inverse laplace of s-2/s^2-4s+13
Answers
Step-by-step explanation:
Method 1: Partial fraction decomposition and table memorization.
s+2s2−4s+13=s+2(s−2)2+9=s−2(s−2)2+32+4(s−2)2+32=s−2(s−2)2+32+433(s−2)2+32s+2s2−4s+13=s+2(s−2)2+9=s−2(s−2)2+32+4(s−2)2+32=s−2(s−2)2+32+433(s−2)2+32
Simply knowing that L[sin(kt)]=ks2+k2L[sin(kt)]=ks2+k2 , L[cos(kt)]=ss2+k2L[cos(kt)]=ss2+k2 , and L[f(t)]=F(s)⟹L[ektf(t)]=F(s−k)L[f(t)]=F(s)⟹L[ektf(t)]=F(s−k) , we get that
L−1[s−2(s−2)2+32+433(s−2)2+32]=e2tcos(3t)+43e2tsin(3t)L−1[s−2(s−2)2+32+433(s−2)2+32]=e2tcos(3t)+43e2tsin(3t) .
Method 2: Mellin’s inverse formula.
If lims→∞F(s)=0lims→∞F(s)=0 , and if cc is a sufficiently large real number (larger than the real parts of all the poles of F(s)F(s) ), then the inverse Laplace transform of F(s)F(s) is 12πilimT→∞∫c+iTc−iTF(s)estds12πilimT→∞∫c−iTc+iTF(s)estds , where TT runs over positive real numbers.
What’s better is, results in complex analysis give us a really easy way to compute this integral. We need only look at each pole, and use Cauchy’s integral formula f(a)=12πi∮γf(z)z−adzf(a)=12πi∮γf(z)z−adz . In our case, we shall assume that, instead of integrating over the vertical line c−iTc−iT to c+iTc+iT , we will integrate over a loop containing the poles. We can close it to the left to get a counterclockwise-oriented loop, and (since lims→∞F(s)=0lims→∞F(s)=0 ), the integral is unaffected by this.
If F(s)=s+2s2−4s+13F(s)=s+2s2−4s+13 , then the poles of FF are the zeroes of s2−4s+13=(s−2)2+32s2−4s+13=(s−2)2+32 , which are 2±3i2±3i . Thus we separate the loop into two loops, one containing 2+3i2+3i and the other containing 2−3i2−3i . We will compute each integral separately, then add them.
For the loop γ1γ1 containing 2+3i2+3i , we take f(z)=z+2z−(2−3i)eztf(z)=z+2z−(2−3i)ezt and a=2+3ia=2+3i in the formula f(a)=12πi∮γ1f(z)z−adzf(a)=12πi∮γ1f(z)z−adz , to get
12πi∮γ1F(z)eztdz=f(a)=a+2a−(2−3i)eat=4+3i6ieat=(12−23i)e(2+3i)t12πi∮γ1F(z)eztdz=f(a)=a+2a−(2−3i)eat=4+3i6ieat=(12−23i)e(2+3i)t .
A similar argument shows that if γ2γ2 is the loop containing 2−3i2−3i , then 12πi∮γ2F(z)eztdz=(12+23i)e(2−3i)t12πi∮γ2F(z)eztdz=(12+23i)e(2−3i)t , therefore, summing the results,
12πi∮γF(s)estds=(12−23i)e(2+3i)t+(12+23i)e(2−3i)t12πi∮γF(s)estds=(12−23i)e(2+3i)t+(12+23i)e(2−3i)t
We may finally use Euler’s formulas to sort this out:
(12−23i)e(2+3i)t+(12+23i)e(2−3i)t=e2t[(12−23i)e3it+(12+23i)e−3it]=e2t[12e3it−23ie3it+12e−3it+23ie−3it]=e2t[12e3it+12e−3it−23ie3it+23ie−3it]=e2t[e3it+e−3it2+43e3it−e−3it2i]=e2t[cos(3t)+43sin(3t)](12−23i)e(2+3i)t+(12+23i)e(2−3i)t=e2t[(12−23i)e3it+(12+23i)e−3it]=e2t[12e3it−23ie3it+12e−3it+23ie−3it]=e2t[12e3it+12e−3it−23ie3it+23ie−3it]=e2t[e3it+e−3it2+43e3it−e−3it2i]=e2t[cos(3t)+43sin(3t)]