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:

    \[\begin{aligned} \boxed{v(t) =V_m\sin(\omega t)} \end{aligned}\]
    image
  • Period = \(2\pi\) radian or \(T\) seconds

  • \[\boxed{f = 1/T} \quad \omega t = 2\pi \quad \boxed{\omega = 2\pi f}\]
    Hz. is periods each second; its frequency must execute Asine wave having a period

Lagging and Leading

  • \[\begin{aligned} &\boxed{v(t) = V_m \sin(\omega t + \theta)}\\ \theta & = \text{Phase angle} \end{aligned}\]
    A more general form of the sinusoid
image
  • \(v(t) = V_m \sin(\omega t + \theta)\) leads \(V_m \sin(\omega t)\) by \(\theta\) rad

  • \(\pm \theta\) indicates leading and lagging \(\Rightarrow\) sinusoidal are out of phase

  • \[\begin{aligned} v = 100 \sin\left(2\pi 1000t - \dfrac{\pi}{6}\right) \Rightarrow 100 \sin(2\pi 1000t - 30^{\circ}) \Leftarrow \text{commonly used} \end{aligned}\]
    NOTE: phase angle is commonly given in degrees, rather than radians
  • 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

\[ \begin{aligned} &\sin (A \pm B)=\sin A \cos B \pm \cos A \sin B \\ &\cos (A \pm B)=\cos A \cos B \mp \sin A \sin B \\ &\sin \left(\omega t \pm 180^{\circ}\right)=-\sin \omega t \\ &\cos \left(\omega t \pm 180^{\circ}\right)=-\cos \omega t \\ &\sin \left(\omega t \pm 90^{\circ}\right)=\pm \cos \omega t \\ &\cos \left(\omega t \pm 90^{\circ}\right)=\mp \sin \omega t \end{aligned} \]
\[\begin{aligned} v_{1} &=V_{m_{1}} \cos \left(5 t+10^{\circ}\right) \\ &=V_{m_{1}} \sin \left(5 t+90^{\circ}+10^{\circ}\right) \\ &=V_{m_{1}} \sin \left(5 t+100^{\circ}\right) \end{aligned}\]
may be written as , since by lags or,

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

    \[ \begin{aligned} z&=x+j y \quad \text { Rectangular form } \\ z&=r / \phi \quad \text { Polar form } \\ z&=r e^{j \phi} \quad \text { Exponential form } \\ r&=\sqrt{x^{2}+y^{2}} \quad \phi=\tan ^{-1} \frac{y}{x} \\ x&=r \cos \phi, \quad y=r \sin \phi \\ z&=x+j y=r / \phi=r(\cos \phi+j \sin \phi) \end{aligned} \]
\[z_{1}+z_{2}=\left(x_{1}+x_{2}\right)+j\left(y_{1}+y_{2}\right)\]
\[z^{*}=x-j y=r \angle-\phi=r e^{-j \phi}\]
\[\sqrt{z}=\sqrt{r} / \phi / 2\]
\[\frac{1}{z}=\frac{1}{r} \angle-\phi\]
\[\frac{z_{1}}{z_{2}}=\frac{r_{1}}{r_{2}} \angle \phi_{1}-\phi_{2}\]
\[z_{1} z_{2}=r_{1} r_{2} \angle \phi_{1}+\phi_{2}\]
\[z_{1}-z_{2}=\left(x_{1}-x_{2}\right)+j\left(y_{1}-y_{2}\right)\]
Addition:
image

Sinusoidal Phasor

image
\(v(t)\) \(\mathrm{V}\)
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.

  • \[\begin{aligned} v(t) & = V_m\cos(\omega t + \phi); \quad \mathrm{V} = V_m \angle \phi \\ \dfrac{dv}{dt} & = \text{Re}(j\omega \mathrm{V}e^{j \omega t}) \end{aligned}\]
    \[\begin{aligned} &\frac{d v}{d t} \qquad \Leftrightarrow \quad j \omega \mathbf{V} \\ &(\text{Time domain}) \qquad (\text{Phasor domain}) \\ & \int v d t \quad \Leftrightarrow \qquad \frac{\mathbf{V}}{j \omega} \\ & (\text{Time domain}) \qquad (\text{Phasor domain}) \end{aligned}\]
    , so phasor domain is known as the frequency domain. However, response depends on
  • 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. \[\mathbf{I}=6 \angle-40^{\circ} \mathrm{A}\]
      \(i=6 \cos \left(50 t-40^{\circ}\right)\)
    2. \[\begin{aligned} v=-4 \sin \left(30 t+50^{\circ}\right) &=4 \cos \left(30 t+50^{\circ}+90^{\circ}\right) \\ &=4 \cos \left(30 t+140^{\circ}\right) \mathrm{V} \end{aligned}\]
      \[\mathbf{V}=4 \angle 140^{\circ} \mathrm{V}\]
      \(v\), Since