Scaling property of Fourier Transform

Question

Problem 4.6(b) from Oppenheim, Wilsky & Nawab (2nd ed) reads:

Given that $x(t)$ has the Fourier transform $X(j\omega)$, express the Fourier transform of $x(3t - 6)$ in terms of $X(j\omega)$.

The way I proceeded, $$ x(3t - 6) = x(3(t-2)) $$

$$ x(t) \longleftrightarrow X(j\omega)$$ $$ \implies x(t-2) \longleftrightarrow X(j\omega)e^{-2j\omega}$$ $$ \implies x(3(t-2)) \longleftrightarrow \frac{1}{3}X(\frac{j\omega}{3})e^{\frac{-2j\omega}{3}}$$ But the answer given at the back of the book is $$ x(3(t-2)) \longleftrightarrow \frac{1}{3}X(\frac{j\omega}{3})e^{-2j\omega}$$

My question is, while applying the scaling property of Fourier Transform, why does scaling happen only to the $X(j\omega)$ term and not the exponential term, when even the latter forms part of the actual $X(j\omega)$ on RHS?

Answer 1

Be careful about the orders of the argument transforms: When you interpret the argument $3t-6$ as $3\cdot(t-2)$, you should first scale (compress) $x(t)$ by $3$ along $t$ axis, and then shift the scaled result by $2$ to the right.

Let $x(t)$ be the original signal whose CTFT is $X(\omega)$, then define $y(t)$ as

$$ y(t) = x(3t) \tag{1}$$ and define $z(t)$ as $$z(t) = y(t-2) \implies z(t) = x(3(t-2)) = x(3t-6) \tag{2}$$

Notice how we replaced $y(t-2)$ by using its definition from the eq. (1).

Now applying CTFT to the above rows yields:

$$ Y(\omega) = \frac{1}{3} X(\omega/3) \tag{3} $$ and $$ Z(\omega) = Y(\omega) e^{-j2\omega} \tag{4} $$

finally insert $Y(\omega)$ from eq. (3) into eq. (4) to get:

$$ \boxed{Z(\omega) = Y(\omega) e^{-j2\omega} = \frac{1}{3} X(\omega/3) e^{-j2\omega} }$$

By the way, you could have also used the alternate approach: first shift $x(t)$ by $6$: $$y(t) = x(t-6) \implies Y(\omega) = X(\omega)e^{-j6\omega}$$ and then scale $y(t)$ by $3$ to obtain $$z(t) = y(3t) = x(3t-6) \implies Z(\omega) = \frac{1}{3}Y(\omega/3) = \frac{1}{3}X(\omega/3)e^{-j2\omega}$$

Answer 2

Fat32's answer is correct and that's the way you should do it. Here I just want to add an explanation of your error, and a way to solve the problem correctly continuing from where you went wrong.

The problem occurs in the $4^{th}$ line of your derivation. The expression on the right-hand side is the Fourier transform of $x(3t-2)$, and not of $x(3(t-2))$. Scaling is related to the independent variable, nothing else. So now you got the Fourier transform of $x(3t-2)$:

$$\mathcal{F}\{x(3t-2)\}=\frac13X\left(\frac{j\omega}{3}\right)e^{-2j\omega/3}\tag{1}$$

Another shifting operation will give you the final result:

$$\mathcal{F}\{x(3t-6)\}=\mathcal{F}\{x(3(t-4/3)-2)\}=\frac13X\left(\frac{j\omega}{3}\right)e^{-2j\omega/3}e^{-4j\omega/3}=\frac13X\left(\frac{j\omega}{3}\right)e^{-2j\omega}\tag{2}$$

Note that I do not recommend this roundabout way, but this was just to make sure you understand your mistake, and you also understand how to arrive at the correct result from how you approached the solution.