8.1 Signals in time, frequency, and phasor domains

In this chapter we consider the behavior of filters and other circuits who’s response depends upon the frequency of the input signal. We limit ourselves to consideration of input signals comprised of DC and sinusoidal components. We say in section 7.x that the steady-state (ie, after any transients have settled down) response of a linear system to a sinusoidal input signal at a certain frequency was an output at the same frequency but, in general, with a different amplitude and phase.

Sinusoids in more detail. Consider a sinusoidal waveform given by

(1)   \begin{equation*} v(t)=V_{m}cos(\frac{2\pi t}{T} + \theta) \end{equation*}

and plotted as a function of time in figure 8.1. Here T is the period of the sinusoid in seconds and \theta is the phase angle in radians. (Note that the argument of the cosine function of eq(1) is in radians, since the units of \frac{2\pi t}{T} and \theta are both radians. The significant of the phase angle can be seen when the waveforms of figure 8.1.a (\theta=0), figure 8.1b (\theta=\frac{\pi}{4}) figure 8.1c (\theta=\frac{\pi}{2}) are compared. The phase angle represents a shift in the waveform to the right by a fraction of a repetition period.  Positive phase angles shift the waveforms to the leftt, while negative angles shift the waveform to the right.

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Equation (1) can also be written as

(2)   \begin{equation*} v(t)=V_{m}cos(2\pi ft + \theta) \end{equation*}

where

(3)   \begin{equation*} f=\frac{1}{T} \end{equation*}

is the oscillation frequency in sec^{-1} or \frac{cycles}{s} or hertz (Hz)

The sinusoidal waveform can also be described as

(4)   \begin{equation*} v(t)=V_{m}cos(\omega t + \theta) \end{equation*}

where

(5)   \begin{equation*} \omega=2\pi f=\frac{2\pi}{T} \end{equation*}

is the radian frequency in \frac{radians}{s}

Figure 8.2 shows plots of eq (2) versus time, from t=0 to t=10 sec for frequencies 0.`1 Hz, 1 Hz, 10 Hz, and 100 Hz, corresponding to periods of 10, 1, 0.1, and 0.01 sec, respectively. In all cases, \theta=0. Over the 1 s of the plot, the waveforms oscillate over 0.1, 1,10, and 100 complete cycles.

The period, frequency, and radian frequency of these sinusoids are summarized below:

 

 

Figure 8.3 shows a  side-by-side time and frequency-domain representations  for these waveforms. The time-domain plots show voltage vs. time, while the frequency-domain plots show peak voltage vs. frequency,

Figure 8.3 time and freq plots. Magnitude and phase vs. frequency

 

Phasor representations of sinusoids. Consider again the sinusoidal signal represented by

(6)   \begin{equation*} v(t)=V_{m}cos(\omega t + \theta) \end{equation*}

Knowing that v(t) is a sinusoid, there are three distinct values, or pieces of information, needed to characterize the waveform: V_{m}, \omega, and \theta. If  this waveform were provided at the input to a linear filter, the output of the filter would be another sinusoid of the same frequency, \omega but having a different amplitude and phase compared to eq (7). Knowing that the frequency of a sinusoid is unchanged when passing through a linear circuit, it is only the amplitude and phased that we need to keep track of.  Phasor notation provides a convenient representation.

The phasor representation of eq (7) depicts the waveform as a vector plotted in the complex plane, as shown in Figure 8.x The phasor is given by

(7)   \begin{equation*} \underline{v}=V_{m}\angle\theta \end{equation*}

and the vector has magnitude V_{m} and angle \theta. The phasor can also be represented in rectangular coordinates as

(8)   \begin{equation*} \underline{v}=V_{m}\cos(\theta)+jV_{m}\cos(\theta) \end{equation*}

Phasor representation and the use of a complex plane representation with magnitude, phase or real, imaginary parts is simply a bookkeeping means of keeping track of the two pieces of information needed to represent the sinusoid of frequency \omega.

 

Examples:

Determine the phasor representations of the following signals:

 

 

Determine the sinusoidal time-domain expression for the signals having phasors xxx if they are all known to be at angular frequency \omega = 500 \frac{rad}{s}

 

 

Complex Impedances

 

Resistor. If the voltage drop across a resistor is given by v(t)=V_{m}cos(\omega t + \theta) then the current is given by i(t)=\frac{V_{m}}{R}cos(\omega t + \theta). We know that the ratio of voltage to current is \frac{v(t)}{i(t)}=R. Similarly, the ratio of the voltage phasor to the current phasor is

\frac{\underline{v}}{\underline{i}}=R

We refer to this ratio as the impedance of the resistor.

The ratio of time-domain voltage to current waveforms is the resistance. The ratio of the voltage phasor to the current phasor os the impedance.

 

Capacitor phasors and impedance. The voltage and current for a capacitor are related by the expression

(9)   \begin{equation*} i(t)=C\frac{dv(t)}{dt} \end{equation*}

Given voltage v(t)=V_{m}cos(\omega t + \theta) the current is

(10)   \begin{equation*} i(t)=C\frac{d}{dt}(V_{m}cos(\omega t + \theta)) \end{equation*}

(11)   \begin{equation*} i(t)=C\frac{d}{dt}(V_{m}cos(\omega t + \theta)) \end{equation*}

(12)   \begin{equation*} i(t)=-\omega C\V_{m}sin(\omega t + \theta) \end{equation*}

since

(13)   \begin{equation*} -sin(x)=cos(x-\frac{\pi}{2})\end{equation*}

we can write

(14)   \begin{equation*} i(t)=\omega C\V_{m}cos(\omega t + \theta -\frac{\pi}{2}) \end{equation*}

The phasor representations for v(t) and i(t) are given by

(15)   \begin{equation*} \underline{v}=V_{m}\angle{\theta} =V_{m}e^{j\theta}\end{equation*}

(16)   \begin{equation*} \underline{I}=\omega C V_{m}\angle{(\theta-\frac{\pi}{2}}) =\omega C V_{m}e^{j(\theta-\frac{\pi}{2}})\end{equation*}

algebraic manipulation of eq (16) gives

(17)   \begin{equation*} \underline{i}=\omega C V_{m}e^{j(\theta-\frac{\pi}{2}})\=\omega C V_{m}e^{j\theta}e^{-j\frac{\pi}{2}} \end{equation*}

since e^{-j\frac{\pi}{2}}=-j we can write

 

The impedance of the capacitor, which is the the ratio of phasor voltage to phasor current for the capacitor is

(18)   \begin{equation*} Z_{c}=\frac{\underline{v}}{\underline{i}}=\frac{1}{\omega C} e^{-j\frac{\pi}{2}} =\frac{1}{j\omega C} \end{equation*}

 

Inductor phasors and impedance.The voltage and current for an inductor  are related by the expression

(19)   \begin{equation*} v(t)=L\frac{di(t)}{dt} \end{equation*}

Given current i(t)=I_{m}cos(\omega t + \theta) the voltage is

(20)   \begin{equation*} v(t)=L\frac{d}{dt}(I_{m}cos(\omega t + \theta)) \end{equation*}

(21)   \begin{equation*} v(t)=L\frac{d}{dt}(I_{m}cos(\omega t + \theta)) \end{equation*}

(22)   \begin{equation*} v(t)=-\omega L\I_{m}sin(\omega t + \theta) \end{equation*}

since

(23)   \begin{equation*} -sin(x)=cos(x-\frac{\pi}{2})\end{equation*}

we can write

(24)   \begin{equation*} v(t)=\omega L\I_{m}cos(\omega t + \theta -\frac{\pi}{2}) \end{equation*}

The phasor representations for v(t) and i(t) are given by

(25)   \begin{equation*} \underline{i}=I_{m}\angle{\theta} =V_{m}e^{j\theta}\end{equation*}

(26)   \begin{equation*} \underline{V}=\omega L I_{m}\angle{(\theta-\frac{\pi}{2}}) =\omega L I_{m}e^{j(\theta-\frac{\pi}{2}})\end{equation*}

algebraic manipulation of eq (26) gives

(27)   \begin{equation*} \underline{i}=\omega L I_{m}e^{j(\theta-\frac{\pi}{2}})\=\omega L I_{m}e^{j\theta}e^{-j\frac{\pi}{2}} \end{equation*}

since e^{-j\frac{\pi}{2}}=-j we can write

 

The impedance of the inductor, which is the the ratio of phasor voltage to phasor current for the inductor is

(28)   \begin{equation*} Z_{L}=\frac{\underline{v}}{\underline{i}}=\omega L} e^{j\frac{\pi}{2}} =j\omega L \end{equation*}

 

The complex impedances of a resistance, R, capacitance, C, and inductance, L,  are given by Z_{R}=R\Omega, Z_{C}=\frac{1}{j\omega C}\Omega, Z_{L}=j\omega L\Omega

 

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Applied Electrical Engineering Fundamentals by David J. McLaughlin is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

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