SHOW: x(at) is a periodic signal with period T/a (a>0)

[VinnyCee]

Homework Statement

If x(t) is a periodic signal with period T, show that x(at), a > 0, is a periodic signal with period $\frac{T}{a}$, and $x\left(\frac{t}{b}\right)$, b > 0, is a periodic signal with period bT.

Homework Equations

HINT: Define $$x_a(t)\,=\,x(at)$$ and $$x_b(t)\,=\,x\left(\frac{t}{b}\right)$$. Show that $$x_a\left(t\,+\,T_a\right)\,=\,x_a(t)\,\forall\,t\,\in\,\mathbb{R}$$ and $$x_b\left(t\,+\,T_b\right)\,=\,x_b(t)\,\forall\,t\,\in\,\mathbb{R}$$, where $$T_a\,=\,\frac{T}{a}$$ and $$T_b\,=\,bT$$.

The Attempt at a Solution

I take the hint, and define

$$x_a(t)\,=\,x(at)$$

Now, I assume that $x_a(t)$ is periodic, with a period $\frac{T}{a}$

$$x_a(t)\,=\,x_a\left(t\,+\,\frac{T}{a}\right)$$

$$x_a\left(t\,+\,\frac{T}{a}\right)\,=\,x\left[a\left(t\,+\,\frac{T}{a}\right)\right]\,=\,x\left(at\,+\,T\right)$$

$$\therefore\,x_a\left(t\,+\,\frac{T}{a}\right)\,=\,x_a(t)\,\forall\,t\,\in\,\mathbb{R}$$


Does this seem right?

 

[VinnyCee]

Anyone verify this?

 

[Avodyne]

Yep, it's right!

 

[HallsofIvy]

No, it's not right!

Homework Statement

If x(t) is a periodic signal with period T, show that x(at), a > 0, is a periodic signal with period $\frac{T}{a}$, and $x\left(\frac{t}{b}\right)$, b > 0, is a periodic signal with period bT.

Homework Equations

HINT: Define $$x_a(t)\,=\,x(at)$$ and $$x_b(t)\,=\,x\left(\frac{t}{b}\right)$$. Show that $$x_a\left(t\,+\,T_a\right)\,=\,x_a(t)\,\forall\,t\,\in\,\mathbb{R}$$ and $$x_b\left(t\,+\,T_b\right)\,=\,x_b(t)\,\forall\,t\,\in\,\mathbb{R}$$, where $$T_a\,=\,\frac{T}{a}$$ and $$T_b\,=\,bT$$.

The Attempt at a Solution

I take the hint, and define

$$x_a(t)\,=\,x(at)$$

Now, I assume that $x_a(t)$ is periodic, with a period $\frac{T}{a}$

You can't assume that- that's what you are trying to prove!

$$x_a(t)\,=\,x_a\left(t\,+\,\frac{T}{a}\right)$$

But from here on you are okay. You aren't using "$x_a(t)= x_a(t + T/a)$" you are using a(x+ T)= a(x) which is your hypothesis.

$$x_a\left(t\,+\,\frac{T}{a}\right)\,=\,x\left[a\left(t\,+\,\frac{T}{a}\right)\right]\,=\,x\left(at\,+\,T\right)$$

$$\therefore\,x_a\left(t\,+\,\frac{T}{a}\right)\,=\,x_a(t)\,\forall\,t\,\in\,\mathbb{R}$$


Does this seem right?