\begin{align*} s_{1}\left ( t\right ) & =A_{c}\left ( 1+k_{a}m\left ( t\right ) \right ) \cos \omega _{c}t\\ s_{2}\left ( t\right ) & =A_{c}\left ( 1+k_{a}m\left ( t\right ) \right ) \cos \omega _{c}t+w\left ( t\right ) \end{align*}
And
\begin{align*} SNR_{c} & =\frac {\left \langle \left ( A_{c}\left ( 1+k_{a}m\left ( t\right ) \right ) \cos \omega _{c}t\right ) ^{2}\right \rangle }{BN_{0}}\\ & =\frac {\frac {A_{c}^{2}}{2}\left \langle \left ( 1+k_{a}m\left ( t\right ) \right ) ^{2}\right \rangle }{BN_{0}}\\ & =\frac {\frac {A_{c}^{2}}{2}\left \langle 1+k_{a}^{2}m^{2}\left ( t\right ) +2k_{a}m\left ( t\right ) \right \rangle }{BN_{0}}\end{align*}
Now assuming \(\left \langle m\left ( t\right ) \right \rangle =0\), the above simplifies to
\[ SNR_{c}=\frac {\frac {A_{c}^{2}}{2}\left ( 1+k_{a}^{2}P_{m}\right ) }{BN_{0}}\]
Hence
\begin{align*} SNR_{i} & =SNR_{c}\frac {B}{B_{T}}\\ & =\frac {\frac {A_{c}^{2}}{2}\left ( 1+k_{a}^{2}P_{m}\right ) }{BN_{0}}\frac {B}{2B}\\ & =\frac {\frac {A_{c}^{2}}{2}\left ( 1+k_{a}^{2}P_{m}\right ) }{2BN_{0}}\end{align*}
Now find \(s_{3}\left ( t\right ) \)
\begin{align*} s_{3}\left ( t\right ) & =A_{c}\left ( 1+k_{a}m\left ( t\right ) \right ) \cos \omega _{c}t+\overset {\text {narrow band noise}}{\overbrace {n\left ( t\right ) }}\\ & =A_{c}\left ( 1+k_{a}m\left ( t\right ) \right ) \cos \omega _{c}t+\left [ n_{I}\left ( t\right ) \cos \omega _{c}t-n_{Q}\left ( t\right ) \sin \omega _{c}t\right ] \\ & =\overset {\text {in phase}}{\overbrace {\left [ A_{c}\left ( 1+k_{a}m\left ( t\right ) \right ) +n_{I}\left ( t\right ) \right ] }}\cos \omega _{c}t-\overset {\text {quadrature}}{\overbrace {n_{Q}\left ( t\right ) }}\sin \omega _{c}t \end{align*}
Now, to find \(s_{4}\left ( t\right ) \), which is the envelope of \(s_{3}\left ( t\right ) .\)
\begin{align*} s_{4}\left ( t\right ) & =\text {envelope}\left ( s_{3}\left ( t\right ) \right ) \\ & =\sqrt {\left ( s_{3}\right ) _{I}^{2}+\left ( s_{3}\right ) _{Q}^{2}}\\ & =\sqrt {\left ( A_{c}\left ( 1+k_{a}m\left ( t\right ) \right ) +n_{I}\left ( t\right ) \right ) ^{2}+n_{Q}^{2}\left ( t\right ) }\end{align*}
Now, assuming \(A_{c}\gg \left \vert n_{I}\left ( t\right ) \right \vert \) and \(A_{c}\gg \left \vert n_{Q}\left ( t\right ) \right \vert \), then the above simplifies to
\[ s_{4}\left ( t\right ) =A_{c}\left ( 1+k_{a}m\left ( t\right ) \right ) +n_{I}\left ( t\right ) \]
now apply the DC blocker, we
obtain
\[ s_{5}\left ( t\right ) =A_{c}k_{a}m\left ( t\right ) +n_{I}\left ( t\right ) \]
\[ SNR_{o}=\frac {\left \langle \left ( A_{c}k_{a}m\left ( t\right ) \right ) ^{2}\right \rangle }{E\left [ n_{I}^{2}\left ( t\right ) \right ] }=\frac {A_{c}^{2}k_{a}^{2}P_{m}}{2BN_{0}}\]
\[ \gamma =\frac {SNR_{o}}{SNR_{c}}=\frac {\frac {A_{c}^{2}k_{a}^{2}P_{m}}{2BN_{0}}}{\frac {\frac {A_{c}^{2}}{2}\left ( 1+k_{a}^{2}P_{m}\right ) }{BN_{0}}}=\frac {k_{a}^{2}P_{m}}{1+k_{a}^{2}P_{m}}\]
We notice, that for Large
\(SNR_{i}\), this detector gives the same result as coherent
detector.
For small \(SNR_{i}\), it is better to use the coherent detector than the envelope detector.