Interference of Waves — Question 6

Question 6

A deep null imposes a small phase-error tolerance

Two equal waves of amplitude $A$ are intended to differ in phase by $\pi$, but the actual difference is $\pi+\epsilon$, with $|\epsilon|\le\pi$. Let $P_0$ be the power of one wave.

1. Find the residual amplitude and $P/P_0$.

2. Find the largest allowed $|\epsilon|$ if $P\le.010P_0$. Compare with a small-error approximation.

Question 6 - Solution

A deep null imposes a small phase-error tolerance

Method and mathematical working

Insert the phase error into the exact equal-amplitude formula before making a small-angle approximation. Use the power requirement to bound the magnitude of the error.$$\begin{aligned}A_{\mathrm{res}}&=2A|\sin(\epsilon/2)|,\\\frac{\overline P}{\overline P_0}&=4\sin^2(\epsilon/2)\le0.01,\\|\epsilon|&\le2\arcsin(0.05).\\[4pt]|\epsilon|_{\max}&=2\arcsin0.05=0.10004\,\mathrm{rad},\qquad P/P_0\simeq\epsilon^2.\end{aligned}$$

(a) Calculation and conclusion

$R^2=2A^2[1+\cos(\pi+\epsilon)]=4A^2\sin^2(\epsilon/2)$. Thus $R=2A|\sin(\epsilon/2)|$ and $P/P_0=4\sin^2(\epsilon/2)$.

(b) Interpretation and limits

The bound is $|\epsilon|\le2\arcsin(.05)=.10004\,\mathrm{rad}$, about $5.732^\circ$. For small error, $P/P_0\approx\epsilon^2$, giving $.100\,\mathrm{rad}$. Near a null, the residual power is second order in phase error.

Physical check

Near the selected null, amplitude is first order in $|\epsilon|$ and leakage power is second order. The bound is local to that null; equivalent nulls repeat every $2\pi$.