7
$\begingroup$

What exactly is a phasor? I was reading about alternating current when I came across the following definition:

A phasor is a vector which rotates about the origin with an angular speed(suppose $\omega$).

Then the book mentions the following statement: Though voltage and current in an AC circuit are represented by phasors-rotating vectors, they are not vectors themselves.

Aren't the 2 statements contradictory?

In my knowledge, a vector quantity is one which follows the law of vector addition(correct me if I'm wrong).

The book even obtains the impedence of an LCR circuit by using phasors and adding them just like vectors. So, what exactly is the difference between the two?

$\endgroup$
5
  • $\begingroup$ You can also have a look at this answer of mine on EE.SE. $\endgroup$ Commented Sep 19, 2016 at 15:03
  • $\begingroup$ It is a (time-dependent) complex number, represented as a (rotating) vector on the complex plane. $\endgroup$ Commented Sep 20, 2016 at 4:12
  • $\begingroup$ @flippiefanus No, a phasor it's not time-dependent: see the answer I linked in the comment above. $\endgroup$ Commented Sep 20, 2016 at 12:49
  • $\begingroup$ OK, the phasor is the part you get after you factor out the time dependence. $\endgroup$ Commented Sep 20, 2016 at 13:06
  • $\begingroup$ @flippiefanus Yes, exactly. If you want instead to consider also the time-varying complex exponential, then you can speak of the analytic signal associated with the real signal (see Example 1 of the Wikipedia article). $\endgroup$ Commented Sep 20, 2016 at 14:12

3 Answers 3

5
$\begingroup$

Think of a combination of the complex plane and ordinary vectors.

A phasor is a complex number,  representing a sinusoidal function whose amplitude (A), angular frequency (ω), and initial phase (θ) are time-invariant.

Image and text from Phasors Wikipedia

Assume you have a network composed of multiple sinusoids (waves). They all have the same frequency, but with differing amplitudes and phases. The only difference in their analytic representations is the complex amplitude (phasor). A linear combination of such functions can be factored into the product of a linear combination of phasors (known as phasor arithmetic) and the time/frequency dependent factor that they all have in common.

enter image description here

enter image description here

When function ${\displaystyle \scriptstyle A\cdot e^{i(\omega t+\theta )}}$ is depicted in the complex plane, the vector formed by its imaginary and real parts rotates around the origin. Its magnitude is $A$, and it completes one cycle every $2π/ω$ seconds. $θ$ is the angle it forms with the real axis at $t = n•2π/ω$, for integer values of n.

$\endgroup$
2
  • $\begingroup$ Just to clarify, phasor is a complex number, and not a vector. So, any physical quantity cannot be phasor but can be represented as a phasor.(right?) $\endgroup$ Commented Sep 19, 2016 at 11:11
  • $\begingroup$ Mathematically, you can consider complex numbers as vectors obeying the vector laws of addition and subtraction considering the real and imaginary parts. With respect to multiplication with real numbers and the vector addition/subtraction laws, complex numbers form a vector space equivalent to the space of translation vectors in the real 2-D plane $\endgroup$
    – freecharly
    Commented Sep 19, 2016 at 19:38
1
$\begingroup$

All vectors follow vector addition laws and multiplication laws. So if you add two phasors they are added like vectors but if you multiply them they are multiplied like simple numbers. Therefore, phasors are like vectors but not vectors. Just like the area vectors which are multiplied like vectors but added like numbers.

$\endgroup$
0
$\begingroup$

Phasors are complex quantities used to partially represent real quantities that vary sinusoidally in time and perhaps in space. All phasors are time-independent. They represent partially the real quantity, and not completely, because they don't have information about the frequency.

To say that phasors are like vectors, is to say that complex numbers are like vectors, which is wrong. First of all, the similarities are for two-dimensional vectors only. Yes, addition and subtraction of two 2D vectors is analogous to addition and subtraction of two complex numbers; and multiplication of a 2D vector by a scalar is analogous to multiplication of a complex number by a real number. But, division of two vectors isn't even defined, however division of two complex numbers is defined; also, you can't just "multiply two 2D vectors, you must specify if it's a dot product or a cross product, however you can "just" multiply two complex numbers.

For electric circuits, a phasor voltage $ \tilde V $ is a complex constant, and it represents the amplitude and phase of the signal, but not its frequency. The signal $v(t)$ is a real-valued function of one real variable (one temporal, $t$), and it represents the real instantaneous value of the signal. The sinor $ v_c(t) $ is a complex-valued function of one real variable (one temporal, $t$), and it represents the complex instantaneous value of the signal. Some relations:

$v(t) = V_m \cos {(\omega t + \phi)} = \Re [\tilde V e^{j \omega t}] = \Re [v_c(t)] \tag*{}$

$\tilde V = V_m e^{j \phi} = V_m \cos {(\phi)} + j V_m \sin {(\phi)} \tag*{}$

$v_c(t) = \tilde V e^{j \omega t} = V_m e^{j \phi} e^{j \omega t} = V_m e^{j (\omega t + \phi)} = V_m \cos {(\omega t + \phi)} + j V_m \sin {(\omega t + \phi)} \tag*{}$

Note: $v(t)=\Re [\tilde V]$ only when $\omega t = \ldots, -4 \pi, -2 \pi, 0, 2 \pi, 4 \pi, \ldots$; in other words, only when $\omega t = 2 \pi k$, where $k$ is any integer.

For long transmission lines (electric circuits with distributed rather than concentrated parameters), a phasor voltage $ \tilde V(x) $ is a complex-valued function of one real variable (one spatial, $x$). The signal $v(x,t)$ is a real-valued function of two real variables (one spatial, $x$; and one temporal, $t$), and it represents the real instantaneous value. The sinor $v_c(x,t)$ is a complex-valued function of two real variables (one spatial, $x$; and one temporal, $t$), and it represents the complex instantaneous value. Some relations:

$v(x,t) = V_m e^{a x} \cos {(\omega t + \beta x + \phi)} = \Re [\tilde V(x) e^{j \omega t}] = \Re [v_c(x,t)] \tag*{}$

$\tilde V(x) = V_m e^{j \phi} e^{a x} e^{j \beta x} = V_m e^{a x} e^{j (\beta x + \phi)} = V_m e^{a x} \cos {(\beta x + \phi)} + j V_m e^{a x} \sin {(\beta x + \phi)} \tag*{}$

$v_c(x,t) = \tilde V e^{j \omega t} = V_m e^{j \phi} e^{a x} e^{j \beta x} e^{j \omega t} = V_m e^{a x} e^{j (\omega t + \beta x + \phi)} = V_m e^{a x} \cos {(\omega t + \beta x + \phi)} + j V_m e^{a x} \sin {(\omega t + \beta x + \phi)} \tag*{}$

For general electromagnetic theory, phasors are complex-valued functions of three real variables (three spatial, $x$, $y$, $z$). For instantaneous electric field vector, $\mathbf E (x,y,z,t)$, its phasor is $\mathbf {\tilde E} (x,y,z)$, and the relation $\mathbf E (x,y,z,t) = \Re [\mathbf {\tilde E} (x,y,z) e^{j \omega t}]$ is satisfied.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.