Phasors — Question 10

Question 10

A nearly closed phasor polygon reveals the direction of an error

A collection of phasors sums exactly to zero. One member is $Z_j=Ae^{i\phi}$. All other members remain fixed while that member acquires either a small fractional amplitude error $\eta$ or a small phase error $\epsilon$.

1. Find the residual phasor to first order for each error separately.

2. Explain how a phase-sensitive residual measurement can distinguish these errors even when their residual magnitudes are equal.

Question 10 - Solution

A nearly closed phasor polygon reveals the direction of an error

Method and mathematical working

Perturb just one term in a sum that originally closes. Subtract its original value to isolate the residual produced by each kind of small error.$$\begin{aligned}\delta Z_{\mathrm{amplitude}}&=\eta Z_j,\\\delta Z_{\mathrm{phase}}&=(e^{i\epsilon}-1)Z_j\approx i\epsilon Z_j.\\[4pt]|\delta Z_{\mathrm{amplitude}}|&=|\eta|A_j,\quad |\delta Z_{\mathrm{phase}}|\simeq|\epsilon|A_j,\\\operatorname{Re}\!\left[(\eta Z_j)^*(i\epsilon Z_j)\right]&=\operatorname{Re}(i\eta\epsilon A_j^2)=0.\end{aligned}$$

(a) Calculation and conclusion

An amplitude change gives $\Delta Z=\eta Ae^{i\phi}=\eta Z_j$. A phase change gives $\Delta Z=Z_j(e^{i\epsilon}-1)\approx i\epsilon Z_j$. Their first-order magnitudes are $A|\eta|$ and $A|\epsilon|$.

(b) Interpretation and limits

The amplitude-error residual lies along the affected phasor, while the phase-error residual is perpendicular to it. Magnitude alone loses this distinction. Near perfect cancellation the resultant phase can change sharply, so reporting only the phase of a very small residual is numerically fragile.

Physical check

The amplitude-error residual is parallel to the perturbed vector, while the first-order phase-error residual is perpendicular. Their different directions provide information that residual magnitude alone loses.