Modifications to Proof for Decreasing Frequency Chirp

Having achieved our proof's goal for the increasing frequency case just above, we now wish to apply a similar analysis to the case of a linearly decreasing frequency chirp. It is not possible to simply substitute $\alpha<0$ in the above analysis, because doing so introduces imaginary numbers in the the $\sqrt{\frac{\pi}{2\alpha}}$ factors, preventing us from using the same real Fresnel properties used above. Thus we will slightly change our definitions and procedures.

We begin by redefining $\alpha$ so that it represents a decreasing frequency characteristic. Formally, we replace the initial reference to $\alpha$ in equation 14 with $-\alpha$ and require that $\alpha$ again be positive so that $-\alpha$ is always negative. Thus our new time domain signal is:

$$y_d(n) = w(n) e^{-i \alpha n^2}.$$ (91)

where the subscript $d$ denotes that this function represents a decreasing frequency chirp as opposed to the original increasing frequency chirp. Proceeding as above, equations 24 and 25 become

$$\Re{Y_{da}(k)} = \int_{-N/2}^{N/2}\cos( -\alpha n^2 -2\pi k n/ K) dn \approx \sum_{n=-(N-1)/2}^{(N-1)/2}\cos(- \alpha n^2 -2\pi k n/ K)$$ (92)

$$\Im{Y_{da}(k)} = \int_{-N/2}^{N/2}\sin( -\alpha n^2 -2\pi k n/ K) dn \approx \sum_{n=-(N-1)/2}^{(N-1)/2}\sin( -\alpha n^2 -2\pi k n/ K).$$ (93)

We presently consider $\Re{Y_{da}(k)}$ (equation 95). We observe that $\cos( -\alpha n^2 -2\pi k n/ N) = \cos( +\alpha n^2 +2\pi k n/ N)$ due to the evenness of the cosine function. Because the sum or integral we are considering is over a window or interval symmetric in the variable $n$, it is arbitrary whether we sum/integrate over $\cos( +\alpha n^2 +2\pi k n/ N)$ or $\cos( +\alpha n^2 -2\pi k n/ N)$. Thus,

$$\Re{Y_{da}(k)} = \int_{-N/2}^{N/2}\cos( -\alpha n^2 -2\pi k n/ K) dn$$ (94)

$$= \int_{-N/2}^{N/2}\cos( \alpha n^2 -2\pi k n/ K) dn$$ (95)

and the rest of the analysis for the real part is identical to that given above, so that for the Hann window case,

$$\Re{Y_{da}(k)} = \Re{Y_a(k)} \approx \sqrt{\frac{\pi}{2\alpha}} \left(\ensuremath{... ...2 k^2}{K^2\alpha}\right) + \sin\left(\frac{\pi^2 k^2}{K^2\alpha}\right)\right).$$ (96)

We consider $\Im{Y_{da}(k)}$ (equation 96) in a similar fashion. Because sine is an odd function, $\sin( -\alpha n^2 -2\pi k n/ N) = -\sin( +\alpha n^2 +2\pi k n/ N)$. Again, since the sum or integral we are considering is over a window or interval symmetric in the variable $n$, it is arbitrary whether we sum/integrate over $\sin( +\alpha n^2 +2\pi k n/ N)$ or $\sin( +\alpha n^2 -2\pi k n/ N)$. Thus,

$$\Im{Y_{da}(k)} = \int_{-N/2}^{N/2}\sin( -\alpha n^2 -2\pi k n/ K) dn$$ (97)

$$= - \int_{-N/2}^{N/2}\sin( \alpha n^2 -2\pi k n/ K) dn$$ (98)

reflecting only a sign change from $\Im{Y_a(k)}$ in the original proof. This gives

$$\Im{Y_{da}(k)} = -\Im{Y_a(k)} \approx \sqrt{\frac{\pi}{2\alpha}} \left(\ensuremath... ...2 k^2}{K^2\alpha}\right) - \cos\left(\frac{\pi^2 k^2}{K^2\alpha}\right)\right).$$ (99)

We may now calculate the phase. To do so, we use the result in equation 91. Since we know that

$$\angle{Y_a(k)} = \arctan\left(\frac{\Im{Y_a(k)}}{\Re{Y_a(k)}}\right) = \frac{\pi}{4}-\frac{\pi^2 k^2}{K^2\alpha},$$ (100)

we may conclude that

$$\angle{Y_{da}(k)} = \arctan\left(\frac{\Im{Y_{da}(k)}}{\Re{Y_{da}(k)}}\right) = \arct... ...{Y_{a}(k)}}{\Re{Y_{a}(k)}}\right) = -\frac{\pi}{4}+\frac{\pi^2 k^2}{K^2\alpha}.$$ (101)

This is an upward-pointing parabola, as we sought. We repeat our conclusion and conditions for this decreasing frequency case:
$\parbox{6.5in}{\begin{eqnarray} \hspace{8cm} \alpha > 0 \nonumbe... ...ase)}\quad \\ \vert k\vert \leq b \cdot k_{\max} \nonumber \end{eqnarray}}$

We note that $\vert Y_{da}(k)\vert$ is identical to $\vert Y_a(k)\vert$ as given in equation 93:

$$\fbox{$ \displaystyle \vert Y_{da}(k)\vert = \vert Y_a(k... ...left(\ensuremath{\frac{2\pi^2 k}{KN\alpha}}\right)\right) $}$$ (102)