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I edited the initial question to make it general and thus applicable in more situations.

If the Fourier Transform of $x(t^2)$ is some function of $\omega^2$. Then what is the Fourier Transform of $x(bt^2)$, for $b>0$?

Is it $\dfrac{1}{b}$ times a function of $\left(\dfrac{\omega}{b}\right)^2$, or $\left(\dfrac{1}{\sqrt{b}}\right)$ times a function of $\left( \dfrac{\omega}{\sqrt{b}} \right)^2$, or neither?

Thank you for your time.

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  • $\begingroup$ i wasn't the one who down-arrowed this question, but Over needs to pose it in a manner that isn't made up $\endgroup$ – robert bristow-johnson May 30 '14 at 22:32
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Let $y(t) = x(t^2)$, and $a$ be such that:

$$ \begin{align} y(at) &= x((at)^2) \\ &= x(a^2t^2) \\ &= x(bt^2) \end{align} $$ In other words, $a^2 = b$.

The Fourier Transform time-scaling property states that given the Fourier Transform pair: $$ \begin{align} y(t) \leftrightarrow Y(w) \end{align} $$ and $a>0$, we have the relation: $$ \begin{align} y(at) &\leftrightarrow \frac{1}{a}Y(\frac{w}{a}) \end{align} $$

Thus, $$ \begin{align} x(bt^2) &\leftrightarrow \frac{1}{a}Y(\frac{w}{a}) \\ &\leftrightarrow \frac{1}{\sqrt{b}}Y(\frac{w}{\sqrt{b}}) \end{align} $$

Now, if we define $$ \begin{align} Y(w) = G(w^2) \end{align} $$ Then, $$ \begin{align} Y(w/a) &= G(w^2/a^2) \\ Y(\frac{w}{\sqrt{b}}) &= G(\frac{w^2}{b}) \end{align} $$

Yielding, $$ \begin{align} x(bt^2) &\leftrightarrow \frac{1}{a}Y(\frac{w}{a}) \\ &\leftrightarrow \frac{1}{\sqrt{b}}G(\frac{w^2}{b}) \\ &\leftrightarrow \frac{1}{\sqrt{b}}G((\frac{w}{\sqrt{b}})^2) \end{align} $$ Or, in prose (1/sqrt(b)) times a function of (w/sqrt(b))^2.

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If $x(t)$ is a time signal, then its Fourier transform can be taken to be $$X(f) = \int_{-\infty}^\infty x(t) \exp(-j2\pi ft)\,\mathrm dt.$$ But what is meant by $x(t)$? Well, $x(t)$ is a real function of a real variable $t$ by which is meant that for each (real number) value of $t$, the function $x(\cdot)$ assigns a real number that we denote by $x(t)$. Usually, $x(t)$ is described by a formula such as $x(t) = \exp(-\pi |t|)$ which says in words that we

take the absolute value of whatever is inside the parentheses, multiply it by $-\pi$ and raise $e$ to the result of the multiplication.

With this, we consider the function $y(t)$ whose value at any $t$ is defined to be the value of $x(t^2)$ where $x(t)$ happens to be $\exp(-\pi |t|)$. Thus, $y(t) = \exp(-\pi t^2)$ whose Fourier transform happens to be $\exp(-\pi f^2)$. This matches what the OP wants: the "function $x(t^2)$ has Fourier transform that is a function of $f^2$."

But now, let's turn the problem around. We are not told what the function $x(t)$ is, but we do know that $y(t) = x(t^2)$ happens to equal $\exp(-\pi t^2)$. How can we deduce that $x(t) = \exp(-\pi |t|)$? We can't because if $x(t)$ were chosen to be $\exp(-\pi t)$ for which the prescription is

take the absolute value of whatever is inside the parentheses, multiply it by $-\pi$ and raise $e$ to the result of the multiplication.

then $x(t^2) = \exp(-\pi t^2)$ also. So, did our $y(t) = \exp(-\pi t^2)$ come from $x(t) = \exp(-\pi |t|)$ or $x(t) = \exp(-\pi t)$?

In short, asking about the Fourier transform of $x(bt^2)$ is futile unless you specify what you mean by $x(t)$.

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