How Do You Solve for x3[t] in Terms of x1[t]?

  • Thread starter Thread starter kougou
  • Start date Start date
  • Tags Tags
    Function
kougou
Messages
80
Reaction score
0

Homework Statement


If x2[t]= x1[t-2]
and x3[t]=x2[2t]

What is x3[t] in terms of x1[t]?

a) x3[t]= x1[2(t-2)]
b) x3[t]= x1[2t-2]

Homework Equations


which one?

The Attempt at a Solution



offcourse, but I am struggling with this concept of shifting. But I believe the answer should be A. since x3[t] is the compression version of x2[t], which is the shift version of x1[t].

So the overall transformation would be x3[2(t-2)], which is not correct.
 
Last edited:
Physics news on Phys.org
I don't see any "attempt at a solution". Have you not tried?
 
HallsofIvy said:
I don't see any "attempt at a solution". Have you not tried?



offcourse, but I am struggling with this concept of shifting. But I believe the answer should be A. since x3[t] is the compression version of x2[t], which is the shift version of x1[t].

So the overall transformation would be x3[2(t-2)], which does not correct.
 
This problem is not so much about "shifting" as it is about understanding functional notation. You are given that x2[t]= x1[t- 2]. That is exactly the same thing as saying that x2[y]= x1[y- 2] or x2[a]= x1[a- 2]. Whatever is in the braces of x2, in the x1 function, you subtract 2 from it. In x3[t]= x2[2t], what is "in the braces of x2" is "2t" so x3[t]= x2[2t]. What do you get when you replace "t" with "2t" in x1?
 
I think this one is easiest to see algebraically. Let's rename some variables to avoid confusion. So let
x2(u) = x1(u - 2)
x3(v) = x2(2v)

Now set v = t. What will u be?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top