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Next: Kirchhoff's Laws Up: Chapter 1: Basic Quantities Previous: Examples: Mechanical and Electrical

Sinusoidal Variables

Sinusoidal variables are of special importance in electrical and electronic systems, not only because they occur frequently in such systems, but also because any periodical signal can be represented as a linear combination of a set of sinusoidal signals of different frequencies, amplitudes, and phase angles (Fourier transform theory).

A sinusoidal variable (voltage or current) can be written as

\begin{displaymath}x(t)=A\;\cos(\omega t + \phi),\;\;\;\mbox{or}\;\;\;
x(t)=A\;\sin(\omega t+\phi+\pi/2) \end{displaymath}

The three parameters $A$, $\omega$ and $\phi$ represent three important elements: Frequency can also be measured by cycles per second. i.e., $f=1/T$ where $T$ is cycle time or period (in seconds). Since one cycle is $2\pi$ radians, we have

\begin{displaymath}\omega=2\pi f=2\pi/T,\;\;\;\;f=1/T=\omega/2\pi,\;\;\;T=1/f=2\pi/\omega \end{displaymath}

Example: (Homework)

A sinusoidal current with a frequency of 60 Hz reaches a positive maximum of 20A at $t=2 \; ms$. Give the expression of this current as a function of time $i(t)$.

Answer

Average Value

The average of a varying current $i(t)$ is the steady value of current $I_{av}$ that in period $T$ would transfer the same charge $Q$:

\begin{displaymath}I_{av}T=Q=\int_0^T i(t) dt,\;\;\;\;\mbox{i.e.}\;\;\;\;
I_{av}=\frac{1}{T}\int_0^T i(t) dt \end{displaymath}

Similarly, the average voltage is defined as:

\begin{displaymath}V_{av}=\frac{1}{T}\int_0^T v(t) dt \end{displaymath}

In general, when $i(t)$ and $v(t)$ are periodic, the time period $T$ is one complete cycle. For a sinusoidal voltage $i(t)=I_p \cos(\omega t)=I_p \cos(2\pi ft)=I_p \cos(2\pi t/T)$ of frequency $f=1/T$, the average over the complete cycle is always zero (the charge transferred during the first half is the opposite to that transferred in the second). We can consider the half-cycle average:
$\displaystyle I_{av}$ $\textstyle =$ $\displaystyle \frac{1}{T/2}\int_{-T/4}^{T/4} i(t) dt
=\frac{2}{T}\int_{-T/4}^{T...
...cos(2\pi t/T)dt
=\frac{2I_p}{T}\frac{T}{2\pi} \sin(2\pi t/T)\vert _{-T/4}^{T/4}$  
  $\textstyle =$ $\displaystyle \frac{I_p}{\pi}[\sin(\pi/2)-\sin(-\pi/2)]
=\frac{2}{\pi}\;I_p=0.637\;I_p$  

Effective or RMS Value

The effective value of a time-varying current $i(t)$ or voltage is the constant value of current $I_{rms}$ or voltage $V_{rms}$ that in period $T$ would transfer the same amount of energy:

\begin{displaymath}W=R I^2_{rms}T=R\int_0^T i^2(t) dt,\;\;\;\;\mbox{or}\;\;\;\;\;
W=\frac{1}{R}V^2_{rms}T=\frac{1}{R}\int_0^T v^2(t) dt \end{displaymath}

i.e.,

\begin{displaymath}I_{rms}=\sqrt{\frac{1}{T}\int_0^T i^2(t) dt},\;\;\;\;\;\;\;\mbox{or}\;\;\;\;\;\;
V_{rms}=\sqrt{\frac{1}{T}\int_0^T v^2(t) dt} \end{displaymath}

As $I_{rms}$ or $V_{rms}$ is the ``square root of the mean of the squared value'', it is also called the root-mean-square (rms) current or voltage.

For a sinusoidal variable $i(t)=I_p \cos(\omega t)$, we have

\begin{displaymath}I^2_{rms} = \frac{1}{T}\int_0^T i^2(t) dt
= \frac{I^2_m}{T}\i...
...rac{I^2_m}{2T}\int_0^T [1+\cos(4\pi t/T)]\; dt=\frac{I^2_m}{2} \end{displaymath}

(trigonometric identity: $\cos^2\alpha=[1+\cos(2\alpha)]/2$) Similarly, we also have

\begin{displaymath}V_{rms}=\frac{V_p}{\sqrt{2}}=0.707 V_p \end{displaymath}

Phasor representation

A sinusoidal voltage or current can be considered as the real (or imaginary) part of a complex variable

    $\displaystyle {\bf V}(t)=V_p e^{j(\omega t+\phi)}=V_p e^{j\omega t} e^{j\phi},\...
...;
v(t)=Re[{\bf V}(t)]=V_p cos(\omega t+\phi)=\sqrt{2}V_{rms} cos(\omega t+\phi)$  
    $\displaystyle {\bf I}(t)=I_p e^{j(\omega t+\psi)}=I_p e^{j\omega t} e^{j\psi},\...
...;
i(t)=Re[{\bf I}(t)]=I_p cos(\omega t+\psi)=\sqrt{2}I_{rms} cos(\omega t+\psi)$  

In the analysis process of an AC circuit, the frequency $\omega=2\pi f$ is always assumed to be the same for all variables. We only need to represent the amplitude $V_{rms}$ or $I_{rms}$ and phase $e^{j\phi}$ of the sinusoidal variables in the analysis, while dropping the time-varying component $e^{j\omega t}$ corresponding to the frequency. We therefore define the phasor representation of a sinusoidal variable $v(t)=Re[V_p e^{j(\omega t+\phi)}]$ or $i(t)=Re[I_p e^{j(\omega t+\phi)}]$ as:

\begin{displaymath}\dot{V}=\frac{V_p}{\sqrt{2}}e^{j\phi}=V_{rms}e^{j\phi},\;\;\;...
...\;\;\;\;\dot{I}=\frac{I_p}{\sqrt{2}}e^{j\phi}=I_{rms}e^{j\phi} \end{displaymath}

Given this phasor representation, we can get the complex variable back by

\begin{displaymath}{\bf V}=\sqrt{2} \dot{V} e^{j\omega t}=V_p e^{j(\omega t+\phi...
...bf I}=\sqrt{2} \dot{I} e^{j\omega t}=I_p e^{j(\omega t+\phi)}
\end{displaymath}

and the corresponding sinusoidal variable can be found as the real part

\begin{displaymath}v(t)=Re[ {\bf V}]=Re[\sqrt{2} \dot{V} e^{j\omega t}]=V_p\cos(...
... I}]=Re[\sqrt{2} \dot{I} e^{j\omega t}]=I_p\cos(\omega t+\phi) \end{displaymath}

Sometimes a phasor can be represented simply by $V$ or $I$ if no confusion will be caused.

A sinusoidal time function can be considered as the real (or imaginary) part of a rotating vector in the complex plane. If two sinusoidal functions have the same frequency $\omega$, i.e., they are rotating at the same rate, their relative positions with respect to each other are fixed independent of the frequency $\omega$. Therefore the vectors can be considered as static instead of rotating if observed from a reference frame rotating at the same frequency as the vectors. An operation of two sinusoids can be carried out based on their phasors, and the resulting phasor can then be converted back to a sinusoidal time function by taking the real part of the phasor now assumed to be rotating.

RotatingVectors.gif

Phasor and the Fourier transform

A signal $x(t)$ can be expressed as a linear combination of infinite frequency components by the Fourier transform:

\begin{displaymath}x(t)=\int_{-\infty}^\infty X(f) e^{j2\pi ft} df
=\int_{-\inft...
... \cos(2\pi ft) df+j\int_{-\infty}^\infty X(f) \sin(2\pi ft) df
\end{displaymath}

When $x(t)$ is real, the second term is zero, and we also have:

\begin{displaymath}\vert X(-f)\vert=\vert X(f)\vert,\;\;\;\angle X(-f)=-\angle X(f) \end{displaymath}

and the corresponding frequency component is
$\displaystyle X(-f)e^{-j2\pi ft}+X(f)e^{j2\pi ft}$ $\textstyle =$ $\displaystyle \vert X(f)\vert e^{-j(2\pi ft+\angle X(f))}+\vert X(f)\vert e^{j(2\pi ft+\angle X(f))}$  
  $\textstyle =$ $\displaystyle 2\vert X(f)\vert\cos(2\pi ft+\angle X(f))$  

and

\begin{displaymath}x(t)=\int_0^\infty 2\vert X(f)\vert\cos(2\pi ft+\angle X(f)) df \end{displaymath}

Same as the phasor, here the complex coefficient $X(f)=\vert X(f)\vert e^{j\angle X(f)}$ also represents the amplitude and phase shift of the corresponding frequency component $\cos(2\pi ft)=\cos(\omega t)$, but not its frequency $f=\omega/2\pi$. In other words, the phasor is actually the same as the Fourier coefficient, both implicitly associated with a certain frequency component.

When a phasor $\dot{V}=V_{rms}e^{j\phi}$ is multiplied by a complex exponential $e^{-j\omega\tau}$, then the corresponding time variable is delayed by $\tau$:

\begin{displaymath}v(t)=Re[\sqrt{2}\dot{V} e^{-j\omega\tau}e^{j\omega t}]=Re[V_pe^{j\phi}e^{j\omega (t-\tau)}]
=V_p\cos(\omega(t-\tau)+\phi) \end{displaymath}

This is actually the time delay property of the Fourier (or Laplace) transform:

\begin{displaymath}\mbox{if}\;\;\;\;\;X(s)={\cal F}[x(t)],\;\;\;\;\;\mbox{then}\;\;\;\;\;
X(s)e^{-j\omega \tau} ={\cal F}[x(t-\tau)] \end{displaymath}


next up previous
Next: Kirchhoff's Laws Up: Chapter 1: Basic Quantities Previous: Examples: Mechanical and Electrical
Ruye Wang 2012-07-02