Deriving equation of a ramp function graph.

  • Thread starter btbam91
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  • #1
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Hello, here is my problem.

http://imageshack.us/a/img543/827/rampfunction.png [Broken]

I'm having a little trouble coming up with the function f(t).

I have something like:

f(t) = [(A/T)*t -A] + [(A/T)*(t-2T) -2A] + A*1(t-3T)

Am I on the right track here?

Thanks!
 
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  • #2
jbunniii
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f(t) = [(A/T)*t -A] + [(A/T)*(t-2T) -2A] + A*1(t-3T)
This is still a linear equation in [itex]t[/itex], so it can't be right: the graph of any linear function is just a straight line. You will have to use a piecewise definition or perhaps a formula that incorporates unit step functions.
 
  • #3
LCKurtz
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Have you studied the unit step function ##u(t)##? Expressing your function with it is the best way if you plan to transform the result. For example, if you want your function to equal g(t) on (0,a) and h(t) on (a,b) and 0 elsewhere you would write it like this:$$
f(t) = g(t)u(t) + (h(t) - g(t))u(t-a) + u(t-b)(0-h(t))$$Notice at each term you take out the old formula and put in the new one.
 
  • #4
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Thanks for the guidance! I will try to come up with a solution now with this additional knowledge!
 
  • #5
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LCKurtz, in trying to follow your formula, I got:

f(t) = [(A/T)*t - A]*u(t) + [((A/T)*t-3A)-((A/T)*t-A)]*u(t-2T)+[(0)-((A/T)*t-3A)]*u(t-3T)

Am I on the right track here?
 
  • #6
LCKurtz
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LCKurtz, in trying to follow your formula, I got:

f(t) = [(A/T)*t - A]*u(t) + [((A/T)*t-3A)-((A/T)*t-A)]*u(t-2T)+[(0)-((A/T)*t-3A)]*u(t-3T)

Am I on the right track here?

That looks OK.
 

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