---
title: "Show g(t) = Ay(Bt) using FT properties"
subject: "Signals and Systems"
chapter: 3
key: "semester_4/signals_and_systems:ft-scaled-convolution"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/ft-scaled-convolution
site: BCT Engineering
---

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

Signals and Systems · Chapter 3: Fourier Transform

- Asked in: 2070 Bhadra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/ft-scaled-convolution
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3

## Question

Given the relationship y(t) = x(t)*h(t) and g(t) = x(3t)*h(3t) and given that x(t) has the Fourier transform X(jω) and h(t) has Fourier transform H(jω), use Fourier transform properties to show that g(t) has the form g(t) = Ay(Bt). Determine the values of A and B.

## Short answer

Using time-scaling and convolution, G(jω) = (1/9)Y(jω/3) = (1/3)ℱ{y(3t)} so g(t) = (1/3)y(3t). Thus A = 1/3, B = 3.

## Answer

**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, $x(at)*h(at) = \frac{1}{|a|}y(at)$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the FT of x(3t)?
   A. 3 X(j3ω)
   B. (1/3) X(j3ω)
   C. X(jω/3)
   D. (1/3) X(jω/3)
   Answer: D. x(at) <-> (1/|a|) X(jω/a). (Key point: Time scaling)

2. What is G(jω) for g = x(3t)*h(3t)?
   A. (1/9) Y(jω/3)
   B. (1/3) Y(jω/3)
   C. (1/9) Y(j3ω)
   D. 3 Y(jω/3)
   Answer: A. Product of two factors of 1/3, each with argument jω/3. (Key point: G(jω))

3. Since y(3t) <-> (1/3) Y(jω/3), what is g(t)?
   A. (1/9) y(3t)
   B. (1/3) y(3t)
   C. 3 y(3t)
   D. (1/3) y(t/3)
   Answer: B. G = (1/3)·[(1/3)Y(jω/3)] = (1/3)F{y(3t)}. (Key point: Compare with y(3t))

4. What are A and B in g(t) = A y(Bt)?
   A. A = 3, B = 1/3
   B. A = 1/9, B = 3
   C. A = 1/3, B = 3
   D. A = 1/3, B = 1/3
   Answer: C. General rule: x(at)*h(at) = (1/|a|) y(at). (Key point: A and B)

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