# How are time resolution and signal bandwidth related?

I am confused by the dual concepts of time-resolution and bandwidth. Often I will hear that a pulse-compressed radar application 'doesnt have enough BW' for some specific time resolution that is sought.

Isnt the maximum time resolution simply the reciprocal of your sampling rate?

How are those concepts related?

• In a zero noise scenario, once you have enough samples to uniquely identify your time-limited signal, you can then interpolate and get sub-sample resolution (e.g. better than 1/Fs). – hotpaw2 Oct 17 '11 at 15:39

Dilip's points in his answer are correct. Speaking more to the context that you referenced of pulse-compression radar, I think you're getting confused by differing meanings of the oft-used word "resolution." In a broad signal processing sense, your time resolution is defined to some extent by your sample rate. But in the specific problem domain of building a radar receiver, you are concerned with being able to identify multiple echoes from distant objects and precisely observe their arrival time. "Resolution" in this context refers to resolving and separating multiple received echoes so that they can be processed independently.

A typical radar receiver uses a sliding cross-correlator to locate echoes from objects that have reflected the transmitted radar signal. The receiver knows the format of the transmitted pulse, so cross-correlation between the RF receiver's output and the transmitted pulse waveform is the optimum scheme for detecting the presence of reflected pulses in AWGN. The correlator output will contain copies of the transmitted pulse waveform's autocorrelation function (which typically has a sinc-like shape) for each received echo, shifted in time based on the range to the target that caused the reflection. In order to discriminate between the targets, their corresponding lobes at the correlator output must be sufficiently separated in time.

A "high-resolution" radar is able to finely discriminate between multiple targets in the range dimension. If your radar has multiple targets at approximately the same range, then their echoes will reach the receiver at nearly the same time. Therefore, their autocorrelation lobes will appear at the output of the correlator at nearly the same time. The ability of the radar to discriminate between the echoes depends upon the time duration of the waveform's autocorrelation lobes; a narrower autocorrelation function (ideally one that looks like an impulse) is better.

This long-winded introduction brings us to the idea of pulse compression. Pulse-compression radar waveforms are typically implemented using linear frequency modulation (also known as "chirping"); instead of transmitting a constant-frequency pulse, the transmitted frequency is swept linearly over the course of the pulse. In practice, the sweep can be done over tens or even hundreds of MHz of spectrum. What's the benefit? An autocorrelation function with nice properties:

$$<s_{c'}, s_{c'}>(t) = T \Lambda \left(\frac{t}{T} \right) \mathrm{sinc} \left[ \pi \Delta f t \Lambda \left( \frac{t}{T}\right) \right] e^{2 i \pi f_0 t}$$

The equation above is borrowed from the Wikipedia article; I'll defer the full explanation to that source. What's important here is the $\Delta f$ term; it refers to the amount of frequency covered by the linear frequency chirp. Since $\Delta f$ is a factor in the argument of the $\mathrm{sinc}$ function, it can easily be seen that by chirping over a larger bandwidth, the main lobe of the pulse's autocorrelation function will be narrower. Narrower lobes are more easily discriminated by the radar receiver, giving such a radar high "resolution" in terms of differentiating between similarly-ranged targets.

Just to wrap up a bit, this sort of finding should make sense. Recall that a wide-sense stationary signal's power spectral density can be defined as the Fourier transform of its autocorrelation function. The ideal autocorrelation function for a radar pulse would be an impulse; separating a bunch of echoes with "zero" width is easier than separating a bunch of more broad lobes. The Fourier transform of an impulse has infinite frequency extent. Qualitatively, it follows that autocorrelation functions with very short time extent would be comparatively wideband in the frequency domain. This is the basis for the oft-used rule of thumb in detection and estimation theory that you need a high-bandwidth signal in order to make high-resolution time-of-arrival measurements.

• Jason thanks for the reply, one Q before some more follow up - let me try to rephrase one question just to make sure I get this right ... What type of signal should I use such that I can resolve (1/fs) seconds? – Spacey Oct 16 '11 at 5:54
• I think you're out of luck. Having resolution of $\frac{1}{f_s}$ would imply that you can discriminate between peaks that are separated by only one sample. That implies that there would be no visible separation between them in your sample stream (it would just look like a two-sample-wide peak). While you may be able to construct a waveform that has an impulsive-enough shape to only be one sample wide at the correlator output, in any practical situation, you're not likely going to rely upon each peak only being one sample wide; it's not likely to be very robust. – Jason R Oct 16 '11 at 14:18
• @Dilip: you're completely right. The ideal autocorrelation function of a PN sequence looks like an impulse surrounded by very small sidelobes. In practice, though, you can't count on seeing an exactly one-sample-wide peak because of finite receiver bandwidth and nonzero timing offset. And yes, they are typically broadband, hence their use in spread spectrum systems. – Jason R Oct 16 '11 at 23:28
• @JasonR The question is trying to establish an upper bound on the maximum time resolution power possible, given a particular sampling rate. (Ignore robustness issues for now - I am talking theoretically). Is the theoretical maximal time resolution available bounded by 1/fs, taking a value of anything between 0 and 1/fs depending on your BW? – Spacey Oct 20 '11 at 0:52
• In a noiseless, interference-free, loss-free (i.e. no gain variations between different signals) environment, you could theoretically resolve two signals separated by $\frac{1}{f_s}$ (i.e. one sample), provided they have suitable autocorrelation functions (like the PN sequences Dilip referenced before). Not very practical, but in theory, you could. – Jason R Oct 20 '11 at 1:27

You don't have usually have a resolution of 1/Fs, because you don't know if it's your signal or not from a single sample. How long does it take to determine that it's your 100 MHz signal and not somebody else's 101 MHz signal? Longer than knowing it's your 100 MHz signal and not somebody else's 110 MHz signal. The closer the enemy signal to be excluded (thus the narrower the bandwidth allowed for your signal), the longer it takes to differentiate friend or foe, which leads to a worse time resolution.

If you could tell from one sample, that implies you're accepting an infinite bandwidth, which is just one corner case of the time-bandwidth dual.