Impulse response functions? I have a very basic model about a small open economy. This is a summary: 


  $$
E_0\sum_{t=0}^{\infty}\theta_t U(c_t, h_t)
$$
  utility function:
  $$
d_t = (1 + r_{t-1})d_{t-1} - y_t + c_t + i_t + \Phi(k_{t-1} - k_t)
$$
  evolution of foreign debt:
  $$
y_t = A_t F(k_t, h_t)
$$
  output:
  $$
k_{t+1} = i_t + (1 - \delta)k_t
$$
  stock of capital evolves according to:
  
  The first order conditions of the household's maximization problem are:
  \begin{align}
\lambda_t = \beta(c_t, h_t)(1 + r_t)E_t\lambda_{t+1}\\
\lambda_t = U_c(c_t,h_t) - \eta_t\beta_c(c_t,h_t)\\
\eta_t = -E_t U(c_{t+1}, h_{t+1}) + E_t \eta_{t+1} \beta(c_{t+1}, h_{t+1}) \\
-U_h(c_t,h_t) + \eta_t\beta_h(c_t,h_t) = \lambda_tA_tF_h(k_t,h_t)\\
\lambda_t[1 + \Phi'(k_{t+1} - k_t)] = \beta(c_t,h_t)E_t\lambda_{t+1}[A_{t+1}F_k(k_{t+1},h_{t+1}) + 1 - \delta + \Phi'(k_{t+2} - k_{t+1})]
\end{align}
  Law of motions of the productivity shock:
  $$
\ln A_{t+1} = \rho \ln A_t + \epsilon_{t+1}; ~~~~ \epsilon_{t+1} \sim { NIID}(0,\sigma_\epsilon^2); ~~~~ t\ge 0
$$
  we parametrize this as 
  \begin{align}
U(c,h) = \frac{[c - \omega^{-1}h^\omega]^{1 - \gamma} - 1}{1 - \gamma}\\
\beta(c,h) = [1 + c - \omega^{-1}h^\omega]^{-\psi_1} \\
F(k,h) = k^{\alpha}h^{1-\alpha} \\
\Phi(x) = \frac{\phi}{2}x^2; ~~~~ \phi&gt;0
\end{align}
  $$
\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline
\gamma &amp; \omega &amp;  \psi_1 &amp; \alpha &amp; \phi &amp; r &amp; \delta &amp; \rho &amp; \sigma_{\epsilon} \\\hline
2 &amp; 1.455 &amp; 0.11 &amp; 0.32 &amp; 0.028 &amp; 0.04 &amp; 0.1 &amp; 0.42 &amp; 0.0129 \\ \hline
\end{array}
$$


I have computed the steady state, but how can I compute the impulse response function?
Start by establishing $ \epsilon _1 = \epsilon , \epsilon_{t&gt;1}=0$ . Now, epsilon will be your impulse.
You can calculate an expression for $A_t,\bar{A}_t\ \forall t$ . From here, you can either:


Work recursively with each variable and provide an analytic function for them, for every moment on time.
Calculate a value if you are simulating it, using the policy functions at each step.



Notice that, if you log-linearized your equations, $\epsilon=1$ would mean a 1% deviation from the steady state.