Exercise 5.9: Minimization of the MSE

Power-spectral densities
with the Wiener filter

Given a stochastic signal $s(t)$ of which only the power-spectral density $\rm (PSD)$ is known:

$${\it \Phi} _s (f) = \frac{\it{\Phi} _{\rm 0} }{1 + ( {f/f_0 } )^2 }.$$

This power-spectral density ${\it \Phi} _s (f)$ is shown in blue in the accompanying diagram.

The average power of $s(t)$ is obtained by integration over the power-spectral density:

$$P_s = \int_{ - \infty }^{ + \infty } {{\it \Phi} _s (f)}\, {\rm d} f = {\it \Phi} _0 \cdot f_0 \cdot {\rm{\pi }}.$$

Additively superimposed on this signal $s(t)$ is white noise $n(t)$ with noise power density ${\it \Phi}_n(f) = N_0/2.$
As an abbreviation, we use $Q = 2 \cdot {\it \Phi}_0/N_0$, where $Q$ could be interpreted as "quality".
Note that $Q$ does not represent a signal–to–noise power ratio.

In this exercise, we want to determine the frequency response $H(f)$ of a filter that minimizes the mean square error $\rm (MSE)$ between the useful signal $s(t)$ and the filter output signal $d(t)$:

$${\rm{MSE}} = \mathop {\lim }\limits_{T_{\rm M} \to \infty } \frac{1}{T_{\rm M} }\int_{ - T_{\rm M} /2}^{T_{\rm M} /2} {\left| {d(t) - s(t)} \right|^2 \, {\rm{d}}t.}$$

Notes:

The exercise belongs to the chapter Wiener–Kolmogorow Filter.
For the optimal frequency response, according to Wiener and Kolmogorov, the following applies in general:

$$H_{\rm WF} (f) = \frac{1}{{1 + {\it \Phi} _n (f)/{\it \Phi} _s (f)}}.$$

The following indefinite integral is given for solving:

$$\int {\frac{1}{a^2 + x^2 }} \, {\rm{d}}x ={1}/{a} \cdot \arctan \left( {{x}/{a}} \right).$$

Questions

	$H(f)$ is a Gaussian low-pass.
	$H(f)$ represents a matched filter.
	$H(f)$ represents a Wiener–Kolmogorov filter.

$H(f = 0) \ = \ $

$H(f = 2f_0)\ = \ $

${\rm MSE}/P_s \ = \ $

$Q_\text{min} \ = \ $

	A filter of the same shape $H(f) = K \cdot H_{\rm WF}(f)$ leads to the same result.
	The output signal $d(t)$ contains more higher frequency components when $Q$ is larger.

Solution

(1) Only the last solution is correct:

The task "Minimization of the mean square error" already points to the Wiener–Kolmogorow filter.
The matched filter, on the other hand, is used to concentrate the signal energy and thereby maximize the S/N ratio for a given time.

(2) For the optimal frequency response, according to Wiener and Kolmogorov, the following applies in general:

$$H(f) = H_{\rm WF} (f) = \frac{1}{{1 + {\it \Phi} _n (f)/{\it \Phi} _s (f)}}.$$

With the given power-spectral density, it is also possible to write for this:

$$H(f) = \frac{1}{{1 + {N_0 }/({{2{\it \Phi} _0 })}\cdot \left[ {1 + ( {f/f_0 } )^2 } \right]}} = \frac{1}{{1 + {1}/{Q}\cdot \left[ {1 + ( {f/f_0 } )^2 } \right]}}.$$

With $Q = 3$ it follows:

$$H( {f = 0} ) = \frac{1}{{1 + {1}/{Q}}} = \frac{Q}{Q + 1} \hspace{0.15cm}\underline {= 0.75},$$

$$H( {f = 2f_0 } ) = \frac{1}{{1 + {5}/{Q}}} = \frac{Q}{Q + 5} \hspace{0.15cm}\underline {= 0.375}.$$

(3) For the filter calculated in subtask (2), taking symmetry into account, the following holds:

$${\rm{MSE = }}\int_{-\infty}^{+\infty} H(f) \cdot {\it \Phi} _n (f) \,\, {\rm{d}}f = \int_{0}^{+\infty} \frac{N_0}{{1 + {N_0 }/({{2{\it \Phi} _0 })}\cdot \left[ {1 + ( {f/f_0 } )^2 } \right]}} \,\, {\rm{d}}f .$$

For this, with $Q = 2 \cdot {\it \Phi}_0/N_0$ and $a^2 = Q + 1$, we can also write:

$${\rm{MSE = }}\int_0^\infty {\frac{{2{\it \Phi} _0 }}{{ Q+1 + ( {f/f_0 })^2 }}} \,\, {\rm{d}}f = 2{\it \Phi} _0 \cdot f_0 \int_0^\infty {\frac{1}{a^2 + x^2 }}\,\, {\rm{d}}x.$$

This leads to the result with the integral given:

$${\rm{MSE}} = \frac{{2{\it \Phi} _0 f_0 }}{{\sqrt {1 + Q} }}\left( {\arctan ( \infty ) - \arctan ( 0 )} \right) = \frac{{{\it \Phi} _0 f_0 {\rm{\pi }}}}{{\sqrt {1 + Q} }}.$$

Normalizing MSE to the power $P_s$ of the signal $s(t)$, we obtain for $Q=3$:

$$\frac{\rm{MSE}}{P_s} = \frac{1}{{\sqrt {1 + Q} }} \hspace{0.15cm}\underline { = 0.5}.$$

(4) From the calculation in subtask (3), the condition $Q \ge 99$ ⇒ $Q_{\rm min} \hspace{0.15cm}\underline{= 99}$ follows directly for ${\rm MSE}/P_s \ge 0.1.$

The larger $Q$, the smaller the mean square error becomes.

(5) Only the second solution is correct:

A frequency response of the same shape as the Wiener–Kolmogorov filter ⇒ $H(f) = K \cdot H_{\rm WF}(f)$ with $K \ne 1$ always leads to large distortions.
This can be illustrated by the noise-free case $(Q \to \infty)$: $d(t) = K \cdot s(t)$ and the optimization task would be extremely poorly solved despite good conditions.
One might incorrectly conclude from the equation

Power-spectral density with the
Wiener–Kolmogorov filter

$${\rm{MSE}} = \int_{ - \infty }^{ + \infty } {H_{\rm WF} (f)} \cdot \it{\Phi} _n (f)\,\,{\rm{d}}f$$

that a filter $H(f) = 2 \cdot H_{\rm WF}(f)$ only doubles the mean squared error.

However, this is not the case, since $H(f)$ is then no longer a Wiener filter and the above equation is no longer applicable.

The second statement is true, as can be seen from the accompanying diagram.

The dots mark the frequency response $H_{\rm WF}(f)$ of the filter for $Q = 3$ and for $Q = 10$, resp.
For larger $Q (= 10)$, high components are attenuated less than for lower $Q (= 3)$.
Therefore, in the case of $Q = 10$, the filter output signal contains more higher frequency components, which are due to the noise $n(t)$.