# Exercise 5.7: Rectangular Matched Filter

Pulse  $g(t)$  and Matched Filter impulse response  $h(t)$

At the input of a lowpass filter with a rectangular impulse response  $h(t)$  the reception signal  $r(t)$  is present, which is additively composed of a pulse-shaped signal component  $g(t)$  and a noise component  $n(t)$.    It holds:

• The pulse  $g(t)$  is rectangular.
• The pulse duration is  $\Delta t_g = 2 \hspace{0.08cm}\rm µ s$.
• The pulse amplitude is  $g_0 = 2 \hspace{0.08cm}\rm V$.
• The center of the pulse is at  $T_g = 3 \hspace{0.08cm}\rm µ s$.
• The noise  $n(t)$  is white and Gaussian distributed.
• The power density is  $N_0 = 4 \cdot 10^{-6} \hspace{0.08cm}\rm V^2\hspace{-0.1cm}/Hz$  with respect to the  $1 \hspace{0.08cm}\rm \Omega$ resistor.

The rectangular impulse response of the filter starts at  $t = 0$.

• The impulse response duration  $\Delta t_h$  is freely selectable.
• The height  $1/\Delta t_h$  of the impulse response is adjusted in each case so that  $H(f = 0) = 1$.

Notes:

• The exercise belongs to the chapter  Matched Filter.
• For the questions  (1)  to  (6):    $\Delta t_h =\Delta t_g = 2 \hspace{0.05cm}\rm µ s$ always applies.

### Questions

1

Which of the statements are true under the assumption  $\Delta t_h =\Delta t_g$?

 The filter is matched to the input pulse  $g(t)$. There is another filter with larger S/N ratio. The filter can be implemented as an integrator over time  $\Delta t_h$.

2

What is the optimal detection time?

 $T_\text{D, opt} \ = \$ $\ \rm µ s$

3

What is the value of the matched filter constant here?

 $K_\text{MF} \ = \$ $\cdot 10^6 \ \rm 1/Vs$

4

What is the S/N ratio at the optimal detection time?

 $\rho_d(T_\text{D, opt}) \ = \$

5

What is the value of the signal component  $d_{\rm S}(t)$  at the optimal time  $T_\text{D, opt}$  and the noise power  $\sigma_d^2$  in front of the detector?

 $d_{\rm S}(T_\text{D, opt}) \ = \$ $\ \rm V$ $\sigma_d^2 \ = \$ $\ \rm V^2$

6

What is the S/N ratio at the detection time  $T_{\rm D} = 3 \hspace{0.08cm}\rm µ s$?

 $\rho_d(T_{\rm D} = 3 \hspace{0.08cm}\rm µ s) \ = \$

7

Which of the following statements are true if  $\Delta t_h =1 \hspace{0.08cm}\rm µ s$  holds?    Note:  In the range from  $0$  to  $1 \hspace{0.05cm}\rm µ s$,  the impulse response thus has the value  $10^6 \ \rm 1/s$.

 Each time point  $T_{\rm D}$  in the range  $3 \hspace{0.08cm}\rm µ s$ ... $4 \hspace{0.05cm}\rm µ s$  leads to the maximum SNR. The signal component  $d_S(T_\text{D, opt})$  is smaller than calculated in subtask  (5). The noise power  $\sigma_d^2$  is larger than calculated in subtask  (5). The S/N ratio is smaller than calculated in subtask  (4).

8

Which of the following statements are true if  $\Delta t_h =3 \hspace{0.08cm}\rm µ s$?    Note:  In the range from  $0$  to  $3 \hspace{0.05cm}\rm µ s$ the impulse response has the value  $0.33 \cdot 10^6 \ \rm 1/s$.

 Each time point  $T_{\rm D}$  in the range  $3 \hspace{0.08cm}\rm µ s$ ... $4 \hspace{0.05cm}\rm µ s$  leads to the maximum SNR. The signal component  $d_S(T_\text{D, opt})$  is smaller than calculated in subtask  (5). The noise power  $\sigma_d^2$  is larger than calculated in subtask  (5). The S/N ratio is smaller than calculated in subtask  (4).

### Solution

#### Solution

(1)  Solutions 1 and 3  are correct:

• For the same pulse duration  $(\Delta t_h =\Delta t_g)$,  there is a matched filter,  even if  $g(t)$  and  $h(t)$  differ in amplitude and temporal position.
• Thus,  there is no other filter with better signal–to–noise power ratio.
• The filter with rectangular impulse response can also be interpreted as an integrator over the time duration  $\Delta t_h$.

(2)  The impulse response of the matched filter is:   $h_{\rm MF} (t) = K_{\rm MF} \cdot g(T_{\rm D} - t).$

• The input impulse  $g(t)$  is non-zero in the range from  $2 \hspace{0.05cm}\rm µ s$  to  $4 \hspace{0.08cm}\rm µ s$,  and in the range from  $-4 \hspace{0.08cm}\rm µ s$  to  $-2 \hspace{0.08cm}\rm µ s$,  when mirrored.
• After shifting by  $4 \hspace{0.08cm}\rm µ s$,  it is achieved that  $g(T_{\rm D} - t)$ is,  like the impulse response  $h(t)$,  between  $0$  and  $2 \hspace{0.08cm}\rm µ s$.
• From this follows:   $T_\text{D, opt}\hspace{0.15cm}\underline{ =4 \hspace{0.08cm}\rm µ s}$.

(3)  With  $\Delta t_h =\Delta t_g = 2 \cdot 10^{-6}\hspace{0.05cm}\rm µ s$  and  $g_0 = 2 \hspace{0.08cm}\rm V$,  we obtain  $K_{\rm MF} = 1/(\Delta t_g \cdot g_0)\hspace{0.15cm}\underline{ =0.25 \cdot 10^{6}\hspace{0.08cm}\rm (1/Vs)}$.

(4)  The energy of the pulse  $g(t)$  is  $E_g = g_0^2 \cdot \Delta t_g = 8 \cdot 10^{-6}\hspace{0.05cm}\rm V^2s$.

• From this it follows for the maximum S/N ratio:
$$\rho _d (T_{{\rm{D, \hspace{0.08cm}opt}}} ) = \frac{2 \cdot E_g }{N_0 } = \frac{{2 \cdot 8 \cdot 10^{ - 6} \;{\rm{V}}^2 {\rm{s}}}}{{4 \cdot 10^{ - 6} \;{\rm{V}}^2 /{\rm{Hz}}}}\hspace{0.15cm}\underline{ = 4}.$$

MF output  $d_{\rm S}(t)$  for subtask  (5)

(5)  The matched filter output pulse  $d_{\rm S}(t)$  is triangular between $2 \hspace{0.05cm}\rm µ s$  and  $6 \hspace{0.05cm}\rm µ s$.

• The maximum  $g_0\hspace{0.15cm}\underline{= 2 \hspace{0.08cm}\rm V}$  is at  $T_\text{D, opt} =4 \hspace{0.05cm}\rm µ s$.
• The noise power is given by:
$$\sigma _d ^2 = \frac{N_0 }{2 \cdot \Delta t_h } \hspace{0.15 cm}\underline{= 1\;{\rm{V}}^2} .$$
• Using these two quantities,  we can calculate the maximum S/N ratio:
$$\rho _d (T_{{\rm{D, \hspace{0.08cm}opt}}} ) = \frac{{d_{\rm S} (T_{{\rm{D, \hspace{0.05cm}opt}}} )^2}}{\sigma _d ^2 } = \frac{({2\;{\rm{V}})^2 }}{{1\;{\rm{V}}^2 }} = 4.$$

(6)  From the above diagram,  one can see that now  $d_{\rm S}(T_\text{D})$  is only half as large,  namely  $1 \hspace{0.08cm}\rm V$.

• Thus for  $T_\text{D} =3 \hspace{0.08cm}\rm µ s$  the S/N ratio is smaller by a factor of  $4$,  i.e.  $\rho _d (T_{{\rm{D}}} )\hspace{0.15 cm}\underline{=1}$.

MF output  $d_{\rm S}(t)$  for subtask  (7)

(7)  Solutions 1, 3 and 4  are correct:

• The diagram shows that now the output pulse  $d_{\rm S}(t)$  is trapezoidal.
• In the range from  $3 \hspace{0.05cm}\rm µ s$  to  $4 \hspace{0.05cm}\rm µ s$  the pulse  $d_{\rm S}(t)$  is constantly equal to  $g_0= 2 \hspace{0.08cm}\rm V$.
• Because of the only half as wide impulse response  $h(t)$,  the frequency response  $H(f)$  is more broadband by a factor of  $2$  and thus the noise power is larger:
$$\sigma_d ^2 = \frac{N_0 }{2} \cdot \int_{ - \infty }^{ + \infty } {h^2 (t)\,{\rm{d}}t} = \frac{N_0 }{2 \cdot \Delta t_h } = 2\;{\rm{V}}^2 .$$
• Thus,  the S/N ratio is now   $\rho_d (T_{{\rm{D, \hspace{0.08cm}opt}}} ) \hspace{0.15cm}\underline{= 2}.$

MF output  $d_{\rm S}(t)$  for subtask  (8)

(8)  Solutions 2 and 4  are correct:

• The matched filter output pulse  $d_{\rm S}(t)$  for  $\Delta t_h = 3 \hspace{0.05cm}\rm µ s$ is sketched on the right.  This is also trapezoidal.
• The optimal detection time is now in the range between  $4 \hspace{0.05cm}\rm µ s$  and  $5 \hspace{0.05cm}\rm µ s$.
• However,  the useful signal  $d_{\rm S}(t)$  is now only one-third as large as when matched:   $d_{\rm S}(T_\text{D, opt}) = 2/3 \hspace{0.08cm}\rm V$.
• For the noise power now applies:
$$\sigma_d ^2 = \frac{N_0 }{2 \cdot \Delta t_h } = \frac{2}{3}\;{\rm{V}}^2 .$$
• Thus,  the noise power is smaller (i.e. more favorable) than with adaptation according to subtask  (5).
• Nevertheless,  the S/N ratio is still worse than in subtask  (7) due to the smaller  $d_{\rm S}(T_{\rm D})$:
$$\rho _d (T_{{\rm{D\hspace{0.15cm},opt}}} ) = \frac{{(2/3\;{\rm{V}})^2 }}{{2/3\;{\rm{V}}^2 }} = {2}/{3}.$$