Time Variance of y(t)=x(2t) System

[Jncik]

Homework Statement

show whether the system

y(t) = x(2t) is time variant or not

Homework Equations

a system is time invariant if a time shift in the input signals results in an identical time shift in the output signal, that is if y[n] is the output of a discrete-time, time invariant system when x[n] is the input, then y[n-n0] is the output when x[n-n0] is applied

The Attempt at a Solution

first of all I have the answer on my book(oppenheim) but I can't understand what he does

what I tried to do is

suppose we apply a signal

x1(t) and we get an output of y1(t) = x1(2t)

now, suppose we apply a signal x2(t) = x1(t-t0)

we get an output of y2(t) = x2(t) = x1(2t - 2t0) = y1(t-t0)

hence it's time invariant

now, the book says it's time variant, and also it creates graphs to prove this point and I don't understand how he proves this..

what am I doing wrong?
thanks in advance

 

[Jncik]

actually now that I'm looking at it better

when we have a system

y(t) = x(2t)

and we say let's input the x(t-2)

will the new system be

y(t) = x(2(t-2))

OR

y(t) = x(2t - 2)

?

for the second part, and I think oppenheims graphs have something to do with the second one

the result is correct

 

[FroChro]

x1(t) and we get an output of y1(t) = x1(2t)

now, suppose we apply a signal x2(t) = x1(t-t0)

we get an output of y2(t) = x2(t) = x1(2t - 2t0) = y1(t-t0)

hence it's time invariant
what am I doing wrong?

y2(t)=x2(2t)=x1(2t-t0)=y1(t-t0/2)
hence it is time variant :P

This is most simply seen if you realize that y in time t is equal to x in time 2*t, which is time origin (t=0) dependent definition.