help with trigonometric derivatives of 2 sine waves.

Discussion in 'Homework Help' started by sian_willcocks, Mar 8, 2016.

  1. sian_willcocks

    Thread Starter New Member

    Mar 8, 2016
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    Hi

    I am currently stuck on a particular question in maths that has completely thrown me.

    I have been given:
    Va = 2sin (314.2t)
    Vb = 2sin (314.2t-70°)

    The task is to determine a single wave from a combination (additive) of 2 waves.

    I have completed the graph and written down the formula for the resulting wave;
    Vc = 3.26sin (314.2t+36°)
    This may not be correct but it must be in the form Vsin (wt-phi)

    I am then asked to use trigonometry to derive an expression for the resultant wave form in the same form?

    This has completely wiped me, the first thing that comes to mind is sina+b=sinacosb+cosasinb but have no idea if I am thinking along the right lines or what to do with it if I am!!

    Please help!!

    Sian.
     
  2. RBR1317

    Active Member

    Nov 13, 2010
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  3. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    Start with this identity:

    [​IMG]
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    Your graphical answer is close, but you've made one big mistake.

    Consider the value at t=0. Will Va be positive, negative, or zero? What about Vb? Based on that, what will Vc be? Now ask whether your expression for Vc will yield the same result.
     
  5. RBR1317

    Active Member

    Nov 13, 2010
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    Not sure what that graph may be, but if it is a phasor diagram then it should look something like the attached image. A phasor diagram will also make clear why your answer is wrong. When carefully drawn, the result can be read directly from the phasor diagram with slide rule accuracy. Not quite as good as a $9 scientific calculator (my Casio fx-260solar), but I got through four years of engineering school with just an inherited Keuffel & Esser.
     
  6. sian_willcocks

    Thread Starter New Member

    Mar 8, 2016
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    I am so confused!I've attached my graph and what I have so far.
     
  7. sian_willcocks

    Thread Starter New Member

    Mar 8, 2016
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    My expression works when I tabulate it so I can't see where I have gone wrong?
     
  8. sian_willcocks

    Thread Starter New Member

    Mar 8, 2016
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    I'm wondering if he wants something along these lines as this is the last thing we did. Though I'm not sure how to get this from the graph.
     
  9. WBahn

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    Mar 31, 2012
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    That just tells me that you didn't check it the way I recommended.

    At t=0, Va has an argument of 0° meaning that it will be zero.
    At t=0, Vb has an argument of -70° meaning that it will be negative.

    Therefore Vc = Va + Vb will be negative.

    But at t=0, YOUR Vc will have an argument of +36° meaning that it will be positive.

    Still don't believe me? Simply run the numbers!

    At t=0, Va = 2*sin(0) = 0.0
    At t=0, Vb = 2*sin(-70°) = -1.879

    So at t=0, Vc = -1.879

    But at t=0, YOUR Vc = 3.26 * sin(36°) = 1.916
     
  10. sian_willcocks

    Thread Starter New Member

    Mar 8, 2016
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    When calculating Vb I get a positive answer of 1.87 as subtracting the minus 70 makes it a positive?

    2 minuses make a positive right?
     
  11. WBahn

    Moderator

    Mar 31, 2012
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    Where do you see two minuses?

    What is (0 - 70°) ?

    You are either adding -70 to 0, or you are subtracting 70 from 0.
     
  12. sian_willcocks

    Thread Starter New Member

    Mar 8, 2016
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    I was given the equation 2sin(wt-phi) to work with though on the sheet I was given Vb is written as 2sin (314.2t-70°).

    As I have been given phi as -70° I used 2sin (314.2t- -70°)
     
  13. WBahn

    Moderator

    Mar 31, 2012
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    How were you given that phi was -70°?

    If you were specifically given Vb = 2sin(314.2t - 70°), on what basis can you change that?

    If Vb = 2sin(wt - phi) and if Vb = 2sin(314.2t - 70°), then phi = +70°.

    If you were separately given that, for Vb, phi = -70° then you have contradictory information and you need to confirm what is what with your instructor.
     
  14. sian_willcocks

    Thread Starter New Member

    Mar 8, 2016
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    I am going to contact my tutor as I am now confused as to whether I have used an incorrect figure. I'm still looking for help on the second part! I though I had this part complete
     
  15. WBahn

    Moderator

    Mar 31, 2012
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    Follow the recommendation made by Alec_t in Post #3. The answer drops out of that in just a couple of lines of simple algebra.
     
  16. sian_willcocks

    Thread Starter New Member

    Mar 8, 2016
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    I will give that a go. I've not used it before or even seen it before so this could be interesting!
     
  17. WBahn

    Moderator

    Mar 31, 2012
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    A good rule is to not use identities that you haven't at least been through the proof of at least once. It's not the end of the world if that isn't the case, but you should want to take whatever opportunities you can to improve your math literacy.

    Are you familiar with the more basic/common trig identities

    <br />
\sin\(A \, \pm \, B\) \; = \; \sin \( A \) \cos \( B \) \, \pm \, \cos \( A \) \sin \( B \)<br />
\.<br />
\cos\(A \, \pm \, B\) \; = \; \cos \( A \) \cos \( B \) \, \mp \, \sin \( A \) \sin \( B \)<br />

    If so, then use them to start with the right hand side and see if you can reduce it to the left hand side.
     
  18. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    Has anyone mentioned this yet? If not here is a challenge...

    Start with the general form for two sine (or cosine) waves with phase shifts pha and phb:
    A*sin(x+pha)+B*sin(x+phb)

    Expand that using known forms, then collect terms. With a little imagination we end up with two rotating vectors 90 degrees out of phase, which have properties where we can easily find both the amplitude and phase shift of any two added waves where they both have their individual phase shifts pha and phb as above.
    It's also almost the same for the two added cosines form.
    Very interesting to try if you havent already :)
     
  19. RBR1317

    Active Member

    Nov 13, 2010
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    I tried to recreate your graph, but got this instead. Since the function definition uses a combination of radian and degree measure, the graph uses a time scale such that 360 time units equals 2π radians. Also note that a negative phase shift moves the waveform to the right because it takes more time to overcome the negative phase shift.
     
  20. sian_willcocks

    Thread Starter New Member

    Mar 8, 2016
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    I have had confirmation from my tutor to use it as it's shown on the sheet, not the minus 70 he gave me so I will be starting again and creating a new graph and I shall go from there.

    I shall be back if I am still completely lost!
     
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