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Find forced response of VC(t) - Can someone check my work?

  1. Nov 14, 2016 #1
    1. The problem statement, all variables and given/known data
    upload_2016-11-13_23-54-31.png
    Find the forced response of VC(t)
    2. Relevant equations


    3. The attempt at a solution
    W=1000
    [tex]\frac{d^2V_C(t)}{dt^2}+\frac{R}{L}\frac{dV_C(t)}{dt}+\frac{V_C}{LC}=\frac{Vin}{LC}[/tex]
    [tex]V_C(t)=Acos(1000t+\theta)[/tex]
    Plug in Vc(t),R,L,C into equation 1

    [tex]-10^6Acos(10^3t+\theta)-10^7Asin(10^3t+\theta)+10^7Acos(10^3t+\theta)=5*10^7cos(10^3t)[/tex]

    Simplifying:
    [tex](9*10^6)Acos(10^3t+\theta)-10^7Asin(10^3t+\theta)=5*10^7cos(10^3t)[/tex]

    Trig ID
    [tex]\sqrt{(9*10^6)^2+(10^7)^2}*Acos(10^3t+tan^-1(-\frac{9*10^6}{-10^7}))[/tex]

    [tex]13.45*10^6Acos(10^3t+41.98°)[/tex]


    Is everything I've done so far correct? How do I find A?
     
  2. jcsd
  3. Nov 14, 2016 #2

    NascentOxygen

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    Staff: Mentor

    Above this line you have an equation, below this line you should also write an equation.

    Above this line your expression contains a ##\theta## term, but its simplified expression below that line has lost the ##\theta## term. Can you restore the missing ##\theta## into that simplified equivalent expression?

    The amplitude of the cosinusoid on the left equates to the amplitude of that on the right, so set them equal and solve to find ##A##.
     
  4. Nov 14, 2016 #3
    Is this correct?
    [tex]\sqrt{(9*10^6)^2+(10^7)^2}*Acos(10^3t+\theta+tan^-1(-\frac{9*10^6}{-10^7}))=5*10^7cos(10^3t)[/tex]
    [tex]13.45*10^6Acos(10^3t+\theta+41.98°)=5*10^7cos(10^3t)[/tex]
     
  5. Nov 14, 2016 #4

    NascentOxygen

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    Staff: Mentor

    I'm not able right now to verify your mathematics, but that is the form that I expect.

    So equate the left side cosine magnitude to that on the right side, and solve for A.
    Similarly, equate the cosine arguments, and solve for θ.​
     
  6. Nov 14, 2016 #5

    NascentOxygen

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    Staff: Mentor

    One trap to be aware of is that tan–1 can have multiple answers. Your calculator will give you only one, but you always need to consider whether the other/s are applicable to your particular problem.
     
  7. Nov 14, 2016 #6
    So
    13.45*10^6 A = 5*10^7
    and
    theta + 41.98 = 0
    ?
     
  8. Nov 15, 2016 #7
    The formula is tan^-1(-A/C)
    if A is 9 and C is -10, wouldn't it be:
    tan^-1(-9/-10)
    tan^-1(9/10)
    =41.98
     
  9. Nov 15, 2016 #8

    NascentOxygen

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    Staff: Mentor

    That's how it's done, yes.
    Geometrically, you need to look at all 4 quadrants when determining the answer to tan –1 (0.9) because there are at least 2 answers.

    Mathematically, you can say there are an infinite number of solutions, but in electronic circuits, we usually consider those phase angles restricted to the range: –180°....+180°
     
  10. Nov 15, 2016 #9

    gneill

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    Just a heads up, I think that somewhere along the line the 9/10 should have been 10/9. I haven't checked closely through the math to see exactly how this happened, but l suspect it would be where the "Trig ID" took place.

    I say that it should have been 10/9 because solving for Vc via another method yields a phase angle of -48.01°, which corresponds to atan(-10/9).
     
  11. Nov 15, 2016 #10
    Can you show the other method?
    My professor wrote this trig identity on this board:
    Acos(x)+Csin(x)=sqrt(A^2+C^2)*cos(x+tan^-1(-A/C))

    but this website: http://myphysicslab.com/springs/single-spring/trig-identity-en.html
    upload_2016-11-15_9-15-15.png
     
  12. Nov 15, 2016 #11

    gneill

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    Staff: Mentor

    I used phasors (which uses complex values for impedances) and standard circuit analysis methods. All the usual analysis methods work with phasors (Ohm's law, KVL, KCL, nodal analysis, mesh analysis, etc.), so it's just like solving a circuit with resistors except it uses complex numbers.
    The trig identity is fine. Your professor's negative sign is merely accounting for the sign of the sine term in the problem rather than making the value of ##a## negative.

    The issue will be with why you found A/C to be 9/10 rather than 10/9.
     
  13. Nov 15, 2016 #12
    upload_2016-11-15_9-30-7.png
    Isn't A = 9*10^6 and C = -10^7?
     
  14. Nov 15, 2016 #13

    gneill

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    Staff: Mentor

    I'm thinking that perhaps the issue is with applying the trig identity correctly. The underlying trig identity being invoked is:

    ##cos(a + b) = cos(a) cos(b) - sin(a) sin(b)##

    and we want to associate cos(a) and sin(a) with the coefficients in your equation. Let's start by getting rid of the large exponents in your equation. Dividing through by ##10^7## and letting ##b = (ω t + θ)## for now yields:

    ##\frac{9}{10} A cos(b) - 1~A sin(b) = 5 cos(ω t)##

    Then the 9/10 and 1 are associated with the cos(a) and sin(a) of the trig identity. That means that the angle a is given by:

    ##a = tan^{-1} \left( \frac{sin(a)}{cos(a)} \right) = tan^{-1} \left( \frac{1}{9/10} \right) = tan^{-1} \left( \frac{10}{9} \right)##

    which is the 10/9 that I found by the different approach.
     
    Last edited: Nov 15, 2016
  15. Nov 15, 2016 #14
    final answer:
    Vc(t)=3.7cos(10^3t-48)
     
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