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?