# Time-bandwidth product and uncertainty on harmonics

For a signal that is not corrupted with noise, how is the time-bandwidth product affecting the uncertainty or error introduced when extracting the fundamental frequency ? I mean, instead of choosing the fundamental frequency of the signal at $f_0$, you choose a harmonic $f_k$ at $k\cdot f_0$ and divide by $k$ to get $f_0$. How is the uncertainty related to $k$ for a given frequency resolution ?

• BT product? What the heck is that?
– jojeck
Jun 19, 2014 at 22:39
• BT product is the time-bandwidth product or bandwidth-time product. The former appellation is more common in literature. Jun 19, 2014 at 22:51
• Oh, you mean $BT$ product - now we know it's math.
– jojeck
Jun 19, 2014 at 22:52
• i don't really get the question. the time-bandwidth product has more to say about how the uncertainty of the frequency measurement (doesn't matter what the frequency actually is, just the uncertainty of it) is related to how long, in time, the signal is. for $T$ seconds (or whatever time units) of data, you get the same uncertainty (in Hz if $T$ is in seconds) of the frequency estimate. doesn't matter if it's $f_0$ or $k \cdot f_0$. same width of uncertainty band of frequencies, same length of time. Jun 20, 2014 at 4:08
• @robertbristow-johnson : it is indeed true that the uncertainty is that of the frequency measurement. R. B. Randall in his book mentions the standard error being inversely proportional to the time-bandwidth product. Doesn't this have an effect if you had to take relative bandwidths of the different harmonics $k\cdot f_0$ ? Jun 20, 2014 at 9:35

If you a-priori assume that $k \cdot f$ is a stationary harmonic of periodic waveform of frequency $f$, that implies that you don't have any separable time locality of that harmonic. But any tighter bandwidth of a frequency estimation might come from the a-priori assumptions regarding the stationarity of the harmonics, not from the data.