Multiple Drug Doses

This project is based on a project by Brian Winkle. https://www.simiode.org/resources/2981/supportingdocs

Inverse Transform Step

Match the expression of the Laplace tranform with known transforms and work backwards.

$$Q(s)=\frac{1}{(s)(s+k)}-e^{-s/2}\cdot \frac{1}{(s)(s+k)}+\frac{2}{(s+k)}$$

We do a little partial fractions:

$$\frac{1}{(s)(s+k)}=\frac{1/k}{s}-\frac{1/k}{s+k}$$
In [31]:
syms s k f(s)
f(s)=1/(s*(s+k))
partfrac(f(s),s)

 
f(s) =
 
1/(s*(k + s))
 
 
ans =
 
1/(k*s) - 1/(k*(k + s))
\begin{align*} Q(s)&=\frac{1}{(s)(s+k)}-e^{-s/2}\cdot \frac{1}{(s)(s+k)}+\frac{2}{(s+k)}\\ &=\frac{1}{k}\cdot \boxed{\frac{1}{s}}-\frac{1}{k}\cdot \boxed{\frac{1}{s+k}}-\frac{1}{k}\cdot \boxed{e^{-s/2}\frac{1}{s}}+\frac{1}{k}\cdot \boxed{e^{-s/2}\frac{1}{s+k}}\\ &\phantom{\text{move over}} +2 \boxed{\frac{1}{s+k}}\\ \end{align*}
\begin{align*} h(t) & \xrightarrow{\mathscr{L}} \frac{1}{s}\\ e^{-kt}& \xrightarrow{\mathscr{L}} \frac{1}{s+k}\\ h\left(t-\frac{1}{2}\right)& \xrightarrow{\mathscr{L}} e^{-\frac{s}{2}}\frac{1}{s}\\ h\left(t-\frac{1}{2}\right)e^{-k\left(t-\frac{1}{2}\right)}& \xrightarrow{\mathscr{L}} e^{-\frac{s}{2}}\frac{1}{s+k}\\ \end{align*}

$$\begin{array}{cccc} &Q(s)&&\\ &\shortparallel&&\\ &\frac{1}{k}\cdot \frac{1}{s}& \xleftarrow{\mathscr{L}}&\frac{1}{k}\cdot h(t)\\ &-\frac{1}{k}\cdot \frac{1}{s+k}&\xleftarrow{\mathscr{L}}&-\frac{1}{k}\cdot e^{-kt}\\ &-\frac{1}{k}\cdot e^{-s/2}\frac{1}{s}&\xleftarrow{\mathscr{L}}&-\frac{1}{k} h\left(t-\frac{1}{2}\right)\\ &+\frac{1}{k}\cdot e^{-s/2}\frac{1}{s+k}&\xleftarrow{\mathscr{L}}&+\frac{1}{k}\cdot h\left(t-\frac{1}{2}\right)e^{-k\left(t-\frac{1}{2}\right)}\\ &+2 \frac{1}{s+k}&\xleftarrow{\mathscr{L}}& +2 e^{-kt}\\ && &\shortparallel\\ &&&q(t)\\ \end{array}$$

Which isn't really as bad as it looks. The function $h(t)$ is really just 1 since we're only concerned with $t>0$. The function $h(t-1/2)$ just means that piece "turns on" at $\ t=1/2$.

\begin{align*} q(t)&=\frac{1}{k}\cdot h(t)-\frac{1}{k}\cdot e^{-kt}-\frac{1}{k} h\left(t-\frac{1}{2}\right)+\frac{1}{k}\cdot h\left(t-\frac{1}{2}\right)e^{-k\left(t-\frac{1}{2}\right)}+2 e^{-kt}\\ &=\frac{1}{k}-\frac{1}{k}\cdot e^{-kt}+2 e^{-kt}+h\left(t-\frac{1}{2}\right)\cdot \left(\frac{1}{k}\cdot e^{-k(t-\frac{1}{2})} -\frac{1}{k} \right) \end{align*}
In [57]:
Qs
qt=ilaplace(Qs, s, t)
pretty(qt)
qtnum=subs(qt, k, -log(1/2))
fplot(qtnum, [0,5])

 
Qs =
 
(1/s - exp(-s/2)/s + 2)/(k + s)
 
 
qt =
 
2*exp(-k*t) + heaviside(t - 1/2)*(exp(-k*(t - 1/2))/k - 1/k) + 1/k - exp(-k*t)/k
 
                                 /    /    /     1 \ \     \
                                 | exp| -k | t - - | |     |
                       /     1 \ |    \    \     2 / /   1 |   1   exp(-k t)
2 exp(-k t) + heaviside| t - - | | ------------------- - - | + - - ---------
                       \     2 / \          k            k /   k       k

 
qtnum =
 
(3479430281589726*exp(-(6243314768165359*t)/9007199254740992))/6243314768165359 + heaviside(t - 1/2)*((9007199254740992*exp(6243314768165359/18014398509481984 - (6243314768165359*t)/9007199254740992))/6243314768165359 - 9007199254740992/6243314768165359) + 9007199254740992/6243314768165359
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