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Signals help needed

  1. Dec 9, 2005 #1
    signals....help needed

    hi,

    can anyone tell me why a discrete time signal x(n-k) where k>0 is shifted right along the X-axis(the time domain) and x(n+k) is shifted left. should'nt it be the reverse? this is for the graphical notation. but for the notation x(n)={...,0,3,2,1,0,1,2,3,0,.....}
    we go left for x(n-k) and go right for x(n+k) which is past samples and future samples respectively. the graphical and short hand notations are really contradictory to me. so plzzzz help me:uhh:
     
  2. jcsd
  3. Dec 9, 2005 #2

    berkeman

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    Staff: Mentor

    It's the same with continuous-time signals, and maybe a little more intuitive there. The -k or +k changes the argument to the function, so that the y value of the function takes on its value later or earlier than it would without the argument offset.

    Take y=sin(x) and y=sin(x-a). Plot y=sin(x) and it crosses the x axis as usual at x=0 and rising. Now set a=PI/4, so y=sin(x-PI/4). We know that sin()=0 when the argument to the sin() is 0, so sin(x-PI/4)=0 when x-PI/4=0, or x=PI/4. So now the first zero crossing of the plot has shifted to the *right* by PI/4, because the *argument* to the sin() function has been shifted to the left by PI/4. Shifting the argument to the function one way has the effect of shifting the plot of the function the other way.

    Hope that helps. It always used to confuse me too, until I understood the difference between shifting the argument and shifting the function. -Mike-
     
  4. Dec 9, 2005 #3
    thanks mike. It's easy when u think about the shifting of sine fuctions. But for a discrete time signal x(n+k) means taking future values of x(n) (n is the sampling variable) whereas x(n-k) means taking past values. and when we plot such signals shouldn't it be that for x(n+k) we take the signal to the right and for x(n-k) we take it to the left? if i'm right we can think of n to be equivalent to the time domain. then n+k would go forward in time and this should take the signal to the right! it's still a bit confusing. but it takes the signal to the left :(
     
  5. Dec 9, 2005 #4

    berkeman

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    Staff: Mentor

    No, the same thing applies to discrete signals. Plot some arbitraty y=x(n). Then plot y=x(n+1). The plot shifts to the left by one clock tick, because where n=0 used to give you the first value of y before, now x=-1 gives you the equivalent of the first value. The plot of y=x(n+1) thus looks like the plot of y=x(n), but shifted left in discrete time by one clock tick.
     
  6. Dec 9, 2005 #5
    OK :) now it's clear . thank's a lot
     
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