The difference here is that SSB signal has transmission bandwidth \(B_{T}=B\) and not \(2B\) as in all the
previous signals. Assume we are working with upper sideband. Analysis is the same for
lower sideband.
\[ s_{1}\left ( t\right ) =k\left [ m\left ( t\right ) \cos \omega _{c}t-\hat {m}\left ( t\right ) \sin \omega _{c}t\right ] \]
Where
\(k\) is a constant. Usually
\(\frac {A_{c}}{2}\) but we will leave it as
\(k\) for now.
\(\hat {m}\left ( t\right ) \) is the
Hilbert transform of
\(m\left ( t\right ) \) \[ s_{2}\left ( t\right ) =k\left [ m\left ( t\right ) \cos \omega _{c}t-\hat {m}\left ( t\right ) \sin \omega _{c}t\right ] +w\left ( t\right ) \]
\begin{align*} SNR_{c} & =\frac {\left \langle \left ( k\left [ m\left ( t\right ) \cos \omega _{c}t-\hat {m}\left ( t\right ) \sin \omega _{c}t\right ] \right ) ^{2}\right \rangle }{BN_{0}}\\ & =\frac {k^{2}\left \langle m^{2}\left ( t\right ) \cos ^{2}\omega _{c}t\right \rangle +k^{2}\left \langle \hat {m}^{2}\left ( t\right ) \sin ^{2}\omega _{c}t\right \rangle -2k^{2}\left \langle m\left ( t\right ) \hat {m}\left ( t\right ) \cos \left ( \omega _{c}t\right ) \sin \left ( \omega _{c}t\right ) \right \rangle }{BN_{0}}\\ & =\frac {k^{2}\left \langle m^{2}\left ( t\right ) \right \rangle \left \langle \cos ^{2}\omega _{c}t\right \rangle +k^{2}\left \langle m^{2}\left ( t\right ) \right \rangle \left \langle \sin ^{2}\omega _{c}t\right \rangle -2k^{2}\left \langle m\left ( t\right ) \hat {m}\left ( t\right ) \cos \left ( \omega _{c}t\right ) \sin \left ( \omega _{c}t\right ) \right \rangle }{BN_{0}}\end{align*}
Assume \(\left \langle m\left ( t\right ) \right \rangle =0\) , we obtain
\begin{align*} SNR_{c} & =\frac {k^{2}\frac {P_{m}}{2}+k^{2}\frac {P_{m}}{2}}{BN_{0}}\\ & =\frac {k^{2}P_{m}}{BN_{0}}\end{align*}
\[ s_{3}\left ( t\right ) =k\left [ m\left ( t\right ) \cos \omega _{c}t-\hat {m}\left ( t\right ) \sin \omega _{c}t\right ] +n\left ( t\right ) \]
Hence
\begin{align*} SNR_{i} & =SNR_{c}\frac {B}{B_{T}}\\ & =\frac {k^{2}P_{m}}{BN_{0}}\frac {B}{B}\\ & =\frac {k^{2}P_{m}}{BN_{0}}\end{align*}
\begin{align*} s_{4}\left ( t\right ) & =\left [ k\left [ m\left ( t\right ) \cos \omega _{c}t-\hat {m}\left ( t\right ) \sin \omega _{c}t\right ] +n\left ( t\right ) \right ] A_{c}^{^{\prime }}\cos \omega _{c}t\\ & =A_{c}^{^{\prime }}km\left ( t\right ) \cos ^{2}\omega _{c}t-A_{c}^{^{\prime }}k\hat {m}\left ( t\right ) \sin \omega _{c}t\cos \omega _{c}t+A_{c}^{^{\prime }}\left [ n_{I}\left ( t\right ) \cos \omega _{c}t-n_{Q}\left ( t\right ) \sin \omega _{c}t\right ] \cos \omega _{c}t\\ & =A_{c}^{^{\prime }}km\left ( t\right ) \left ( \frac {1}{2}+\frac {1}{2}\cos 2\omega _{c}t\right ) -A_{c}^{^{\prime }}k\hat {m}\left ( t\right ) \frac {1}{2}\left ( \sin \left ( 0\right ) +\sin \left ( 2\omega _{c}t\right ) \right ) \\ & +A_{c}^{^{\prime }}\left [ n_{I}\left ( t\right ) \cos ^{2}\omega _{c}t-n_{Q}\left ( t\right ) \sin \omega _{c}t\cos \omega _{c}t\right ] \\ & =\frac {1}{2}A_{c}^{^{\prime }}km\left ( t\right ) +\frac {1}{2}A_{c}^{^{\prime }}km\left ( t\right ) \cos 2\omega _{c}t-A_{c}^{^{\prime }}k\hat {m}\left ( t\right ) \frac {1}{2}\sin \left ( 2\omega _{c}t\right ) \\ & +A_{c}^{^{\prime }}\left [ n_{I}\left ( t\right ) \left ( \frac {1}{2}+\frac {1}{2}\cos 2\omega _{c}t\right ) -n_{Q}\left ( t\right ) \frac {1}{2}\left ( \sin \left ( 0\right ) +\sin \left ( 2\omega _{c}t\right ) \right ) \right ] \\ & =\frac {1}{2}A_{c}^{^{\prime }}km\left ( t\right ) +\frac {1}{2}A_{c}^{^{\prime }}km\left ( t\right ) \cos 2\omega _{c}t-\frac {1}{2}A_{c}^{^{\prime }}k\hat {m}\left ( t\right ) \sin \left ( 2\omega _{c}t\right ) \\ & +\frac {A_{c}^{^{\prime }}}{2}n_{I}\left ( t\right ) +\frac {A_{c}^{^{\prime }}}{2}n_{I}\left ( t\right ) \cos 2\omega _{c}t-\frac {A_{c}^{^{\prime }}}{2}n_{Q}\left ( t\right ) \sin 2\omega _{c}t \end{align*}
After low pass filter, we obtain
\[ s_{5}\left ( t\right ) =\frac {1}{2}A_{c}^{^{\prime }}km\left ( t\right ) +\frac {A_{c}^{^{\prime }}}{2}n_{I}\left ( t\right ) \]
Hence,
\begin{align*} SNR_{o} & =\frac {\left \langle \left ( \frac {1}{2}A_{c}^{^{\prime }}km\left ( t\right ) \right ) ^{2}\right \rangle }{E\left ( \left [ \frac {A_{c}^{^{\prime }}}{2}n_{I}\left ( t\right ) \right ] ^{2}\right ) }\\ & =\frac {\frac {1}{4}\left ( A_{c}^{^{\prime }}\right ) ^{2}k^{2}P_{m}}{\frac {1}{4}\left ( A_{c}^{^{\prime }}\right ) ^{2}N_{0}B}\\ & =\frac {k^{2}P_{m}}{N_{0}B}\end{align*}
Hence
\begin{align*} \frac {SNR_{o}}{SNR_{i}} & =\frac {\frac {k^{2}P_{m}}{N_{0}B}}{\frac {k^{2}P_{m}}{BN_{0}}}\\ & =1 \end{align*}
Hence
\begin{align*} \gamma & =\frac {SNR_{o}}{SNR_{c}}\\ & =\frac {\frac {k^{2}P_{m}}{N_{0}B}}{\frac {k^{2}P_{m}}{BN_{0}}}\\ & =1 \end{align*}