Horizontal expansion/compression of a function - proof

Question

How can I prove that $y(\alpha t)$ with $\alpha>1$ is a compression of $y(t)$ along x axis?

I tried to do it in this way: given $x(t)$, when I consider $t_1$ I have $x(t_1)=x_1$. Now I consider $y(\alpha t)$ with $\alpha>1$ where $y(t)=x(t)$, thus:

$$y(\alpha t°=t_1)=x(t_1) \iff t°=t_1/\alpha$$

Since $\alpha>1$, it follows that $t°<t_1$. But this could mean also that $y(\alpha t)$ is the function $x(t)$ shifted to the left side.

Thank you for your time.

Answer 1

Let's compare the graphs of the two functions $y$ and $z$, where $y$ is your original function and $z(t)=y(\alpha t)$. Consider any point $P$ on the graph of $y$. So the coordinates of $P$ are $(t,y(t))$ for some number $t$. If you compress the graph by a factor $\alpha$ in the horizontal direction, then $P$ gets moved to the point $Q$ whose coordinates are $(t/\alpha, y(t))$ --- the same vertical coordinate $y(t)$, but the horizontal coordinate $t$ is reduced by a factor $\alpha$.

Now notice that $Q=(t/\alpha,y(t))$ is a point on the graph of $z$, because $z(t/\alpha)=y(\alpha\cdot(t/\alpha))=y(t)$. This shows that, if you compress the graph of $y$ as in the question, you get points that are on the graph of $z$.

A similar argument shows that every point on the graph of $z$ is the image, under horizontal compression, of a point on the graph of $y$.

Answer 2

Consider a finite signal:

$$ x(t) = \begin{cases} f, & -t_1\leq t \leq t_1 \\[2ex] 0, & \text{otherwise} \end{cases} $$

Now, we apply a scaling factor $\alpha$ to the variable $t$:

$$ \begin{align} x(\alpha t) &= \begin{cases} f, & -t_1\leq \alpha t \leq t_1 \\[2ex] 0, & \text{otherwise} \end{cases}\\[2ex] x(\alpha t) &= \begin{cases} f, & -\frac{t_1}{\alpha}\leq t \leq \frac{t_1}{\alpha} \\[2ex] 0, & \text{otherwise} \end{cases} \end{align} $$

As you can see, there is no time shift associated. If $|\alpha|>1$, there is a compression, if $0<|\alpha|<1$, there is an expansion. This is easily verifiable comparing plots. If $\alpha<0$, the same proerties apply, along with an inversion of the signal.

Now, for a delay of $T_d$ on the signal:

$$ \begin{align} x(t-T_d) &= \begin{cases} f, & -t_1\leq t-T_d \leq t_1 \\[2ex] 0, & \text{otherwise} \end{cases}\\[2ex] x(t-T_d) &= \begin{cases} f, & -t_1+T_d\leq t \leq t_1+T_d \\[2ex] 0, & \text{otherwise} \end{cases} \end{align} $$

For this function, when would the time scaled equal the time shifted?

$$ \begin{align} \alpha t &= t-T_d\\ \alpha t -t & =-T_d\\ t(1-\alpha)&=T_d \end{align} $$

$$t=\frac{T_d}{(1-\alpha)}$$