Interference — Question 7

Question 7

Phase jitter washes out a designed interference minimum

At a microphone two equal-amplitude tones have pressure amplitudes $P$ and relative phase $\pi+\epsilon$. The phase error is slowly and uniformly distributed over $[-a,a]$. Average first over carrier cycles and then over phase error.

1. Derive the averaged mean-square pressure as a function of $a$.

2. For $a=\pi/2$, compare the result with the sum of the two separate mean-square pressures. State the small-$a$ limit.

Question 7 - Solution

Phase jitter washes out a designed interference minimum

(a) Method and calculation

For fixed error, the combined mean-square pressure is$$\langle p^2\rangle_{\mathrm{cycle}}=P^2[1+\cos(\pi+\epsilon)]=P^2(1-\cos\epsilon).$$Uniform phase averaging gives$$\overline{\langle p^2\rangle}=P^2\left[1-\frac1{2a}\int_{-a}^a\cos\epsilon\,d\epsilon\right]=P^2\left(1-\frac{\sin a}{a}\right).$$

(b) Interpretation and further calculation

The separate mean-square pressures sum to $P^2$. At $a=\pi/2$, the ratio is$$1-\frac2\pi=0.36338.$$For small $a$, $\sin a/a=1-a^2/6+O(a^4)$, so the residual is $P^2a^2/6$. The null degrades quadratically with small uniformly distributed phase error.

Physical check

The averaging convention matters: $P$ is a peak pressure amplitude, so each isolated sinusoid has mean square $P^2/2$.