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I find myself confused how two properties of the ambiguity function relate to one another. The properties in question are (presented here from wikipedia, but similar formulas can be found in books by Levanon, Mahafza, etc...)

  1. The maximum value of the ambiguity function is in point $(0, 0)$ $$ \tag{1} |\chi(\tau, f)|^2 \le |\chi(0, 0)|^2 $$

  2. The volume under the ambiguity function is invariant $$ \tag{2} \intop_{-\infty}^{\infty}\intop_{-\infty}^{\infty}|\chi(\tau, f)|^2 d\tau df = |\chi(0, 0)|^2 = E^2 $$

From these 2 properties doesn't it naturally follow that all the volume of the ambiguity function is completely concentrated at point $(0, 0)$? And the only valid form for the ambiguity function to satisfy $(2)$ would be $$ \tag{3} |\chi(\tau, f)|^2 = E^2 \cdot \delta(f) \cdot \delta(\tau) $$ Obviously, this can't be the case. But I can't seem to grasp how $(2)$ could be satisfied otherwise.

UPDATE:

My logic for this reasoning is this following. If someone could walk me through where it should fall apart, that would be helpful. From my perspective:

  1. The first property states that the ambiguity function at point $|\chi(0, 0)|^2$ will equal $E^2$ (or 1, if you normalize it), and at all other points it will be smaller.
  2. The second property states that its entire volume, integrated over $f$ and $\tau$ will equal $E^2$ (or 1, if you normalize it).
  3. Since $|\chi(\tau, f)|^2$ is non-negative, if at any point other than $(0, 0)$ the ambiguity function is non-zero, its volume will be greater than $E^2$, which contradicts the second property.
  4. Hence, the only possible form for the ambiguity function from these two properties is a delta-function.
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  • $\begingroup$ The short answer is no. Your conclusions do not follow. $\endgroup$ Commented Jul 12, 2023 at 20:24
  • $\begingroup$ @AnonSubmitter85 Sorry for the later reply. I've updated my question. If you could walk me through where my logic falls apart, that would be helpful. $\endgroup$
    – Narrava
    Commented Jul 22, 2023 at 9:58
  • $\begingroup$ Why does $3$ follow? If I have a cylinder with a hight of $100(\mathrm{ft})$, does that mean it must have a volume $>100(\mathrm{ft}^3)$? If the the radius of this cylinder were $\delta$, then wouldn't its volume be $0$? $\endgroup$ Commented Jul 24, 2023 at 17:41
  • $\begingroup$ I see your point. Although in the case of an AF, the dimensions will remain the same. Moreover, when integrating over a cylinder, it would be incorrect to write $h = V$, $100 (\mathrm{ft}) = 100(\mathrm{ft}^3)$. By that logic, wouldn't it be incorrect to write $\iint|\chi(\tau, f)|^2 d\tau df = |\chi(0, 0)|^2$, unless we integrate over a $\delta$-function and use its sifting property. On the other hand, the dimensions are the same here. Granted, I've only seen this direct equality on wikipedia, Levanon and Mahafza don't write it out explicitly. $\endgroup$
    – Narrava
    Commented Jul 26, 2023 at 9:47
  • $\begingroup$ I'm not sure what you mean by the radius being $\delta$. Radius is a scalar value, and a $\delta$ is a type of function. Concerning this question, through its sifting property, the $\delta$ would restrict integration to one dimension less. So, two deltas in $f$ and $\tau$ would transform the double integral into a scalar value, as written in property $2$. $\endgroup$
    – Narrava
    Commented Jul 26, 2023 at 9:48

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