Describing a simple signal with unit step functions...

Discussion in 'Homework Help' started by Davezor, May 25, 2018.

  1. Davezor

    Thread Starter New Member

    May 25, 2018
    5
    0
    Hello there, I'm having trouble describing a simple triangle signal in terms of unit step functions only (turning on and off).
    I get the part where you describe the signal with ramps, but when you have to transform the signal to
    unit steps only, I get stuck. I just can't transform my equation to match the right answer in the text book.

    When I insert t*u(t) instead of ramp functions I simply cannot transform my equation to match the answer, not even close and I tried several ways.
    Can someone point me in the right direction? How should I start?

    [​IMG]
     
  2. danadak

    Well-Known Member

    Mar 10, 2018
    1,430
    292
    Seems to be a lot of resources on web for this -

    Google "representation of signals in terms of unit step function and ramp function"

    Regards, Dana.
     
  3. WBahn

    Moderator

    Mar 31, 2012
    22,857
    6,821
    x(t) = t*u(t) IS a ramp function. It is a function that is zero for t < 0 and x(t) = t for t >= 0. It is NOT in term of unit step functions only.

    The easiest (from a plug-and-chug no-thinking-required perspective) way to build up a signal like this is to identify each transition point and at each point add two terms, each multiplied by u(t-t0) at that time. The first term turns off the prior function and is simply the same function that was turned on at the last transition point. The second term is the function being turned on.

    So say you want a train of steps that first go to 10 V at t = 5 s and then to -14 V at t = 20 s and then back to 0 V at t = 21 s.

    The two transition points are t1 = 5 s, t2 = 20 s, and t3 = 21 s.

    The waveform for t < t1 is simply x(t) = 0 V.

    The waveform between t1 and t2 is x(t) = 10 V.

    The waveform for t2 < t < t3 is x(t) = -14 V.

    The waveform for t > t3 is x(t) = 0 V.

    So now crank the recipe:

    x(t) = 0 V - 0 V u(t - 5 s) + 10 V u(t - 5 s) - 10 V u(t - 20 s) + -14 V u(t - 20 s) - -14 V u(t - 21 s) + 0 V u(t - 21 s)

    Now you can go about simplifying it quite a bit.

    x(t) = 10 V u(t - 5 s) - 24 V u(t - 20 s) + 14 u(t - 21 s)
     
  4. MrAl

    AAC Fanatic!

    Jun 17, 2014
    5,117
    1,109
    Hello,

    To get an intuitive feel for how this works try graphing the effect of each term in the solution separately, then adding or subtracting graphically as the solution demands. You'll see how the individual ramps combine to form the final graphical solution and that will show you how this works.
    It's a matter of turning 'on' a ramp and then counter acting it with another ramp that turns 'on' and subtracts from the first to get the first ramp to stop acting or act differently, then possibly using more ramps like that to form the rest of the signal. The description is rather vague, but when you see the individual ramps graphed it makes a lot more sense. There are only four individual ramps for this problem so it's not that hard to graph. The final signal level is zero so all the ramps have to cancel out at the end of the triangular pulse even though the individual ones may continue forever.
    The unit step function is just used to turn 'on' a ramp at a given time, and it may be either added or subtracted from the previous ramp(s).
     
  5. Davezor

    Thread Starter New Member

    May 25, 2018
    5
    0
    Thanks to everyone. :) Not sure how I missed the point here but now I understand it, thanks again. :)
     
Loading...