WSS Process On Causal LTI System

Consider a real LTI system with a WSS process $x(t)$ as input and WSS process $y(t)$ as output. Base on the WSS correlation properties，we get these equations

$egin{align*}
&Time-Domain  &:&R_{yy}( au) &= h( au)*h(- au)*R_{xx}( au)\
&Frequency-Domain &:&S_{yy}(jomega) &= H(jomega)H^*(jomega)S_{xx}(jomega)
end{align*}$

The way we get $x(t)$ from white noise is no different. Let the input be a white noise with PSD $W_{xx}(jomega)=1$，which means that its auto-correlation is $delta$. Then the system can be seen to be a modeling filter denoted by $m(t)$ in time-domain and $M_{xx}(jomega)$ in frequency-domain.

This can be summarized as the following equations

$egin{align*}
&Time-Domain  &:&R_{xx}( au) &= m_{xx}( au)*m_{xx}(- au)\
&Frequency-Domain &:&S_{xx}(jomega) &= M_{xx}(jomega)M_{xx}^*(jomega)
end{align*}$

Now, to think of a system which is the cascade of the filter $m_{xx}( au)$ and $m_{xx}(- au)$.

The filter $m_{xx}( au)$ can be decomposed into the sum of an even part $m_e( au)$, and an odd part $m_o( au)$

$m_{xx}( au) = m_e( au)+m_o( au)$

where

$egin{align*}
m_e( au)&= frac{1}{2}(m_{xx}( au)+m_{xx}(- au))\
m_o( au)&= frac{1}{2}(m_{xx}( au)-m_{xx}(- au))\
end{align*}$

If the filter $m_{xx}( au)$ is causal, in order that $m_{xx}( au)=0$ for $ au<0$, we require that

$m_o( au) = left{egin{matrix}
m_e( au), & au >0 \
-m_e( au), & au<0
end{matrix} ight. =sgn( au)m_e( au)$

Then the causal impulse response may be written in terms of the even function alone

$egin{align*}
&m_{xx}( au) &= m_e( au)+sgn( au)m_e( au)\
&m_{xx}(- au) &= m_e( au)-sgn( au)m_e( au)
end{align*}$

For example

In the frequency domain, the frequency response function $M_{xx}(jomega)$ can also be expressed in terms of the even function alone

$egin{align*}
M_{xx}(jomega) &= mathcal{F}Big{m_e( au)Big}+mathcal{F}Big{sgn( au)m_e( au)Big}\
&= mathcal{F}Big{m_e( au)Big}+frac{1}{2pi}mathcal{F}Big{sgn( au)Big}otimes mathcal{F}Big{m_e( au)Big}qquad convolution theorem\
&= M_e(jomega) + jleft[frac{1}{piomega}otimes M_e(jomega) ight]\
&= M_e(jomega) + jwidehat{M}_e(jomega) qquad widehat{M}_e(jomega) means Hilbert Transform of M_e(jomega)
end{align*}$

The frequency response function $M_{xx}^*(jomega)$ can be derived with the same argument.

$displaystyle{M_{xx}^*(jomega) = M_e(jomega) - jwidehat{M}_e(jomega)}$

Thus

$egin{align*}
S_{xx}(jomega)&=M_{xx}(jomega)M_{xx}^*(jomega)\
&=Big{M_e(jomega)+jwidehat{M}_e(jomega)Big}Big{M_e(jomega)-jwidehat{M}_e(jomega)Big}\
&=M_e^2(jomega)+widehat{M}_e^2(jomega)
end{align*}$

Back to the WSS process, $S_{xx}(jomega)$ is the PSD of $x(t)$. For real WSS process, the PSD should meet 3 condictions:even, real, non-negative. These condictions can be easily varified on $M_e^2(jomega)+widehat{M}_e^2(jomega)$.

1. $M_e^2(jomega)+widehat{M}_e^2(jomega)$ is real, because it is the sum of square
2. $M_e^2(jomega)+widehat{M}_e^2(jomega)$ is non-negative, because it is the sum of square
3. The first term is the square of FT of real even function, so that $M_e(jomega)$ is real and even. The second term is the Hilbert transform of the real even function $M_e(jomega)$. According to the Hilbert transform duality, $widehat{M}_e(jomega)$ is odd, which means that $widehat{M}_e^2(jomega)$ is even. With these understanding, it is evident that $M_e^2(jomega)+widehat{M}_e^2(jomega)$ is even.

Reference :

MIT Open course 2.161 Signal Processing: Continuous and Discrete: Determining a System's Causality from its Frequency Response

Alan V. Oppenheim: Signals, Systems and Inference, Chapter 11: Wiener Filtering