Aufgaben:Exercise 3.7Z: Partial Fraction Decomposition

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Pole-zero diagrams

In the graph,  four two-port networks are given by their pole–zero diagrams  $H_{\rm L}(p)$.

• They all have in common that the number  $Z$  of zeros is equal to the number  $N$  of poles.
• The constant factor in each case is  $K=1$.

In the special case  $Z = N$  the residue theorem cannot be applied directly to compute the impulse response  $h(t)$.

Rather,  a  "partial fraction decomposition"  corresponding to

$$H_{\rm L}(p) =1- H_{\rm L}\hspace{0.05cm}'(p)

\hspace{0.05cm}$$

must be made beforehand. Then,

$$h(t) = \delta(t)- h\hspace{0.03cm}'(t)

\hspace{0.05cm}$$ 

holds for the impulse response.  $h\hspace{0.03cm}'(t)$  is the inverse Laplace transform of  $H_{\rm L}\hspace{0.05cm}'(p)$,  where the condition  $Z' < N'$  is satisfied.

Two of the four configurations given are so-called  "all-pass filters".

• This refers to two-port networks for which the Fourier spectrum satisfies the condition  $|H(f)| = 1$   ⇒   $a(f) = 0$ .
• In Exercise 3.4Z it is given how the poles and zeros of such an all-pass filter must be positioned.

Furthermore,  in this exercise the  $p$–transfer function

$$H_{\rm L}^{(5)}(p) =\frac{p/A}{\left (\sqrt{p/A}+\sqrt{A/p} \right )^2}

\hspace{0.05cm}$$

⇒   "configuration $(5)$" will be examined in more detail,  which can be represented by one of the four pole–zero diagrams given in the graph if the parameter  $A$  is chosen correctly.

Please note:

• The exercise belongs to the chapter  Inverse Laplace Transform.
• In particular, reference is made to the page  Partial fraction decomposition.

Questions

1 Which of the sketched two-port networks are all-pass filters?

	Configuration  $(1)$,
	configuration  $(2)$,
	configuration  $(3)$,
	configuration  $(4)$.

2 Which two-port network has the transfer function  $H_{\rm L}^{(5)}(p)$?

	Configuration  $(1)$,
	configuration  $(2)$,
	configuration  $(3)$,
	configuration  $(4)$.

3 Compute the function  $H_{\rm L}\hspace{0.01cm}'(p)$  after a partial fraction decomposition for configuration  (1).  Enter the function value for  $p = 0$.

$H_{\rm L}\hspace{0.01cm}'(p = 0) \ = \ $

4 Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for configuration  $(2)$.  Which statements are true here?

	$H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
	$H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
	The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

5 Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for configuration  $(3)$.  Which statements are true here?

	$H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
	$H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
	The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

6 Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for configuration  $(4)$.  Which statements are true here?

	$H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
	$H_{\rm L}\hspace{0.01cm}'(p)$  has the same poles as  $H_{\rm L}(p)$.
	The constant factor of  $H_{\rm L}\hspace{0.01cm}'(p)$  is  $K' = 8$.

Solution

Solution

(1)  The suggested solutions 1 and 2 are correct:

• According to the criteria given in exercise 3.4Z, there is always an all-pass filter at hand if there is a corresponding zero  $p_{\rm o} = + A + {\rm j} \cdot B$  in the right $p$–half-plane for each pole  $p_{\rm x} = - A + {\rm j} \cdot B$  in the left half-plane.
• Considering  $K = 1$  the attenuation function is then  $a(f) = 0 \ \rm Np$   ⇒   $|H(f)| = 1$.
• The following can be seen from the graph on the information page:   The configurations  $(1)$ and  $(2)$ satisfy exactly these symmetry properties.

(2)  The suggested solution 4 is correct:

• The transfer function  $H_{\rm L}^{(5)}(p)$  is also described by configuration  $(4)$  as the following calculation shows:

$$H_{\rm L}^{(5)}(p) \hspace{0.25cm} = \hspace{0.2cm} \frac{p/A}{(\sqrt{p/A}+\sqrt{A/p})^2}

=\frac{p/A}{{p/A}+2+ {A/p}}
 =  \hspace{0.2cm}\frac{p^2}{p^2 + 2A \cdot p + A^2} = \frac{p^2}{(p+A)^2
}= H_{\rm L}^{(4)}(p)
\hspace{0.05cm}.$$

• The double zero is at  $p_{\rm o} = 0$ and the double pole at  $p_{\rm x} = -A = -2$.

(3)  The following holds for configuration  $(1)$ :

$$H_{\rm L}(p) =\frac{p-2}{p+2}=\frac{p+2-4}{p+2}= 1 - \frac{4}{p+2}=1- H_{\rm L}\hspace{-0.05cm}'(p)

\hspace{0.3cm} \Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = \frac{4}{p+2}

\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\hspace{0.15cm}\underline{H_{\rm L}\hspace{-0.05cm}'(p =0)
=2}
\hspace{0.05cm}.$$

(4)  Similarly, the following is obtained for configuration  $(2)$:

$$H_{\rm L}(p) =\frac{(p-2 - {\rm j} \cdot 2)(p-2 + {\rm j} \cdot 2)}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}=

 \frac{p^2 -4\cdot p  +8 }{p^2 +4\cdot p  +8}=
   \hspace{0.2cm}\frac{p^2 +4\cdot p  +8 -8\cdot p}{p^2 +4\cdot p
+8} =1- \frac{8\cdot p}{p^2 +4\cdot p  +8}=1- H_{\rm L}\hspace{-0.05cm}'(p)$$

$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 8

\cdot \frac{p}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
\hspace{0.05cm}.$$

Thus, the suggested solutions 2 and 3 are correct in contrast to statement 1:

• While  $H_{\rm L}(p)$  has two conjugate complex zeros,
•  $H_{\rm L}\hspace{0.01cm}'(p)$  only has a single zero at  $p_{\rm o}\hspace{0.01cm}' = 0$.

(5)  The following applies for configuration  $(3)$ :

$$H_{\rm L}(p) =

 \frac{p^2 }{p^2 +4\cdot p  +8}=\frac{p^2 +4\cdot p  +8 -4\cdot p  -8 }{p^2 +4\cdot p  +8}
= 1- H_{\rm L}\hspace{-0.05cm}'(p)$$

$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = 4

\cdot \frac{p+2}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
\hspace{0.05cm}.$$

• The zero of  $H_{\rm L}\hspace{0.01cm}'(p)$  is now at  $p_{\rm o}\hspace{0.01cm}' = -2$.
• The constant is  $K\hspace{0.01cm}' = 4$   ⇒   only suggested solution 2 is correct here.

(6)  Finally, the following holds for configuration  $(4)$:

$$H_{\rm L}(p) = \frac{p^2 }{(p+2)^2}=\frac{p^2 +4\cdot p +4 -4\cdot p -4 }{p^2 +4\cdot p +4}

 = 1- \frac{4\cdot p  +4 }{p^2 +4\cdot p  +4}

\hspace{0.3cm}\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 4

\cdot \frac{p+1}{(p+2)^2}
\hspace{0.05cm}.$$

Suggested solution 2 is also correct here. In general, it can be said that:

• The partial fraction decomposition changes the number and position of the zeros.
• On the contrary, the poles of $H_{\rm L}\hspace{0.01cm}'(p)$ are always identical to those of $H_{\rm L}(p)$.