Complex to polar form

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Homework Statement



Show that the solution x(t) = Ge^(iwt), where G is in general complex, can be written in the form x(t) = Dcos(wt - [tex]\delta[/tex]).

D(w) and [tex]\delta[/tex](w) are real functions of w.

Homework Equations



z = Ae^(i[tex]\phi[/tex])

The Attempt at a Solution



So I know I should start by writing G in polar form. I am confused though as to how to go to polar form with just the G. Is it simply just Ge^(i[tex]\phi[/tex]). Then, I could use Euler's formula to write:

Ge^(i[tex]\phi[/tex]) = Gcos([tex]\phi[/tex]) + iGsin([tex]\phi[/tex]).

I am not sure where this gets me. Any help on where to go from here or if this is even correct would be much appreciated.
 

Answers and Replies

  • #2
tiny-tim
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hi w3390! :smile:

(have a delta: δ and a rho: ρ and a phi: φ and an omega: ω :wink:)

Homework Statement



Show that the solution x(t) = Ge^(iwt), where G is in general complex, can be written in the form x(t) = Dcos(wt - [tex]\delta[/tex]).

D(w) and [tex]\delta[/tex](w) are real functions of w.

but that's obviously not true …

the RHS is real, but x isn't :confused:
 
  • #3
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Why isn't x real?
 
  • #4
tiny-tim
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it's the product of two complex numbers … it's very unlikely to be real
 
  • #5
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I am confused then because my question was from a test prep sheet from my professor. Should I perhaps only consider the real part of x(t)?
 
  • #6
tiny-tim
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dunno :redface:

maybe :smile:
 
  • #7
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What I'm saying is:

x(t) = Ge^(i[tex]\phi[/tex])

x(t) = G[cos([tex]\omega[/tex]t - [tex]\delta[/tex]) + i*sin([tex]\omega[/tex]t - [tex]\delta[/tex])

Then taking only the real part of this:

x(t) = Gcos([tex]\omega[/tex]t - [tex]\delta[/tex]).

From here, I can compare to the given solution of x(t) = Dcos([tex]\omega[/tex]t - [tex]\delta[/tex]) and say that G = D.

Does this make sense?
 
  • #8
tiny-tim
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Hi w3390! :smile:

(what happened to that δ φ and ω i gave you? :confused:)

I don't understand where your second line came from …

x(t) = G[cos([tex]\omega[/tex]t - [tex]\delta[/tex]) + i*sin([tex]\omega[/tex]t - [tex]\delta[/tex])
 
  • #9
HallsofIvy
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Write [itex]G= re^{i\theta}[/itex]. Then [itex]Ge^{i\omega t}= r e^{i(\omega t+ \theta)}= r (cos(\omega t+ \theta)+ i sin(\omega t+ \theta)[/itex].
 

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