An LTI system has a wide-sense stationary (WSS) input signal with zero mean. Its output is:

Analysis:

The output of filter is Y(t) = h(t) * X(t), where h(t) is impulse response of the filter. Let the output mean be μ_y.

μ_y = E[Y(t)] = E[h(t) * X(t)]

\( {{\rm{μ }}_{\rm{Y}}} = {\rm{E}}\left[ {\mathop \smallint \limits_{ - \infty }^\infty {\rm{h}}\left( {\rm{\tau }} \right){\rm{X}}\left( {{\rm{t}} - {\rm{\tau }}} \right){\rm{d\tau }}} \right]\)

\({{\rm{μ }}_{\rm{Y}}} = \mathop \smallint \limits_{ - \infty }^\infty {\rm{h}}\left( {\rm{\tau }} \right) \cdot {\rm{E}}\left[ {{\rm{X}}\left( {{\rm{t}} - {\rm{\tau }}} \right)} \right]{\rm{d\tau }} \)

Now, if X(t) is WSS, a shift in time does not affect its mean, i.e.

\({\rm{E}}\left[ {{\rm{X}}\left( {\rm{t}} \right)} \right] = {\rm{E}}\left[ {{\rm{X}}\left( {{\rm{t}} - {\rm{\tau }}} \right)} \right] = {{\rm{μ }}_{\rm{X}}}\) .

\({\rm{E}}\left[ {{\rm{Y}}\left( {\rm{t}} \right)} \right] = \mathop \smallint \limits_{ - \infty }^\infty {\rm{h}}\left( {\rm{\tau }} \right){{\rm{μ }}_{\rm{X}}}{\rm{d\tau }} \)

\(= {{\rm{μ }}_{\rm{X}}}\mathop \smallint \limits_{ - \infty }^\infty {\rm{h}}\left( {\rm{\tau }} \right){\rm{d\tau }}\)

\(\because {\mathop \smallint \limits_{ - \infty }^\infty {\rm{h}}\left( {\rm{\tau }} \right){\rm{d\tau }} = {\rm{H}}\left( 0 \right)}\)

\({{\rm{μ }}_{\rm{Y}}} = {{\rm{μ }}_{\rm{X}}} \cdot {\rm{H}}\left( 0 \right)\)

Since H(0) will be some constant, we conclude that the output will also have the same nature as the input signal, i.e. WSS and zero mean.