Differential equation proof

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Homework Statement


Given a function g(t)=acosωt + bsinωt, where a and b are constants, show that g(t) is the real part of the complex function: keeiωt for some k and Φ
Remark: the complex expression ke is called a phasor. If we know that g(t) has the form kcos(ωt+Φ) then we need know only the constants k and Φ-the amplitude and the phase- to know the function g. Hence we can use the phasor ke as a notation for the function g(t)=keeiωt

Homework Equations


Euler's formula eiωt= cosωt +isinωt

The Attempt at a Solution


Not really sure where to start here except for expanding using euler;s as the first step. any help would be greatly appreciated.
 

Answers and Replies

  • #2
haruspex
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Not really sure where to start here except for expanding using euler;s as the first step.
So do that, for both exponential terms.
 
  • #3
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okay so then I get k(cosΦt +isinΦt)(cosωt+isinωt)
but what should this be equal too?
 
  • #4
haruspex
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okay so then I get k(cosΦt +isinΦt)(cosωt+isinωt)
but what should this be equal too?
How do you get φt terms?
 
  • #5
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umm I'm not sure. I didn't know we would be using that symbol. What does φ represent?
 
  • #6
haruspex
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umm I'm not sure. I didn't know we would be using that symbol. What does φ represent?
It is a constant. You need to find the values of k and φ which make the real part match g().
 
  • #7
haruspex
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umm I'm not sure. I didn't know we would be using that symbol. What does φ represent?
I think perhaps you did not understand my question. In your post #3, you had terms like sin(φt). I don't understand how you got those. Please post your working.
 
  • #8
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Sorry I'm kind of lost. Where did φ come from?
 
  • #9
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In my third post I only haveΦ k ω and t
 
  • #11
haruspex
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no φ
Looks like a font issue. Are you saying Φ and φ look very different to you? In the font that comes up on my iPad they are almost the same, so I may have used the wrong one.
 
  • #12
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yes they look very different. I think thats where the confusion was.
 
  • #13
haruspex
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yes they look very different. I think thats where the confusion was.
Ok, so please reread my posts with that in mind.
 
  • #14
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But in regards to your earlier quastion. I used euler's formula to obtain sin(Φt) eiΦt=cosΦt +isin(Φt)
right?
 
  • #15
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that sin (Φt) is not apart of that calculation
 
  • #16
haruspex
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But in regards to your earlier quastion. I used euler's formula to obtain sin(Φt) eiΦt=cosΦt +isin(Φt)
right?
According to your initial post, it is e, not eiφt.
 
  • #17
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oops, my mistake, than diregard the t's
 
  • #18
haruspex
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oops, my mistake, than diregard the t's
Ok. So multiply out the corrected version of the expression in your post #3.
 
  • #19
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k(cosΦ +isinΦ)(cosω+isinω)
 
  • #20
haruspex
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k(cosΦ +isinΦ)(cosω+isinω)
No, the other exponential term does have a t factor: eiωt.
 
  • #21
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k(cosΦ +isinΦ)(cosωt+isinωt)
 
  • #23
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kcosΦcosωt +kisinΦcosωt + kisinωtcosΦ -ksinΦsinωt
 
  • #24
haruspex
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kcosΦcosωt +kisinΦcosωt + kisinωtcosΦ -ksinΦsinωt
Right. What part of that is relevant in this question?
 
  • #25
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The part that does not have i's in them? the real parts
 
  • #26
haruspex
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The part that does not have i's in them? the real parts
Yes. What values of k and φ make the real part match g?
 
  • #27
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i'm not sure I want to say 0 or π/2 for Φ but that doesnt seem right.
 
  • #28
haruspex
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i'm not sure I want to say 0 or π/2 for Φ but that doesnt seem right.
Remember that t is a variable, so the functions are to produce the same value for every value of t. This means that the cos(ωt) term must look the sameinboth functions, and the sin term must look the same too. That givesyou two equations for the coefficients.
 
  • #29
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well if I equate this to g(t)=acosωt+bsinωt than I would get Φ=π/2 and b for the sin's and Φ=0 and a for the cos's or is my algebra wrong on this?
 
  • #30
haruspex
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ωT)
well if I equate this to g(t)=acosωt+bsinωt than I would get Φ=π/2 and b for the sin's and Φ=0 and a for the cos's or is my algebra wrong on this?
Compare the expression for g with the real part of the complex expression in post #23. Look at the two cos(ωt) terms. If they are to be exactly the same, they must have the same coefficient. That gives you an equation relating a and b to k and φ. Do the same with the two sin(ωt) terms.
 
  • #31
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sorry I am pretty lost here, do you mind working it out for me?
 
  • #32
haruspex
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sorry I am pretty lost here, do you mind working it out for me?
No, I will not do that, and no-one else on this forum should either.
Take the expression you had in post #23 and throw away the imaginary parts. Post what that leaves.
You should have the sum (or difference) of two terms, one with a cos(ωt) in it and the other with a sin(ωt). Call this expression R(t).
Similarly, g(t) has a cos(ωt) term and a sin(ωt) term. We need to arrange that R and g are the same function.
In order that R(t) and g(t) are to be identical functions, the cos term in one must be the same as the cos term in the other. Write that as an equation: (cos term extracted from g(t), complete with its coefficient there) = (cos term from R(t), complete with its coefficient there).
 

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