Examples of The Inverse Laplace Transform

Example 1: Find the inverse Laplace transform of$$F(s)=\frac{3}{s}-\frac{5}{s+1}+\frac{6}{s^{2}+4}$$ Solution: The inverse transform is given by where Table $ 2 $ has been consulted for the inverse of each term.

Example 2: Find $ f(t) $ given that Solution: Unlike in the previous example where the partial fractions have been provided, we first need to determine the partial fractions. Since there are three poles, we let where $ A, B $, and $ C $ are the constants to be determined. We can find the constants using two approaches. METHOD I Residue method: METHOD 2 Algebraic method: Multiplying both sides of Eq. (1) by $ s(s+2)(s+3) $ gives or Equating the coefficients of like powers of $ s $ gives Thus $ A=2, B=-8, C=7 $, and Eq. (1) becomes By finding the inverse transform of each term, we obtain

Example 3: Calculate $ v(t) $ given that Solution: While the previous example is on simple roots, this example is on repeated roots. Let METHOD I Residue method: METHOD 2 Algebraic method: Multiplying Eq. (2) by $ s(s+1)(s+2)^{2} $, we obtain or Equating coefficients, Solving these simultaneous equations gives $ A=1, B=-14, C=22 $, $ D=13 $, so that Taking the inverse transform of each term, we get

Example 4: Find the inverse transform of the frequency-domain function: Solution: In this example, $ H(s) $ has a pair of complex poles at $ s^{2}+8 s+25=0 $ or $ s=-4 \pm j 3 $. We let We now determine the expansion coefficients in two ways. METHOD I Combination of methods: We can obtain $ A $ using the method of residue, Although $ B $ and $ C $ can be obtained using the method of residue, we will not do so, to avoid complex algebra. Rather, we can substitute two specific values of $ s $ [say $ s=0,1 $, which are not poles of $ F(s) $ ] into Eq. (3). This will give us two simultaneous equations from which to find $ B $ and $ C $. If we let $ s=0 $ in Eq. (3), we obtain or Since $ A=2 $, Eq. (4) gives $ C=-10 $. Substituting $ s=1 $ into Eq. (3) gives or But $ A=2, C=-10 $, so that Eq. (15.11.3) gives $ B=-2 $ METHOD 2 Algebraic method: Multiplying both sides of Eq. $ (3) $ by $ (s+3)\left(s^{2}+8 s+25\right) $ yields Equating coefficients, That is, $ B=-2, C=-10 $. Thus Taking the inverse of each term, we obtain It is alright to leave the result this way. However, we can combine the cosine and sine terms as To obtain Eq. (6) from Eq. (5), we apply Eq. (A). Next, we determine the coefficient $ A $ and the phase angle $ \theta $ : Thus,