Sinusoidal Signals & Phasors

Demonstrative Video

Introduction

Complete response $\Rightarrow$ natural response and forced response.
Natural response $\Rightarrow$ short-lived transient response of a circuit to a sudden change in its condition.
Forced response $\Rightarrow$ long-term steady-state response of a circuit to any independent sources
Only forced response we have considered is that due to dc sources.
Another very common forcing function is the sinusoidal waveform.
voltage available at household electrical sockets as well as the voltage of power lines connected to residential and industrial areas.
Assuming transient response is of little interest, and the steady-state response of a circuit (a television set, a toaster, or a power distribution network) to a sinusoidal voltage or current is needed.
We will analyze such circuits using a powerful technique that transforms integrodifferential equations into algebraic equations

Sinusoidal Currents and Voltages

A sinusoid is a signal that has the form of the sine or cosine function.
Consider a sinusoidally varying voltage:

$\begin{aligned} \boxed{v(t) =V_m\sin(\omega t)} \end{aligned}$ $\begin{aligned} V_m& = \text{amplitude}\\ \omega & = \text{angular frequency (radian/sec)}\\ \omega t & = \text{argument} \end{aligned}$
Period = $2\pi$ radian or $T$ seconds
Asine wave having a period $T$ must execute $1/T$ periods each second; its frequency $f$ is $1/T$ Hz. $\boxed{f = 1/T} \quad \omega t = 2\pi \quad \boxed{\omega = 2\pi f}$

Lagging and Leading

A more general form of the sinusoid $\begin{aligned} &\boxed{v(t) = V_m \sin(\omega t + \theta)}\\ \theta & = \text{Phase angle} \end{aligned}$

$v(t) = V_m \sin(\omega t + \theta)$ leads $V_m \sin(\omega t)$ by $\theta$ rad
$\pm \theta$ indicates leading and lagging $\Rightarrow$ sinusoidal are out of phase
NOTE: phase angle is commonly given in degrees, rather than radians $\begin{aligned} v = 100 \sin\left(2\pi 1000t - \dfrac{\pi}{6}\right) \Rightarrow 100 \sin(2\pi 1000t - 30^{\circ}) \Leftarrow \text{commonly used} \end{aligned}$
Two sinusoidal waves whose phases are to be compared must:
1. Both be written as sine waves, or both as cosine waves.
2. Both be written with positive amplitudes.
3. Each have the same frequency.

Converting Sines to Cosines

sine and cosine are essentially the same function, but with a $90^{\circ}$ phase difference
$\sin \omega t = \cos(\omega t - 90^{\circ})$
Multiplies of $\pm 360^{\circ}$ from argument without function change

$\begin{aligned} &\sin (A \pm B)=\sin A \cos B \pm \cos A \sin B \\ &\cos (A \pm B)=\cos A \cos B \mp \sin A \sin B \\ &\sin \left(\omega t \pm 180^{\circ}\right)=-\sin \omega t \\ &\cos \left(\omega t \pm 180^{\circ}\right)=-\cos \omega t \\ &\sin \left(\omega t \pm 90^{\circ}\right)=\pm \cos \omega t \\ &\cos \left(\omega t \pm 90^{\circ}\right)=\mp \sin \omega t \end{aligned}$

$\begin{aligned} v_{1} &=V_{m_{1}} \cos \left(5 t+10^{\circ}\right) \\ &=V_{m_{1}} \sin \left(5 t+90^{\circ}+10^{\circ}\right) \\ &=V_{m_{1}} \sin \left(5 t+100^{\circ}\right) \end{aligned}$ $\text{leads}~v_{2}=V_{m_{2}} \sin \left(5 t-30^{\circ}\right) \quad \text{by} ~ 130^{\circ}$ or, $v_{1}$ lags $v_{2}$ by $230^{\circ}$ , since $v_{1}$ may be written as $v_1 = V_{m1} \sin(5t-260^{\circ})$

Phasors

Sinusoids are easily expressed in terms of phasors, which are more convenient to work with than sine and cosine functions
A phasor is a complex number that represents the amplitude and phase of a sinusoid.
Complex number

$\begin{aligned} z&=x+j y \quad \text { Rectangular form } \\ z&=r / \phi \quad \text { Polar form } \\ z&=r e^{j \phi} \quad \text { Exponential form } \\ r&=\sqrt{x^{2}+y^{2}} \quad \phi=\tan ^{-1} \frac{y}{x} \\ x&=r \cos \phi, \quad y=r \sin \phi \\ z&=x+j y=r / \phi=r(\cos \phi+j \sin \phi) \end{aligned}$

Addition: $z_{1}+z_{2}=\left(x_{1}+x_{2}\right)+j\left(y_{1}+y_{2}\right)$ Subtraction: $z_{1}-z_{2}=\left(x_{1}-x_{2}\right)+j\left(y_{1}-y_{2}\right)$ Multiplication: $z_{1} z_{2}=r_{1} r_{2} \angle \phi_{1}+\phi_{2}$ Division: $\frac{z_{1}}{z_{2}}=\frac{r_{1}}{r_{2}} \angle \phi_{1}-\phi_{2}$ Reciprocal: $\frac{1}{z}=\frac{1}{r} \angle-\phi$ Square Root: $\sqrt{z}=\sqrt{r} / \phi / 2$ Complex Conjugate: $z^{*}=x-j y=r \angle-\phi=r e^{-j \phi}$

Sinusoidal Phasor

$v(t)$	$\mathrm{V}$
Instantaneous or time-domain	frequency or phasor domain
time-dependent	not
real no complex term	generally complex

In phasor representation, the frequency (or time) factor $e^{j\omega t}$ is suppressed because $\omega$ is constant.
However, response depends on $\omega$ , so phasor domain is known as the frequency domain. $\begin{aligned} v(t) & = V_m\cos(\omega t + \phi); \quad \mathrm{V} = V_m \angle \phi \\ \dfrac{dv}{dt} & = \text{Re}(j\omega \mathrm{V}e^{j \omega t}) \end{aligned}$ $\begin{aligned} &\frac{d v}{d t} \qquad \Leftrightarrow \quad j \omega \mathbf{V} \\ &(\text{Time domain}) \qquad (\text{Phasor domain}) \\ & \int v d t \quad \Leftrightarrow \qquad \frac{\mathbf{V}}{j \omega} \\ & (\text{Time domain}) \qquad (\text{Phasor domain}) \end{aligned}$
Advantage of phasor: removal of time differentiation and integration, and summing sinusoids of the same frequency

Problem

Transform these sinusoids to phasors:
1. $i=6 \cos \left(50 t-40^{\circ}\right) \mathrm{A}$
2. $v=-4 \sin \left(30 t+50^{\circ}\right) \mathrm{V}$
Solution:
1. $i=6 \cos \left(50 t-40^{\circ}\right)$ has the phasor $\mathbf{I}=6 \angle-40^{\circ} \mathrm{A}$
2. Since $-\sin A=\cos \left(A+90^{\circ}\right)$ , $\begin{aligned} v=-4 \sin \left(30 t+50^{\circ}\right) &=4 \cos \left(30 t+50^{\circ}+90^{\circ}\right) \\ &=4 \cos \left(30 t+140^{\circ}\right) \mathrm{V} \end{aligned}$ The phasor form of $v$ is $\mathbf{V}=4 \angle 140^{\circ} \mathrm{V}$