Answered: Design a 1st order chebysehev circuit…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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Question

Design a 1st order chebysehev circuit with a band-pass with a cutoff frequency of 100 Hz to 100 kHz. Analyze AC / DC

Expert Solution

Step 1

Given the minimum cut-off frequency, f_pis 1 Hz and the maximum cut-off frequency, f_s is 1 kHz.

The number of the order for Chebyshev is determined by the following expression,

$N \geq \frac{\cos h^{- 1} \sqrt{\frac{10^{0.1 α_{s}} - 1}{10^{0.1 α_{p}} - 1}}}{\cos h^{- 1} (\frac{2 {πf}_{s}}{2 {πf}_{p}})} 1 = \frac{\cos h^{- 1} \sqrt{\frac{10^{0.1 α_{s}} - 1}{10^{0.1 α_{p}} - 1}}}{\cos h^{- 1} (\frac{2 π (100 k H z)}{2 π (100 H z)})} \cos h^{- 1} \sqrt{\frac{10^{0.1 α_{s}} - 1}{10^{0.1 α_{p}} - 1}} = 7.6 \sqrt{\frac{10^{0.1 (2)} - 1}{10^{0.1 α_{p}} - 1}} = \cos h (7.6) [Consider α_{s} = 2 dB] α_{p} \approx 0$

Step 2

The following parameter is determined as:

$ε = \sqrt{10^{0.1 α_{p}} - 1} = 0, μ = ε^{- 1} + \sqrt{ε^{- 2} + 1} = 1$

The minor axis is determined as:

$\begin{array}{rcl} a & = & 2 {πf}_{p} [\frac{μ^{\frac{1}{N}} - μ^{- \frac{1}{N}}}{2}] \\ = & 0 \end{array}$

The major axis is determined as:

$\begin{array}{rcl} b & = & 2 {πf}_{p} [\frac{μ^{\frac{1}{N}} + μ^{- \frac{1}{N}}}{2}] \\ = & 2 π (1 H z) [\frac{μ^{\frac{1}{1}} + μ^{- \frac{1}{1}}}{2}] \\ = & 2 π \end{array}$

Step 3

The location for the poles is determined by the following expression,

$p_{k} = a \cos ϕ_{k} + j b \sin ϕ_{k} k = 1, 2, \dots$

Here,

$\begin{array}{rcl} ϕ_{k} & = & [\frac{2 k - 1}{2 N}] π k = 1, 2, \dots N \\ = & [\frac{2 (1) - 1}{2}] π \\ = & \frac{π}{2} \\ = & π \end{array}$

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