Phasors — Question 6

Question 6

Two phase-controlled measurements recover an unordered amplitude pair

Two unknown positive amplitudes $A,B$ give resultant amplitude $7\,\mathrm{mm}$ when in phase and $5\,\mathrm{mm}$ when in quadrature.

1. Find $A$ and $B$ up to interchange.

2. What phase measurement in the quadrature setting would tell which source has amplitude $A$, if $A$ defines zero phase and $B$ leads by $\pi/2$?

Question 6 - Solution

Two phase-controlled measurements recover an unordered amplitude pair

Method and mathematical working

Use the in-phase measurement for the sum of amplitudes and the quadrature measurement for the sum of their squares. Subtract to recover their product.$$\begin{aligned}A+B&=7,\qquad A^2+B^2=25,\\2AB&=49-25=24,\qquad z^2-7z+12=0.\\[4pt](z-3)(z-4)&=0\Rightarrow\{A,B\}=\{3,4\}\,\mathrm{mm},\\A=3,\ B=4&:\quad\arg(A+iB)=\arctan(4/3)=0.92730\,\mathrm{rad}.\end{aligned}$$

(a) Calculation and conclusion

In phase, $A+B=7$. In quadrature, $A^2+B^2=25$. Hence $2AB=49-25=24$, so $AB=12$. The roots of $z^2-7z+12=0$ are $3$ and $4\,\mathrm{mm}$.

(b) Interpretation and limits

The quadrature phasor is $A+iB$, with angle $\phi=\arctan(B/A)$. A phase $.9273\,\mathrm{rad}$ selects $A=3,B=4$; $.6435\,\mathrm{rad}$ selects $A=4,B=3$. Magnitude-only measurements cannot label the two sources.

Physical check

The roots are $3$ and $4$, but the two measurements are symmetric under exchanging source labels. Additional phase information is required to assign either amplitude to a named source.