- #1
JasonMech
- 5
- 0
Hi guys,
I need some help. I have the following modulation equations:
[tex]\Re:-\zeta a+\lambda a\cos 2\delta-\eta \cos (\delta+\phi)=0[/tex]
and
[tex]\Im:\omega a+\lambda a\sin 2\delta-\eta \sin (\delta+\phi)=0[/tex]
which, with proper manipulation (i.e. by squaring and adding and/or diving them with each other) should give me an expression for the phase:
[tex]\delta=\arctan \frac{\left(\lambda-\omega\right)\sin\phi-\zeta\cos\phi}{\left(\lambda+\omega\right)\cos\phi-\zeta\sin\phi}[/tex]
I just can't see the link... That is, I cannot reach this expression! By dividing the imaginary part with the real part of the modulation eqns:
[tex]\tan \left(\delta+\phi\right)=\frac{\omega+\lambda\sin 2\delta}{-\zeta+\lambda\cos 2\delta}.[/tex]
Then, we should obtain an expression for the phase $\delta$ in terms of $\phi$ and through some manipulation we should arrive at [tex]\delta=\arctan \left[...\right][/tex]
Can you solve this challenge? (This is NOT classical homework although it might seem so!)
I need some help. I have the following modulation equations:
[tex]\Re:-\zeta a+\lambda a\cos 2\delta-\eta \cos (\delta+\phi)=0[/tex]
and
[tex]\Im:\omega a+\lambda a\sin 2\delta-\eta \sin (\delta+\phi)=0[/tex]
which, with proper manipulation (i.e. by squaring and adding and/or diving them with each other) should give me an expression for the phase:
[tex]\delta=\arctan \frac{\left(\lambda-\omega\right)\sin\phi-\zeta\cos\phi}{\left(\lambda+\omega\right)\cos\phi-\zeta\sin\phi}[/tex]
I just can't see the link... That is, I cannot reach this expression! By dividing the imaginary part with the real part of the modulation eqns:
[tex]\tan \left(\delta+\phi\right)=\frac{\omega+\lambda\sin 2\delta}{-\zeta+\lambda\cos 2\delta}.[/tex]
Then, we should obtain an expression for the phase $\delta$ in terms of $\phi$ and through some manipulation we should arrive at [tex]\delta=\arctan \left[...\right][/tex]
Can you solve this challenge? (This is NOT classical homework although it might seem so!)
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