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Sketching a Signal Graph Based on a Step Function

Date: 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 

          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
  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 

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.
Associated Topics:
High School Functions

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