# Matched Filter Properties

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## Applet Description

The applet is intended to illustrate the properties of the  »matched filter«  $({\rm MF})$.

• This is used to optimally detection of amplitude and/or location of a known waveform in a highly noisy environment.
• Or more generally speaking:  The matched filter – sometimes also referred to as  »optimal filter«  or as  »correlation filter«  – is used to detect the existence of such a signal.
Block diagram of the  "matched filter receiver"

The graphic shows the so-called  »matched filter receiver«:

• This can decide with the greatest possible certainty – in other words:   with the maximum signal–to–noise–ratio  $($ $\rm SNR)$  – whether a pulse-shaped signal component  $g(t)$  disturbed by additive noise  $n(t)$  is present or not.
• One application is radar technology,  where the pulse shape  $g(t)$  is known,  but not when the pulse was sent and with what strength and delay it arrives.
• However,  the matched filter is also used as a receiver filter in digital transmission systems  $($or at least as part of them$)$  to minimize the error probability of the system.

All parameters,  times and frequencies are to be understood as normalized quantities and thus dimensionless.

• For the  »input pulse«  $g(t)$,  "rectangular",  "Gaussian"  and  "exponential"  can be set,  each described by the pulse amplitude  $A_g$,  the equivalent pulse duration  $\Delta t_g$  and the shift  $\tau_g$  with respect to the time–symmetrical case.  Further information in the section  »Further information about the considered input pulses«.
• For the  »receiver filter«  you can choose between the alternatives  "slit low-pass"  $($rectangular-in-time$)$,  "Gaussian low-pass",  "first order lowpass" and  "extremely acausal filter".  The respective impulse responses  $h(t)$,  are shown, characterized by their height  $A_h$,  the equivalent duration  $\Delta t_h$  and the shift  $\tau_h$.  More information in the section  »Further information on the considered impulse responses«.
• Further input parameters are the detection time  $T_{\rm D}$  and the also normalized noise power density  $N_0$  at the receiver input.

The following are output as numerical values

1. The energy  $E_g$  of the input pulse  $g(t)$,  useful signal value  $d_{\rm S} (T_{\rm D})$  at the filter output as well as the noise variance  $\sigma_d^2$  at the filter output,
2. the signal–to–noise ratio  $\rho_{d} (T_{\rm D})$  at the filter output and the corresponding dB specification  $10 \cdot \lg \ \rho_{d} (T_{\rm D})$,
3. the maximum value for this is  $10 \cdot \lg \ \rho_{\rm MF}$.

If the input configuration meets the matched filter conditions,  then:   $10 \cdot \lg \ \rho_{d} (T_{\rm D,\ opt}) = 10 \cdot \lg \ \rho_{\rm MF}$.

## Theoretical Background

### Detailed description of the underlying model

The following conditions apply to the individual components:

• Let the useful component  $g(t)$  of the received signal  $r(t)=g(t)+n(t)$  be pulse-shaped and thus  "energy-limited".
• That means:   The integral over  $\big [g(t)\big ]^2$  from  $–∞$  to  $+∞$  yields the finite value  $E_g$.
• Let the noise signal  $n(t)$  be  "white Gaussian noise"  with  (one–sided)  noise power density  $N_0$.
• The signal  $d(t)$  is additively composed of two components:  The component  $d_{\rm S}(t)$  is due to the  "$\rm S$"ignal  $g(t)$,  the component  $d_{\rm N}(t)$  is due to the  "$\rm N$"oise  $n(t)$.
• The receiver,  consisting of a linear filter   ⇒   frequency response  $H_{\rm MF}(f)$  and the  "decision maker",  is to be dimensioned
so that the instantaneous S/N ratio at the output is maximized:
$$\rho _d ( {T_{\rm D} } ) = \frac{ {d_{\rm S} ^2 ( {T_{\rm D} } )} }{ {\sigma _d ^2 } }\mathop = \limits^{\rm{!} }\hspace{0.1cm} {\rm{Maximum} }.$$
• Here,  $σ_d^2$  denotes the  variance  ("power")  of the signal  $d_{\rm N}(t)$,  and  $T_{\rm D}$  denotes the (suitably chosen)  "detection time".

### Matched filter optimization

Let be given an energy-limited useful signal  $g(t)$  with the corresponding spectrum  $G(f)$.

• Thus,  the filter output signal at detection time  $T_{\rm D}$  for any filter with impulse response  $h(t)$  and frequency response  $H(f) =\mathcal{ F}\{h(t)\}$  can be written as follows
(ignoring noise   ⇒   subscript  $\rm S$  for "signal"):
$$d_{\rm S} ( {T_{\rm D} } ) = g(t) * h(t) = \int_{ - \infty }^{ + \infty } {G(f) \cdot H(f) \cdot {\rm{e}}^{ {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f \hspace{0.05cm}\cdot \hspace{0.05cm}T_{\rm D} }\hspace{0.1cm} {\rm{d}}f} .$$
• The  "noise component"  $d_{\rm N}(t)$  of the filter output signal  (subscript  $\rm N$  for "noise")  stems solely from the white noise  $n(t)$  at the input of the receiver.  For its variance  (power)  applies independently of the detection time  $T_{\rm D}$:
$$\sigma _d ^2 = \frac{ {N_0 } }{2} \cdot \int_{ - \infty }^{ + \infty } {\left| {H(f)} \right|^{\rm{2} }\hspace{0.1cm} {\rm{d} }f} .$$
• Thus,  the optimization problem at hand is:
$$\rho _d ( {T_{\rm D} } ) = \frac{ {\left| {\int_{ - \infty }^{ + \infty } {G(f) \cdot H(f) \cdot {\rm{e} }^{ {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f \hspace{0.05cm}\cdot \hspace{0.05cm}T_{\rm D} }\hspace{0.1cm} {\rm{d} }f} } \right|^2 } }{ {N_0 /2 \cdot \int_{ - \infty }^{ + \infty } {\left| {H(f)} \right|^{\rm{2} }\hspace{0.1cm} {\rm{d} }f} } } \stackrel{!}{=} {\rm{Maximum} }.$$

$\text{Here first without proof:}$    One can show that this quotient becomes largest for the following frequency response  $H(f)$:

$$H(f) = H_{\rm MF} (f) = K_{\rm MF} \cdot G^{\star} (f) \cdot {\rm e}^{- {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f \hspace{0.05cm}\cdot \hspace{0.05cm}T_{\rm D} } .$$
• Thus,  for the signal–to–noise power ratio at the matched filter output  $($independent of the dimensionally constant  $K_{\rm MF})$,  we obtain:
$$\rho _d ( {T_{\rm D} } ) = { {2 \cdot E_g } }/{ {N_0 } }.$$
• $E_g$  denotes the energy of the input pulse,  which can be calculated using  Parseval's theorem  in both the time and frequency domains:
$$E_g = \int_{ - \infty }^{ + \infty } {g^2 (t)\hspace{0.1cm}{\rm{d} }t} = \int_{ - \infty }^{ + \infty } {\left \vert {G(f)} \right\vert ^{\rm{2} }\hspace{0.1cm} {\rm d}f} .$$

$\text{Example 1:}$   A rectangular pulse  $g(t)$  with amplitude  $\rm 1\hspace{0.05cm}V$,  duration  $0.5\hspace{0.05cm} \rm ms$  and unknown position is to be found in a noisy environment.

• Thus the pulse energy  $E_g = \rm 5 · 10^{–4} \hspace{0.05cm}V^2s$.
• Let the noise power density be  $N_0 = \rm 10^{–6} \hspace{0.05cm}V^2/Hz$.

The best result   ⇒   the  maximum S/N ratio  is obtained with the matched filter:

$$\rho _d ( {T_{\rm D} } ) = \frac{ {2 \cdot E_g } }{ {N_0 } } = \frac{ {2 \cdot 5 \cdot 10^{-4}\, {\rm V^2\,s} } }{ {10^{-6}\, {\rm V^2/Hz} } } = 1000 \hspace{0.3cm}\Rightarrow\hspace{0.3cm} 10 \cdot {\rm lg}\hspace{0.15cm}\rho _d ( {T_{\rm D} } ) = 30\,{\rm dB}.$$

The matched filter criterion given above is now derived step by step.  If you are not interested in this, please skip to the next section  "Interpretation of the matched filter".

$\text{Derivation of the matched filter criterion:}$

$(1)$  The Schwarz inequality with the two  (generally complex)  functions  $A(f)$  and  $B(f)$:

$$\left \vert {\int_a^b {A(f) \cdot B(f)\hspace{0.1cm}{\rm{d} }f} } \right \vert ^2 \le \int_a^b {\left \vert {A(f)} \right \vert^{\rm{2} } \hspace{0.1cm}{\rm{d} }f} \cdot \int_a^b {\left\vert {B(f)} \right \vert^{\rm{2} } \hspace{0.1cm}{\rm{d} }f} .$$

$(2)$  We now apply this equation to the signal–to–noise ratio:

$$\rho _d ( {T_{\rm D} } ) = \frac{ {\left \vert {\int_{ - \infty }^{ + \infty } {G(f) \cdot H(f) \cdot {\rm{e} }^{ {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f \hspace{0.05cm}\cdot \hspace{0.05cm}T_{\rm D} } \hspace{0.1cm}{\rm{d} }f} } \right \vert^2 } }{ {N_0 /2 \cdot \int_{ - \infty }^{ + \infty } {\left \vert {H(f)} \right \vert^{\rm{2} }\hspace{0.1cm} {\rm{d} }f} } }.$$

$(3)$  Thus,  with  $A(f) = G(f)$  and  $B(f) = H(f) · {\rm e}^{ {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f \hspace{0.05cm}\cdot \hspace{0.05cm}T_{\rm D} }$  the following bound is obtained:

$$\rho_d ( {T_{\rm D} } ) \le \frac{1}{ {N_0 /2} } \cdot \int_{ - \infty }^{ + \infty } {\left \vert {G(f)} \right \vert^{\rm{2} } }\hspace{0.1cm}{\rm{d} }f .$$

$(4)$  We now tentatively set for the filter frequency response:

$$H(f) = H_{\rm MF} (f) = K_{\rm MF} \cdot G^{\star} (f) \cdot {\rm{e} }^{- {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f \hspace{0.05cm}\cdot \hspace{0.05cm}T_{\rm D} }.$$

$(5)$  Then,  from the above equation  $(2)$,  we obtain the following result:

$$\rho _d ( {T_{\rm D} } ) = \frac{ {\left \vert K_{\rm MF}\cdot {\int_{ - \infty }^{ + \infty } {\left \vert {G(f)} \right \vert ^{\rm{2} }\hspace{0.1cm} {\rm{d} }f} } \right \vert ^2 } }{ {N_0 /2 \cdot K_{\rm MF} ^2 \cdot \int_{ - \infty }^{ + \infty } {\left \vert {G(f)} \right \vert ^{\rm{2} }\hspace{0.1cm} {\rm{d} }f} } } = \frac{1}{ {N_0 /2} } \cdot \int_{ - \infty }^{ + \infty } {\left \vert {G(f)} \right \vert ^{\rm{2} }\hspace{0.1cm} {\rm{d} }f} .$$

$\text{This means:}$

• With the approach  $(4)$  for the matched filter $H_{\rm MF}(f)$,  the maximum possible value is indeed obtained in the above estimation.
• No other filter  $H(f) ≠ H_{\rm MF}(f)$  can achieve a higher signal–to–noise power ratio.
• The matched filter is optimal with respect to the maximization criterion on which it is based.
q.e.d.

### Interpretation of the matched filter

In the last section,  the frequency response of the matched filter was derived as follows:

$$H_{\rm MF} (f) = K_{\rm MF} \cdot G^{\star} (f) \cdot {\rm{e} }^{- {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f \hspace{0.05cm}\cdot \hspace{0.05cm}T_{\rm D} } .$$

By  Fourier inverse transformation  the corresponding impulse response is obtained:

$$h_{\rm MF} (t) = K_{\rm MF} \cdot g(T_{\rm D} - t).$$

These two functions can be interpreted as follows:

• The  "matched filter"  is matched by the term  $G^{\star}(f)$  to the spectrum of the pulse  $g(t)$  which is to be found  –  hence its name.
• The  "constant"  $K_{\rm MF}$  is necessary for dimensional reasons.
• If  $g(t)$  is a voltage pulse,  this constant has the unit "Hz/V".  The frequency response  $H_{\rm MF} (f)$  is therefore dimensionless.
• The  "impulse response"  $h_{\rm MF}(t)$  results from the useful signal  $g(t)$  by mirroring   ⇒   from  $g(t)$  becomes $g(–t)$ $]$   as well as a shift by  $T_{\rm D}$  to the right.
• The  "earliest detection time"  $T_{\rm D}$  follows for realizable systems from the condition  $h_{\rm MF}(t < 0)\equiv 0$   $($"causality",  see book  Linear and Time-Invariant Systems$)$.
• The  "useful component"  $d_{\rm S} (t)$  of the filter output signal is equal in shape to the  energy auto-correlation function   $\varphi^{^{\bullet} }_{g} (t )$  and shifted with respect to it by  $T_{\rm D}$.  It holds:
$$d_{\rm S} (t) = g(t) * h_{\rm MF} (t) = K_{\rm MF} \cdot g(t) * g(T_{\rm D} - t) = K_{\rm MF} \cdot \varphi^{^{\bullet} }_{g} (t - T_{\rm D} ).$$

$\text{Please note:}$  For an energy-limited signal  $g(t)$,  one can only specify the  energy ACF

$$\varphi^{^{\bullet} }_g (\tau ) = \int_{ - \infty }^{ + \infty } {g(t) \cdot g(t + \tau )\,{\rm{d} }t} .$$

Compared to the ACF definition of a power-limited signal  $x(t)$, viz.

$$\varphi _x (\tau ) = \mathop {\lim }_{T_{\rm M} \to \infty } \frac{1}{ {T_{\rm M} } }\int_{ - T_{\rm M} /2}^{+T_{\rm M} /2} {x(t) \cdot x(t + \tau )\hspace{0.1cm}\,{\rm{d} }t} ,$$

the division by the measurement duration  $T_{\rm M}$  and the boundary transition  $T_{\rm M} → ∞$  are omitted in the calculation of the energy ACF.

$\text{Example 2:}$  We assume that the rectangular pulse is between   $\rm 2\hspace{0.08cm}ms$   and   $\rm 2.5\hspace{0.08cm}ms$   and the detection time  $T_{\rm D} =\rm 2\hspace{0.08cm}ms$  is desired.

Under these conditions:

• The matched filter impulse response  $h_{\rm MF}(t)$  must be constant in the range from   $t_1 (= 4 - 2.5) =\rm 1.5\hspace{0.08cm}ms$   to  $t_2 (= 4 - 2) =\rm 2\hspace{0.08cm}ms$.
• For  $t < t_1$  as well as for  $t > t_2$  it must not have any components.
• The magnitude frequency response  $\vert H_{\rm MF}(f)\vert$  is  $\rm sinc$–shaped here.
• The magnitude of the impulse response  $h_{\rm MF}(t)$  is not important for the S/N ratio, because  $\rho _d ( {T_{\rm D} } )$  is independent of  $K_{\rm MF}$.

### Further information about the considered input pulses

All data are without consideration of the delay  $\tau_g$.

(1)  Rectangular Pulse

• The pulse  $g(t)$  has constant height  $A_g$  in the range  $\pm \Delta t_g/2$  and is zero outside.
• The spectral function  $G(f)=A_g\cdot \Delta t_g \cdot {\rm si}(\pi\cdot \Delta t_g \cdot f)$  has zeros at equidistant distances $1/\Delta t_g$.
• The pulse energy is  $E_g=A_g^2\cdot \Delta t_g$.

(2)  Gaussian Pulse

• The pulse  $g(t)=A_g\cdot {\rm e}^{-\pi\cdot(t/\Delta t_g)^2}$  is infinitely extended.  The maximum is  $g(t= 0)=A_g$.
• The smaller the equivalent time duration  $\Delta t_g$,  the broader and lower the spectrum   $G(f)=A_g \cdot \Delta t_g \cdot {\rm e}^{-\pi\hspace{0.05cm}\cdot\hspace{0.05cm}(f\hspace{0.05cm}\cdot\hspace{0.05cm} \Delta t_g)^2}$.
• The pulse energy is  $E_g=A_g^2\cdot \Delta t_g/\sqrt{2}$.

(3)  Exponential Pulse

• The pulse is identically zero for  $t<0$  and infinitely extended for positive times   ⇒   $g(t)=A_g\cdot {\rm e}^{-t/\Delta t_g}$.
• $g(t)$  is (strongly) asymmetric   ⇒   the spectrum   $G(f)=A_g \cdot \Delta t_g/( 1 + {\rm j} \cdot 2\pi \cdot f \cdot \Delta t_g)$  is complex-valued;
• The pulse energy is  $E_g=A_g^2\cdot \Delta t_g/2$.

### Further information on the considered impulse responses

The different receiver filters  $H(f)$  are described by their impulse responses  $h(t)$.

Similar to the input pulses  $g(t)$,  these are characterized by the pulse height  $A_h$, the equivalent pulse duration   $\Delta t_h$  and the delay  $\tau_h$  compared to the symmetrical case.  The following short descriptions are always valid for   $\tau_h= 0$.

(1)  Slit lowpass  ⇒   "rectangular impulse response"

• The impulse response  $h(t)$  has constant height  $A_h$  in the range  $\pm \Delta t_h/2$  and is zero outside.
• The frequency response  $H(f)=K \cdot {\rm si}(\pi\cdot \Delta t_g \cdot f)$  has zeros at equidistant intervals $1/\Delta t_h$.
• For white noise, the noise variance at the filter output is:  $\sigma_d^2= N_0/2 \cdot A_h^2 \cdot \Delta t_h$.

(2)  Gaussian lowpass  ⇒   "Gaussian impulse response"

• The impulse response  $h(t)=A_h\cdot {\rm e}^{-\pi\cdot(t/\Delta t_h)^2}$  ist is infinitely extended.  The maximum is  $h(t= 0)=A_h$.
• The smaller the equivalent time duration  $\Delta t_h$,  the broader and lower the frequency response  $H(f)=K \cdot {\rm e}^{-\pi\hspace{0.05cm}\cdot\hspace{0.05cm}(f\hspace{0.05cm}\cdot\hspace{0.05cm} \Delta t_h)^2}$.
• For white noise, the noise variance at the filter output is:  $\sigma_d^2= N_0/2 \cdot A_h^2 \cdot \Delta t_h/\sqrt{2}$.

(3)  First order lowpass  ⇒   "Exponentially decaying impulse response"

• The impulse response is identically zero for  $t<0$  and infinitely extended for positive times   ⇒   $h(t)=A_h\cdot {\rm e}^{-t/\Delta t_h}$.
• $h(t)$  is causal and (strongly) asymmetric.  The frequency response $H(f)=A_g \cdot \Delta t_g/( 1 + {\rm j} \cdot 2\pi \cdot f \cdot \Delta t_g)$  is complex-valued.
• For white noise, the noise variance at the filter output is:  $\sigma_d^2= N_0/4 \cdot A_h^2 \cdot \Delta t_h$.

(4)  Lowpass 4"  ⇒   "Impulse response mirror image of  (3)"

• The impulse response is identically zero for  $t>0$  and infinitely extended for negative times   ⇒   $h(t)=A_h\cdot {\rm e}^{t/\Delta t_h}$  für  $t<0$.
• The frequency response $H(f)$  is conjugate complex to the frequency response of the first order lowpass filter.
• The noise variance at the filter output is exactly the same for white noise as for the first order lowpass:  $\sigma_d^2= N_0/4 \cdot A_h^2 \cdot \Delta t_h$.

## Exercises

• First, select the number  $(1,\ 2, \text{...} \ )$  of the task to be processed.  The number  $0$  corresponds to a "Reset":  Same setting as at program start.
• A task description is displayed.  The parameter values are adjusted.  Solution after pressing "Show Solution".
• Both the input signal  $x(t)$  and the filter impulse response  $h(t)$  are normalized, dimensionless and energy-limited ("time-limited pulses").
• All times, frequencies, and power values are to be understood normalized, too.

(1)  Let the input pulse be Gaussian with  $A_g=1,\ \Delta t_g=1,\ \tau_g=1$.   Which setting leads to the "Matched Filter"?  What value has  $10 \cdot \lg \ \rho_{\rm MF}$  with  $N_0=0.01$?

•  The Matched Filter must also have a Gaussian shape and it must hold:  $\Delta t_h=\Delta t_g=1,\ \tau_h =\tau_g=1$   ⇒   $T_{\rm D} = \tau_h +\tau_g=2$.
•  The (instantaneous) signal-to-noise power ratio at the Matched Filter output is  $\rho _{\rm MF} = { {2 \cdot E_g } }/{ {N_0 } } \approx 141.4$  ⇒   $10 \cdot \lg \ \rho _{\rm MF} \approx 21.5$  dB.
•  With no other filter than the Matched Filter this  $\rm SNR$  (or an even better one)  can be achieved   ⇒   $10 \cdot \lg \ \rho _{d} \le 10 \cdot \lg \ \rho _{\rm MF}$.

(2)  The "Matched Filter" on rectangular input pulse with  $A_g=1,\ \Delta t_g=1,\ \tau_g=0$  is a rectangular-in-time low–pass   ⇒   rectangular impulse response.
What value has  $10 \cdot \lg \ \rho_{\rm MF}$  with  $N_0=0.01$?  Interpret all the graphs shown and the numerical results in different ways

•  The MF parameters are  $A_h=A_g=1, \ \Delta t_h=\Delta t_g=1,\ \tau_h =\tau_g=0$   ⇒   $T_{\rm D} = \tau_h +\tau_g=0$   ⇒   $\rho _{\rm MF} = 200$   ⇒   $10 \cdot \lg \ \rho _{\rm MF} \approx 23$  dB.
•  The pulse energy is the integral over  $g^2(t)$   ⇒   $E_g = A_g^2 \cdot \Delta t_g=1$   ⇒   $\rho _{\rm MF} = 2 \cdot E_g /N_0 =200$.   $T_{\text{D, opt} }=0$  is implicitly considered here.
•  Another equation is  $\rho_d (T_{\rm D}) =d_{\rm S}^2 (T_{\rm D})/\sigma_d^2$.  The noise variance can, for example, be calculated as the integral over  $h^2(t)$   ⇒   $\sigma_d^2= N_0 \cdot \Delta t_h/2 = 0.005$.
•  The useful detection signal  $d_{\rm S} (t)= g(t) * h(t)$  has a triangular shape with the maximum  $d_{\rm S} (T_{\rm D, \ opt} = 0 )= 1$   ⇒   $\rho_d (T_{\rm D, \ opt} = 0 ) = 200= \rho _{\rm MF}$.

(3)  The settings of  $(2)$  continue to apply, with the exception of  $N_0=0.02$  instead of  $N_0=0.01$.  What changes can be seen?

•  The only difference is twice the noise variance  $\sigma_d^2= 0.01$   ⇒   $\rho_d (T_{\rm D, \ opt} = 0 ) = 100= \rho _{\rm MF}$   ⇒   $10 \cdot \lg \rho_{\rm MF} =20$ dB.

(4)  The settings of  $(3)$  continue to apply, except  $T_{\rm D} = 0.1$  instead of  $T_{\rm D, \ opt} = 0$.  What is the effect of this non-optimal detection time?

•  Now the useful signal value  $d_{\rm S} (T_{\rm D} = 0.1 )= 0.9$  is smaller   ⇒   $\rho_d (T_{\rm D} = 0.1 ) =0.9^2/0.01= 81< \rho _{\rm MF}$.  There is a degradation of nearly  $1$ dB.
•  For the further tasks the optimal detection time  $T_{\rm D, \ opt}$  is assumed, if not explicitly stated otherwise.

(5)  The settings of  $(3)$  apply again except for a lower impulse response  $A_h = 0.8$  instead of  $A_h = 1$.  Interpret the changes.

•  With  $A_h \ne A_g$  it is also a Matched Filter as long as  $h(t)$  is equal in shape to  $g(t)$    ⇒   $\rho _{\rm MF} = { {2 \cdot E_g } }/{ {N_0 } } =100$   ⇒   $10 \cdot \lg \rho _{\rm MF} =20$ dB.
•  The equation  $\rho_d (T_{\rm D}=0) =d_{\rm S}^2 (T_{\rm D}=0)/\sigma_d^2$  leads to the same result, since  ${d_{\rm S}}^2 (T_{\rm D})$  and  $\sigma_d^2$  are compared to  $(3)$  each reduced by a factor  $0. 8^2$.

(6)  Compared to  $(5)$  now the height of the input pulse  $g(t)$  is increased from  $A_g = 1$  to  $A_g = 1. 25$.  Does  $h(t)$  describe a Matched Filter?  What is the SNR  $\rho_{\rm MF}$?

•  Again, this is a Matched Filter, since  $h(t)$  and  $g(t)$  are equal in shape.  With  $E_g = 1.25^2$:     $\rho _{\rm MF} = { {2 \cdot 1.25^2 } }/{ 0.02 } =156.25$  ⇒  $10 \cdot \lg \rho_{\rm MF} \approx 21.9$ dB.
•  The higher value  $21.9$ dB compared to  $(5)$  is related to the fact that for the same noise variance  $\sigma_d^2= 0.0064$  the useful detection sample is again  ${d_{\rm S}} (T_{\rm D}) = 1$.

(7)  We continue from the rectangle–rectangle combination with  $A_h=A_g=1,\ \ \Delta t_h=\Delta t_g=1,\ \tau_h=\tau_g=0,\ N_0 =0.02,\ T_{\rm D}=0$.
Interpret the results after varying the equivalent pulse duration  $\Delta t_h$  of  $h(t)$  in the range  $0.6$ ... $1.4$.  Use the graph representation over  $\Delta t_h$.

•  As expected the optimum is obtained for the equivalent pulse duration  $\Delta t_h=\Delta t_g=1$.  Then  $10 \cdot \lg \ \rho_d (T_{\rm D, \ opt} = 0 ) =20$ dB  $\big(= 10 \cdot \lg \rho_{\rm MF}\big)$.
•  If  $\Delta t_h<\Delta t_g=1$, the useful detection signal is trapezoidal.  For  $\Delta t_h=0.6$:   $d_{\rm S} (T_{\rm D}=0)= 0.6$,  $\sigma_d^2\approx0.006$   ⇒   $10 \cdot \lg \ \rho_d (T_{\rm D, \ opt} = 0 ) \approx 17.8$ dB.
•  Also for  $\Delta t_h>1$  the useful detection signal is trapezoidal, but now still  $d_{\rm S} (T_{\rm D}=0)= 1$.  The noise variance  $\sigma_d^2$  increases continuously with  $\Delta t_h$.
•  For  $\Delta t_h=1.4$:     $\sigma_d^2=0.014$   ⇒   $10 \cdot \lg \ \rho_d (T_{\rm D, \ opt} = 0 ) \approx 18. 5$ dB.  Compared to the Matched Filter  $(\Delta t_h=1)$  the degradation is approx.  $1.5$ dB.

(8)  Now interpret the results for different  $\Delta t_g$  of the input pulse  $g(t)$  in the range  $0.6$ ... $1.4$.  Use the graph representation over  $\Delta t_g$.

•  Note:   The blue curve  $10 \cdot \lg \ \rho_d (T_{\rm D,\ opt} )$  is the difference between  $20\cdot \lg \ \big [{K \cdot d_{\rm S}} (T_{\rm D,\ opt}) \big ]$    (purple curve)  and  $20\cdot \lg \ \big [K \cdot \sigma_d \big ]$  (green curve).
•  For the considered parameter set  and  $K=10$  the "green term"  $20\cdot \lg \ \big [K \cdot \sigma_d \big ] = 0$ dB  for all  $\Delta t_g$   ⇒   the blue and the purple curves are identical.
•  The blue curve  $10 \cdot \lg \ \rho_d (T_{\rm D,\ opt} )$  increases from  $15.6$ dB  $($for  $\Delta t_g = 0. 6)$  to  $20$ dB  $($for  $\Delta t_g = 1)$  continuously and then remains constant for  $\Delta t_g > 1$.
•  But, the setting  $(\Delta t_g = 1.4,\ \Delta t_h = 1)$  does not yield a "Matched Filter".   Rather, with  $\Delta t_h = \Delta t_g = 1.4$:    $10 \cdot \lg \ \rho_{\rm MF}=10 \cdot \lg \ (2 \cdot E_g/N_0) \approx 21.5$ dB.
•  On the other hand the plot over  $\Delta t_h$  with the default setting  $(\Delta t_g = 1.4,\ \Delta t_h = 1)$  shows a monotonic increase of the blue curve   $10 \cdot \lg \ \rho_d (T_{\rm D,\ opt} )$.
•  For  $\Delta t_h = 0.6$  this gives  $10 \cdot \lg \ \rho_d (T_{\rm D,\ opt} )\approx 17.8$ dB,  and for  $\Delta t_h = 1. 4$  against  $10 \cdot \lg \ \rho_d (T_{\rm D,\ opt} )\approx 21.5$ dB  $=10 \cdot \lg \ \rho_{\rm MF}$.

(9)  We consider the exponential pulse  $g(t)$  and the first order low–pass, where  $A_h=A_g=1,\ \Delta t_h=\Delta t_g=1,\ \tau_h=\tau_g=0,\ N_0 =0.02,\ T_{\rm D}=1$.
Does this setting meet the Matched Filter criteria?  Justify your answers with as many arguments as possible.

•  No!   Here  $h(t)=g(t)$.  In a Matched Filter configuration, the impulse response should be  $h(t)={\rm const.}\cdot g(T_{\rm D}-t)$.
•  The useful detection signal  $d_{\rm S}(t)$  does not have a symmetric shape around the maximum. For the Matched Filter,  $d_{\rm S}(T_{\rm D}-t) = d_{\rm S}(T_{\rm D}+t)$  would have to hold.
•  Despite  $\Delta t_h=\Delta t_g$  the SNR  $10 \cdot \lg \ \rho_d (T_{\rm D,\ opt}) \approx 14. 3$ dB   is now less than  $10 \cdot \lg \ \rho _{\rm MF} = 10 \cdot \lg \ 2 \cdot E_g/N_0 \approx 17$ dB.

(10)  With all other settings being the same, what changes with the "extremely acausal filter"?  Does the setting meet the Matched Filter criteria?  Reason.

•  Now here  $h(t)=g(-t)$  and the useful detection signal  $d_{\rm S}(t)$  is symmetric around  $t=0$.  It makes sense to choose  $T_{\rm D} = 0$  here.
•  This gives  $10 \cdot \lg \ \rho_d (T_{\rm D,\ opt}) =10 \cdot \lg \ d_{\rm S}^2 (T_{\rm D,\ opt})/\sigma_d^2 = 17$ dB   –   the same value as for  $10 \cdot \lg \ \rho _{\rm MF} = 10 \cdot \lg \ 2 \cdot E_g/N_0 = 17$ dB.
•  The useful detection signal  $d_{\rm S}(t)$  is of the same shape as the energy ACF of the input pulse  $g(t)$.  The Matched Filter focuses the energy around the time  $T_{\rm D,\ opt}$.

(11)  With which rectangular pulse  $g(t)$  can one achieve the same  $\rho _{\rm MF}=50$  as in task  $(10)$?
With  $(A_h=A_g=1,\ \ \Delta t_h=\Delta t_g=0.5)$   or with   $(A_h=A_g=0.5,\ \ \Delta t_h=\Delta t_g=1)$ ?

•  From the equation  $\rho _{\rm MF} = 2 \cdot E_g/N_0$  it is already clear that the SNR depends only on the energy  $E_g$  of the input pulse and not on its shape.
•  The exponential pulse with   $(A_g=1,\ \Delta t_g=1)$   has the energy  $E_g=0.5$  ⇒   $\rho _{\rm MF}=50$.  As well as the rectangular pulse with  $(A_g=1,\ \Delta t_g=0.5)$.
•  In contrast, the rectangular pulse with  $(A_g=0.5,\ \Delta t_g=1)$  has a smaller energy   ⇒   $E_g=0. 25$   ⇒   $\rho _{\rm MF}=25$   ⇒   $10 \cdot \lg \ \rho _{\rm MF} = 14$ dB.

## Applet Manual

(A)     Selection  of the input pulse  $g(t)$:
Rectangle | Gaussian | Exponential

(B)   Parameter selection of the input pulse:
amplitude  $A_g$,     equivalent pulse duration  $\Delta t_g$,
shift  $\tau_g$

(C)     Selection  of the receiver filter impulse response  $h(t)$:
rectangular-in-time | Gaussian low-pass |
1st order low-pass | extremely acausal filter

(D)     Parameterwahl der Impulsantwort:
amplitude  $A_h$,     equivalent pulse duration  $\Delta t_h$,
shift  $\tau_g$,     detection time  $T_{\rm D}$

(E)     Sketches of  $g(t)$  $($red$)$  and  $h(t)$  $($blue$)$

(F)     Sketch of the convolution result  $d_{\rm S}= g(t)\star h(t)$

(G)     Input of the one-sided noise power density  $N_0$

(H)     Selection whether in area  (K)  the simulation results
should be displayed numerically or graphically

( I )     Display area for  $h(t)^2$

(J)     Display area for  $|H(f)|^2$

(K)     Area for numerical or graphical output

(L)     Area for exercises:  Task selection

(M)     Area for exercises:  Questions, sample solution

This interactive calculation tool was designed and implemented at the  Institute for Communications Engineering  at the  Technical University of Munich.

• The first version was created in 2006 by Markus Elsberger  as part of his bachelor thesis with “FlashMX – Actionscript” (Supervisor: Günter Söder).
• Last revision and English version 2020/2021 by  Carolin Mirschina  in the context of a working student activity.

The conversion of this applet to HTML 5 was financially supported by  Studienzuschüsse  ("study grants")  of the TUM Faculty EI.  We thank.