Aufgaben:Exercise 1.8Z: Cosine-Square Low-Pass Filter

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Concerning the cosine-square low-pass filter

When studying digital systems a dirac-shaped input signal $x(t) = T \cdot \delta(t)$  is often assumed such that  $X(f) = T$  holds.

The output spectrum $Y(f)$  is then of equal form compared with the overall frequency response of the transmitter and receiver filter:

$$H(f) = H_{\rm S}(f) \cdot H_{\rm E}(f).$$

This frequency response is often assumed to be  $\cos^2$-shaped (see graph):

• For  $f \cdot T \ge 1$ ,  $H(f) = 0$ holds.
• In the inner region the following holds:  $H(f) = \cos^2(f \cdot T \cdot {\pi}/{ 2} ) .$

Note that the equivalent bandwidth is $\Delta f = 1/{\Delta t}$ . Thus, for the equivalent ${\Delta t}$  of the impulse response one also obtains  $T$  and the following holds:

$$y(t) = T \cdot h(t) = {\rm si}(\pi \cdot {t}/{T} )\cdot \frac {\cos(\pi \cdot t / T )}{1 - (2 \cdot t/T )^2}.$$

Also, note that the output signal $y(t)$  is dimensionless in cotrast to the impulse response  $h(t)$ . This signal can also be represented as follows by applying trigonomic transformations:

$$y(t) = {\pi}/{4} \cdot {\rm si}(\pi \cdot {t}/{T} )\cdot \big[ {\rm si}\big(\pi \cdot \left({t}/{T}+ 0.5 \right) \big)+ {\rm si}\big(\pi \cdot \left({t}/{T}- 0.5 \right) \big)\big].$$

For each of the following tasks select the equation that is easier to handle.

For subtask  (3)  it shall be assumed that the signal $s(t)$  is a rectangular pulse in the middle between the two frequency responses $H_{\rm S}(f)$  and  $H_{\rm S}(f)$ . Consequently, it must hold:

$$H_{\rm E}(f) = {\rm si }(\pi f T ) .$$

Please note:

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

Questions

1

Compute the output signal at times $t = 0$  and  $t = T$.

$y(t = 0) \ = \ $

$y(t = T) \ = \ $

2

Compute the output signal at times $t = 0.5 T$  and  $t = 1.5 T$.

$y(t = 0.5 T) \ = \ $

$y(t = 1.5 T) \ = \ $

3

Compute $y(t)$  for large $t$-values. Suitable approximations are allowed and encouraged.  What is the signal value at $t = 10.75 T$?

$y(t = 10.75 T) \ = \ $ $ \ \cdot \ 10^{-6}$

4

State the required receiver frequency response $H_{\rm E}(f)$  for  $H_{\rm S}(f)= {\rm si}(πfT)$ .  What are the values for the given frequencies?

$H_{\rm E}(f=0) \ = \ $

$H_{\rm E}(f=0.5/T) \ = \ $

$H_{\rm E}(f=1/T) \ = \ $

Solution

Solution

(1)  From the first equation on the information page it directly follows that  $y(t = 0) = 1$  and  $y(t = T) = y(t = 2T) = \text{...} =0$ due to the $\rm si$–function.

• From the second equation these results are obtained, for example

$$y(t = 0) = {\pi}/{4} \cdot {\rm si}(0)\cdot \left[ {\rm si}(\pi/2)+ {\rm si}(-\pi/2)\right] {\pi}/{2} \cdot {\rm si}(\pi/2) = {\pi}/{2} \cdot \frac{{\rm sin}(\pi/2)}{\pi/2} \hspace{0.15cm}\underline{= 1},$$

$$y(t = T) \hspace{0.15cm}=\hspace{0.15cm}{\pi}/{4} \cdot {\rm si}(\pi)\cdot \left[ {\rm si}(3\pi/2)+ {\rm si}(\pi/2)\right] \hspace{0.15cm}\underline{= 0}.$$

Output signal of the cosine-square low-pass filter

(2)  The second representation is more suitable for the computation of these signal values:

$$y(t = T/2) = {\pi}/{4} \cdot {\rm si}(\pi/2)\cdot \big[ {\rm si}(\pi)+ {\rm si}(0)\big].$$

• Considering  ${\rm si}(0) = 1$  and  ${\rm si}(\pi) = 0$  one obtains thus:

$$y(t = T/2) = {\pi}/{4} \cdot {\rm si}(\pi/2)= {\pi}/{4} \cdot \frac{{\rm sin}(\pi/2)}{\pi/2} \hspace{0.15cm}\underline{ = 0.5}.$$

• Analogously, the following is obtained for  $t = 1.5T$:

$$y(t = 1.5T) = {\pi}/{4} \cdot {\rm si}(3\pi/2)\cdot \left[ {\rm si}(2\pi)+ {\rm si}(\pi)\right] \hspace{0.15cm}\underline{ = 0}.$$

• Here,   ${\rm si}(\pi) = {\rm si}(2\pi) = 0$  is considered.
• Also, at times  $t/T = 2.5, 3.5,\text{ ... }$ ,   $y(t) = 0$ holds as shown in the above graph.

(3)  For large values of  $t$  the following holds by approximation  (neglecting the "1" in the denominator):

$$y(t)= {\rm si}(\pi \cdot {t}/{T} )\cdot \frac {\cos(\pi \cdot t / T )}{1 - (2 \cdot t/T )^2} \approx \frac {\sin(\pi \cdot t / T )\cdot \cos(\pi \cdot t / T )}{ - (\pi \cdot t/T )(2 \cdot t/T )^2} = - \frac {\sin(2\pi \cdot t / T )}{ 8 \pi \cdot( t/T )^3}.$$

• This takes into account that $\sin(\alpha) \cdot \cos(\alpha) = \sin(2\alpha)/2$ .
• Hence at time $t = 10.75 T$  it holds that:

$$\sin(2\pi \cdot t / T ) = \sin (21.5\pi)= \sin (1.5\pi) = -1\hspace{0.3cm} \Rightarrow \hspace{0.3cm} y(t = 10.75 T) = \frac {1 }{ 8 \pi \cdot( 10.75 )^3} \hspace{0.15cm}\underline{= 32 \cdot 10^{-6}}.$$

Searched-for receiver frequency response

(4)  The receiver frequency response is the following for  $|f \cdot T| \le 1$:

$$H_{\rm E}(f) = \frac{H(f)}{H_{\rm S}(f)}= \frac{\cos^2(\pi f T /2)}{{\rm si}(\pi fT)}.$$

• This function curve is shown in the graph. The following holds or the interpolation values:

$$H_{\rm E}(f = 0) = \frac{\cos^2(0)}{{\rm si}(0)} \hspace{0.15cm}\underline{ = 1},$$

$$H_{\rm E}(f = {0.5}/T \hspace{-0.15cm} = \hspace{-0.15cm} \frac{\cos^2(\pi/4)}{{\rm si}(\pi/2)}= (\sqrt{2} / 2)^2 \cdot \frac{\pi}{2} = \hspace{-0.15cm} \frac{\pi}{4}\hspace{0.15cm}\underline{ \approx 0.785},$$

$$H_{\rm E}(f = {1}/{T}) = \frac{\cos^2(\pi/2)}{{\rm si}(\pi)} = "0/0"\hspace{0.15cm}\underline{= 0}.$$

This result takes into account that in the whole frequency domain  $H_{\rm S}(f) \ge H(f) $  holds. Actually, the last calculated value should be determined by a limiting process in a mathematically exact way. We did that too, but not in this sample solution.