Bode plot of an unstable system? I am a bit confused on how to sketch a bode plot for an unstable system? (being a/all  pole(s) lies on RHP). 
I tried plotting it in matlab,  but it doesn't resemble the output i was expecting  using "the rule of thumb" -  rule (poles => -20dB/decade and zero => +20dB/decade). 
so i was wondering if the method differs for a stable system and unstable system??
 A: You can construct the Bode plot of an unstable system in the sense that you can plot $\log|G(j\omega)|$ and $\arg G(j\omega)$, but, since the system is unstable, these quantities might not be informative (with some exceptions).
Let me be more specific.

Theorem 1. Let $G$ be the transfer function of a BIBO-stable linear dynamical system which is excited with a sinusoidal input,
 \begin{equation}\label{eq:freq:sin_input}
  u(t) = A\sin(\omega t).
 \end{equation}
 Then, at large times, the system response can be approximated by
 \begin{equation}
  x_{\infty}(t) = A R_{\infty} \sin(\omega t + \phi_0),
 \end{equation}
 with 
  \begin{align}
   R_{\infty} {}={}& |G(j\omega)|,
   \\
   \phi_0     {}={}& \arg G(j\omega).
  \end{align}
Proof. You can find a sketch of the proof here.

As you see, the theorem states explicitly that the system must be known to be BIBO-stable for $|G(j\omega)|$ and $\phi_0$ to be its amplitude ratio and phase lag respectively.

As a counterexample take $G(s) = \frac{1}{s-1}$ where $|G(j\omega)|$ is not, as one could expect, infinite.
The above theorem can be extended to the case of systems with a pole at zero (which are not BIBO stable):

Theorem 2. Suppose that a dynamical system has the transfer function 
 \begin{equation}\label{eq:thm_freq_resp_zero_pole:system}
  G(s) = \frac{H(s)}{s},
 \end{equation}
 where $H$ has all its poles in the open left complex plane.
 If the system is excited with  a sinusoidal input 
of the form $u(t) = A\sin(\omega t)$, $A,\omega>0$, then at large times, the output
can be approximated by 
\begin{equation}
 x_{\infty}(t) {}={}
\tfrac{A}{\omega}H(0)
 {}+{} AR_{\infty} \sin(\omega t + \phi_0),
\end{equation}
with 
 \begin{align}
  R_{\infty} {}={}& |G(j\omega)|
  \\
  \phi_0 {}={}& \arg G(j\omega).
 \end{align}
Proof. Use the partial fractions expansion of $H(s)/s$.

Note that the above theorem does not apply if you have a double pole at zero.
Another exception is the case where the system has simple pairs of complex conjugate poles on the imaginary axis, $\pm j b$, $b>0$, and $b\neq \omega$. Then, the system response is a sum of two frequencies ($\omega$ and $b$). This can be extended to the case where the system has $N$ such pairs, $\pm j b_\nu$, all different from one another and $\omega \neq b_{\nu}$ for all $\nu$. The pertinent result is a bit lengthy, so I'll skip it. 
