Aufgaben:Exercise 1.8: Variable Edge Steepness

• Return to book

Trapezoidal low-pass filter (red) and raised-cosine low-pass filter (green)

Two low-pass filters with variable edge steepnesses are compared with each other. For frequencies $|f| ≤ f_1$ ,  $H(f) = 1$  holds in both cases. In contrast, all frequencies $|f| ≥ f_2$  are suppressed entirely.

In the range $f_1 ≤ |f| ≤ f_2$  the frequency responses are defined by the following equations:

• Trapezoidal low-pass filter  $\rm (TLP)$:

$$H(f) = \frac{f_2 - |f|}{f_2 - f_1} ,$$

• Raised-cosine low-pass filter  $\rm (RCLP)$:

$$H(f) = \cos^2 \left(\frac{|f|- f_1}{f_2 - f_1} \cdot\frac{\pi}{2} \right).$$

Alternative system parameters for both low-pass filters are

• the equivalent bandwidth $Δf$ defined by the equal-area rectangle, and also
• the roll-off factor (in frequency domain):

$$r=\frac{f_2 - f_1}{f_2 + f_1} .$$

Throughout the task, $Δf = 10 \ \rm kHz$  and  $r = 0.2$ hold true.

The impulse responses are with the equivalent impulse duration $Δt = 1/Δf = 0.1 \ \rm ms$:

$$h_{\rm TTP}(t) = \frac{1}{\Delta t} \cdot {\rm si}(\pi \cdot \frac{t}{\Delta t} )\cdot {\rm si}(\pi \cdot r \cdot \frac{t}{\Delta t} ),$$

$$h_{\rm CRTP}(t) = \frac{1}{\Delta t} \cdot {\rm si}(\pi \cdot \frac{t}{\Delta t} )\cdot \frac {\cos(\pi \cdot r \cdot t / \Delta t )}{1 - (2 \cdot r \cdot t/\Delta t )^2}.$$

Please note:

• The exercise belongs to the chapter  Some Low-Pass Functions in Systems Theory.
• In particular, reference is made to the pages  Trapezoidal low-pass filter  and  Raised-cosine low-pass filter.
• You can check your results with the interactive applet  Frequency response and impulse response.

Questions

1

What is the equation for the equivalent bandwidth $Δf$?  It holds that

	$Δf = f_2 - f_1$,
	$Δf = f_1 + f_2$,
	$Δf = (f_2 + f_1)/2$.

2

Determine the low-pass filter parameters $f_1$  and $f_2$  for  $Δf = 10 \ \rm kHz$  and  $r = 0.2$.

$f_1 \ = \ $ $\ \rm kHz$

$f_2 \ = \ $ $\ \rm kHz$

3

Which statements are true for the impulse response of the trapezoidal low-pass filter if  $r = 0.2$  is assumed?

	$h(t)$  has zeros at  $±\hspace{0.03cm}n · Δt \ (n = 1, 2, \text{...})$.
	$h(t)$  has additional zeros at other times.
	$h(t)$  would decay faster with  $r = 0$ .
	$h(t)$  would decay faster with  $r = 1$ .

4

Which statements are true for the impulse response of the raised-cosine low-pass filter if  $r = 0.2$  is assumed?

	$h(t)$  has zeros at  $±\hspace{0.03cm}n · Δt \ (n = 1, 2, \text{...})$.
	$h(t)$  has additional zeros at other times.
	$h(t)$  would decay faster with  $r = 0$ .
	$h(t)$  would decay faster with  $r = 1$ .

Solution

Solution

(1)  Approach 2 is correct:

• For both low-pass filters, the integral over $H(f)$  is equal to $f_1 + f_2$.
• Thus, due to $H(f = 0 = 1)$  Approach 2 is correct:   $\Delta f = f_1 + f_2.$

(2)  Substituting the relation found in  (1)  into the defining equation of the roll-off factor the following is obtained:

$${f_2 - f_1} = r \cdot \Delta f = {2\,\rm kHz}, \hspace{0.5cm} {f_2 + f_1} = {10\,\rm kHz}.$$

• By addition or subtraction of both equations the so-called "corner frequencies" result in

$$f_1 \underline{= 4 \ \rm kHz},$$

$$f_2 \underline{= 6 \ \rm kHz}.$$

(3)  Suggestions 1 and 4 are correct:

• The first $\rm sinc$–function of $h_{\rm TTP}(t)$  causes zeros at an interval of $\Delta t$  (see also the equation on the information page).
• The second  $\rm sinc$–function causes zeros at multiples of $5 · \Delta t$.
• There are no additional zeros since these coincide exactly with the zeros of the first  $\rm sinc$–function.
• The special case $r = 0$  corresponds to the ideal rectangular low-pass filter with  $\rm sinc$–shaped impulse response. This decays extremely slowly.
• The  $\rm si^2$–shaped impulse response of the triangular low-pass filter  $($special case for $r = 1)$  decays asymptotically with  $1/t^2$, i.e. faster than with $r = 0.2$.

(4)  Suggestions 1, 2 and 4 are correct here:

• The impulse response $h_{\rm CRTP}(t)$  of the raised-cosine low-pass filter also has zeros at an interval of $\Delta t$ due to the  $\rm sinc$–function.
• The cosine function has zeros at the following times:

$${\cos(\pi \cdot r \cdot {t}/{ \Delta t} )} = 0 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}r \cdot {t}/{ \Delta t} = \pm 0.5, \pm 1.5, \pm 2.5, \text{...} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {t}/{ \Delta t} = \pm 2.5, \pm 7.5, \pm 12.5, ... $$

$$\Rightarrow \hspace{0.3cm} {t}/{ \Delta t} = \pm 2.5, \pm 7.5, \pm 12.5, \text{...}. $$

• However, the zero of the numerator at $t / \Delta t = 2.5$  is nullified by the denominator which also vanishes.
• By contrast, the other zeros at $7.5, 12.5,\text{...} $  remain.
• Here,  $r = 0$  also results in the rectangular low-pass filter and thus in the $\rm sinc$–shaped impulse response.
• In contrast, the impulse response of the cosine-square low-pass filter  $($special case for  $r = 1)$  decays extremely fast.
• This is studied in detail in  Exercise 1.8Z .