The Laplace transform of t is 1/s^2. In this post, we will find the Laplace transform of t by definition and by the Laplace transform formulas for integrals and derivatives.
Laplace transform of t formula
We know that the Laplace transform of integral powers of t is given by the formula:
L{tn} = n!/sn+1.
Putting n=1, we get the Laplace transform formula of t.
L{t} = 1!/s1+1 = 1/s2.
What is the Laplace transform of t?
Answer: The Laplace transform of t is 1/s2.
Proof:
The definition of the Laplace transform of f(t) is given as follows:
L{f(t)} = $\int_0^\infty$ f(t) e-st dt.
Step 1: Put f(t) = t in the above definition of Laplace transform.
So the Laplace transform of t by definition is equal to
L{t} = $\int_0^\infty$ te-st dt
Step 2: We use the integrating by parts formula, we obtain that
L{t} = $[$ t ∫e-st dt – $\int \{\frac{d}{dt}(t) \int e^{-st} dt\} dt$ $]_0^\infty$
= $\left[ t \cdot \dfrac{e^{-st}}{-s} – \int \dfrac{e^{-st}}{-s} dt \right]_0^\infty$
= $\left[ \dfrac{te^{-st}}{-s} – \dfrac{e^{-st}}{s^2} \right]_0^\infty$
= $\lim\limits_{t \to \infty}\left[ \dfrac{te^{-st}}{-s} – \dfrac{e^{-st}}{s^2} \right]$ $-\big( 0 – \dfrac{1}{s^2} \big)$
= $ 0-0 + \dfrac{1}{s^2}$
= $\dfrac{1}{s^2}$.
Thus, the Laplace transform of t is 1/s2 and this is obtained by the definition of the Laplace transform.
Also Read:
Laplace transform of sin at: | a/(s2+a2) |
Laplace transform of cos at: | s/(s2+a2) |
Laplace transform of e-t: | 1/(s+1) |
Laplace transform of 1: | 1/s |
Find The Laplace Transform of t
We know that L{t} =1/s2.
Method 1: Using the multiplication by tn Laplace transform formula, we will now find the Laplace transform of t. This formula says that
L{tnf(t)} = (-1)n $\dfrac{d^n}{ds^n}[F(s)]$, where F(s)=L{f(t)}.
Put n=1 and f(t)=1. Then F(s)=L{f(t)}=1/s.
So L{t} = (-1)1 $\dfrac{d}{ds}[\dfrac{1}{s}]$
= $- \dfrac{-1}{s^2}$
= 1/s2.
Method 2: Using the Laplace transform formula for integrals, we will now find the Laplace transform of t. The formula says that
$L\{\int_0^t f(u) du \}=\frac{1}{s}F(s)$, where F(s)=L{f(t)}.
Put f(u)=1. Then F(s)=L{f(u)}=1/s.
Therefore, $L\{\int_0^t du \}=\dfrac{1}{s} \cdot \dfrac{1}{s}$
⇒ L{t} = 1/s2.
FAQs
Answer: The Laplace transform of 0 is 0.
Answer: 1/s2 is the Laplace transform of t.