Unit Step Function Integral

Thread Starter

p75213

Joined May 24, 2011
70
Can somebody help me with understanding how they have graphed the integral? From the integral I get the following:

10r(t)=0, t<=0
=10t, t>=0

-20r(t-2)=0, t<=2
=-20t+40, t>=2

10r(t-4)=0, t<=4
=10t-40. t>=4

I can't see how the graph has been drawn from these equations???
 

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Zazoo

Joined Jul 27, 2011
114
Can somebody help me with understanding how they have graphed the integral? From the integral I get the following:

10r(t)=0, t<=0
=10t, t>=0

-20r(t-2)=0, t<=2
=-20t+40, t>=2

10r(t-4)=0, t<=4
=10t-40. t>=4

I can't see how the graph has been drawn from these equations???
For t<0, the ramp functions are all zero, so:
f(t)=0

For 0<t<2, only the first ramp function is non-zero, so:
f(t)=10t. (the left side of the triangle)

For 2<t<4, both the first and second ramp function are non-zero, so:
f(t) = (10t) + (-20t+40) = -10t +40. (the right side of triangle)

For t>4, all three ramp functions are non-zero, so:
f(t) = (10t) + (-20t+40) + (10t-40) = 0
 
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