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Homework Help: Fundamental Shift and Scaling of Signals

  1. Dec 11, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm confused on whether or not two functions would be equivalent or not. Let's say x(t) is a triangle with height 1, width 1. The hypotenuse of it is the function t (with a slope of 1). I'm told that x((t+2) / 4) then is making it 4 times as wide and left-shifting the back, pointy end (just to help you visualize what the signal looks like, lol) by 2. What, then, is x((t/4) + 2)? It can't be exactly the same thing! Also, if we split up the first function into x((t/4) + (2/4)), would those be equivalent? I'm so confused...

    2. Relevant equations

    3. The attempt at a solution
    Given Above
  2. jcsd
  3. Dec 12, 2012 #2
    Start with:

    x(t) = (t), 0<=t<=1

    When you make a variable substitution like this one:


    always keep in mind you are replacing the 't' in the original x(t) by the whole expression:

    x((t+2)/4) = ((t+2)/4)

    I placed brackets around the t in the original expression to make sure the (t+2)/4 correctly replaces the t.

    We can get to the same expression by following these steps:

    (i) x(t) = t

    (ii) x(t/4) = t/4 = y(t)

    (iii) y(t+2) = (t+2)/4

    This means we start with x(t), the triangle, then fatten the function in (ii) then left shift by two units in (iii)

    This is *not* the same as x((t/4) + 2)

    We get this one by following these steps:

    (i) x(t) = t

    (ii) x(t+2) = t+2 = y(t)

    (iii) y(t/4) = t/4 + 2

    This means we start with x(t), the triangle, then we left shift by two units in (ii) and then fatten that by four in (iii)

    Notice the change in order here means we have two different functions!

    How do I know what order to do these things? In the end I need to replace the 't' in the x(t) function with the final expression.


    With x(t/4+2) I have to find a sequence that will replace t one operation at a time and result in an equivalent replacement of t by 't/4+2'

    start with (t)
    replace t by (t+2) --> (t+2)
    replace t by (t/4) --> (t/4+2)

    check -- that's what I wanted.

    I can't do this:

    start with (t)
    replace t by (t/4) --> t/4
    replace t by (t+2) --> (t+2)/4 = t/4 + 0.5

    not the same!
  4. Dec 12, 2012 #3
    Yes it would. In both cases you would be replacing the 't's in x(t) by an equivalent expression.
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