Dear all,
In wideband digital phase array radars, how the true-time delays are realized? If the true-time delays are realized by the digital signal processing, for example, the signal is s(n) = [1 2 3 4], and the sample interval is 80 ps (the corresponding sampling rate is 12.5 Gbps ), is the minimum time-delay we can realize for s(n) is 80 ps? And the delayed signal is s1(n) = [4 1 2 3]? Does it mean that the time-delay precision is related with the sampling rate? And are there any other parameters related with the time-delay precision? Thanks very much for answering this question.
Dear Sir,
In radar, the time-delay precision is related to the bandwidth of the waveform, and the signal/noise ration. Basically,
- The resolution power (the ability of the radar to distinguish two objects close to each other, one more distant than the other), is related to the bandwidth. The resolution power is proportional to the inverse of the bandwidth.
- The resolution precision can be increased by searching the maximum of the echo, given good signal/noise. Hence, the resolution precision is proportional to the signal/noise ratio and to the inverse of the bandwidth.
- Note that bandwidth here is RF bandwidth, not IF bandwidth. Between RF and IF there is usually a mixer, and depending on the kind of the radar, the IF bandwidth is usually lower than the RF bandwidth. When A/D conversion is done, the sampling is done on the IF, and the A/D convertor sampling period does not determine the resolution precision nor the resolution power.
- An example of the latter is chirp radar, where the waveform sent out is a chirp, eg a relatively long pulse of constantly increasing frequency. The IF is obtained by mixing the return signal with the transmit signal. The IF can be relatively narrow band, while the RF can be wideband.
Best Regards,
Henri.
