Realization of a filter based on its transfer function How can we check whether the filter is realizable given its transfer function and What are the parameters the realization depends on?
Here is an example:

Show that a filter with transfer function
$H(ω)=\frac{2(10^5)}{ω^2 + 10^{10}} e^{-jωt_0 }$
is unrealizable. Can this filter be made approximately realizable by
choosing a sufficiently large t0.

--Modern digital and analog communication systems (3rd edition) - B.P. Lathi
Note: The question is of general nature. The example quoted above serves only as to make it clear, what is being asked. No one has to provide a solution for this problem.
A transfer function is called realizable if it can be implemented by a causal and stable system. The given frequency response is continuous and doesn't have any impulses, so the corresponding system is stable.

The transfer function (as a function of $s$) is given by

$$H(s)=\frac{a}{b-s^2}e^{-st_0}\tag{1}$$

The given frequency response is obtained form $(1)$ by substituting $s=j\omega$. Since $b&gt;0$, $H(s)$ has two real-valued poles at $\pm\sqrt{b}$, and, consequently, the corresponding impulse response is two-sided (i.e., non-causal).

Partial fraction expansion of $(1)$ gives

$$H(s)=\frac{a}{2\sqrt{b}}\left[\frac{1}{\sqrt{b}-s}+\frac{1}{\sqrt{b}+s}\right]e^{-st_0},\qquad -\sqrt{b}&lt;\textrm{Re}\{s\}&lt;\sqrt{b}\tag{2}$$

From $(2)$ it is straightforward to obtain the impulse response

$$\begin{align}h(t)=\mathcal{L}^{-1}\{H(s)\}&amp;=\frac{a}{2\sqrt{b}}\left[e^{\sqrt{b}(t-t_0)}u(-(t-t_0))+e^{-\sqrt{b}(t-t_0)}u(t-t_0)\right]\\&amp;=\frac{a}{2\sqrt{b}}e^{-\sqrt{b}|t-t_0|}\tag{3}\end{align}$$

No matter how large you choose $t_0$, the impulse response will never be causal. However, since the system is stable, the impulse response decays for $|t|\to\infty$, so you can approximate the actual impulse response well by sufficiently shifting it to the right (i.e., by choosing some large $t_0$), and truncating the non-causal part.

Also take a look at this related question and its answer.