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Homework Help: Complex waveform/harmonics question

  1. Sep 1, 2009 #1
    Really stuck with this one any help appreciated
    1. The problem statement, all variables and given/known data
    a voltage amplifier ideally should have the relationship Vo=100vi
    but in practice is Vo=vi(98+ (2/.001) vi)

    determine the expression for the output voltage when a 10mv peak 800hz sinusoidal input signal is applied

    2. Relevant equations


    3. The attempt at a solution

    right i know that the equation needs to have the fundamental and second harmonic in

    i have so far

    vo=10mv(98+(2/.001)10mv vo=1.18v

    vo=0.1+98sin(1600pi t)

    iknow at 1600hz 1 cycle will be 1.25ms
    and at 3200hz 1 cyle will be 3.125ms

    the example in the lesson is in a different format and i cant relate the steps to this example

    i dont think im far off with the expression but cant justify the steps in getting that.

    any help appreciated
  2. jcsd
  3. Sep 1, 2009 #2


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    You should not replace vi by 10mV. This is a dc voltage.
    vi should be [tex]10x10^{-3}sin(1600\pi t)[/tex]
    Notice that the output voltage has a term that depends on vi squared. What is the square of a sinusoid?
  4. Sep 2, 2009 #3
    grateful for the reply.

    i foujnd this on the net
    "From basic trigonometry the square of a sinusoidal input is a DC shifted sinusoid of twice the input frequency"

    this is the second harmonic i believe Vi=10mv sin (1600pi t)?
    i know i need to have a cos wave something like 1/2cos(3200pi t) ?

    but i havnt had enough in th elesson to feel confident of what im writing. ive done the lesson 3 times now and each time get a little more from it

    thanks ian
    Last edited: Sep 2, 2009
  5. Sep 2, 2009 #4


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    The amplitude of the second harmonic is not 10mV.
    Remember your trigonometry lessons. What is the expression for the sine of the double arc?
    Remember also that [tex]sin^2 x + cos^2 x = 1[/tex].
    Subtracting one expression from the other, what do you get?
  6. Sep 2, 2009 #5
    Sorry man im lost, i have just started this open learning course after being out of education for 15 years i have managed to answer th efirst 5 questions of my assignment fine as the lessons had enough examples for me to practice on however the lesson im on with now only has 1 in a different format to what the question above is asking, so im sorry if i sound dumb

    thanks for trying

  7. Sep 2, 2009 #6


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    By the rules of the forum, we are not allowed to give you the answers, only to point you to the right direction.
    You must do your work and show it. Then we can correct you.
  8. Sep 3, 2009 #7
    Yeah i understand that but sadly i dont understand it enough il keep on at it and try again

  9. Sep 5, 2009 #8


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    Science Advisor

    a voltage amplifier ideally should have the relationship Vo=100vi
    but in practice is Vo=vi(98+ (2/.001) vi)

    determine the expression for the output voltage when a 10mv peak 800hz sinusoidal input signal is applied

    I see nothing in the question about second harmonics.
    10mV IS the peak amplitude of the input signal.

    OK, ideally the amplifier should have a gain of 100. (Vout = 100 * Vin)
    So the output should be 10 mV * 100 or 1 volt peak.

    But in practice has an output voltage of Vin * (98 + 2000*Vin) = 1.18 volts peak.

    And this is exactly what you got.

    "peak" is the amplitude of half a sinewave. The ideal gain and the 800 hz are irrelevant to the final answer.
  10. Sep 5, 2009 #9


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    If you perform the multiplication, you get vi squared. The square of a sinusoid corresponds to a dc plus a sinusoid with the double of the frequency of the input.
  11. Sep 6, 2009 #10
    hello there right i think im on to something now my final expression is

    Vo=100+980 sin (1600pi t)-100 cos(3200pi t)

    100 is the dc term
    1600hz is the fundamental
    and 3200hz is the second harmonic

    it now asks for the percentage second harmonic distortion present in the output

    need to check this out and repost my workings

    thanks for replies and pointing me in the right direction
  12. Sep 6, 2009 #11


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    It is still wrong. What is the coefficient of [tex]vi^2[/tex]?
    You must use:
    [tex]cos^2 x+sin^2 x = 1[/tex]
    [tex]cos^2 x-sin^2 x=cos 2x[/tex]
  13. Mar 20, 2010 #12
    i need help too i'm getting a second harmonic at 1600hz at an amplitude of 10v as 2/1*10^-3 is 2000, takes me to 20(1/2(1-cos(2*1600pi t)))
  14. Mar 27, 2010 #13
    okay i'm getting an answer of dc term 10v, fundamental of 0.98v at 800hz and a secondary of 10v at 1600hz, but that would say my amplifier was Vo=10000Vi why am i getting this answer? please help
  15. Apr 1, 2010 #14
    Hi there, I think I've got the answer. Just need it checking!

    Vo = 0.1 + 0.98sin(1600 pi t) - 0.1cos (3200 pi t)

    Thanks for any reply.
  16. Apr 2, 2010 #15
    No the answer comes out as 10+0.98sin(1600pi t)-10cos(3200pi t) as the second part of the equation is 2/k which is 2/1*10^-3 = 2000 multiply that by 10mV or 0.01 gives 20 multiply that by half gives you 10
  17. Apr 5, 2010 #16


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    You have a sign error. The answer is
    0.1 + 0.98 cos (1660 pi t) + 0.1 cos (3200 pi t)

    Notice that the amplitudes sum to more than 1. I think that there is an error in the original statement that should be

    vo = vi (98 + 2/0.01 vi)
  18. Apr 5, 2010 #17


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    In reality the amplitude of the wave must be squared. 10 mV squared is 0.0001.
  19. Apr 5, 2010 #18
    I have the answer now it's 10 + 0.98 sin (1600 pi t) + 10 cos(3200 pi t) as we're working in millivolts the dc term is 10 mV, the fundamental is 0.98v at 800hz and the second harmonic is 10mV at 1600hz I've had this checked and it is correct
  20. Apr 5, 2010 #19


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    If you are working in millivolts, the amplitude of the fundamental sinusoid should be 98 and not 0.98.
  21. Apr 5, 2010 #20
    Yes it is but in the equation the fundamental is worked out in volts whereas the second harmonic and dc term are multiplied by 2/0.001 to change it to work out in millivolts
  22. Apr 8, 2010 #21
    Hi Cel

    Thanks for your reply.

    I've been looking at this question for a few days now and I still get a negative sign at latter end of the expression. Here's how I've got there.


    A voltage amplifier ideally shouldm have the input-output realationship of v(0) = 100 v(i) but relationship in practice is :

    v(0) = v(i) (98 + 2/k v(i) )

    (a) Determine the expression for the output voltage when a 10 mV (peak), 800 Hz sinusoidal signal is applied to the amplifier.


    v (i) = 0.01 sin (1600 pi *t)

    Substitute into original practical expression:

    v (0) = 0.01 sin (1600 pi *t) x (98 + 2/0.001 * 0.01 sin (1600 pi *t) )

    = 0.01 sin (1600 pi *t) x (98 +2000 * 0.01 sin (1600 pi *t) )

    = 0.01 sin (1600 pi *t) x (98 + 20 sin (1600 pi *t) )

    = 0.98 sin (1600 pi *t) + 0.2 sin^2 (1600 pi*t)

    Using trig identity sin ^2(x) = 1/2 (1 - cos 2x):

    = 0.98 sin (1600 pi *t) + 0.2 [ 1/2 (1 - cos 2 * 1600 pi *t)]

    = 0.98 sin (1600 pi *t) + 0.1 - 0.1 cos(3200 pi*t)

    = 0.1 + 0.98 sin (1600 pi *t) - 0.1 cos(3200 pi*t)

    Thanks in advance for any reply.
    Last edited: Apr 8, 2010
  23. Apr 8, 2010 #22


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    You are right. I have made the calculations with the cosine (that is the usual waveform) and found a plus sign.
    With a sine waveform you get a minus sign,
  24. Apr 8, 2010 #23
    Thanks again CEL,

    I thought i was losing it.
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