Answered: 5 x 10°s s2 + 10°s + 1014 A,(s) =

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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Question

What are Amid in dB, f_H , f_L , the BW in Hz, and
the Q for the amplifier described by What type of amplifier is this?

Expert Solution

Step 1

The given transfer function is

$\begin{array}{rcl} A_{v} (s) & = & - \frac{5 \times 10^{6} s}{s^{2} + 10^{5} s + 10^{14}} \\ = & (- 50) \frac{10^{5} s}{s^{2} + 10^{5} s + 10^{14}} . . . . . (1) \end{array}$

The above transfer function passes only a particular band of frequency. Hence it is a Bandpass filter.

The general expression of the bandpass filter is given by

$A_{v} (s) = A_{m i d} (\frac{\frac{ω_{o} s}{Q}}{s^{2} + \frac{ω_{o} s}{Q} + ω_{o}^{2}}) . . . . . . (2)$

Here,

$A_{m i d}$ is mid-band gain.

$ω_{o}$ is cut-off frequency.

$Q$ is the quality factor.

Step 2

On comparing equation (1) and (2), we get

$|A_{m i d}| = 50$

The gain in dB is given by

$\begin{array}{rcl} A_{m i d (d B)} & = & 20 \log (100) \\ = & 40 dB \end{array}$

$\begin{array}{rcl} ω_{o}^{2} & = & 10^{14} \\ ω_{0} & = & 10^{7} \end{array}$

The quality factor is given by

$\begin{array}{rcl} \frac{ω_{o} s}{Q} & = & 10^{5} s \\ Q & = & \frac{10^{7}}{10^{5}} \\ = & 100 \end{array}$

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