Convolution Theorem

1. Using the Convolution theorem, evaluate \[ L^{-1}\left\{ \frac{1}{(s+a)(s+b)} \right\} \]

Solution: Let \( F(s) = \frac{1}{s+a} \) and \( G(s) = \frac{1}{s+b} \).

Then, \[ L^{-1}\{F(s)\} = e^{-at} = f(t) \] and \[ L^{-1}\{G(s)\} = e^{-bt} = g(t). \]

By the Convolution theorem, \[ L^{-1}\{F(s)G(s)\} = \int_{0}^{t} f(u)g(t-u) \, du \] i.e., \[ L^{-1}\left\{ \frac{1}{s+a} \cdot \frac{1}{s+b} \right\} = \int_{0}^{t} e^{-au} e^{-b(t-u)} \, du = \int_{0}^{t} e^{-au} e^{-bt} e^{bu} \, du = e^{-bt} \int_{0}^{t} e^{(b-a)u} \, du \] \[ = e^{-bt} \left[ \frac{e^{(b-a)u}}{b-a} \right]_{0}^{t} = \frac{e^{-bt}}{b-a} \left[ e^{(b-a)t} - e^{0} \right] = \frac{e^{-bt}}{b-a} \left[ e^{bt} e^{-at} - 1 \right] \] \[ = \frac{1}{b-a} \left[ e^{-at} - e^{-bt} \right]. \]

Therefore, \[ L^{-1}\left\{ \frac{1}{(s+a)(s+b)} \right\} = \frac{1}{a-b} \left[ e^{-bt} - e^{-at} \right]. \]

2. Using the Convolution theorem, evaluate \[ L^{-1}\left\{ \frac{s}{(s+2)(s^2+9)} \right\} \]

Solution: Let \( F(s) = \frac{s}{s^2+9} \) and \( G(s) = \frac{1}{s+2} \).

Then, \[ L^{-1}\{F(s)\} = L^{-1}\left\{ \frac{s}{s^2+9} \right\} = \cos 3t = f(t) \] and \[ L^{-1}\{G(s)\} = L^{-1}\left\{ \frac{1}{s+2} \right\} = e^{-2t} = g(t). \]

By the Convolution theorem, \[ L^{-1}\{F(s)G(s)\} = \int_{0}^{t} f(u)g(t-u) \, du \] i.e., \[ L^{-1}\left\{ \frac{s}{s^2+9} \cdot \frac{1}{s+2} \right\} = \int_{0}^{t} \cos 3u \cdot e^{-2(t-u)} \, du = e^{-2t} \int_{0}^{t} e^{2u} \cos 3u \, du \] \[ = e^{-2t} \left[ \frac{e^{2u}}{2^2 + 3^2} \left( 2\cos 3u + 3\sin 3u \right) \right]_{0}^{t} \] \[ = \frac{e^{-2t}}{13} \left[ e^{2t} \left( 2\cos 3t + 3\sin 3t \right) - e^{0} \left( 2\cos 0 + 3\sin 0 \right) \right] \] \[ = \frac{1}{13} \left[ 2\cos 3t + 3\sin 3t - 2e^{-2t} \right]. \]

Therefore, \[ L^{-1}\left\{ \frac{s}{(s+2)(s^2+9)} \right\} = \frac{1}{13} \left[ 2\cos 3t + 3\sin 3t - 2e^{-2t} \right]. \]

Model Problems

Use the Convolution theorem to find the inverse Laplace transforms of the following:

1. \[ L^{-1}\left\{ \frac{1}{(s^2 + a^2)^2} \right\} \]
2. \[ L^{-1}\left\{ \frac{1}{(s-2)(s+2)^2} \right\} \]
3. \[ L^{-1}\left\{ \frac{s}{(s^2 + a^2)^2} \right\} \]
4. \[ L^{-1}\left\{ \frac{s^2}{(s^2 + a^2)(s^2 + b^2)} \right\} \]
5. \[ L^{-1}\left\{ \frac{1}{(s^2 + 1)(s^2 + 9)} \right\} \]
6. \[ L^{-1}\left\{ \frac{1}{s^2(s^2 + a^2)} \right\} \]
7. \[ L^{-1}\left\{ \frac{1}{s^2(s^2 + 1)^2} \right\} \]
8. \[ L^{-1}\left\{ \frac{1}{(s+1)(s^2 + 1)} \right\} \]
9. \[ L^{-1}\left\{ \frac{1}{s^2(s-a)} \right\} \]
10. \[ L^{-1}\left\{ \frac{1}{s^2(s^2 + 1)} \right\} \]

Answers

1. \[ \frac{1}{2a^3} (\sin at - at \cos at) \]
2. \[ \frac{1}{16} \left[ e^{2t} - (4t + 1)e^{-2t} \right] \]
3. \[ \frac{1}{2a} t \sin at \]
4. \[ \frac{a \cos at - b \sin bt}{a^2 - b^2} \]
5. \[ \frac{1}{24} (3 \sin t - \sin 3t) \]
6. \[ \frac{1}{a^3} (at - \sin at) \]
7. \[ e^{-t}(t + 2) + t - 2 \]
8. \[ \frac{1}{2} (\sin t - \cos t + e^{-t}) \]
9. \[ \frac{1}{a^2} \left[ e^{at} - at - 1 \right] \]
10. \[ t - \sin t \]