If the Fourier transform of the function $f(x, y)$ is $F(m, n)$, then the Fourier transform of the function $f(2x, 2y)$ is :
- $\cfrac{1}{4} F(m/2,n/2)$
- $\cfrac{1}{4}F(2m,2n)$
- $\cfrac{1}{4} F(m,n)$
- $\cfrac{1}{4} F(m/4,n/4)$
My attempt: somewhere it explained as following:
$f(x) =x$
$f(k) = \cfrac{1}{2 \pi}\int x e^{ikx} dx$
Let $f(2x) = 2x $
$f(k') = \cfrac{1}{2 \pi}\int 2 x e^{ikx} dx= 2 f(k)$
Similarly for $y$.
$f(m',n') = 4 f(m,n)$
But, I didn't get this. Official key is given option $(1)$.
Can you explain it, please?
I will try to make my arguments as heuristic and intuitive as possible, at the expense of absolute rigour. Think of it logically - what is the relation of the function $f(2x)$ to $f(x)$? $f(2x)$ is $f(x)$ compressed by a factor of 2. If we think of $x$ as time, this means that it will take half the time to reach an "identical" point on each function. In effect, the period will be halved. Since the period and frequency are reciprocals, the frequency will double. This means that $m \mapsto 2m$. For the function to take a parameter of $2m$ and returm $m$, $2m$ must be multiplied by $1/2$. Therefore the associated frequency function for $f(2x)$ is $F(1/2m)$.
To determine the factor by which the frequency changes, recall $F(\omega)=\int_{-\infty}^{+\infty}f(x)e^{2\pi i x \omega} dx$ Replace $x$ with $2x$. Then, $F(\omega)=\int_{-\infty}^{+\infty}f(2x)e^{4\pi i x \omega} dx$. Let $2x=u$. Then, $\frac{du}{dx} = 2$ or $dx = \frac{du}{2}$. Substituting gives $F(\omega)=\frac{1}{2} \int_{-\infty}^{+\infty}f(u)e^{2\pi i u \omega} du$. But this integral is the same as the defining integral for the fourier transform for $f(x)$ with the $x$'s replaced with $u$'s! Therefore, $F(\omega)=\frac{1}{2}F(\omega ')$. But we know $\omega ' = m/2$, So $F(\omega)$ for $f(2x)$ is $\frac{1}{2}F(m/2)$. By similar argument, for $f(2y)$, $F(\omega)=\frac{1}{2}F(n/2)$. Combining the functions gives the fourier transform, for $f(2x,2y)$, as $\frac{1}{4}F(m/2,n/2)$.