Sketching a Signal Graph Based on a Step FunctionDate: 08/03/2006 at 10:17:08 From: Annuar Subject: Draw a signal graph How do we sketch the graph of x(t) = 3(t+3)u(t+3) - 6tu(t) + 3(t-3)u(t-3) where u(t) is a step function. The difficult part is that its a function multiplied by a function plus the addition operations. It involves a unit step function and linear line. How do we multiply and add those lines? I was told by my lecturer that its the graph of a triangle. I only get the half of the graph between -3 and 0. For the other half I can't figure it out because of the -6t. Date: 08/03/2006 at 12:06:51 From: Doctor Peterson Subject: Re: Draw a signal graph Hi, Annuar. I'll assume that your "unit step function" is the one defined here, though it may be a variant: Heaviside Step Function http://mathworld.wolfram.com/HeavisideStepFunction.html 0 x<0 H(x) = 1/2 x=0 1 x>0 You can treat this in a piecewise fashion. Since your function uses u(t), u(t+3), and u(t-3), you will need to consider four intervals, in each of which each of these functions has a constant value: -3 0 3 <------------+----------+----------+------------->t H(t+3): 0 1/2 1 1 1 H(t): 0 0 1/2 1 1 H(t-3): 0 0 0 1/2 1 Now consider each of the four open intervals, and each of the three boundaries, separately. For instance, in the leftmost interval, all three H functions are 0, so x(t) = 3(t+3)u(t+3) - 6tu(t) + 3(t-3)u(t-3) = 3(t+3)*0 - 6t*0 + 3(t-3)*0 = 0 The other intervals will be a little more interesting. Once you have done a couple of these, you may start to see a pattern that will make it easier. If you need more help, please write back and show me how far you got. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 08/09/2006 at 10:05:48 From: Annuar Subject: Thank you (Draw a signal graph) I want to thank you for your help. It helped me a lot to figure out the signal. I never thought to do it by intervals. For t < -3, similar to you I've got 0. For -3 < t < 0 I've got 3t + 9 which is the first half portion of the signal. For 0 < t < 3 I've got -3t + 9 which is the other half portion of the signal, and for t > 3 I've got 0. Finally, I've got the triangle signal! Guess what? Thanks to your help now I can do the reverse--that is to find the equation in unit step function form for a given signal. It's really a big help to me, so once again, thank you. |
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