Frequency Content Implied by $\alpha$

The single parameter $\alpha$ was chosen for notational simplicity to represent the rate of linear frequency increase in the chirp signal. We now consider the frequencies actually represented. Since $\alpha n^2$ represents the argument of our analytic signal, the frequencies contained correspond to $\ensuremath{\frac{\Delta (\alpha n^2)}{\Delta n}}= 2\alpha(n-0.5)$.⁴

Since we are in discrete time, the factor $\alpha$ needs to incorporate the sampling rate and the $2\pi$ factor required for representing discrete time frequency in radians per sample. Thus, to consider the actual signal frequency in Hz, we use $a = \ensuremath{\frac{F}{2\pi}}\alpha$ (or $\alpha = \ensuremath{\frac{2\pi}{F}}a$) where $F$ is the sampling frequency and $a$ is the actual signal frequency in Hz. Playing an analogous role to $\alpha$, $a$ is a positive constant equal to half the linear frequency increase in Hz per sample. To consider the actual frequencies of the signal, we thus have $2a(n-0.5)$.

We note that the mathematics require that the frequencies correspond to those not observed at actual time domain samples, but rather the frequencies at points in time between those samples (at $(n-0.5)$). This will not present a problem, as we will eventually use an integral approximation in our analysis whose midpoint rule requires us to use points half way between samples as bounds. This integral has an interval of $-N/2$ to $N/2$ and thus contains frequencies which are $2a(-N/2)$ to $2a(N/2)$ or, simplified, $-a N$ to $a N$. We may calculate the frequency bin numbers corresponding to these values as

$$\fbox{$\displaystyle k_{\max} = \pm\ensuremath{\frac{\alpha K N}{4\pi}}$}$$ (15)

where $K$ is the length of the optionally zero padded FFT. If no zero padding is used, $K=N$ and we clearly have $k_{\max} = \pm \ensuremath{\frac{\alpha N^2}{4\pi}}$. We note that in any representation, $y$ has a ``center frequency'' of zero (DC). In the time domain, this occurs at $n=0$. (Cases of other center frequencies are also discussed below.)

We presently note that $\alpha$ must not be so large as to cause frequencies greater than the Nyquist frequency, namely $\ensuremath{\frac{F}{2}}$. Since the maximum frequencies occur at the edges of the time window, we have then, that $\vert a N\vert< \ensuremath{\frac{F}{2}}$. Thus $a<\ensuremath{\frac{F}{2N}}$ and, recalling our comment about $\alpha$ and $a$ above, $\alpha<\ensuremath{\frac{\pi}{N}}$. These constraints will be important as we progress in the proof, and so we repeat them here:

$$\fbox{$ \displaystyle a < \ensuremath{\frac{F}{2N}}$}$$ (16)

$$\fbox{$ \displaystyle \alpha < \ensuremath{\frac{\pi}{N}}$}$$ (17)

As our proof progresses, we will see that a lower limit on $\alpha$ is also required for model validity. This is addressed in section 3.4.1.