Answered: 5 x 10's A, (S) = s+ 2007

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 , and the BW in Hz
for the amplifier described by What type of amplifier is this?

Expert Solution

Step 1

The given transfer function is

$A_{v} (s) = \frac{5 \times 10^{3} s}{s + 200 π} . . . . . . . (a)$

This is a low pass filter with a single-pole transfer function.

The general expression for low pass filter is given by

$A_{v} (s) = \frac{A_{o} ω_{H}}{s + ω_{H}} . . . . . . . (b)$

Here,

$A_{o}$ is low-frequency gain

$ω_{H}$ is upper-cut off frequency

Step 2

On comparing equation (a) and (b), we get

$ω_{H} = 200 π$

The upper-cut frequency is given by

$\begin{array}{rcl} ω_{H} & = & 2 π f_{H} \\ 200 π & = & 2 π f_{H} \\ f_{H} & = & 100 Hz \end{array}$

The amplifier works as a low pass filter, so the lower cut off frequency is $f_{L} = 0 Hz$

The low-frequency gain is given by

$\begin{array}{rcl} |A_{o}| s ω_{H} & = & 2 π \times 10^{6} s \\ |A_{o}| & = & \frac{2 π \times 10^{6}}{ω_{H}} \\ = & \frac{2 π \times 10^{6}}{200 π} \\ = & 10^{4} \end{array}$

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