Solve f(t) with Heaviside Functions: Find b

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SUMMARY

The function f(t) can be expressed using Heaviside functions as f(t) = 7H(t) + bH(t-5). To determine the value of b, the equation f(5) = 15 leads to the calculation -34 = 7b, resulting in b = -34/7. However, it is crucial to note that the function must be defined on the entire real line, necessitating the inclusion of conditions for x < 0. A corrected version of the function is f(x) = 7H(x) + 8H(x-5), which accurately represents the function's behavior across its domain.

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Uniman
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Consider the function

Then f(t) can be expressed with Heaviside functions as

Determine b.

Workdone so far...

f(t) = 7H(t) + bH(t-5)
when t = 5
f(5) = 49 + 7b
15 = 49 + 7b
-34 = 7b
b = -34/7
http://www.chegg.com/homework-help/questions-and-answers/-q3136934#
Is the answer correct...
 
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Uniman said:
Consider the function

Then f(t) can be expressed with Heaviside functions as

Determine b.

Workdone so far...

f(t) = 7H(t) + bH(t-5)
when t = 5
f(5) = 49 + 7b
15 = 49 + 7b
-34 = 7b
b = -34/7
Is the answer correct...

As it is written your function cannot be written as a combination of Heaviside steps, since such a combination is defined on the whole real line, but the domain of your function is just the positive real line. So you either to use Heaviside steps restricted to the \(x>0\) of you need to specify that your function is zero foir \(x<0\).

Anyway what you probably want is \(f(x)=7H(x)+8H(x-5)\) and then to check that this is correct at the steps (this is \(0\) up to \(0\), \(7\) from \(0\) to \(5\) and \(7+8=15\) past \(x=5\) )

Also remove the link from your post.

CB
 

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