Exercise 5.8Z: Matched Filter for Rectangular PSD

Spectrum $G(f)\ \bullet\!\!\!-\!\!\!-\!\!\!-\!\!\circ\, \ g(t)$;
power-sprectral density ${\it \Phi}_n (f)$

The interference power-spectral density effective on a system can be assumed to be constant in range:

$$\it{\Phi} _n \left( f \right) = \left\{ \begin{array}{l} N_0 /2 \\ N_1 /2 \\ \end{array} \right.\quad \begin{array}{*{20}c} \rm{for} \\ \rm{for} \\\end{array}\quad \begin{array}{*{20}c} {\left| f \right| \le f_{\rm N} ,} \\ {\left| f \right| > f_{\rm N} .} \\\end{array}$$

Here, let the power-spectral density $N_1$ in the outer region $|f| > f_{\rm N}$ always be much smaller than $N_0$.
For example, use the following values:

$$N_0 = 2 \cdot 10^{ - 6} \;{\rm{V}}^{\rm{2}} /{\rm{Hz}},\quad N_1 = 2 \cdot 10^{ - 8} \;{\rm{V}}^{\rm{2}}/ {\rm{Hz}}.$$

Such an interference signal $n(t)$ occurs, for example, when the dominant interference source contains only components below the frequency limit $f_{\rm N}$. Due to the unavoidable thermal noise, also for $|f| > f_{\rm N}$ interference power-spectral density is ${\it \Phi}_n(f) \ne 0$.

Further, it holds:

Let the spectrum $G(f)$ of the useful signal $g(t)$ also be rectangular according to the above diagram.
Therefore, $g(t)$ has the following curve with $\Delta f = 2 \cdot f_{\rm G}$:

$$g(t) = G_0 \cdot \Delta f \cdot {\mathop{\rm sinc}\nolimits} \left( { \Delta f \cdot t} \right).$$

Let the frequency response $H_{\rm E}(f)$ of the receiver filter (German: "Empfangsfilter" ⇒ subscript "E") be optimally matched to the spectrum $G(f)$ and the interference power-spectral density ${\it \Phi}_n(f)$.
That is, let $H_{\rm E}(f) = H_{\rm MF}(f)$ ⇒ "Matched Filter".
Let the detection time be simplified $T_{\rm D} = 0$ (acausal system description).

Notes:

The exercise belongs to the chapter Matched Filter.

For numerical calculations always use the numerical values

$$G_0 = 10^{ - 4} \;{\rm{V/Hz}}{\rm{, }}\quad \Delta f = 10\;{\rm{kHz}}.$$

Questions

Solution

(1) Solutions 1 and 3 are correct:

For all frequencies $|f| > f_{\rm G}$ at which the useful signal $d_{\rm S}(t)$ has spectral components $(G_d(f) \ne 0)$,
the interference power-spectral density is ${\it}\Phi_n(f) = N_0/2$.
Thus, the frequency response of the matched filter is, assuming $T_{\rm D} = 0$:

$$H_{\rm MF} (f) = K_{\rm MF} \cdot G(f).$$

In this case, the optimal frequency response $H_{\rm MF}(f)$, just like $G(f)$, is rectangular with width $\Delta f$.
Thus, for the useful component of the matched filter output signal $d(t)$ holds:

$$d_{\rm S}(t)\quad \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \quad G(f) \cdot H_{\rm MF} (f).$$

The product of two rectangular functions of equal width again yields a rectangular function.
It further follows that the output pulse of the matched filter is also $\rm sinc$–shaped.

(2) With white noise one obtains:

$$\rho _d = \frac{1}{N_0 /2}\int_{ - \infty }^{ + \infty } {\left| {G(f)} \right|^2 \, {\rm{d}}f.}$$

The integral yields the value $G_0^2 \cdot \Delta f$. It follows that:

Regarding subtask (3)

$$\rho _d = \frac{G_0 ^2 \cdot \Delta f }{N_0 /2} = \frac{ 10^{ - 8}\,(\rm V/Hz)^2 \;\cdot10^4 \;{\rm{Hz}} }{10^{ - 6}\,\rm V^2/Hz} = 10^2 \quad \Rightarrow \quad 10\lg \rho _d \hspace{0.15cm}\underline { = 20\;{\rm{dB}}}.$$

(3) In general, the SNR for colored interference is:

$$\rho _d = 2 \cdot \int_0^\infty \frac{\left| {G(f)} \right|^2 }{{\it \Phi}_n (f)} \, {\rm{d}}f.$$

As can be seen from the accompanying qualitative diagram, that the integrand is piecewise constant for the given frequency responses.
Thus, with $f_{\rm G} = 5 \; \rm kHz$ and $f_{\rm N} = f_{\rm G}/2 = 2.5 \; \rm kHz$, we obtain:

$$\rho _d = 2 \cdot 2.5\;{\rm{kHz}}\left( { \frac{10^{ - 2}}{\rm{Hz}} + \frac{1}{{{\rm{Hz}}}} } \right) = 5.05 \cdot 10^3 \quad \Rightarrow \quad 10\cdot\lg \rho _d \hspace{0.15cm}\underline {= 37.03\;{\rm{dB}}}.$$

Interpretation:

The matched filter frequency response $H_{\rm MF}(f)$ has exactly the same shape as the integrand sketched above.
If the constant $K_{\rm MF}$ is chosen (arbitrarily) so that $H_{\rm MF}(f) = 1$ in the range $f_{\rm N} \le |f| \le f_{\rm G}$, then for low frequencies $(|f| < f_{\rm N})$: $H_{\rm MF}(f) = 0.01$. This means: The matched filter favors those frequencies that are only slightly affected by the interference ${\it \Phi}_n(f)$.
If instead we would use a filter $H(f)$, which gives equal weight to all frequencies up to and including $f_{\rm G}$ (purple curve in the sketch below),
the following ratios would result:

$$d_{\rm S}( {T_{\rm D} } ) = G_0 \cdot 2 \cdot f_{\rm G} = 1\;{\rm{V}}, \quad \sigma _d ^2 = 10^{ - 6} \frac{{{\rm{V}}^{\rm{2}} }}{{{\rm{Hz}}}} \cdot f_{\rm G} + 10^{ - 8} \frac{{{\rm{V}}^{\rm{2}} }}{{{\rm{Hz}}}} \cdot ( {f_{\rm G} - f_{\rm N} } ) = 2.5 \cdot 1.01 \cdot 10^{ - 3} \;{\rm{V}}^{\rm{2}}$$

$$ \Rightarrow \hspace{0.3cm} \rho _d = \frac {d_{\rm S}( {T_{\rm D} } )^2}{\sigma _d ^2} = \frac{1 \;{\rm{V}}^{\rm{2}}}{2.525 \cdot 10^{ - 3} \;{\rm{V}}^{\rm{2}}} = 396 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}10 \cdot {\rm lg} \, \rho _d = 25.98 \, {\rm dB}.$$

The signal–to–noise ratio is thus about $11\ \rm dB$ worse than when using the matched filter for colored interference.

	Applicable is the "matched filter for white noise".
	The matched filter output pulse is triangular.
	The matched filter output pulse is $\rm sinc$–shaped.
	The matched filter output pulse is $\rm sinc^2$–shaped.