### Reordering the Transfer FunctionTo reorder the transfer function for the Black disk, we use the formula to match the form \( \frac{C}{\tau s + 1} \), where \( C \) is a constant.Given transfer function:\[\frac{\Omega(s)}{Vt(s)} = \frac{Kt}{RaJs + Kt \times Kb}\]#### Values for Black Disk- \( Ra = 25 \)- \( Kt = 0.0281 \)- \( Kb = 0.0312 \)- \( J = 3.248 \times 10^{-5} \)#### Correct AnswerThe reordered transfer function should be:\[\frac{32.0531}{0.9265s + 1}\]#### Steps to Find the Answer1. **Substitute Values**: Insert the given values of \( Ra \), \( Kt \), \( Kb \), and \( J \) into the transfer function.2. **Simplify**: Perform algebraic manipulation to simplify the expression to the desired form.3. **Compare & Adjust**: Match the coefficients to achieve the form \( \frac{C}{\tau s + 1} \).By following these steps, the correct transfer function form is obtained.

Given a transfer function with below valuesRa = 25 Kt = 0.0281 Kb = 0.0312 J = 3.248 × 10-5 Reorder…

Reorder the transfer function (shown below) using the values for the Black disk to the form Here C is just a constant. Q(s) Vt(s) = Kt Rajs +Kt*Kb Values for Black disk Below Ra = 25 Kt = 0.0281 Kb = 0.0312 J = 3.248 x 10-5 Had the answer correct which is = 32.0531 0.9265 + 1 Just step by step how to get the answer C TS + 1

Reorder the transfer function (shown below) using the values for the Black disk to the form Here C is just a constant. Q(s) Vt(s) = Kt Rajs +Kt*Kb Values for Black disk Below Ra = 25 Kt = 0.0281 Kb = 0.0312 J = 3.248 x 10-5 Had the answer correct which is = 32.0531 0.9265 + 1 Just step by step how to get the answer C TS + 1

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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