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Differential equation proof

  • Thread starter Dusty912
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  • #1
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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
 
  • #10
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no φ
 
  • #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)
 
  • #22
haruspex
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k(cosΦ +isinΦ)(cosωt+isinωt)
Right. Now multiply out.
 
  • #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
 

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