 Solution

Use the projected baseline along the source offset, $b_p$, and $\alpha =10^{\prime }=2.909\times 10^{-3}\mathrm{rad}$. The first bandwidth null occurs at $\Delta \nu =c/(b_p\alpha )$. A worst-case $b_p=25.46\mathrm{m}$ gives $4.05\mathrm{GHz}$. That is a zero of the response, not a good usable bandwidth. A quantitative maximum needs an allowed attenuation.

For example, requiring a response of at least $0.9$ gives $\mathrm{sinc}x=0.9$ at $x\simeq 0.787$. Hencefor that worst projection. A strictly east-west offset in a sides-aligned square instead samples at most $18\mathrm{m}$, giving a $5.73\mathrm{GHz}$ first null and a $1.43\mathrm{GHz}$ ten-percent-loss bandwidth. The original two-antenna geometry supplies an additional $\cos \theta$ projection; these examples use its maximum magnitude. Thus a bandwidth of order $1\mathrm{GHz}$ is a reasonable conservative estimate with a stated small-loss criterion, rather than a unique geometry-free limit.

To use a much wider total band, channelize the signals before correlation or before irreversible frequency averaging. Correlate sufficiently narrow channels, track the reference delay and phase at each frequency, and combine frequency-aware visibilities or images. This preserves sensitivity while avoiding field-dependent averaging losses in radio interferometry. Merely adding a phase correction after a broad channel has decorrelated cannot recover the lost signal; nor can one tracking delay cancel every sky direction simultaneously.^[ref]