The function g(t) in (21) can be expressed in a more

Chapter 7, Problem 42E

(choose chapter or problem)

The function g(t) in (21) can be expressed in a more

convenient form as follows:

Show that for each \(n=0,1,2, \ldots\)

\(g(t)=e^{-\alpha t}\left[\frac{e^{(n+1) \alpha T}-1}{e^{\alpha T}-1}\right]\) for \(n T<t<(n+1) T\)

[Hint: Use the fact that \(1+x+x^{2}+\ldots+x^{\prime \prime}=\left(x^{n+1}-1\right) /(x-1)\)

Let \(v=1-(n+1) T\). Show that when \(n T<1<(n+1) T\), then \(-T<v<0\) and

(22) \(g(t)=\frac{e^{-\alpha v}}{e^{\alpha T}-1}-\frac{e^{-\alpha t}}{e^{\alpha T}-1}\)

Use the facts that the first term in (22) is periodic with period T and the second term is independent of n to sketch the graph of g(t) in (22) for \(\alpha=1\) and T = 2.

Equation transcription:

Text transcription:

n=0,1,2, \ldots

g(t)=e^{-\alpha t}\left[\frac{e^{(n+1) \alpha T}-1}{e^{\alpha T}-1}\right]

n T<t<(n+1) T

1+x+x^{2}+\ldots+x^{\prime \prime}=\left(x^{n+1}-1\right) /(x-1)

v=1-(n+1) T

n T<1<(n+1) T

-T<v<0

g(t)=\frac{e^{-\alpha v}}{e^{\alpha T}-1}-\frac{e^{-\alpha t}}{e^{\alpha T}-1}

\alpha=1