 2+sRC)( 2+sRC ( 2+sRC ) 2 −1)v1−( R F R)(1 ( 2+sRC ) 2 −1)v1

Substituting values of v0 and v2 in equation (1):

{ v 1( 1 R V + 1 R +sC)− 1 R( 2+sRC ( 2+sRC ) 2 −1 ) v 1}=−(1 R V )( R F R)(1 ( 2+sRC ) 2 −1)v11RV+1R+sC−2+sRCR[ ( 2+sRC ) 2−1]=−(1 R V )( R F R)(1 ( 2+sRC ) 2 −1)

Now simplifying more:

1[ R VR ( 2+sRC ) 2−R R V]{R( 2+sRC)2−RRV( 2+sRC)2−RV+sRRVC( 2+sRC)2−sRRVC−2RV−sRRVC}={( 1 R V )( R F R)( 1 ( 2+sRC ) 2 −1)}

{R(4+4sRC+ s 2 R 2 C 2)−R+RV(4+4sRC+ s 2 R 2 C 2)−3RV+sRVRC(4+4sRC+ s 2 R 2 C 2)−2sRRVC}=−RF4R+4sR2C+s2R2C2−R+4RV+4sRRVC+s2R2RVC2−3RV+4sRRVC+4s2R2RVC2+s3R3RVC3−2sRRVC=−RF{3R+4sR2C+s2R3C2+RV+5s2R2RVC2+s3R3RVC3+6sRRVC}=−RF Evaluating the frequency of oscillation, considering s=jω :

{3R+j4ωR2C+ω2R3C2+RV+5ω2R2RVC2+jω3R3RVC3+j6ωRRVC}=−RF{3R+ω2R3C2+RV−5ω2R2RVC2+j(4ω R 2C− ω 3 R 3 R V C 3+6ωR R VC)}=−RF

Now equating imaginary part to 0:

4ωR2C−ω3R3RVC3+6ωRRVC=0ωRC(4R−ω2R2RVC2+6RV)=04R−ω2R2RVC2+6RV=06RV+4R=ω2R2RVC2

Now simplifying:

ω2=1R2C2( 2 R V +4 R V +4R R V )ω2=1R2C2(2+4R[ R V +R R R V ])ω2=1R2C2(2+ 4R R R V )

Squaring on the both sides:

ω=1 R 2 C 2 ( 2+ 4R R R V )ω=1RC2+ 4R R R V

Since, the frequency is given as:

2πf=1RC2+4RRRV

Hence, the required frequency of oscillation is

f0=12πRC2+4RRRV

b.

To determine

The range in the frequency of the oscillation.

b.

Expert Solution

Answer to Problem 15.31P

The range of frequency of oscillation:

19.45≤f0≤22.66kHz

Explanation of Solution

Given:

The circuit is given for the phase shift oscillator:

MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL), Chapter 15, Problem 15.31P , additional homework tip 3

R=25kΩ,C=0.001μF and 15≤RV≤30kΩ

The expression for the frequency of the oscillation is given as:

f0=12πRC2+4RRRV

Substitute the known values in the above expression:

f0=12π( 25× 10 3 )( 0.001× 10 −6 )2+4( ( 25× 10 3 25× 10 3 ×15× 10 3 25× 10 3 +15× 10 3 ))=12π( 25× 10 −6 )2+4( 25× 10 3 9.375× 10 3 )=1157.08× 10 −62+4( 2.667)

More simplifying:

f0=1157.08× 10 −612.667=3.599157.08× 10 −6=22657.464=22.66kHz

If, RV=30kΩ , the frequency is given as:

f0=12πRC2+4RR□RV

Substitute the known values:

f0=12π( 25× 10 3 )( 0.001× 10 −6 )2+4( ( 25× 10 3 25× 10 3 ×30× 10 3 25× 10 3 +30× 10 3 ))=12π( 25× 10 −6 )2+4( 25× 10 3 13.636× 10 3 )=1157.08× 10 −62+4( 1.833)

More simplifying:

f0=1157.08× 10 −69.333=3.055157.08× 10 −6=0.01945×10−6=19.45kHz

Hence, the required range of frequency of oscillation:

19.45≤f0≤22.66kHz

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Answer 4

Question

Chapter 15, Problem 15.31P

a.

To determine

The expression for the frequency of the oscillation in the terms of C, R and R_v.

a.

Expert Solution

Answer to Problem 15.31P

The frequency of oscillation:

f0=12πRC2+4RRRV

Explanation of Solution

Given:

The circuit is given for the phase shift oscillator:

MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL), Chapter 15, Problem 15.31P , additional homework tip 1

Redrawing the given circuit

MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL), Chapter 15, Problem 15.31P , additional homework tip 2

Nodal analysis at the node v1 :

v1−v0RV+v1−v2R+v11 sC=0v1(1 R V +1R+sC)−v2R=v0RV...........(1)

Nodal analysis at the node v2 :

v2−v1RV+v2−v3R+v21 sC=0v2+v2−v3+v2sRC=v1v1=v2(2+sRC)−v3..........(2)

Nodal analysis at the node v3 :

v3−v2RV+v3−0R+v31 sC=0−v2+v2−v3+v3(sRC)=0v3(2+sRC)=v2v3=v22+sRC........(5)

Now, solving the equation 2 and 3:

v1=v2(2+sRC)−v22+sRC=v2(2+sRC−1 2+sRC)v1=v2( ( 2+sRC ) 2 −1 2+sRC)v2=v1( 2+sRC ( 2+sRC ) 2 −1).........(4)

Nodal analysis at the inverting terminal:

0−v0RF+0−v3R=0v0RF=−v3Rv0=−RFRv3

From equation (3):

v0=−( R F R)(1 2+sRC)v2−( R F R)(1 2+sRC)v2

From equation 4:

v0=−( R F R)(1 2+sRC)( 2+sRC ( 2+sRC ) 2 −1)v1−( R F R)(1 ( 2+sRC ) 2 −1)v1

Substituting values of v0 and v2 in equation (1):

{ v 1( 1 R V + 1 R +sC)− 1 R( 2+sRC ( 2+sRC ) 2 −1 ) v 1}=−(1 R V )( R F R)(1 ( 2+sRC ) 2 −1)v11RV+1R+sC−2+sRCR[ ( 2+sRC ) 2−1]=−(1 R V )( R F R)(1 ( 2+sRC ) 2 −1)

Now simplifying more:

1[ R VR ( 2+sRC ) 2−R R V]{R( 2+sRC)2−RRV( 2+sRC)2−RV+sRRVC( 2+sRC)2−sRRVC−2RV−sRRVC}={( 1 R V )( R F R)( 1 ( 2+sRC ) 2 −1)}

{R(4+4sRC+ s 2 R 2 C 2)−R+RV(4+4sRC+ s 2 R 2 C 2)−3RV+sRVRC(4+4sRC+ s 2 R 2 C 2)−2sRRVC}=−RF4R+4sR2C+s2R2C2−R+4RV+4sRRVC+s2R2RVC2−3RV+4sRRVC+4s2R2RVC2+s3R3RVC3−2sRRVC=−RF{3R+4sR2C+s2R3C2+RV+5s2R2RVC2+s3R3RVC3+6sRRVC}=−RF Evaluating the frequency of oscillation, considering s=jω :

{3R+j4ωR2C+ω2R3C2+RV+5ω2R2RVC2+jω3R3RVC3+j6ωRRVC}=−RF{3R+ω2R3C2+RV−5ω2R2RVC2+j(4ω R 2C− ω 3 R 3 R V C 3+6ωR R VC)}=−RF

Now equating imaginary part to 0:

4ωR2C−ω3R3RVC3+6ωRRVC=0ωRC(4R−ω2R2RVC2+6RV)=04R−ω2R2RVC2+6RV=06RV+4R=ω2R2RVC2

Now simplifying:

ω2=1R2C2( 2 R V +4 R V +4R R V )ω2=1R2C2(2+4R[ R V +R R R V ])ω2=1R2C2(2+ 4R R R V )

Squaring on the both sides:

ω=1 R 2 C 2 ( 2+ 4R R R V )ω=1RC2+ 4R R R V

Since, the frequency is given as:

2πf=1RC2+4RRRV

Hence, the required frequency of oscillation is

f0=12πRC2+4RRRV

b.

To determine

The range in the frequency of the oscillation.

b.

Expert Solution

Answer to Problem 15.31P

The range of frequency of oscillation:

19.45≤f0≤22.66kHz

Explanation of Solution

Given:

The circuit is given for the phase shift oscillator:

MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL), Chapter 15, Problem 15.31P , additional homework tip 3

R=25kΩ,C=0.001μF and 15≤RV≤30kΩ

The expression for the frequency of the oscillation is given as:

f0=12πRC2+4RRRV

Substitute the known values in the above expression:

f0=12π( 25× 10 3 )( 0.001× 10 −6 )2+4( ( 25× 10 3 25× 10 3 ×15× 10 3 25× 10 3 +15× 10 3 ))=12π( 25× 10 −6 )2+4( 25× 10 3 9.375× 10 3 )=1157.08× 10 −62+4( 2.667)

More simplifying:

f0=1157.08× 10 −612.667=3.599157.08× 10 −6=22657.464=22.66kHz

If, RV=30kΩ , the frequency is given as:

f0=12πRC2+4RR□RV

Substitute the known values:

f0=12π( 25× 10 3 )( 0.001× 10 −6 )2+4( ( 25× 10 3 25× 10 3 ×30× 10 3 25× 10 3 +30× 10 3 ))=12π( 25× 10 −6 )2+4( 25× 10 3 13.636× 10 3 )=1157.08× 10 −62+4( 1.833)

More simplifying:

f0=1157.08× 10 −69.333=3.055157.08× 10 −6=0.01945×10−6=19.45kHz

Hence, the required range of frequency of oscillation:

19.45≤f0≤22.66kHz

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Answer 5

Chapter 15, Problem 15.31P

a.

To determine

The expression for the frequency of the oscillation in the terms of C, R and R_v.

a.

Expert Solution

Answer to Problem 15.31P

The frequency of oscillation:

f0=12πRC2+4RRRV

Explanation of Solution

Given:

The circuit is given for the phase shift oscillator:

MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL), Chapter 15, Problem 15.31P , additional homework tip 1

Redrawing the given circuit

MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL), Chapter 15, Problem 15.31P , additional homework tip 2

Nodal analysis at the node v1 :

v1−v0RV+v1−v2R+v11 sC=0v1(1 R V +1R+sC)−v2R=v0RV...........(1)

Nodal analysis at the node v2 :

v2−v1RV+v2−v3R+v21 sC=0v2+v2−v3+v2sRC=v1v1=v2(2+sRC)−v3..........(2)

Nodal analysis at the node v3 :

v3−v2RV+v3−0R+v31 sC=0−v2+v2−v3+v3(sRC)=0v3(2+sRC)=v2v3=v22+sRC........(5)

Now, solving the equation 2 and 3:

v1=v2(2+sRC)−v22+sRC=v2(2+sRC−1 2+sRC)v1=v2( ( 2+sRC ) 2 −1 2+sRC)v2=v1( 2+sRC ( 2+sRC ) 2 −1).........(4)

Nodal analysis at the inverting terminal:

0−v0RF+0−v3R=0v0RF=−v3Rv0=−RFRv3

From equation (3):

v0=−( R F R)(1 2+sRC)v2−( R F R)(1 2+sRC)v2

From equation 4:

v0=−( R F R)(1 2+sRC)( 2+sRC ( 2+sRC ) 2 −1)v1−( R F R)(1 ( 2+sRC ) 2 −1)v1

Substituting values of v0 and v2 in equation (1):

{ v 1( 1 R V + 1 R +sC)− 1 R( 2+sRC ( 2+sRC ) 2 −1 ) v 1}=−(1 R V )( R F R)(1 ( 2+sRC ) 2 −1)v11RV+1R+sC−2+sRCR[ ( 2+sRC ) 2−1]=−(1 R V )( R F R)(1 ( 2+sRC ) 2 −1)

Now simplifying more:

1[ R VR ( 2+sRC ) 2−R R V]{R( 2+sRC)2−RRV( 2+sRC)2−RV+sRRVC( 2+sRC)2−sRRVC−2RV−sRRVC}={( 1 R V )( R F R)( 1 ( 2+sRC ) 2 −1)}

{R(4+4sRC+ s 2 R 2 C 2)−R+RV(4+4sRC+ s 2 R 2 C 2)−3RV+sRVRC(4+4sRC+ s 2 R 2 C 2)−2sRRVC}=−RF4R+4sR2C+s2R2C2−R+4RV+4sRRVC+s2R2RVC2−3RV+4sRRVC+4s2R2RVC2+s3R3RVC3−2sRRVC=−RF{3R+4sR2C+s2R3C2+RV+5s2R2RVC2+s3R3RVC3+6sRRVC}=−RF Evaluating the frequency of oscillation, considering s=jω :

{3R+j4ωR2C+ω2R3C2+RV+5ω2R2RVC2+jω3R3RVC3+j6ωRRVC}=−RF{3R+ω2R3C2+RV−5ω2R2RVC2+j(4ω R 2C− ω 3 R 3 R V C 3+6ωR R VC)}=−RF

Now equating imaginary part to 0:

4ωR2C−ω3R3RVC3+6ωRRVC=0ωRC(4R−ω2R2RVC2+6RV)=04R−ω2R2RVC2+6RV=06RV+4R=ω2R2RVC2

Now simplifying:

ω2=1R2C2( 2 R V +4 R V +4R R V )ω2=1R2C2(2+4R[ R V +R R R V ])ω2=1R2C2(2+ 4R R R V )

Squaring on the both sides:

ω=1 R 2 C 2 ( 2+ 4R R R V )ω=1RC2+ 4R R R V

Since, the frequency is given as:

2πf=1RC2+4RRRV

Hence, the required frequency of oscillation is

f0=12πRC2+4RRRV

b.

To determine

The range in the frequency of the oscillation.

b.

Expert Solution

Answer to Problem 15.31P

The range of frequency of oscillation:

19.45≤f0≤22.66kHz

Explanation of Solution

Given:

The circuit is given for the phase shift oscillator:

MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL), Chapter 15, Problem 15.31P , additional homework tip 3

R=25kΩ,C=0.001μF and 15≤RV≤30kΩ

The expression for the frequency of the oscillation is given as:

f0=12πRC2+4RRRV

Substitute the known values in the above expression:

f0=12π( 25× 10 3 )( 0.001× 10 −6 )2+4( ( 25× 10 3 25× 10 3 ×15× 10 3 25× 10 3 +15× 10 3 ))=12π( 25× 10 −6 )2+4( 25× 10 3 9.375× 10 3 )=1157.08× 10 −62+4( 2.667)

More simplifying:

f0=1157.08× 10 −612.667=3.599157.08× 10 −6=22657.464=22.66kHz

If, RV=30kΩ , the frequency is given as:

f0=12πRC2+4RR□RV

Substitute the known values:

f0=12π( 25× 10 3 )( 0.001× 10 −6 )2+4( ( 25× 10 3 25× 10 3 ×30× 10 3 25× 10 3 +30× 10 3 ))=12π( 25× 10 −6 )2+4( 25× 10 3 13.636× 10 3 )=1157.08× 10 −62+4( 1.833)

More simplifying:

f0=1157.08× 10 −69.333=3.055157.08× 10 −6=0.01945×10−6=19.45kHz

Hence, the required range of frequency of oscillation:

19.45≤f0≤22.66kHz

Answer 6

Chapter 15, Problem 15.31P

a.

To determine

The expression for the frequency of the oscillation in the terms of C, R and R_v.

a.

Expert Solution

Answer to Problem 15.31P

The frequency of oscillation:

f0=12πRC2+4RRRV

Explanation of Solution

Given:

The circuit is given for the phase shift oscillator:

MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL), Chapter 15, Problem 15.31P , additional homework tip 1

Redrawing the given circuit

MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL), Chapter 15, Problem 15.31P , additional homework tip 2

Nodal analysis at the node v1 :

v1−v0RV+v1−v2R+v11 sC=0v1(1 R V +1R+sC)−v2R=v0RV...........(1)

Nodal analysis at the node v2 :

v2−v1RV+v2−v3R+v21 sC=0v2+v2−v3+v2sRC=v1v1=v2(2+sRC)−v3..........(2)

Nodal analysis at the node v3 :

v3−v2RV+v3−0R+v31 sC=0−v2+v2−v3+v3(sRC)=0v3(2+sRC)=v2v3=v22+sRC........(5)

Now, solving the equation 2 and 3:

v1=v2(2+sRC)−v22+sRC=v2(2+sRC−1 2+sRC)v1=v2( ( 2+sRC ) 2 −1 2+sRC)v2=v1( 2+sRC ( 2+sRC ) 2 −1).........(4)

Nodal analysis at the inverting terminal:

0−v0RF+0−v3R=0v0RF=−v3Rv0=−RFRv3

From equation (3):

v0=−( R F R)(1 2+sRC)v2−( R F R)(1 2+sRC)v2

From equation 4:

v0=−( R F R)(1 2+sRC)( 2+sRC ( 2+sRC ) 2 −1)v1−( R F R)(1 ( 2+sRC ) 2 −1)v1

Substituting values of v0 and v2 in equation (1):

{ v 1( 1 R V + 1 R +sC)− 1 R( 2+sRC ( 2+sRC ) 2 −1 ) v 1}=−(1 R V )( R F R)(1 ( 2+sRC ) 2 −1)v11RV+1R+sC−2+sRCR[ ( 2+sRC ) 2−1]=−(1 R V )( R F R)(1 ( 2+sRC ) 2 −1)

Now simplifying more:

1[ R VR ( 2+sRC ) 2−R R V]{R( 2+sRC)2−RRV( 2+sRC)2−RV+sRRVC( 2+sRC)2−sRRVC−2RV−sRRVC}={( 1 R V )( R F R)( 1 ( 2+sRC ) 2 −1)}

{R(4+4sRC+ s 2 R 2 C 2)−R+RV(4+4sRC+ s 2 R 2 C 2)−3RV+sRVRC(4+4sRC+ s 2 R 2 C 2)−2sRRVC}=−RF4R+4sR2C+s2R2C2−R+4RV+4sRRVC+s2R2RVC2−3RV+4sRRVC+4s2R2RVC2+s3R3RVC3−2sRRVC=−RF{3R+4sR2C+s2R3C2+RV+5s2R2RVC2+s3R3RVC3+6sRRVC}=−RF Evaluating the frequency of oscillation, considering s=jω :

{3R+j4ωR2C+ω2R3C2+RV+5ω2R2RVC2+jω3R3RVC3+j6ωRRVC}=−RF{3R+ω2R3C2+RV−5ω2R2RVC2+j(4ω R 2C− ω 3 R 3 R V C 3+6ωR R VC)}=−RF

Now equating imaginary part to 0:

4ωR2C−ω3R3RVC3+6ωRRVC=0ωRC(4R−ω2R2RVC2+6RV)=04R−ω2R2RVC2+6RV=06RV+4R=ω2R2RVC2

Now simplifying:

ω2=1R2C2( 2 R V +4 R V +4R R V )ω2=1R2C2(2+4R[ R V +R R R V ])ω2=1R2C2(2+ 4R R R V )

Squaring on the both sides:

ω=1 R 2 C 2 ( 2+ 4R R R V )ω=1RC2+ 4R R R V

Since, the frequency is given as:

2πf=1RC2+4RRRV

Hence, the required frequency of oscillation is

f0=12πRC2+4RRRV

Answer 7

Chapter 15, Problem 15.31P

a.

To determine

The expression for the frequency of the oscillation in the terms of C, R and R_v.

Answer 8

a.

Expert Solution

Answer to Problem 15.31P

The frequency of oscillation:

f0=12πRC2+4RRRV

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

The circuit is given for the phase shift oscillator:

MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL), Chapter 15, Problem 15.31P , additional homework tip 1

Redrawing the given circuit

MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL), Chapter 15, Problem 15.31P , additional homework tip 2

Nodal analysis at the node v1 :

v1−v0RV+v1−v2R+v11 sC=0v1(1 R V +1R+sC)−v2R=v0RV...........(1)

Nodal analysis at the node v2 :

v2−v1RV+v2−v3R+v21 sC=0v2+v2−v3+v2sRC=v1v1=v2(2+sRC)−v3..........(2)

Nodal analysis at the node v3 :

v3−v2RV+v3−0R+v31 sC=0−v2+v2−v3+v3(sRC)=0v3(2+sRC)=v2v3=v22+sRC........(5)

Now, solving the equation 2 and 3:

v1=v2(2+sRC)−v22+sRC=v2(2+sRC−1 2+sRC)v1=v2( ( 2+sRC ) 2 −1 2+sRC)v2=v1( 2+sRC ( 2+sRC ) 2 −1).........(4)

Nodal analysis at the inverting terminal:

0−v0RF+0−v3R=0v0RF=−v3Rv0=−RFRv3

From equation (3):

v0=−( R F R)(1 2+sRC)v2−( R F R)(1 2+sRC)v2

From equation 4:

v0=−( R F R)(1 2+sRC)( 2+sRC ( 2+sRC ) 2 −1)v1−( R F R)(1 ( 2+sRC ) 2 −1)v1

Substituting values of v0 and v2 in equation (1):

{ v 1( 1 R V + 1 R +sC)− 1 R( 2+sRC ( 2+sRC ) 2 −1 ) v 1}=−(1 R V )( R F R)(1 ( 2+sRC ) 2 −1)v11RV+1R+sC−2+sRCR[ ( 2+sRC ) 2−1]=−(1 R V )( R F R)(1 ( 2+sRC ) 2 −1)

Now simplifying more:

1[ R VR ( 2+sRC ) 2−R R V]{R( 2+sRC)2−RRV( 2+sRC)2−RV+sRRVC( 2+sRC)2−sRRVC−2RV−sRRVC}={( 1 R V )( R F R)( 1 ( 2+sRC ) 2 −1)}

{R(4+4sRC+ s 2 R 2 C 2)−R+RV(4+4sRC+ s 2 R 2 C 2)−3RV+sRVRC(4+4sRC+ s 2 R 2 C 2)−2sRRVC}=−RF4R+4sR2C+s2R2C2−R+4RV+4sRRVC+s2R2RVC2−3RV+4sRRVC+4s2R2RVC2+s3R3RVC3−2sRRVC=−RF{3R+4sR2C+s2R3C2+RV+5s2R2RVC2+s3R3RVC3+6sRRVC}=−RF Evaluating the frequency of oscillation, considering s=jω :

{3R+j4ωR2C+ω2R3C2+RV+5ω2R2RVC2+jω3R3RVC3+j6ωRRVC}=−RF{3R+ω2R3C2+RV−5ω2R2RVC2+j(4ω R 2C− ω 3 R 3 R V C 3+6ωR R VC)}=−RF

Now equating imaginary part to 0:

4ωR2C−ω3R3RVC3+6ωRRVC=0ωRC(4R−ω2R2RVC2+6RV)=04R−ω2R2RVC2+6RV=06RV+4R=ω2R2RVC2

Now simplifying:

ω2=1R2C2( 2 R V +4 R V +4R R V )ω2=1R2C2(2+4R[ R V +R R R V ])ω2=1R2C2(2+ 4R R R V )

Squaring on the both sides:

ω=1 R 2 C 2 ( 2+ 4R R R V )ω=1RC2+ 4R R R V

Since, the frequency is given as:

2πf=1RC2+4RRRV

Hence, the required frequency of oscillation is

f0=12πRC2+4RRRV

Answer 13

b.

To determine

The range in the frequency of the oscillation.

b.

Expert Solution

Answer to Problem 15.31P

The range of frequency of oscillation:

19.45≤f0≤22.66kHz

Explanation of Solution

Given:

The circuit is given for the phase shift oscillator:

MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL), Chapter 15, Problem 15.31P , additional homework tip 3

R=25kΩ,C=0.001μF and 15≤RV≤30kΩ

The expression for the frequency of the oscillation is given as:

f0=12πRC2+4RRRV

Substitute the known values in the above expression:

f0=12π( 25× 10 3 )( 0.001× 10 −6 )2+4( ( 25× 10 3 25× 10 3 ×15× 10 3 25× 10 3 +15× 10 3 ))=12π( 25× 10 −6 )2+4( 25× 10 3 9.375× 10 3 )=1157.08× 10 −62+4( 2.667)

More simplifying:

f0=1157.08× 10 −612.667=3.599157.08× 10 −6=22657.464=22.66kHz

If, RV=30kΩ , the frequency is given as:

f0=12πRC2+4RR□RV

Substitute the known values:

f0=12π( 25× 10 3 )( 0.001× 10 −6 )2+4( ( 25× 10 3 25× 10 3 ×30× 10 3 25× 10 3 +30× 10 3 ))=12π( 25× 10 −6 )2+4( 25× 10 3 13.636× 10 3 )=1157.08× 10 −62+4( 1.833)

More simplifying:

f0=1157.08× 10 −69.333=3.055157.08× 10 −6=0.01945×10−6=19.45kHz

Hence, the required range of frequency of oscillation:

19.45≤f0≤22.66kHz

Answer 14

b.

To determine

The range in the frequency of the oscillation.

Answer 15

b.

Expert Solution

Answer to Problem 15.31P

The range of frequency of oscillation:

19.45≤f0≤22.66kHz

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given:

The circuit is given for the phase shift oscillator:

MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL), Chapter 15, Problem 15.31P , additional homework tip 3

R=25kΩ,C=0.001μF and 15≤RV≤30kΩ

The expression for the frequency of the oscillation is given as:

f0=12πRC2+4RRRV

Substitute the known values in the above expression:

f0=12π( 25× 10 3 )( 0.001× 10 −6 )2+4( ( 25× 10 3 25× 10 3 ×15× 10 3 25× 10 3 +15× 10 3 ))=12π( 25× 10 −6 )2+4( 25× 10 3 9.375× 10 3 )=1157.08× 10 −62+4( 2.667)

More simplifying:

f0=1157.08× 10 −612.667=3.599157.08× 10 −6=22657.464=22.66kHz

If, RV=30kΩ , the frequency is given as:

f0=12πRC2+4RR□RV

Substitute the known values:

f0=12π( 25× 10 3 )( 0.001× 10 −6 )2+4( ( 25× 10 3 25× 10 3 ×30× 10 3 25× 10 3 +30× 10 3 ))=12π( 25× 10 −6 )2+4( 25× 10 3 13.636× 10 3 )=1157.08× 10 −62+4( 1.833)