Aufgaben:Exercise 5.3: 1st order Digital Filter

• Return to book

First order digital filter

We consider the filter arrangement shown on the right with coefficients  $a_0$,  $a_1$  and  $b_1$,  each of which can take values between  $0$  and  $1$. 

• Let the input signal  $x(t)$  be a single Dirac delta impulse with unit weight  "1"   ⇒   $x(t) = \delta(t)$,  which corresponds to the following discrete-time representation:

$$\left\langle {\hspace{0.05cm}x_\nu } \hspace{0.05cm}\right\rangle = \left\langle \hspace{0.05cm}1,\;0,\;0,\;0,\;\text{...} \hspace{0.05cm}\right\rangle .$$

• Due to this special input sequence,  the sequence  $\left\langle {\hspace{0.05cm}y_\nu \hspace{0.05cm}} \right\rangle$  at the filter output simultaneously describes the discrete-time impulse response  $\left\langle {\hspace{0.05cm}h_\nu \hspace{0.05cm}} \right\rangle$  of the filter.  The spacing of the samples here is  $T_{\rm A} = 1 \hspace{0.05cm} \rm µ s$.

Note:

• The exercise belongs to the chapter  Digital Filters.
• The HTML5/JavaScript applet  "Digital Filters"  illustrates the subject matter of this chapter.

Questions

1

Which of the following statements are true?

	The special case  $b_1 = 1$  leads to a non-recursive filter.
	With  $a_0 = 1$,  $a_1 = 0$  and  $b_1 = 0$:    $y(t) = x(t)$  is true.
	With  $a_0 = 0$,  $a_1 = 0.5$  and  $b_1 = 0$:   $y(t)$  is undistorted with respect to  $x(t)$.

2

Let now  $a_0 = 1$,  $a_1 = 0$  and  $b_1 = 0.6$.  Calculate the output sequence  $\left\langle {y_\nu } \right\rangle$.  What output value  $y_3$  occurs at time  $t = 3 \cdot T_{\rm A}$? 

$y_3 \ = \ $

3

Let  $a_0 = 1$,  $a_1 = 0$  and  $b_1 = 0.6$.  On which range  $0$, ... , $M \cdot T_{\rm A}$  is the impulse response limited to if values smaller than  $0.001$  are neglected?

$M \ = \ $

4

Let  $a_0 = 1$  and  $b_1 = 0.6$.  Given the result from  (2),  calculate the output value  $y_3$  for  $a_1 = -0.5$.

$y_3 \ = \ $

Solution

Solution

(1)  Solutions 2 and 3  are correct:

• The filter is non-recursive if the feedback is omitted:   $b_1 = 0$.
• If additionally  $a_0 = 1$  and  $a_1 = 0$,  the sequences  $\left\langle {x_\nu } \right\rangle$  and  $\left\langle {y_\nu } \right\rangle$  and thus of course the signals  $x(t)$  and  $y(t)$  are equal.
• With  $a_0 = 0$  and  $a_1 = 1$,    $y(t) = x(t-T_{\rm A})$  is delayed by  $T_{\rm A}$  with  $a_1 = 0.5$  additionally attenuated.
• However,  delay and damping do not result in distortion.

(2)  At time  $\nu = 0$:    $y_{\nu} = x_{\nu} = 1$. 

• For all further time points:  $\nu$,    $x_{\nu} = 0$  and thus:

$$y_\nu = b_1 \cdot y_{\nu - 1} = {b_1 }^\nu .$$

• In particular,  $y_3 = b_1^3 = 0.6^3\hspace{0.15cm}\underline{= 0.216}$.

(3)  According to the problem definition must be valid:  

$$y_{M + 1} = {b_1} ^{M + 1} < 0.001.$$

• This leads to the result:

$$M + 1 \ge \frac{{\lg \ \left( {0.001} \right)}}{{\lg \ \left( {0.6} \right)}} = \frac{ - 3}{ - 0.222} \approx 13.51\quad \Rightarrow \quad \hspace{0.15cm} \underline{M = 13}.$$

• Checking the values of  $y_{13} \approx 0.0013$  and  $y_{14} \approx 0.0008$  confirms this result.

(4)  Due to the linearity of the present filter,  the same result is obtained

• if the filter is not changed compared to subtask  (2)    $(a_1 = 0)$
• and the input sequence   $\left\langle {x_\nu } \right\rangle = \left\langle {1,\; - 0.5,\;0,\;0,\;\text{...} } \right\rangle$  is considered.

One then obtains in general for  $\nu \gt 0$:

$$y_\nu = {b_1} ^\nu + a_1 \cdot {b_1} ^{\nu - 1} = \left( {b_1 + a_1 } \right) \cdot {b_1} ^{\nu - 1} .$$

• With  $b_1 = 0.6$  and  $a_1 = -0.5$,  this gives  $y_\nu = 0.1\cdot {0.6} ^{\nu - 1}$,  and thus the sequence   $\left\langle {y_\nu } \right\rangle = \left\langle {1,\;0.1,\;0.06,\;0.036,\;\text{...} } \right\rangle .$

• The value we are looking for is  $y_3\hspace{0.15cm}\underline{= 0.036}$.