Transfer function of double exponential smoothing?
I would use the $\mathcal{Z}$-transform. Let me change the notation slightly by using $n$ as the discrete time index, lower case letters for time domain quantities, $x[n]$ as the observed input data sequence, and $y[n]$ as the output sequence, which is an estimate of $x[n+1]$. With this notation, the time domain equations are

$$\tag{1}\begin{align}
l[n]&amp;=ax[n]+(1-a)(l[n-1]+t[n-1])\\
t[n]&amp;=b(l[n]-l[n-1])+(1-b)t[n-1]\\
y[n]&amp;=l[n]+t[n]
\end{align}$$

Taking the $\mathcal{Z}$-transform of these 3 equations gives

$$\begin{align}\tag{2}
L(z)&amp;=aX(z)+(1-a)z^{-1}(L(z)+T(z))\\
T(z)&amp;=b(1-z^{-1})L(z)+(1-b)z^{-1}T(z)\\
Y(z)&amp;=L(z)+T(z)
\end{align}$$

From the second equation in (2) we can express $T(z)$ in terms of $L(z)$:

$$T(z)=\frac{b(1-z^{-1})}{1-(1-b)z^{-1}}L(z)\tag{3}$$

Plugging (3) into the first equation of (2), we can express $L(z)$ in terms of $X(z)$:

$$L(z)\left[1-(1-a)z^{-1}\left(1+\frac{b(1-z^{-1})}{1-(1-b)z^{-1}}\right)\right]=aX(z)$$

from which, after some algebra, you get

$$L(z)=\frac{a\left[1-(1-b)z^{-1}\right]}{1+[a(1+b)-2]z^{-1}+(1-a)z^{-2}}X(z)\tag{4}$$

Finally, plugging (3) into the last equation of (2) gives

$$Y(z)=\left[1+\frac{b(1-z^{-1})}{1-(1-b)z^{-1}}\right]L(z)=\frac{1+b-z^{-1}}{1-(1-b)z^{-1}}L(z)$$

and combining with (4) results in a relation between the output $Y(z)$ and the input $X(z)$:

$$Y(z)=\frac{a(1+b-z^{-1})}{1+[a(1+b)-2]z^{-1}+(1-a)z^{-2}}X(z)=H(z)X(z)\tag{5}$$

where $H(z)$ is the desired transfer function, which is of course a second order IIR filter.