adding harmonics calculating lag

Discussion in 'Homework Help' started by ninjaman, Dec 7, 2013.

  1. ninjaman

    Thread Starter Member

    May 18, 2013
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    hello
    I want to learn about adding harmonics to the fundamental and calculating lag. I also have to find the voltage after 20milliseconds by simulation and calculating.

    what would this come under. I know its not phasor diagrams as they only work on one frequency. I have three, the fundamental third and fifth harmonics. the fifth I have to add a lag to, graph and show voltages. then calculate the voltage after 20milliseconds.

    do I use Vsin(wt+a)?

    voltage for the fundamental would be 100v
    w would be 2*pi*50
    then add "a"? on my excel graph I have "a" as time. I found the periodic time of the first wave which was 1/50 = 0.02seconds. then I broke that up so the graph had from 0 up to 0.02
    form what I have read on some sites the "a" is an angle?

    any help would be great, a website that showed how to calculate the lag

    thanks
    simon
     
  2. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    When signal is represented in the form V*sin(wt+a):
    V is magnitude of the amplitude
    w is angular frequency
    a is phase angle

    a, the phase angle, is how you introduce the lag.
     
  3. MikeML

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    Oct 2, 2009
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    but a is in radians; not degrees. a = 2∏*degrees/360, so 70 deg would be 1.221730476 radians.
     
  4. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    So, so far we have:
    Magnitude of the amplitude=100 volts
    Fundamental frequency f=50 Hz
    Third harmonic, 3f=3(50 Hz)=150 Hz
    Fifth harmonic, 5f=5(50 Hz)=250 Hz

    So.
    For fundamental frequency: Signal1=100*sin(2*pi*50*t+0)
    For third harmonic: Signal3=100*sin(2*pi*150*t+0)
    For fifth harmonic: Signal5=100*sin(2*pi*250*t+a)

    Since your fifth harmonic will be delayed, you can calculate the delay.
    This is from wiki:
    When φ is non-zero, the entire waveform appears to be shifted in time by the amount φ/ω seconds. A negative value represents a delay, and a positive value represents an advance.
    http://en.wikipedia.org/wiki/Sine_wave

    Since you want a delay, then the fifth harmonic will look like this:
    Signal5=100*sin(2*pi*250*t-a)
    * minus sign shows that you are delaying the signal by a
    * to find a=(delay in seconds)(w)=(delay in seconds)(2*pi*50)
     
    Last edited: Dec 7, 2013
  5. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    ω
    You were taught wrong.

    sin(anything), the units of "anything" are always in radians (my HP calculator cant be wrong) ;)

    btw=the units of the expression ωt are in radians, because ω is radians per second.
     
  6. shteii01

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    Feb 19, 2010
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    My mistake. I found better explanation.
    I will delete the earlier post.
     
  7. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    How many threads is Ninjaman going to start about this same question? I think we are now up to four...
     
  8. shteii01

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    Feb 19, 2010
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    Well, the fact that he is using Excel for this... is telling.
     
  9. atferrari

    AAC Fanatic!

    Jan 6, 2004
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    Sure not less than three...

    The OP insists in having doubts if amplitudes should be added or not. Of course he has the right to have doubts.

    That is a basic concept to be acquired so, using Excel, would have been a straight procedure for him.

    Enough.
     
  10. WBahn

    Moderator

    Mar 31, 2012
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    What's wrong with using Excel? What other tool -- besides another spreadsheet -- that someone is likely to have sitting on their computer should he be using?
     
  11. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    It depends how you approach the topic. If you did not include that last qualifier, I would have said MatLab or some other mathematics program/package even in demo form.

    With that last qualifier, I still say MatLab, but it is less simple to get for "free".
     
  12. ninjaman

    Thread Starter Member

    May 18, 2013
    306
    1
    hello

    ok I have just a few more questions.....

    1. with the sine at 30degrees and consine at 2004degrees......no im kidding.

    I think I have this now, I would like to thank everyone for their help. I don't understand fourier series and most of this is quite confusing. I have some sites to read about sine and cosine, then maybe onto fourier series.
    I don't understand excel that well and still struggling getting the results I wanted.
    what I did to get the results in excel was subtract 70 from 250, which is 70 degrees away from 250 hertz. that how much I don't understand it.
    I know how to show the circuit in multisim and should be able to get the correct results.
    thanks again
    all the best
    simon
     
  13. shteii01

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    Feb 19, 2010
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    Ouch. Mixing units is bad, very bad.
     
  14. WBahn

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    Mar 31, 2012
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    Oh, there are definitely more suitable programs -- I pointed that out many a post ago -- and MatLab or Mathematica rank right up there. But unless they are something that you are likely to use on a recurring basis, the learning curve probably isn't worth it -- the learning curve with Excel for stuff like this is pretty tame (and that's not saying that it's horrendous with MatLab for this kind of stuff, either).

    I've done reasonably sophisticated stuff in MatLab, but even so I tend to use Excel for this kind of stuff because I use MatLab so infrequently that each time I do I have to take the time to reteach myself all the basics. Now, I'll admit that this is someone of a self-fulfilling prophecy because there are lots of things that I use Excel for that I could/should use MatLab for and the fact that I don't ensures that I will get too rusty at it and put off using it even more until I finally get to something that is just too much of a pain in Excel.

    None-the-less, Excel is a surprisingly capable package and most people just barely scratch the surface of its capabilities.
     
  15. WBahn

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    Mar 31, 2012
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    I absolutely agree, but also disagree somewhat.

    The sin(), log(), and other such functions are transcendental functions and, as such, their arguments have to be dimensionless and they return a value that is dimensionless. The easiest, though not most rigorous, way to see this is to consider that the functions could be defined in terms of a power series. For instance, just taking the first two terms of the sine function, we have

    sin(θ) ≈ θ - (θ^3)/6

    Let's say that θ has dimensions of "bobs". The first term then has dimensions of "bobs" while the second term has dimensions of "bobs-cubed". These cannot be added unless a "bob" is the same as a "bob-cubed". The only way this will be the case is if "bob" is identically equal to 1 -- i.e., if θ is dimensionless.

    That does not mean that something like

    y = sin(30°)

    is incorrect. Yes, the argument has units, but the units are stated and, most important, they are compatible with a dimensionless unit, namely radians. So it is a simple matter to convert from degrees to radians since we know that a full circle consists of 2∏ radians or 360°. Thus, writing

    y = sin(ωt + 30°)

    is perfectly reasonable, valid, and common.

    This is really nothing more than saying that something like

    y = 6ft + 3in

    is perfectly valid and meaningful. I can add the two values because they have compatible units -- both have units of length. But I cannot perform the arithmetic until I have converted them such that the units are not only compatible, but identical. The easiest way to understand this is to understand that addition and subtraction of units is meaningless. So we need to be able to factor out the units so that we can get at the numbers that can be added.

    y = 6ft + 3in

    y = 6ft*12in/ft + 3in

    y = 72in + 3in

    y = (72 + 3)in

    y = 75in

    An example that is common, but neither reasonable nor valid, is

    <br />
y = e^{10t}<br />

    The exponent has units of time. The exponential function is a transcendental function and has to have a dimensionless argument. When this isn't ignored completely, it is patched by writing off to the side something like "t in ms".

    It shouldn't matter what t is in. The units should take care of themselves -- and will take care of themselves if they are properly tracked. So this should be

    <br />
y = e^{10\frac{1}{ms}\cdot t}<br />

    Now no one has to be told that "t is in ms". If someone wants to use a value of t that is in μs, they can. All the necessary information is right there; the equation is self-contained and self-consistent.

    More importantly, it will remain intact as the expression, which is likely part of a larger expression, is manipulated -- be it multiplied, integrated, differentiated, transformed, or you name it.
     
  16. WBahn

    Moderator

    Mar 31, 2012
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    250 Hz - 70°

    Sorry, can't be done. That's like trying to subtract the height of a desk from a gallon of milk. It's meaningless.
     
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