Show g(t) = Ay(Bt) using FT properties

Given: $y(t) = x(t) * h(t)$, so by the convolution property $Y(j\omega) = X(j\omega)H(j\omega)$.

Time-scaling property: $x(at) \leftrightarrow \frac{1}{|a|}X\left(\frac{j\omega}{a}\right)$.

Step 1: transforms of $x(3t)$ and $h(3t)$.

$$x(3t) \leftrightarrow \frac{1}{3}X\left(\frac{j\omega}{3}\right), \qquad h(3t) \leftrightarrow \frac{1}{3}H\left(\frac{j\omega}{3}\right)$$

Step 2: transform of $g(t) = x(3t) * h(3t)$. Convolution becomes multiplication:

$$\begin{aligned} G(j\omega) &= \frac{1}{3}X\left(\frac{j\omega}{3}\right)\cdot\frac{1}{3}H\left(\frac{j\omega}{3}\right) \\ &= \frac{1}{9}\,Y\left(\frac{j\omega}{3}\right) \end{aligned}$$

Step 3: compare with the transform of $y(3t)$.

$$y(3t) \leftrightarrow \frac{1}{3}Y\left(\frac{j\omega}{3}\right) \;\Rightarrow\; Y\left(\frac{j\omega}{3}\right) = 3\,\mathcal{F}\{y(3t)\}$$

So

$$G(j\omega) = \frac{1}{9}\cdot 3\,\mathcal{F}\{y(3t)\} = \frac{1}{3}\,\mathcal{F}\{y(3t)\}$$

Taking the inverse transform:

$$g(t) = \frac{1}{3}\,y(3t)$$

Answer: $g(t) = A\,y(Bt)$ with $A = \dfrac{1}{3}$ and $B = 3$.

Time-domain check:

$$\begin{aligned} g(t) &= \int x(3\tau)h(3t - 3\tau)\,d\tau \quad (\text{let } \lambda = 3\tau,\ d\tau = d\lambda/3) \\ &= \frac{1}{3}\int x(\lambda)h(3t - \lambda)\,d\lambda = \frac{1}{3}y(3t) \end{aligned}$$

Both methods agree. In general, $x(at)*h(at) = \frac{1}{|a|}y(at)$.