MHB Solve f(t) with Heaviside Functions: Find b

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The function f(t) is expressed using Heaviside functions as f(t) = 7H(t) + bH(t-5), with the goal of determining the value of b. The calculation shows that at t = 5, f(5) equals 15, leading to the equation 15 = 49 + 7b, which simplifies to b = -34/7. However, a key point raised is that the function cannot be represented as a combination of Heaviside steps without addressing its domain, which is limited to positive real numbers. It is suggested to redefine the function as f(x) = 7H(x) + 8H(x-5) to ensure it is valid across the intended intervals. The discussion emphasizes the importance of correctly specifying the function's domain when using Heaviside functions.
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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