Step 1

Let the time delay is T,

So the ramp function will be

\(f(t)=(t-T)u(t-T)\)

Where

\(u(t-T)=\begin{cases}1 & t>T\\0 & t

Step 2

Let the Laplace transform of f(t) is F(s)

So the Laplace transform is

\(F(s)=\int_0^\infty f(t)e^{-st}dt\)

\(\Rightarrow F(s)=\int_T^\infty T(t-T)e^{-st}dt\)

let

x=t-T

\(\Rightarrow dx=dt\)

So we have

\(F(s)=\int_0^\infty xe^{-s(x+T)}dx\)

\(\Rightarrow F(s)=e^{-sT}\int_0^\infty xe^{-sx}dx\)

\(\Rightarrow F(s)=\frac{e^{-sT}}{s^2}\)

Let the time delay is T,

So the ramp function will be

\(f(t)=(t-T)u(t-T)\)

Where

\(u(t-T)=\begin{cases}1 & t>T\\0 & t

Step 2

Let the Laplace transform of f(t) is F(s)

So the Laplace transform is

\(F(s)=\int_0^\infty f(t)e^{-st}dt\)

\(\Rightarrow F(s)=\int_T^\infty T(t-T)e^{-st}dt\)

let

x=t-T

\(\Rightarrow dx=dt\)

So we have

\(F(s)=\int_0^\infty xe^{-s(x+T)}dx\)

\(\Rightarrow F(s)=e^{-sT}\int_0^\infty xe^{-sx}dx\)

\(\Rightarrow F(s)=\frac{e^{-sT}}{s^2}\)