Eqn for phase / modulation eqns

[JasonMech]

Hi guys,

I need some help. I have the following modulation equations:
$$\Re:-\zeta a+\lambda a\cos 2\delta-\eta \cos (\delta+\phi)=0$$
and
$$\Im:\omega a+\lambda a\sin 2\delta-\eta \sin (\delta+\phi)=0$$
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:
$$\delta=\arctan \frac{\left(\lambda-\omega\right)\sin\phi-\zeta\cos\phi}{\left(\lambda+\omega\right)\cos\phi-\zeta\sin\phi}$$
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:
$$\tan \left(\delta+\phi\right)=\frac{\omega+\lambda\sin 2\delta}{-\zeta+\lambda\cos 2\delta}.$$

Then, we should obtain an expression for the phase $\delta$ in terms of $\phi$ and through some manipulation we should arrive at $$\delta=\arctan \left[...\right]$$
Can you solve this challenge? (This is NOT classical homework although it might seem so!)

 

[berkeman]

(Thread moved from ME to Calculus & Analysis to get better math views...)

 

[hunt_mat]

Why don't you expand?
$\sin 2x=2\sin x\cos x$ and so forth and you have to write sin and cos in terms of tan.

 

[JasonMech]

Hi hunt_mat,

I have already, along other approaches, done that. With your proposal, i.e. invoking the relations:

$$\sin 2\delta=2\sin\delta\cos\delta,\quad \cos 2\delta=1-2\sin^2\delta$$

and

$$\tan\delta=\frac{\sin\delta}{\cos\delta},\quad \tan(\delta+\phi)=\frac{\tan\delta+\tan\phi}{1-\tan\delta\tan\phi},$$

and substituting these into:

$$\tan \left(\delta+\phi\right)=\frac{\omega+\lambda\sin 2\delta}{-\zeta+\lambda\cos 2\delta}.$$

one obtains:

$$\tan\delta+\tan\phi=\frac{\omega+2\lambda \tan \delta}{-\zeta+\lambda\left(1-2\frac{\tan^2\delta}{\cos^2\delta}\right)}\left(1-\tan\delta\tan\phi\right)$$

This expression can be isolated wrt. \phi with some additional algebraic manipulations:

$$\phi=\arctan\left\{\frac{\omega+2\lambda \tan \delta -\left[-\zeta+\lambda\left(1-2\frac{\tan ^2\delta}{\cos^2\delta}\right)\tan \delta\right]}{\left(\omega+2\lambda \tan \delta\right)\tan \delta-\zeta+\lambda\left(1-2\frac{\tan ^2\delta}{\cos^2\delta}\right)}\right\}.$$

However, I am interested in an expression for the variable \delta in terms of \phi.

I have difficulties isolating \delta in the aforementioned eqn. If you have some ideas that would be highly appreciated.

 

[hunt_mat]

First I did what you did and isolate the $a$ and the $\eta$ and divide through by each other to obtain:
$$\tan \left(\delta+\phi\right)=\frac{\omega+\lambda\sin 2\delta}{-\zeta+\lambda\cos 2\delta}$$
Then I turned everything I could see into tan, so
$$\sin 2\delta =\frac{2\tan\delta}{1+\tan^{2}\delta},\quad\cos 2\delta =\frac{1-\tan^{2}\delta}{1+\tan^{2}\delta}$$
doing this lead me to this equation:
$$\frac{\omega (1+\tan^{2}\delta)+2\lambda\tan\delta}{\lambda -\zeta -(\lambda +\zeta)\tan^{2}\delta}=\frac{\tan\delta +\tan\phi}{1-\tan\delta\tan\phi}$$
From here, cos multiply, rearrange and cross your fingers.

 

[JasonMech]

Have you gotten from:

$$\frac{\omega (1+\tan^{2}\delta)+2\lambda\tan\delta}{\lambda -\zeta -(\lambda +\zeta)\tan^{2}\delta}=\frac{\tan\delta +\tan\phi}{1-\tan\delta\tan\phi}$$

to:

$$\delta=\arctan \frac{\left(\lambda-\omega\right)\sin\phi-\zeta\cos\phi}{\left(\lambda+\omega\right)\cos\phi-\zeta\sin\phi}$$

?

I didnt think about using the trig id:

$$\sin 2\delta =\frac{2\tan\delta}{1+\tan^{2}\delta},\quad\cos 2\delta =\frac{1-\tan^{2}\delta}{1+\tan^{2}\delta}$$

before, so that's a nice comment. I don't know whether you with "cos multiply" meant both in the numerator and denominator or only numerator and whether it should be only on the right hand side of the eqn or on both sides of the eqn. All approaches are valid I guess. I tried all of them of which the firstmentioned writes:

$$\frac{\sin \delta +\cos \delta \tan \phi}{\cos \delta -\sin \delta \tan \phi}=\frac{\omega \left(\cos \delta +\frac{\sin ^2\delta}{\cos \delta}\right)+2\lambda \sin \delta}{\left(\lambda-\zeta\right)\cos \delta -\left(\zeta+\lambda\right)\frac{\sin ^2\delta}{\cos \delta}}$$

which is not, to me, near the sought solution:

$$\delta=\arctan \frac{\left(\lambda-\omega\right)\sin\phi-\zeta\cos\phi}{\left(\lambda+\omega\right)\cos\phi-\zeta\sin\phi}$$

Solving:

$$\frac{\omega (1+\tan^{2}\delta)+2\lambda\tan\delta}{\lambda -\zeta -(\lambda +\zeta)\tan^{2}\delta}=\frac{\tan\delta +\tan\phi}{1-\tan\delta\tan\phi}$$

wrt. $\delta$ and multiplying with $\cos \phi$ (note that this approach is different than the one you proposed; here I solve and subsequently multiply with $\cos \phi$ and not $\cos \delta$) in Maple yields:

$$\delta=\arctan \left\{\frac{\left(\zeta+\lambda\right)\sin \phi +\omega \cos \phi}{-\left(\zeta+\lambda\right)\cos \phi+\omega \sin \phi}\right\}$$

which differs with change in sign and (more importantly) the variables are attached to different trigonometric functions. Thus, I am very curious if you got the correct result? And of course, I am dying to get a hint... :-)

 

[hunt_mat]

It was a typo, cos multiply was meant to be cross multiply. That is as far as I got as I only wanted to get a feel of the problem to give you a hint of how to solve your problem.

My idea was to get an equation for $\tan\delta$ first.