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

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

\begin{equation} \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) \end{equation}

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  • $\begingroup$ I would say - WolframAlpha answers: wolframalpha.com/input/… $\endgroup$
    – georg
    Aug 12 '16 at 16:52
  • $\begingroup$ 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? $\endgroup$
    – Gina
    Aug 12 '16 at 16:59
  • $\begingroup$ The first job is not difference. $\endgroup$
    – georg
    Aug 12 '16 at 17:13
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Notice:

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

So, we get:

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