# Laplace Transformation

I have the following expressions in the frequency domain and I want to transform them back to time domain. Are the following two correct?

$$\mathcal{L}^{-1}_s\left[\frac{a}{b} \left(\frac{1}{s+\frac{1}{c}}\right)\right](t)= \frac{a}{b}e^{-t/c}$$

$$\mathcal{L}^{-1}_s\left[\frac{a}{b} \left(\frac{1}{\left(s+\frac{1}{c}\right)\left(s+\frac{1}{d}\right)}\right)\right](t)= \frac{a}{b}\left(e^{-t/c}+e^{-t/d}\right)$$

• I would say - WolframAlpha answers: wolframalpha.com/input/… Aug 12 '16 at 16:52
• What's the difference? I thought that since a,b are constants does't really matter. So if I have $\frac{aw}{bz}$ instead of $\frac{a}{b}$ the inverse transform would be different?
– Gina
Aug 12 '16 at 16:59
• The first job is not difference. Aug 12 '16 at 17:13

• $$\mathcal{L}_s^{-1}\left[\text{C}\right]_{(t)}=\text{C}\cdot\mathcal{L}_s^{-1}\left[1\right]_{(t)}=\text{C}\delta(t)$$
• $$\mathcal{L}_s^{-1}\left[\frac{1}{s+\text{C}}\right]_{(t)}=e^{-\text{C}t}$$
1. $$\mathcal{L}_s^{-1}\left[\frac{\text{a}}{\text{b}}\left(\frac{1}{s+\frac{1}{\text{C}}}\right)\right]_{(t)}=\frac{\text{a}}{\text{b}}\cdot\mathcal{L}_s^{-1}\left[\frac{1}{s+\frac{1}{\text{C}}}\right]_{(t)}=\frac{\text{a}}{\text{b}}\cdot e^{-\frac{t}{\text{C}}}$$
2. $$\mathcal{L}_s^{-1}\left[\frac{\text{a}}{\text{b}}\left(\frac{1}{\left(s+\frac{1}{\text{C}}\right)\left(s+\frac{1}{\text{d}}\right)}\right)\right]_{(t)}=\frac{\text{a}}{\text{b}}\cdot\mathcal{L}_s^{-1}\left[\frac{1}{\left(s+\frac{1}{\text{C}}\right)\left(s+\frac{1}{\text{d}}\right)}\right]_{(t)}=$$ $$\frac{\text{a}}{\text{b}}\cdot\mathcal{L}_s^{-1}\left[\frac{1}{\left(\frac{1}{\text{C}}-\frac{1}{\text{d}}\right)\left(s+\frac{1}{\text{d}}\right)}+\frac{1}{\left(\frac{1}{\text{d}}-\frac{1}{\text{C}}\right)\left(s+\frac{1}{\text{C}}\right)}\right]_{(t)}=$$ $$\frac{\text{a}}{\text{b}}\cdot\left(\mathcal{L}_s^{-1}\left[\frac{1}{\left(\frac{1}{\text{C}}-\frac{1}{\text{d}}\right)\left(s+\frac{1}{\text{d}}\right)}\right]_{(t)}+\mathcal{L}_s^{-1}\left[\frac{1}{\left(\frac{1}{\text{d}}-\frac{1}{\text{C}}\right)\left(s+\frac{1}{\text{C}}\right)}\right]_{(t)}\right)=$$ $$\frac{\text{a}}{\text{b}}\cdot\left(\frac{1}{\frac{1}{\text{C}}-\frac{1}{\text{d}}}\cdot\mathcal{L}_s^{-1}\left[\frac{1}{s+\frac{1}{\text{d}}}\right]_{(t)}+\frac{1}{\frac{1}{\text{d}}-\frac{1}{\text{C}}}\cdot\mathcal{L}_s^{-1}\left[\frac{1}{s+\frac{1}{\text{C}}}\right]_{(t)}\right)=$$ $$\frac{\text{a}}{\text{b}}\cdot\left(\frac{1}{\frac{1}{\text{C}}-\frac{1}{\text{d}}}\cdot e^{-\frac{t}{\text{d}}}+\frac{1}{\frac{1}{\text{d}}-\frac{1}{\text{C}}}\cdot e^{-\frac{t}{\text{C}}}\right)$$