Inverse Laplace Transform Notes

Inverse Laplace Transform

Definition

If LF(t)=f(s)L{F(t)} = f(s), then F(t)=L−1f(s)F(t) = L^{-1}{f(s)}.

Examples

  1. Find L−1as2+a2L^{-1}{ a \over s^2+a^2}

    • Since Lcos⁡at=ss2+a2L{\cos{at}} = {s \over s^2 + a^2}, we have L−1ss2+a2=cos⁡atL^{-1}{ s \over s^2+a^2} = \cos{at}.
    • Find L−11s2+a2L^{-1}{ 1 \over s^2+a^2}
    • Since Lsin⁡at=as2+a2L{\sin{at}} = {a \over s^2 + a^2}, we have L−1as2+a2=sin⁡atL^{-1}{ a \over s^2+a^2} = \sin{at}.
    • Therefore, L−11s2+a2=sin⁡ataL^{-1}{ 1 \over s^2+a^2} = {\sin{at} \over a}.

  2. Find L−11snL^{-1}{ 1 \over s^n}

    • Since Ltn=Γ(n+1)sn+1L{t^n} = {\Gamma(n+1) \over s^{n+1}}, we have L−1Γ(n+1)sn+1=tnL^{-1}{\Gamma(n+1) \over s^{n+1}} = t^n.
    • Thus, L−11sn+1=tnΓ(n+1)L^{-1}{ 1 \over s^{n+1}} = {t^n \over \Gamma(n+1)}.

Properties of Inverse Laplace Transform

  1. Linearity:
    • L−1c<em>1f</em>1(s)+c<em>2f</em>2(s)=c<em>1L−1f</em>1(s)+c<em>2L−1f</em>2(s)L^{-1}{c<em>1f</em>1(s) + c<em>2f</em>2(s)} = c<em>1L^{-1}{f</em>1(s)} + c<em>2L^{-1}{f</em>2(s)}
    • Example:
      • L−15s2+a2+7s2=5L−11s2+a2+7L−11s2=5sin⁡ata+7tL^{-1}{ 5 \over s^2+a^2} + {7 \over s^2} = 5L^{-1}{ 1 \over s^2+a^2} + 7L^{-1}{ 1 \over s^2} = 5{\sin{at} \over a} + 7t

First Translation or Shifting Property

Finding the formula directly
  • If LF(t)=f(s)L{F(t)} = f(s), then LeatF(t)=f(s−a)L{e^{at}F(t)} = f(s-a).
  • If L−1f(s)=F(t)L^{-1}{f(s)} = F(t), then L−1f(s−a)=eatF(t)=eatL−1f(s)L^{-1}{f(s-a)} = e^{at}F(t) = e^{at}L^{-1}{f(s)}.
  • L−1f(s−a)=eatL−1f(s)L^{-1}{f(s-a)} = e^{at}L^{-1}{f(s)}
Finding the formula by proof
  • If L−1f(s)=F(t)L^{-1}{f(s)} = F(t), then show that L−1f(s−a)=eatF(t)L^{-1}{f(s-a)} = e^{at}F(t)
  • Proof:
    • f(s)=LF(t)=∫0∞e−stF(t)dtf(s) = L{F(t)} = \int_0^{\infty} e^{-st} F(t) dt
    • f(s−a)=∫<em>0∞e−(s−a)tF(t)dt=∫</em>0∞e−steatF(t)dt=LeatF(t)f(s-a) = \int<em>0^{\infty} e^{-(s-a)t} F(t) dt = \int</em>0^{\infty} e^{-st} e^{at} F(t) dt = L{e^{at}F(t)}
    • L−1f(s−a)=eatF(t)L^{-1}{f(s-a)} = e^{at}F(t)
    • L−1f(s−a)=eatL−1f(s)L^{-1}{f(s-a)} = e^{at}L^{-1}{f(s)}
Example
  • Find L−11(s+10)2L^{-1}{ 1 \over (s+10)^2}
    • Using the property, L−1f(s−a)=eatL−1f(s)L^{-1}{ f(s-a) } = e^{at} L^{-1}{ f(s) }
    • L−11(s+10)2=e−10tL−11s2=e−10tt=te−10tL^{-1}{ 1 \over (s+10)^2} = e^{-10t} L^{-1}{ 1 \over s^2} = e^{-10t}t = te^{-10t}
  • Find L−11(s2−4s+8)L^{-1}{ 1 \over (s^2 - 4s + 8)}
    • L−11(s2−4s+8)=L−11(s−2)2+4=e2tL−11s2+4=12e2tsin⁡2tL^{-1}{ 1 \over (s^2 - 4s + 8)} = L^{-1}{ 1 \over (s-2)^2 + 4} = e^{2t} L^{-1}{ 1 \over s^2 + 4} = {1 \over 2}e^{2t}\sin{2t}
  • Find L−1s−7(s−7)2+52L^{-1}{ s-7 \over (s-7)^2 + 5^2}
    • L−1s−7(s−7)2+52=e7tL−1ss2+52=e7tcos⁡5tL^{-1}{ s-7 \over (s-7)^2 + 5^2} = e^{7t} L^{-1}{ s \over s^2 + 5^2} = e^{7t}\cos{5t}
  • Find L−11(s+5)2+16L^{-1}{ 1 \over (s+5)^2 + 16}
    • L−11(s+5)2+16=e−5tL−11s2+16=14e−5tsin⁡4tL^{-1}{ 1 \over (s+5)^2 + 16} = e^{-5t} L^{-1}{ 1 \over s^2 + 16} = {1 \over 4} e^{-5t} \sin{4t}
      *Find L−13s+1(s+5)2+16L^{-1}{ 3s+1 \over (s+5)^2 + 16}

Second Translation or Shifting Property

Finding the formula directly

  • If LF(t)=f(s)L{F(t)} = f(s) and

    G(t)={F(t−a),amp;tgt;a 0,amp;tlt;aG(t) = \begin{cases} F(t-a), &amp; t &gt; a \ 0, &amp; t &lt; a \end{cases}

    then LG(t)=e−asf(s)L{G(t)} = e^{-as}f(s)

  • If L−1f(s)=F(t)L^{-1}{f(s)} = F(t), then

    L−1e−asf(s)=G(t)={F(t−a),amp;tgt;a 0,amp;tlt;aL^{-1}{e^{-as}f(s)} = G(t) = \begin{cases} F(t-a), &amp; t &gt; a \ 0, &amp; t &lt; a \end{cases}

Finding the formula by proof

  • If LF(t)=f(s)L{F(t)} = f(s) and

    G(t)={F(t−a),amp;tgt;a 0,amp;tlt;aG(t) = \begin{cases} F(t-a), &amp; t &gt; a \ 0, &amp; t &lt; a \end{cases}

    then LG(t)=e−asf(s)L{G(t)} = e^{-as}f(s)

  • If L−1f(s)=F(t)L^{-1}{f(s)} = F(t), then show that

    L−1e−asf(s)=G(t)={F(t−a),amp;tgt;a 0,amp;tlt;aL^{-1}{e^{-as}f(s)} = G(t) = \begin{cases} F(t-a), &amp; t &gt; a \ 0, &amp; t &lt; a \end{cases}

Proof
  • f(s)=∫0∞e−stF(t)dtf(s) = \int_0^{\infty} e^{-st} F(t) dt
  • e−asf(s)=∫<em>0∞e−ase−stF(t)dt=∫</em>0∞e−s(t+a)F(t)dte^{-as}f(s) = \int<em>0^{\infty} e^{-as} e^{-st} F(t) dt = \int</em>0^{\infty} e^{-s(t+a)} F(t) dt
  • Let z=t+az = t + a, then dt=dzdt = dz
  • e−asf(s)=∫a∞e−szF(z−a)dze^{-as}f(s) = \int_a^{\infty} e^{-sz} F(z-a) dz
  • e−asf(s)=∫<em>0ae−sz(0)dz+∫</em>a∞e−szF(z−a)dze^{-as}f(s) = \int<em>0^a e^{-sz} (0) dz + \int</em>a^{\infty} e^{-sz} F(z-a) dz
  • e−asf(s)=∫<em>0∞e−szG(z)dz=∫</em>0∞e−stG(t)dt=LG(t)e^{-as}f(s) = \int<em>0^{\infty} e^{-sz} G(z) dz = \int</em>0^{\infty} e^{-st} G(t) dt = L{G(t)}
  • L−1e−asf(s)=G(t)={F(t−a),amp;tgt;a 0,amp;tlt;aL^{-1}{e^{-as}f(s)} = G(t) = \begin{cases} F(t-a), &amp; t &gt; a \ 0, &amp; t &lt; a \end{cases}
Example

  • Find L−1e−3π2sss2+1L^{-1}{ e^{- {3\pi \over 2}s } {s \over s^2 + 1} }

  • We know that if L−1f(s)=F(t)L^{-1}{f(s)} = F(t), then

    L−1e−asf(s)={F(t−a),amp;tgt;a 0,amp;tlt;aL^{-1}{e^{-as}f(s)} = \begin{cases} F(t-a), &amp; t &gt; a \ 0, &amp; t &lt; a \end{cases}

  • L−1ss2+1=cos⁡t=F(t)L^{-1}{ s \over s^2 + 1} = \cos{t} = F(t)

  • L−1e−3π2sss2+1={cos⁡(t−3π2),amp;tgt;3π2 0,amp;tlt;3π2=cos⁡(t−3π2)U(t−3π2)L^{-1}{ e^{- {3\pi \over 2}s } {s \over s^2 + 1} } = \begin{cases} \cos(t - {3\pi \over 2}), &amp; t &gt; {3\pi \over 2} \ 0, &amp; t &lt; {3\pi \over 2} \end{cases} = \cos(t - {3\pi \over 2}) U(t - {3\pi \over 2})

  • Where U(t−a)={1,amp;tgt;a 0,amp;tlt;aU(t-a) = \begin{cases} 1, &amp; t &gt; a \ 0, &amp; t &lt; a \end{cases} is the Heaviside unit step function.

  • Find L−1e−7s(s−4)5L^{-1}{ e^{-7s} \over (s-4)^5 }

  • We know that if L−1f(s)=F(t)L^{-1}{f(s)} = F(t), then

    L−1e−asf(s)={F(t−a),amp;tgt;a 0,amp;tlt;aL^{-1}{e^{-as}f(s)} = \begin{cases} F(t-a), &amp; t &gt; a \ 0, &amp; t &lt; a \end{cases}

  • L−11(s−4)5=e4tL−11s5=e4tt424=F(t)L^{-1}{ 1 \over (s-4)^5 } = e^{4t} L^{-1}{ 1 \over s^5 } = e^{4t} {t^4 \over 24} = F(t)

  • L−1e−7s(s−4)5={124(t−7)4e4(t−7),amp;tgt;7 0,amp;tlt;7=124(t−7)4e4(t−7)U(t−7)L^{-1}{ e^{-7s} \over (s-4)^5 } = \begin{cases} {1 \over 24} (t-7)^4 e^{4(t-7)}, &amp; t &gt; 7 \ 0, &amp; t &lt; 7 \end{cases} = {1 \over 24} (t-7)^4 e^{4(t-7)} U(t-7)
    Find L−1(3s+1)e−7s(s+5)2+16L^{-1}{(3s+1)e^{-7s} \over (s+5)^2 + 16}

Change of Scale Property

Formula
  • If L−1f(s)=F(t)L^{-1}{f(s)} = F(t), then L−1f(ks)=1kF(tk)L^{-1}{f(ks)} = {1 \over k} F({t \over k})
  • We know that if LF(t)=f(s)L{F(t)} = f(s), then LF(at)=1af(sa)L{F(at)} = {1 \over a} f({s \over a})
  • If L−1f(s)=F(t)L^{-1}{f(s)} = F(t), then L−11af(sa)=F(at)L^{-1}{ {1 \over a} f({s \over a}) } = F(at)
  • L−1f(sa)=aF(at)L^{-1}{ f({s \over a}) } = a F(at)
  • Putting 1a=k{1 \over a} = k, we get L−1f(ks)=1kF(tk)L^{-1}{f(ks)} = {1 \over k} F({t \over k})
Finding the formula by proof
  • If L−1f(s)=F(t)L^{-1}{f(s)} = F(t), then show that L−1f(ks)=1kF(tk)L^{-1}{f(ks)} = {1 \over k} F({t \over k})
Proof
  • Given that f(s)=∫0∞e−stF(t)dtf(s) = \int_0^{\infty} e^{-st} F(t) dt
  • f(ks)=∫0∞e−kstF(t)dtf(ks) = \int_0^{\infty} e^{-kst} F(t) dt
  • Suppose z=ktz = kt, then dz=kdtdz = k dt
  • f(ks)=∫<em>0∞e−szF(zk)1kdz=1k∫</em>0∞e−szF(zk)dzf(ks) = \int<em>0^{\infty} e^{-sz} F({z \over k}) {1 \over k} dz = {1 \over k} \int</em>0^{\infty} e^{-sz} F({z \over k}) dz
  • f(ks)=1k∫0∞e−stF(tk)dt=1kLF(tk)f(ks) = {1 \over k} \int_0^{\infty} e^{-st} F({t \over k}) dt = {1 \over k} L{ F({t \over k}) }
  • L−1f(ks)=1kF(tk)L^{-1}{f(ks)} = {1 \over k} F({t \over k})
Example
  • Find L−1s3s2+16L^{-1}{ s \over 3s^2 + 16 }
  • We know that if L−1f(s)=F(t)L^{-1}{f(s)} = F(t), then L−1f(ks)=1kF(tk)L^{-1}{f(ks)} = {1 \over k} F({t \over k})
  • L−1ss2+42=cos⁡2tL^{-1}{ s \over s^2 + 4^2 } = \cos{2t}
  • L−1s3s2+16=13L−13s(3s)2+42=1313cos⁡(2t3)=13cos⁡(2t3)L^{-1}{ s \over 3s^2 + 16 } = {1 \over \sqrt{3}} L^{-1}{ {\sqrt{3}s } \over (\sqrt{3}s)^2 + 4^2 } = {1 \over \sqrt{3}} {1 \over \sqrt{3}} \cos( {2t \over \sqrt{3}} ) = {1 \over 3} \cos( {2t \over \sqrt{3}} )