# Exercise 3.4: Attenuation and Phase Response

Pole-zero diagram and definition of some auxiliary quantities

We assume the sketched pole–zero diagram, i.e. the values

$$K = 5, \hspace{0,2cm}Z = 1, \hspace{0,2cm}N = 2,$$
$$p_{\rm o}= 1,\hspace{0,2cm}p_{\rm x1}= -3 + 3{\rm j},\hspace{0,2cm}p_{\rm x2}= -3 - 3{\rm j}\hspace{0.05cm} .$$

Thus,  the  $p$–transfer function is:

$$H_{\rm L}(p)= K \cdot \frac {p - p_{\rm o }} {(p - p_{\rm x 1})(p - p_{\rm x 2})} \hspace{0.05cm} .$$

Considering the substitution  $p = {\rm j} \cdot 2 \pi f$,  the conventional transfer function can be specified,  which is also called  "frequency response":

$$H(f) = H_{\rm L}(p)\Bigg |_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm}2\pi \it f}} = {\rm e}^{-a(f)\hspace{0.05cm}}\cdot {\rm e}^{- \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}b(f)} \hspace{0.05cm}.$$

From this equation it can also be seen the relationship between

• the transfer function  $H(f)$,
• the attenuation function  $a(f)$  and
• the phase function  $b(f)$.

The attenuation and phase values can be determined as follows for a frequency  $f$  indirectly specified by the point  $p = {\rm j} \cdot 2 \pi f$ :

$$a(f)\hspace{0.15cm}{\rm in}\hspace{0.15cm}{\rm Np} \hspace{0.25cm} = \hspace{0.2cm} -{\rm ln} \hspace{0.1cm} K + {\rm ln} \hspace{0.1cm} |R_{\rm x1}|+{\rm ln} \hspace{0.1cm} |R_{\rm x1}|- {\rm ln} \hspace{0.1cm} |R_{{\rm o} }|\hspace{0.05cm} ,$$
$$b(f)\hspace{0.15cm}{\rm in}\hspace{0.15cm}{\rm rad} \hspace{0.25cm} = \hspace{0.2cm} \phi_{\rm x1}+ \phi_{\rm x2}-\phi_{\rm o} \hspace{0.05cm} .$$

The corresponding magnitudes  $|R_{\rm o}|$,  $|R_{\rm x1}|$  and  $|R_{\rm x2}|$ as well as the angles  $\phi_{\rm o}$,  $\phi_{\rm x1}$  and  $\phi_{\rm x2}$  can be taken from the graph .

### Questions

1

Compute  $H(f)$.  What is its magnitude at very large frequencies?

 $|H(f → ∞)| \ = \$

2

Compute the magnitude frequency response and the attenuation value for  $f → 0$.

 $|H(f = 0)| \ = \$ $a(f = 0) \ = \$ $\ \rm Np$

3

Compute the attenuation value at  $f =4/(2 \pi)$  in neper  $\rm(Np)$  and decibel  $\rm(dB)$  according to the described approach.

 $a(f = 2/ \pi)\ = \$ $\ \rm Np$ $a(f = 2/ \pi)\ = \$ $\ \rm dB$

4

Compute the phase value at frequency  $f = 4/(2 \pi)$ according to the described approach.

 $b(f = 2/ \pi)\ = \$ $\ \rm Grad$

### Solution

#### Solution

(1)  The  $p$–transfer function is:

$$H_{\rm L}(p)= K \cdot \frac {p - p_{\rm o }} {(p - p_{\rm x 1})(p - p_{\rm x 2})} \hspace{0.05cm} .$$
• The conventional transfer function  (the frequency response)  is obtained via the substitution  $p = {\rm j} \cdot 2 \pi f$:
$$H(f)= K \cdot \frac {{\rm j \hspace{0.05cm}2\pi \it f} - p_{\rm o }} {({\rm j \hspace{0.05cm}2\pi \it f} - p_{\rm x 1})({\rm j \hspace{0.05cm}2\pi \it f} - p_{\rm x 2})} = {\rm e}^{-a(f)\hspace{0.05cm}}\cdot {\rm e}^{- \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}b(f)} \hspace{0.05cm} .$$
• In the limiting case  $f → \infty$,  the following is obtained for the magnitude,  attenuation and phase:
$$\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} H(f)= \frac{K}{{\rm j \hspace{0.05cm}2\pi \it f}}\hspace{0.15cm}\Rightarrow \hspace{0.15cm}\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} |H(f)|\hspace{0.15cm}\underline {= 0} \hspace{0.05cm} \Rightarrow \hspace{0.15cm}\lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} a(f)= \infty,\hspace{0.1cm} \lim_{f \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} b(f)\underline {= {\pi}/{2}\hspace{0.1cm}(+90^\circ)} \hspace{0.01cm}.$$

(2)  From the general equation in subtask  (1)  the following is obtained with the limit process  $f → 0$:

Magnitude  (⇒ Betrag)  $|H(f)|$,
attenuation  (⇒ Dämpfung)  $a(f)$  and phase  $b(f)$
$$|H(f=0)|= -\frac {K \cdot p_{\rm o }} {p_{\rm x 1}\cdot p_{\rm x 2}} = \frac {5 \cdot 1}{ (-3 + 3{\rm j})\cdot (-3 + 3{\rm j})}= \frac {5 }{18}\hspace{0.15cm}\underline {\approx 0.278} \hspace{0.05cm} ,$$
$$a(f=0)=- {\rm ln} \hspace{0.1cm}\hspace{0.15cm}\underline { |H(f=0)|= 1.281\,{\rm Np }} \hspace{0.05cm} .$$

The screen capture of the Flash–module  "Causal Systems"  summarizes the results of this exercise:

• middle axis (blue):   magnitude $|H(f)|$,   ⇒   here labeled with  $|Y(f)|$,
• left axis (red):   attenuation  $a(f)$,
• right axis (green):   phase  $b(f)$.
• black point:   values for  $2\pi f = 4.$

Pole-zero diagram and some auxiliary quantities

(3)  According to the detailed description in the  theory part,  the following holds for the attenuation function:

$$a(f)= -{\rm ln} \hspace{0.1cm} K + {\rm ln} \hspace{0.1cm} |R_{\rm x1}|+{\rm ln} \hspace{0.1cm} |R_{\rm x2}|- {\rm ln} \hspace{0.1cm} |R_{{\rm o} }|\hspace{0.05cm} .$$
• Furthermore,  the additional unit  "neper"  $\rm (Np)$  must be taken into account.
• The attenuation at  $f = 2/\pi$  is searched-for.  For this, we set  $p = {\rm j} \cdot 2 \pi f = 4$  and determine the following distances:
$$R_{\rm o} = 1 - 4 \cdot {\rm j}, \hspace{0.2cm}|R_{\rm o}| \hspace{0.25cm} = \hspace{0.2cm} \sqrt{1^2 + 4^2}= 4.123, \hspace{1.15cm} {\rm ln} \hspace{0.1cm}|R_{\rm o}| \hspace{0.25cm} = \hspace{0.2cm}1.417\,{\rm Np }\hspace{0.05cm},$$
$$R_{\rm x1} = -3 - 1 \cdot {\rm j}, \hspace{0.2cm}|R_{\rm x1}| \hspace{0.25cm} = \hspace{0.2cm} \sqrt{3^2 + 1^2}= 3.162,\hspace{0.5cm} {\rm ln} \hspace{0.1cm}|R_{\rm x1}| \hspace{0.25cm} = \hspace{0.2cm}1.151\,{\rm Np }\hspace{0.05cm},$$
$$R_{\rm x2} = -3 - 7 \cdot{\rm j}, \hspace{0.2cm}|R_{\rm x2}| \hspace{0.25cm} = \hspace{0.2cm} \sqrt{3^2 + 7^2}= 7.616,\hspace{0.5cm} {\rm ln} \hspace{0.1cm}|R_{\rm x2}| \hspace{0.25cm} = \hspace{0.2cm}2.030\,{\rm Np }\hspace{0.05cm}.$$

$$\Rightarrow \hspace{0.3cm}a(f = \frac{4}{2\pi})= -{\rm ln} \hspace{0.1cm} 5 + 1.151+ 2.030- 1.417\hspace{0.15cm}\underline{=0.155\,{\rm Np }} \hspace{0.05cm}.$$

This is equivalent to   $0.155\ {\rm Np} \cdot 8.686 \ {\rm dB/Np} \hspace{0.15cm} \underline{= 1.346 \ {\rm dB}}$.

(4)  The following holds for the phase function according to the description in the theory section due to  $K > 0$ :

$$b(f ={2}/\pi) = \phi_{\rm x1} + \phi_{\rm x2}-\phi_{\rm o}\hspace{0.05cm},$$
$$\phi_{\rm x1} ={\rm arctan}\hspace{0.15cm}(1/3) = 18.4^\circ\hspace{0.05cm}, \hspace{0.5cm}\phi_{\rm x2} = {\rm arctan}\hspace{0.15cm}(7/3) = 66.8^\circ\hspace{0.05cm},\hspace{0.5cm} \phi_{\rm o} = {\rm arctan}\hspace{0.15cm}(-1/4) = 180^\circ - 76^\circ = 104^\circ$$
$$\Rightarrow \hspace{0.3cm}b(f ={2}/\pi) = 18.4^\circ + 66.8^\circ - 104^\circ \hspace{0.15cm} \underline{= -18.8^\circ} \hspace{0.05cm}.$$