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Vector Coordinate Conversion & Phasors

  1. Jul 3, 2009 #1
    I would attempt to solve these questions with "relevant equations", but my questions simply derive from an attempt of understanding class notes. If someone could attempt to help my understanding, that would be great. I've attached the PDF file that contains my corresponding questions.

    Questions:
    I. Pg. 2 (from the PDF file): Reading the first slide on this page, I understand the table containing various dot products. However, when I look at the second slide (on this page 2), I am slightly confused. Some of my questions include:

    1. Why are each element in the matrix dotted products?
    *** I know how to take dot products: multiply the first element of each vector together, multiply the second elements of each vector, and so on until the elements [for each unit direction] is exhausted, then sum them all. Or one could take the magnitude of each vector and multiply these to the angle difference between the two.***

    ----------------------------------------------------------------------------------------------------------------------------
    II. Pg. 7 (from the PDF file): My question come from the second slide of this page (pg. 7/22):

    1. What applications are phasors used for? I've looked online, and found that phasors are sinusoidal functions where [tex]A, \omega, \theta[/tex] doesn't change with time (time-invariant). But I still don't know how this is useful.
    2. What is the "instantaneous form"?
    3. What is [tex]v_x[/tex] (which is also shown on the diagram on the top right)?
    4. How does applying Kirchoffs Current law at the node containing [tex]v_x[/tex] provide the first equation (sum of two fractions = 0)?
    ***I know from physics, KCL tells us the amount of charge entering a particular system must equal the amount of energy exiting the system- So it is conservative.***
    5. Can someone explain to me why -90degrees accounts for the phase shift between voltage and current in the capacitor?
     

    Attached Files:

    Last edited: Jul 3, 2009
  2. jcsd
  3. Jul 3, 2009 #2
    I) You're trying to convert cartesians to cylindrical coordinates, right? Just trace through the matrix multiplication carefully. Ask yourself what are the components of Ax after multiplication. If you don't get it, maybe you don't really understand that first slide...

    If all this is v new to you, you're probably better off looking for a textbook on vector calculus. These notes probably suffice as a recap, but it's going to be tough to learn from scratch using those.

    II) I'm not sure what Vx is either, so I won't try to answer lest I mess things up.
     
  4. Jul 3, 2009 #3
    I understand what they are after dot product multiplication. However, I don't understand why in the first matrix each element in the matrix are dot products. For instance, in the (1,1) entry of the matrix we are dotting r with x (I know by the table it produces the cosine).

    thanks,


    Jeff
     
  5. Jul 3, 2009 #4
    Alright, assuming you are now in cylindrical polars, you want to convert to cartesians to get Ax. To do that you will need to find the x component in each component of the cylindrical coordinates. To find the component, you take the dot product. Thus you get

    (r.x)Ar + (phi.x)Aphi + (z.x) Az
    [all are unit vectors]

    which is what is being represented in the matrix multiplication (and hence the matrix is built that way)
     
  6. Jul 3, 2009 #5
    I think I just understood the concept of this vector coordinate conversion. But how does this differ from the typical conversion where, [tex]x = r cos(\phi), y = rsin(\phi) and z = z [/tex] (#1). When the calculations are done, one method is not equivalent to the other. For instance [tex]A_x = cos(\phi)\hat{r} + (-sin(\phi))\hat{\phi}[/tex], which implies [tex]x \neq A_x[/tex].
     
    Last edited: Jul 3, 2009
  7. Jul 3, 2009 #6
    They are actually equivalent.

    Note that for describing a point in space, the phi component is usually 0, since you only need r and z to define a point properly, if you choose r to point along the vector you want. So it simplifies to the familiar x = r cos (phi)
     
  8. Jul 3, 2009 #7
    I'm sure you are correct, but I don't understand what you just said. I always thought a point in 3 dimensional space is usually defined by three variables, that is; in cartesian (x,y,z); in cylindrical [tex]r, \phi, z[/tex]; in spherial also three variables. But if I interpret what you said to how I am thinking [which is probably where my error arises], if [tex]\phi = 0[/tex], then,

    [tex] A_x = cos(0) - sin(0) = 1 \neq r = rcos(0) = x[/tex] since r can have any value. So by my interpretation, they are not equal.
     
  9. Jul 3, 2009 #8
    It should be
    [tex]
    A_x = A_r cos(\phi)\hat{r} + A_\phi (-sin(\phi))\hat{\phi}
    [/tex]

    I hadn't spotted that earlier. Remember you only have unit vectors there!

    Yes in cylindrical polars you have 3 variables, a point in space in cylindrical polars IS defined by 3 variables, but we can usually DEFINE phi = 0 for a single point in space. (You can choose the r vector to point anywhere, see)
     
  10. Jul 3, 2009 #9

    Got it. So we're letting r not only be the radius, but also act as [tex]\phi[/tex].
    [tex]
    A_x = A_r cos(\phi)\hat{r} + A_\phi (-sin(\phi))\hat{\phi} = A_r cos(0)\hat{r} + A_\phi (-sin(0))\hat{\phi} = A_r\hat{r} = r = rcos(0) = rcos(\phi)
    [/tex]

    Thanks,

    Jeffrey Levesque
     
  11. Jul 4, 2009 #10
    Can someone check over the second paragraph to see whether my understanding for taking the dot product between [tex]\hat{z} \cdot \hat{R}[/tex] is correct?

    I continued onto the next page, and it seems the notes didn't include a table like they did for the Cartesian-Cylindrical Coordinate conversion. So I was trying to think of some ideas how to go about this, but I'm affraid I do not know. In the last table, there was only one angle that varied which was [tex]\phi[/tex], which allowed us to determine the cosine and sine functions respectively. However, for the Spherical-Cartesian (and vice-versa), there are two angles [tex]\theta, \phi[/tex]. Would that mean in order to construct a similar table (as pg. 2), one of the angles would have to be fixed? In short, I would like to know how [tex](\hat{x}\cdot \hat{R}) = cos(\phi) = (\hat{y} \cdot \hat{\theta}); (\hat{y} \cdot \hat{R}) = sin(\phi) = -(\hat{x} \cdot \hat{\theta}), (\hat{z} \cdot \hat{\phi}) = 1; (\hat{r} \cdot \hat{R}) = sin(\theta)[/tex], along with other dot products.

    This is what I think: If I vary [tex]\phi[/tex] from 0degrees to 90 degrees, and hold [tex]\theta[/tex] fixed, then I can produce a table (just like the cylindrical-cartesian table). Unfortunately, I'm affraid I didn'tunderstand the original table to begin with, and my lack of understanding transfers over to this table also. I am not sure how to take the dot product for the following: [tex]\hat{z} \cdot \hat{R}[/tex] which changes to [tex]\hat{z} \cdot \hat{x}[/tex] when [tex]\phi[/tex] = 0 degrees and [tex]\phi[/tex] = 90 degrees. I mean I think the answer for both is 0, but I can quite understand. I'm guessing because [tex]\hat{z}[/tex] and [tex]\hat{x}[/tex] are perpendicular to one another, the cosine is 0, which causes the dot product to be 0 (and [tex]\phi[/tex] has no effect on the dot product for this case). I think this reasoning actually makes sense, and should be correct, but I feel hesistant to conclude that my reasoning is right. This is similar to the previous table (Cylindrical-Cartesian), for the [tex]\hat{r} \cdot \hat{z}[/tex] column. Where entries are both 0 for both 0 degrees and 90 degrees. Can someone tell me if I am wrong. But if this is correct, how do you take the dot product between [tex]\hat{x} \cdot \hat{\theta}[/tex], or any other dot products involving [tex]\theta[/tex]

    thanks so much,


    Jeff
     
    Last edited: Jul 4, 2009
  12. Jul 4, 2009 #11
    In addition to my last question [above, which I am still not sure about], I was wondering if anyone knows what a phasor is- I think more specifically, my question resides on pg. 7 of the attached slide.

    thanks once again,


    JL
     
  13. Jul 4, 2009 #12
    Short answer (sorry, haven't got much time now): Phasor is just a *vector*. I think it's easier to understand how they work rather than try to define them (at least for me) Have you learnt interference before? Young's double slit? They almost always use phasors to try to explain constructive/destructive interference.
     
  14. Jul 4, 2009 #13
    I don't know any of that stuff. I've only taken basic physics courses, learning basic kinematics, basic electrical concepts [coulomb, right hand rule, and such], basic circuit concepts [KCL], and some basic interference- that is if sound waves overlap, they cancel each other out. I guess, If I could get more information simply on the premise of the variables I asked about- I would be greatly thankful.

    Also, could you [or anyone else], help me understand how to produce tables- for vector conversion of the spherical conversion [to Cylindrical and Cartesian]? Because on the slides, they show me the matrix conversion, but I don't know how to derive them.


    Thanks so much,


    Jeffrey Levesque
    **Happy 4th of July.
     
  15. Jul 5, 2009 #14
    I actually just figured out how to produce the remaining vector transformations- how to produce the tables and its matrix for the spherical coordinates. I guess for the phasors, I will try to borrow a book on circuitry tomorrow, but if anyone could help- that would be great.

    Thanks,

    JL
     
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