(a) Explanation Given information: Given transfer function is − T ( f ) = β ( 5 × 10 4 ) ( 1 + j f 10 3 ) ( 1 + j f 5 × 10 4 ) 2 The phase of transfer function is − 180 ° Calculation: Phase angle of transfer function is − ϕ ( f ) = − tan − 1 ( f 10 3 ) − 2 tan − 1 ( f 5 × 10 4 ) It is known that − 2 tan − 1 ( A ) = tan − 1 ( 2 A 1 − A 2 ) ( 1 ) tan − 1 ( A ) + tan − 1 ( B ) = tan − 1 ( A + B 1 − A B ) ( 2 ) Value of frequency for phase angle − 180 ° is − ϕ ( f ) = − tan − 1 ( f 10 3 ) − 2 tan − 1 ( f 5 × 10 4 ) ϕ ( f ) = − 180 ° − 180 ° = − ( tan − 1 ( f 10 3 ) + 2 tan − 1 ( f 5 × 10 4 ) ) ( tan − 1 ( f 10 3 ) + 2 tan − 1 ( f 5 × 10 4 ) ) = 180 ° tan − 1 ( f 10 3 ) + tan − 1 ( 2 f 5 × 10 4 1 − ( (b) To determine Value of ß for the given values.

MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL)

4th Edition

ISBN: 9781266368622

Author: NEAMEN

Publisher: MCG

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Chapter 12, Problem 12.72P

(a)

To determine

Value of frequency for the given values.

(b)

To determine

Value of ß for the given values.

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Chapter 12, Problem 12.71P

Chapter 12, Problem 12.73P

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Chapter 12 Solutions

MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL)

Ch. 12 - (a) The open-loop gain of an amplifier is A=5104...Ch. 12 - (a) Consider a general feedback system with...Ch. 12 - (a) A feedback amplifier has an open-loop...Ch. 12 - (a) Consider the circuit shown in Figure...Ch. 12 - (a) The closed-loop gain of a feedback amplifier...Ch. 12 - The gain factors in a feedback system are A=5105...Ch. 12 - Prob. 12.3TYU Ch. 12 - An ideal series-shunt feedback amplifier is shown...Ch. 12 - Consider the ideal shunt-series feedback amplifier...Ch. 12 - An ideal series-series feedback amplifier is shown...

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Value of frequency for the given values.

Solution Summary: The author explains the value of frequency for the given values. The phase angle of transfer function is -180°.

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Value of frequency for the given values.

Value of ß for the given values.

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Chapter 12 Solutions