Chapter 14, Problem 1P

Plot the following waveform versus time showing one clear. Complete cycle. Then determine the derivative of the waveform using Eq. (14.1) and sketch one complete cycle of the derivative directly under the Original waveform. Compare the magnitude of the derivative at various points versus the slope of the original sinusoidal function.

$v=4\sin 62.8t$

To determine

To plot:

The graph of the waveform versus time for one complete cycle. Calculate the derivative of the waveform and then draw one complete cycle of it. Also, compare the magnitude of the derivative at various points versus the slope of the original sinusoidal function.

Explanation of Solution

Given:

The given equation is $v=4\sin 62.8t$.

Calculation:

Let us consider the sinusoidal expression of voltage as:

  $v\left(t\right)=4\sin 62.8t$

Standard sinusoidal equation is:

  $v\left(t\right)=V_m\sin \left(\omega t\pm \theta \right)$

Comparing, we get

  $\omega =62.8\text{rad/sec}$

Value of frequency will be

  $\begin{array}{l}\omega =2\pi f\\ f=\frac{\omega }{2\pi }\\ \text{Substitute the}\text{value}\text{of}\text{ω,}\\ f=\frac{62.8}{2\pi }\\ f=10\text{Hz}\end{array}$

Voltage at $t=0$

  $\begin{array}{l}v\left(t\right)=4\sin 62.8t\\ v\left(0\right)=4\sin 62.8\left(0\right)\\ v\left(0\right)=0\text{V}\end{array}$

Voltage at $t=0.02$

  $\begin{array}{l}v\left(t\right)=4\sin 62.8t\\ v\left(0.02\right)=4\sin 62.8\left(0.02\right)\\ v\left(0.02\right)=3.80\text{V}\end{array}$

Voltage at $t=0.04$

  $\begin{array}{l}v\left(t\right)=4\sin 62.8t\\ v\left(0.04\right)=4\sin 62.8\left(0.04\right)\\ v\left(0.04\right)=2.355\text{V}\end{array}$

Voltage at $t=0.06$

  $\begin{array}{l}v\left(t\right)=4\sin 62.8t\\ v\left(0.06\right)=4\sin 62.8\left(0.06\right)\\ v\left(0.06\right)=-2.35\text{V}\end{array}$

Voltage at $t=0.08$ :

  $\begin{array}{l}v\left(t\right)=4\sin 62.8t\\ v\left(0.08\right)=4\sin 62.8\left(0.08\right)\\ v\left(0.08\right)=-3.80\text{V}\end{array}$

Voltage at $t=0.1$

  $\begin{array}{l}v\left(t\right)=4\sin 62.8t\\ v\left(0.1\right)=4\sin 62.8\left(0.1\right)\\ v\left(0.1\right)=0\text{V}\end{array}$

Graph showing voltage versus time

Derivative of standard sinusoidal equation is

  $\frac{d}{dt}v\left(t\right)=\omega V_m\cos \left(\omega t\pm \theta \right)$

Value of derivative of voltage at $t=0$

  $\begin{array}{l}\frac{d}{dt}v\left(t\right)=\omega V_m\cos \left(\omega t\pm \theta \right)\\ \frac{d}{dt}v\left(0\right)=\left(62.8\right)\left(4\right)\cos \left(62.8\times 0\right)\\ \frac{d}{dt}v\left(0\right)=251.2\text{V}\end{array}$

Value of derivative of voltage at $t=0.02$

  $\begin{array}{l}\frac{d}{dt}v\left(t\right)=\omega V_m\cos \left(\omega t\pm \theta \right)\\ \frac{d}{dt}v\left(0.02\right)=\left(62.8\right)\left(4\right)\cos \left(62.8\times 0.02\right)\\ \frac{d}{dt}v\left(0.02\right)=77.77\text{V}\end{array}$

Value of derivative of voltage at $t=0.04$

  $\begin{array}{l}\frac{d}{dt}v\left(t\right)=\omega V_m\cos \left(\omega t\pm \theta \right)\\ \frac{d}{dt}v\left(0.04\right)=\left(62.8\right)\left(4\right)\cos \left(62.8\times 0.04\right)\\ \frac{d}{dt}v\left(0.04\right)=-203.04\text{V}\end{array}$

Value of derivative of voltage at $t=0.06$

  $\begin{array}{l}\frac{d}{dt}v\left(t\right)=\omega V_m\cos \left(\omega t\pm \theta \right)\\ \frac{d}{dt}v\left(0.06\right)=\left(62.8\right)\left(4\right)\cos \left(62.8\times 0.06\right)\\ \frac{d}{dt}v\left(0.06\right)=-203.50\text{V}\end{array}$

Value of derivative of voltage at $t=0.08$

  $\begin{array}{l}\frac{d}{dt}v\left(t\right)=\omega V_m\cos \left(\omega t\pm \theta \right)\\ \frac{d}{dt}v\left(0.08\right)=\left(62.8\right)\left(4\right)\cos \left(62.8\times 0.08\right)\\ \frac{d}{dt}v\left(0.08\right)=77.01\text{V}\end{array}$

Value of derivative of voltage at $t=0.1$

  $\begin{array}{l}\frac{d}{dt}v\left(t\right)=\omega V_m\cos \left(\omega t\pm \theta \right)\\ \frac{d}{dt}v\left(0.1\right)=\left(62.8\right)\left(4\right)\cos \left(62.8\times 0.1\right)\\ \frac{d}{dt}v\left(0.1\right)=251.2\text{V}\end{array}$

Graph of derivative of voltage directly under the original waveform: