Phasors + Wave equation

Discussion in 'Homework Help' started by Arausio, Jan 3, 2016.

  1. Arausio

    Thread Starter New Member

    Jun 18, 2015
    22
    0
    Hello guys, I need some of your expertise to help me understand the notion of phasors and their representation in the wave equation. From school the notes show that a phasor is V_{0}e^{\alpha} and this is shown in the page marked phasors below. However, reading from wikipedia a phasor for a harmonic signal such as a(t)=Acos(\omega t + \phi) is
    Re\left \{ b(t) \right \} where b(t) is an imaginary number,
    b(t)=Ae^{\mathit{j (\omega t+ \phi)}}=Ae^{\mathit{j \phi}}e^{\mathit{j \omega t}}. Now, the real part of \widehat{A}=Ae^{\textit{j} \phi} multiplied by cos(\omega t) gives us the original a(t) back. I.e. Re\left \{ \widehat{A} \right \}=Acos(\phi);<br />
a(t)=Acos(\phi)*cos(\omega t)
    Moreover, looking at the KFUPM Open Courseware I also found that a phasor should have a j multiplied in the exponent. So, what is the proper representation? Whom should I believe and what exactly should I leave in my notes?
    2016-01-02_231836.jpg 2016-01-02_232013.jpg 2016-01-03_111446.jpg
     
  2. Papabravo

    Expert

    Feb 24, 2006
    10,145
    1,791
    Phasors are a convenient way to represent periodic functions of time such as AC voltage and current. They represent real things that you can observe and measure, but they use algebra of complex numbers, to keep track of things like magnitude and phase.

    Point #1: A complex number can be represented in multiple ways:
    1. In Cartesian Coordinates as a + jb, where a and b are real numbers an j is the imaginary unit aka √(-1).
    2. In Polar Coordinates as a magnitude and a phase, for example 6.3 ∠ 30°. In this case the magnitude is 6.3 and the phase angle is 30°
    Point #2: You can easily convert a Cartesian Form to a Polar Form
    1. If we have √3 + j in Cartesian Coordinates, this is equivalent to 2 ∠ 30°
    2. If we have 2 ∠ 60° in Polar Coordinates, this is equivalent to 1 + j√3
    Point #3: A phasor is a representation of a periodic sinewave that has a magnitude and a phase. It is a function of time, but time is not shown explicitly in the representation. For purposes of calculation the algebra of phasors does not require the explicit use of time t.

    Point #4: A complex exponential may also be used to represent a periodic sinewave due to the following identity:

    e ^ {j \phi} = cos(\phi)+j sin(\phi)

    Point #5: The explicit time dependence is encapsulated in the following term:

    e ^ {j \omega t} = cos(\omega t) + j sin(\omega t)

    This is a unit vector the rotates around the origin in a counter clockwise direction.

    If it helps you to visualize what is going on, you can think of A e^{j \phi} as the initial position of the phasor at time t = 0. Also the magnitude of e ^ {j \omega t} is 1 and the initial angle is 0° at t = 0.

    In summary:
    1. a + jb can be treated as a phasor
    2. M ∠ ∅ can be treated as a phasor
    3. A e^{j \phi}e^{j \omega t} can be treated as a phasor
    Here is a reference for Phasor Algebra

    https://en.wikiversity.org/wiki/Phasor_algebra
     
    Last edited: Jan 3, 2016
    Arausio likes this.
  3. Arausio

    Thread Starter New Member

    Jun 18, 2015
    22
    0
    Papabravo,
    Thank you for your post it helped me a lot! Now, what I'm thinking is trying to decode the professor's trail of thought going from phasors to the wave equation. So, on his phasors page he represented phasor without the j which is incorrect (from what I understood). However, in regards to \widetilde{V}(z)=V_{0}e^{\gamma z}=V_{0}e^{\alpha z}e^{\textit{j}\beta z} where \gamma=\alpha +\textit{j} \beta(phasor) the V_{0}e^{\gamma z} represents only the phasor component where the exponential looks like V_{0}e^{\textit{j} \phi}=V_{0}e^{\alpha z}e^{\textit{j}\beta z} after plugging in for gamma. Sorry if I'm asking easy questions it's just I'm still a little confused on the components in the wave equation which constitute the attenuation and oscillating constants and what represents the phasor in it.
     
  4. Papabravo

    Expert

    Feb 24, 2006
    10,145
    1,791
    The difference is fairly subtle. Any exponential terms without a j in them, can be thought of as contributing to the magnitude rather than the phase . In the simple case of a periodic waveform you have only the phase angle changing with time. In the more general case you have both the magnitude and the phase changing with time. In your equation replace z with t and you have:

    V_0 e^{\gamma t}=V_0 e^{\alpha t} e^{j \beta t}

    Now using the alternative phasor notation in polar form:
    1. The magnitude is V_0 e^ {\alpha t}. The magnitude starts at V0 and α must be strictly ≤ 0 or it will diverge to ∞. With α < 0 you have a damped response.
    2. The remaining term shows the explicit time dependence of the phase angle.
    Let me sum up:
    1. With α < 0, you have a damped sinusoidal response.
    2. With α = 0, you have periodic solution at a single frequency determined by β.
    3. With α > 0, you have a divergent solution which goes off to ∞.
    Got it?

    Extra Credit: Examine the derivative of a phasor, substitute it back into the original equation, and see if it makes sense.
     
    Last edited: Jan 3, 2016
    Arausio likes this.
  5. WBahn

    Moderator

    Mar 31, 2012
    17,757
    4,799
    There are several waves of linking a phasor to a sinusoidal signal and all that matters is that you are consistent in how you do it.

    The key relationship is

    <br />
Ae^{j \( \omega t+\phi \)} \; = \; A\cos(\omega t+\phi) \, + \, j \, A\sin(\omega t+\phi)<br />

    The two most common ways mapping the phasor to the sinusoidal signal are to exploit one of the following relationships:

    <br />
A\cos(\omega t+\phi) \; = \; Re \{ Ae^{j \( \omega t+\phi \)} \} \; = \; Re \{ Ae^{j \phi}e^{j \omega t} \}<br />
    or
    <br />
A\sin(\omega t+\phi) \; = \; Im \{ Ae^{j \( \omega t+\phi \)} \} \; = \; Im \{ Ae^{j \phi}e^{j \omega t} \}<br />

    Since the e^{j \omega t} is a multiplying term that does not change, we essentially factor it out of all of our work and work with just the Ae^{j \phi} as our "phasor".

    The "phasor" is just a convenient way of keeping track of the magnitude (A) and the phase (φ) of the complex exponential. This is because if you add, subtract, multiply, or divide complex exponentials, the magnitude and phase combine the same way that the components of the phasors do.
     
    Last edited: Jan 3, 2016
  6. Papabravo

    Expert

    Feb 24, 2006
    10,145
    1,791
    Yup! What he said.
     
    Arausio likes this.
  7. Arausio

    Thread Starter New Member

    Jun 18, 2015
    22
    0
    Hello guys, thank you for the help. I have another phasor representation straight from the board. It seems a bit incorrect to me... since \gamma = \alpha +\textit{j} \beta
    V(t)=V_{0}^{+}e^{\textit{j}\phi }e^{\textit{j}\gamma t}e^{\textit{j}\omega t} where the phasor component is V_{0}^{+}e^{\textit{j}\phi }e^{\textit{j}\gamma t}
    Anyway, what I understood from all this is the following (please do correct me if I'm wrong):
    1) Phasor consists of a complex amplitude + phase angle
    2) The wave eqn. \widetilde{V}(z)=V_{0}e^{\gamma z}=V_{0}e^{\alpha z}e^{\textit{j}\beta z} is the phasor where the frequency component \omega z of the exponential is omitted
    2a) Magnitude of this phasor is V_{0}e^{\alpha z} = (magnitude A in previous posts). Damping occurs when {\alpha<0}. The angle changes due to the 2nd term e^{\textit{j}\beta z} which is just like e^{\textit{j}\phi}
    2b) So, this corresponds to the notion of phasors in polar (short-hand) form which is [​IMG]
    2016-01-03_180347.jpg

    P.S. I know you have been trying to lead me to the answer; however, nobody has yet said for sure if the professor's note regarding phasors is incorrect. I mean all I care about is writing the CORRECT form in my own notes that's all... Therefore, please do point it out if possible.
     
    Last edited: Jan 3, 2016
  8. WBahn

    Moderator

    Mar 31, 2012
    17,757
    4,799
    Phasors have no time-domain component -- so there should not be a 't' anywhere in one. This is not to say that it is impossible to construct a problem in which a time-dependent phasor exists, but it would be very unusual.

    Are you sure that your middle exponential isn't spatial -- meaning that it would be e^(jϒx) or some other spatial parameter instead of time?
     
  9. Arausio

    Thread Starter New Member

    Jun 18, 2015
    22
    0
    I'm not sure exactly. I mean the course name is fields and waves and the topic of review was phasors leading in to the telegraph equations and the solution to the wave eqn. So, my doubts are approaching infinity in regards to phasors )) what would be the correct form for the solution to the telegraph eqns?? Also, acording to Papabravo,
     
    Last edited: Jan 3, 2016
  10. WBahn

    Moderator

    Mar 31, 2012
    17,757
    4,799
  11. Arausio

    Thread Starter New Member

    Jun 18, 2015
    22
    0
    Ok, so is my reasoning correct about phasors as they pertain to the telegraph equations? I mean wiki is ok but that was not my question in this thread....
     
  12. Papabravo

    Expert

    Feb 24, 2006
    10,145
    1,791
    In many problems where phasors are used, there is no explicit time dependence in either the Cartesian Form or the Polar Form, and no need for it to solve many types of problems. In these problems the magnitude is constant and the frequency is also constant and the phasor represents a sinusoidal quantity which is a function of time. Adding a time dependence to the magnitude of a phasor, does not affect the algebra of phasors at all. Using the exponential form of the solution shows explicitly what is going on with both the magnitude and the phase of the solution to the equation with respect to time. You would only do this if the time dependence was important to the problem at hand. For example if the frequency of the solution was important you, would probably not use phasors in the Cartesian or Polar Forms since they do not necessarily have have frequency information.
     
    Arausio likes this.
  13. Arausio

    Thread Starter New Member

    Jun 18, 2015
    22
    0
    Papabravo,
    So, is my 1-->2b in Post #7 correct?
    From what you're saying it is fine if a phasor has a time element in the exponential for these problems? Then, does \phi \equiv \beta z where z represents the time variable?
     
  14. Papabravo

    Expert

    Feb 24, 2006
    10,145
    1,791
    Yes, that appears to be correct.
     
    Arausio likes this.
  15. WBahn

    Moderator

    Mar 31, 2012
    17,757
    4,799
    I'm guessing here a bit, but it looks like you are trying to describe both the temporal and the spatial behavior of the wave, in which case your starting point is probably something like:

    <br />
V(t,z) \; = \; V_{\(0,0\)}^+ e^{j \gamma z} e^{j \omega t}<br />

    In which case your phasor is

    <br />
\widetilde{V} \; = \; V_{\(0,0\)}^+ e^{j \gamma z}<br />

    If

    <br />
\gamma \; = \; \alpha \, + \, j \beta<br />

    then

    <br />
\widetilde{V} \; = \; V_{\(0,0\)}^+ e^{j \(\alpha \, + \, j \beta\) z}<br />
\,<br />
\widetilde{V} \; = \; V_{\(0,0\)}^+ e^{j \alpha z} e^{-\beta z}<br />

    In this formulation, alpha is a spatial frequency and beta is a spatial damping factor.
     
    Arausio likes this.
  16. MrAl

    Well-Known Member

    Jun 17, 2014
    2,438
    492
    Hi,

    Note sure what is being asked, so just a guess, but maybe the form:
    v(x,t)=RealPart(V(x)*e^jwt

    and then using that (and the form for the current) in the wave equation.
    Then we can write:
    d2V(x)/dx2=(1/vp2)*jw2*V(x)

    where the four 2's in the above are used to show powers of 2 so dx2 is dx squared for example.
     
  17. Arausio

    Thread Starter New Member

    Jun 18, 2015
    22
    0
    Another thing that I want to ask is if we look at WBahn's, key relationship in Post #5 taking the Re (real part) of both sides would mean we end up with
    Acos(\omega t+\phi)=Re\left \{ Ae^{j \phi}e^{j \omega t} \right \}=Acos(\phi)cos(\omega t) for the real component and taking the Im (imaginary part) of both sides Asin(\omega t+\phi)=Im\left \{ Ae^{j \phi}e^{j \omega t} \right \}=Asin(\phi)sin(\omega t) for the imaginary component. I just want to verify that I'm correct in the trigonometric variant


     
  18. Papabravo

    Expert

    Feb 24, 2006
    10,145
    1,791
    Last edited: Jan 5, 2016
  19. WBahn

    Moderator

    Mar 31, 2012
    17,757
    4,799
    Nope. Work it out -- and you don't need any trig identities to do it. Just put each exponential in rectangular form and multiply them out collecting real and imaginary parts.

    Then compare that to the trig identities for the sine and cosine of the sum of two angles.

    What you will have discovered is that you no longer need to memorize the basic trig identities -- they are trivially easy to derive using complex exponentials.
     
Loading...