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Draw heaviside and ramp function

  1. Sep 12, 2013 #1
    1. The problem statement, all variables and given/known data
    Plot this signal

    3. The attempt at a solution

    I tried to draw each single function, but do not know the way to plot for the whole signal like the bottom graph
  2. jcsd
  3. Sep 13, 2013 #2


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    Try breaking the problem down into 2 parts--a graph resulting from only the u-components, and another from only the r-components. Post what you get.
  4. Sep 16, 2013 #3
    Only for u function

    Only for r function

    What things I have to do next to get the correct graph ?
  5. Sep 16, 2013 #4


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    Your u-function looks right. Your r-function has a problem with the rightmost segment. Fix that and then just add them together. It may be easier to start with the r-function and then add the offsets from the u-function. You may even be able to do it visually if your x-axes are the same scale.
  6. Sep 16, 2013 #5
    Let me ask some question first :
    Both of graph is plotted using wolframalpha, for the u-functions :
    The 3u(t+3) function is correct, but for -2u(t-4), i think it should have a line starts from (2;-2) but the graph only shows a line starts from (4;-1), and where is the line for -2u(t-2). Where does the line start from (2;1) to (4;1) come from ?
  7. Sep 16, 2013 #6


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    Wolfram alpha would be good for checking your work, but you should first try to do the graphs manually.

    Why would you think that? The function -2u(t-4) "activates" at t=4.

    (4,-1) is correct.

    That would be from the third time interval (I3), below:

    Realize that y(t)=3u(t+3)-2u(t-2)-2u(t-4) is composed of 4 time intervals:

    I1: -∞ < t < -3 [nothing is happening because no u-function is "active" --for amplitude, think 0]
    I2: -3 <= t < 2 [3u(t+3) is the only "active" --for amplitude, think 3]
    I3: 2 <= t < 4, [3u(t+3) AND -2u(t-2) are both "active" --for amplitude, think 3-2]
    I4: 4 <= t < ∞, [3u(t+3) AND -2u(t-2) AND -2u(t-4) are all "active" --for amplitude, think 3-2-2 ]
    Last edited: Sep 16, 2013
  8. Sep 17, 2013 #7
    In the definition of unit step function, the function-2u(t-4) should be
    -2 if t-4>0 and 0 if t-4<0

    so it should start from (4;-2), am I right ?
  9. Sep 17, 2013 #8


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    Yes. Technically its: -2 if t-4 >= 0 and 0 if t-4 < 0
  10. Sep 17, 2013 #9
    Another example x(t)=2u(t)-u(t-1)-u(t+1)
    For 2u(t) = 2 if t>0 and 0 if t<0, a line will start from (0;2)
    For -1u(t-1) = -1 if t>1 and 0 if t<1, line starts from (+1;-1)
    For -1u(t+1) = -1 if t>-1 and 0 if t<-1. line starts from (-1;-1)

    I tried to plot all thing above in a graph, but does not know how to combine them into one. The correct graph have lines start from (-1;-1) to (0;-1) and (0;1) to (1;1).

    I am stucking at see the way 2u(t) represent in the graph.
    Last edited: Sep 17, 2013
  11. Sep 17, 2013 #10


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    You don't need any special tools to solve these kinds of problems. A pencil and some graph paper will do the job.

    Not really. (-∞;0) to (-1;0) to (-1;-1) to (0;-1) and (0;1) to (1;1) to (1;0) to (∞;0)

    2u(t) is an additive term in the composite graph. You won't "see it" in the graph as you would if it stood alone. It is responsible for the 2-unit jump from (0;-1) to (0;1).

    Before jumping all over this and other examples, did you understand what I wrote in post #6? If not, what is not clear? I am trying to understand what you don't understand about the OP so please focus on post #6 for now.
  12. Sep 18, 2013 #11
    Thank you, I got the #6. Now come to the ramp function, I have tried to seperate time intervals but it seems there is nothing like the amplitude for heaviside, it doesn't work.
  13. Sep 18, 2013 #12


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    What doesn't work? The time intervals for the combined ramp functions can be obtained from inspection:

    yramp(t) = -r(t+2) + 2r(t+0) -r(t-3)

    You know, maybe you are still trying to do too much all at once. Why not just take two of the function terms and add them together:

    yramp-first-2-terms(t) = -r(t+2) + 2r(t+0)

    What would that look like?
  14. Sep 18, 2013 #13
    Let me do it :
    -∞<t<-2 : No signal, 0 for amplitude.
    -2<t<0 : -r(t+2) is active at t=-2, -1 for amplitude.
    0<t<3 : -r(t+2) and 2r(t) are both active, 2-1 for amplitude
    3<t<+∞ : -r(t+2) and 2r(t) and -r(t-3) are all active, 2-1-1 for amplitude.

    I have no idea what the graph would look like, the u-functions only have lines paralled with t-axis.
    Last edited: Sep 18, 2013
  15. Sep 18, 2013 #14


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    So jumping ahead (against my advice :frown:), you are now working on yramp(t) = -r(t+2) + 2r(t+0) -r(t-3).


    False. r(t) is the unit ramp function, with amplitude 0 for t<0 and a "ramping" output for t >= 0 (output starts at zero and increases linearly with a slope of 1).

    -r(t+2) starts ramping at t=-2 with a slope of -1.

    "amplitude" is wrong. You are thinking about the behavior of the u-function.

    Again, the "constant amplitude" concept only applies to u-functions.

    You started out this post with a collection of horizontal and diagonal lines. Those diagonal lines are the the result of the r-functions. By the way, the red dash-dot-dot-dashed line is not shown correctly (it should start at t=3).
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