Complex number exponential subtraction

  • Thread starter Exulus
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  • #1
Exulus
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Hi all,

Im having a bit of trouble with a question. I have to convert:

[tex]Ke^{j\delta} - Ke^{j\psi}[/tex]

Into the form

[tex]re^{j\theta}[/tex]

This is the second part of the question, the first part was an addition instead of subtraction which i managed by using this formula:

[tex]z_1 + z_2 = K(e^{j\delta} + e^{j\psi}) = Ke^{j(\delta + \psi)/2}(e^{j(\delta - \psi)/2} + e^{-j(\delta - \psi)/2}) = 2K\cos((\delta - \psi)/2).e^{j(\delta + \psi)/2}[/tex]

I cant really see where to go with the subtraction though...is it maybe to do with a sin rule? To be honest i dont fully understand the formula above but it was given to us...Ive fiddled around with the maths for a while but its totally headbanging :( Hoping someone can help!
 

Answers and Replies

  • #2
HallsofIvy
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Since you already have a formula for sums, how about using the fact that [itex]e^{i(\psi+ \pi)}= e^{i\psi}e^{i\pi}= -e^{i\psi}[/itex]? In other words, replace [itex]\psi[/itex] with [itex]\psi+ \pi[/itex] in your formula.
 
  • #3
dextercioby
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Okay.It's not difficult.Jst follow the same pattern using the half angle

[tex] D=z_{1}-z_{2}=Ke^{j(\frac{\delta+\psi}{2})}[e^{j(\frac{\delta-\psi}{2})}-e^{j(\frac{\psi-\delta}{2})}]=Ke^{j(\frac{\delta+\psi}{2})}[e^{j(\frac{\delta-\psi}{2})}-e^{-j(\frac{\delta-\psi}{2})}]=[2jK\sin(\frac{\delta-\psi}{2})]e^{j(\frac{\delta+\psi}{2})} [/tex]

Okay?

Daniel.
 
  • #4
Exulus
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brilliant thanks for your help! :)
 
  • #5
ph23ms
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URGENT: Can anyone help me with this?

Can anyone tell me what difference it makes to the subtraction/addition formulae when the base factors are not the same? i.e

[TEX]D=z_{1}-z_{2}=Ke^{j(\delta)}-Le^{j(\psi)}[/TEX]

I have fiddled with the above formulae, but cant seem to get anything sensible looking from it... And I need to solve this quite urgently!!!

Thanks in advance...
 
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