Aufgaben:Exercise 3.3: p-Transfer Function

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Considered two-port network

Any linear time-invariant system that can be realized by a circuit of discrete time-constant components  (resistances  $R$,  capacitances  $C$,  inductances  $L$,  amplifier elements, etc.)  is causal and has a fractional–rational  $p$–transfer function of the form

$$H_{\rm L}(p)= \frac {A_Z \cdot p^Z +\text{ ...} + A_1 \cdot p + A_0} {B_N \cdot p^N + \text{ ...} + B_1 \cdot p + B_0}= \frac {Z(p)}{N(p)} \hspace{0.05cm} .$$

• All coefficients  $A_Z$, ... ,  $A_0$,  $B_N$, ... ,  $B_0$  are real.
• $Z$  denotes the degree of the numerator polynomial  $Z(p)$.
• $N$  denotes the degree of the denominator polynomial  $N(p)$ .

An equivalent representation form of the above equation is:

$$H_{\rm L}(p)= K \cdot \frac {\prod\limits_{i=1}^Z p - p_{\rm o i}} {\prod\limits_{i=1}^N p - p_{\rm x i}}= K \cdot \frac {(p - p_{\rm o 1})(p - p_{\rm o 2})\cdot \text{ ...} \cdot (p - p_{{\rm o} \hspace{-0.03cm} Z})} {(p - p_{\rm x 1})(p - p_{\rm x 2})\cdot \text{ ...} \cdot (p - p_{{\rm x} \hspace{-0.03cm} N})} \hspace{0.05cm} .$$

The  $Z + N + 1$  parameters mean:

• $K = A_Z/B_n$  is a constant factor.  If  $Z = N$ applies, then this is dimensionless.
• The solutions of the equation  $Z(p) = 0$  yield the  $Z$ zeros  $p_{{\rm o}1}$, ... , $p_{{\rm o}Z}$  of  $H_{\rm L}(p)$.
• The zeros of the denominator polynomial  $N(p)$  yield the  $N$  poles  $p_{{\rm x}1}$, ... , $p_{{\rm x}N}$  of the transfer function.

These characteristics are to be determined for the circuit shown in the diagram with the following components:

$$R = 50\,\,{\rm \Omega}\hspace{0.05cm},\hspace{0.2cm} L = 10\,\,{\rm µ H}\hspace{0.05cm},\hspace{0.2cm}C = 25\,\,{\rm nF}$$

Additionally, the Fourier frequency response  $H(f)$  is to be determined which arises as a result from  $H_{\rm L}(p)$  by the substitution  $p= {\rm j } \cdot 2\pi f$ .

Please note:

• The exercise belongs to the chapter  Laplace Transform and p-Transfer Function.

• The following are the auxiliary quantities used in this exercise:

$$A = \frac{R}{2L}\hspace{0.05cm},\hspace{0.2cm} B = \frac{1}{\sqrt{LC}}\hspace{0.05cm} .$$

Questions

1

Determine the  $p$–transfer function.  What asymptotic values are obtained for  $p → 0$  and  $p → \infty$?

$H_L(p → 0) \ = \ $

$H_L(p → ∞) \ = \ $

2

Find the frequency response  $H(f)$ from  $H_{\rm L}(p)$  by setting  $p= {\rm j } \cdot 2\pi f$ .  Which of the following statements are true?

	It is a band-pass filter.
	It is a band-stop filter.
	Without exact knowledge of  $R$,  $L$ and  $C$  it is not possible to make a statement.

3

Compute the auxiliary quantities  $A$  and  $B$  for  $R = 50 \ \rm \Omega$,  $L = 10 \ µ\rm H$,  $C = 25 \ \rm nF$.

$A \ = \ $ $\ \cdot \ 10^6 \ \rm 1/s$

$B \ = \ $ $\ \cdot \ 10^6 \ \rm 1/s$

4

Express  $H_{\rm L}(p)$  in pole–zero form.  How many zeros  $(Z)$  and poles  $(N)$  are there?  What is the constant factor  $K$?

$Z \ = \ $

$N \ = \ $

$K \ = \ $

5

Compute the zeros  $p_\text{o1}$ (in the upper half-plane) and  $p_\text{o2}$ (in the lower half-plane).  Consider the unit  $\rm 1/ µs$.

${\rm Re}\{p_\text{o1}\} \ =\ $ $\ \rm 1/ µ s$

${\rm Im}\{p_\text{o1}\} \ = \ $ $\ \rm 1/ µ s$

${\rm Re}\{p_\text{o2}\} \ =\ $ $\ \rm 1/ µ s$

${\rm Re}\{p_\text{o2}\} \ = \ $ $\ \rm1/ µ s$

6

Compute the poles  $p_\text{x1}$  and  $p_\text{x2}$.  Let  $|p_\text{x2}| > |p_\text{x1}|$ hold.

${\rm Re}\{p_\text{x1}\} \ =\ $ $\ \rm 1/ µ s$

${\rm Im}\{p_\text{x1}\} \ =\ $ $\ \rm 1/ µ s$

${\rm Re}\{p_\text{x2}\} \ =\ $ $\ \rm 1/ µ s$

${\rm Im}\{p_\text{x2}\} \ =\ $ $\ \rm 1/ µ s$

7

How can you change the position of the poles without changing the zeros?

	Change of  $R$;   $L$ and $C$ unchanged.
	Change of  $L$;   $R$ and $C$ unchanged.
	Change of  $C$;   $L$ and $R$ unchanged.

8

How must the auxiliary quantity  $A$  be changed  $(B$ unchanged$)$ so that a double pole occurs  (critically-damped case)?

$A \ =\ $ $\ \rm \cdot 10^6\ 1/s$

Solution

Solution

(1)  According to the voltage divider principle,  the following can be written for the $p$–transfer function:

$$H_{\rm L}(p)= \frac {pL +{1}/{(pC)}} {R + pL + {1}/{(pC)}}= \frac { p^2 \cdot{LC}+1} {p^2 \cdot{LC} + p \cdot{RC}+ 1} \hspace{0.05cm} .$$

• The two desired limit processes yield:

$$\underline {H_{\rm L}(p \rightarrow 0)= 1, \hspace{0.2cm}H_{\rm L}(p \rightarrow \infty)= 1} \hspace{0.05cm} .$$

1. From this it follows that it can be neither a low-pass filter nor a high-pass filter.
2. Both at very low and very high frequencies,   $y(t)=x(t)$ holds.

(2)  Suggested solution 2  is correct:

• Replacing  $p$  by  ${\rm j } \cdot 2\pi f$ the following is obtained:

$$H(f)= \frac {1 - (2\pi f)^2 \cdot LC} {1 - (2\pi f)^2 \cdot LC + {\rm j} \cdot 2\pi f \cdot RC} \hspace{0.05cm} .$$

• So,  there is always a frequency at which the numerator is zero, namely the resonance frequency of  $L$  and  $C$.
• For this frequency  $f_0 = 1 \ \rm MHz/2\pi$  the series connection of  $L$  and  $C$  acts like a short circuit.
• From this it follows:  Regardless of the values of  $R$,  $L$  and  $C$ it is a  $\rm band–stop \:filter$.

(3)  The following holds according to the information sheet:

$$A = \frac{R}{2L}= \frac{50\,{\rm \Omega}}{2 \cdot 10\,{\rm \mu H}} = \frac{50\,{\rm \Omega}}{2 \cdot 10^{-5 }\,{\rm \Omega s}}\hspace{0.15cm} \underline {= 2.5} \cdot 10^6 \, \,{1}/{\rm s}\hspace{0.05cm},$$

$$ B = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{10^{-5 }\,{\rm \Omega s} \cdot 25 \cdot 10^{-9 }\,{\rm s/\Omega }}}\hspace{0.15cm} \underline {= 2.0} \cdot 10^6 \, \,{1}/{\rm s}\hspace{0.05cm} .$$

(4)  Using  $A=R/(2L)$  and  $B^2 = 1/(LC)$  the following is obtained from the  $p$–transfer function determined in subtask  (1) :

$$H_{\rm L}(p)= \frac { p^2 + {1}/(LC)} {p^2 + p \cdot{R}/{L} +{1}/(LC)} = \frac { p^2 + B^2} {p^2 + 2A \cdot p + B^2} \hspace{0.05cm} .$$

• The numerator polynomial  $Z(p)$  and the denominator polynomial  $N(p)$  are each quadratic   ⇒   $\underline {Z = N = 2}$.
• The constant factor is  $\underline {K = 1}$.

(5)  Solving the equation  $p^2 + B^2 = 0$  leads to the result  $p = \pm {\rm j} \cdot B$  and thus to the zeros

$${\rm Re}\{ p_{\rm o1}\} \underline {= 0}\hspace{-0.3cm} \hspace{1cm}{\rm Im}\{ p_{\rm o1}\} \underline {=+2.5} \cdot 10^6 \, {1}/{{\rm s}} \hspace{0.05cm},$$

$$ {\rm Re}\{ p_{\rm o2}\}\hspace{0.15cm} \underline { = 0}\hspace{-0.3cm} \hspace{1cm}{\rm Im}\{ p_{\rm o2}\} \underline {=-2.5} \cdot 10^6 \, {1}/{{\rm s}} \hspace{0.05cm}.$$

• The normalization of the variable  $p$  and all poles and zeros to the unit  $( \rm 1/µ s)$  simplifies the numerical evaluation,  especially in the time domain.
• If the unit is dispensed with altogether,  all  $t$–values are obtained in microseconds.

(6)  If the denominator polynomial is set  $N(p) = 0$,  the following conditional equation arises as a result:

$$p^2 + 2A \cdot p + B^2 = 0 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm x1,\hspace{0.05cm}2}= -A \pm \sqrt{A^2 - B^2} \hspace{0.05cm},$$

$${\rm Mit}\hspace{0.2cm}A = 2.5 \cdot 10^6 \cdot {1}/{\rm s}\hspace{0.05cm},\hspace{0.2cm} \sqrt{A^2 - B^2}= 1.5 \cdot 10^6 \cdot {1}/{{\rm s}}\hspace{0.05cm}:$$

$${\rm Re}\{ p_{\rm x1}\}\hspace{0.15cm} \underline {= -1} \cdot 10^6 \cdot {1}/{{\rm s}}\hspace{0.15cm} \underline {= -1} \cdot {1}/{{\rm \mu s}}, \hspace{0.2cm}{\rm Im}\{ p_{\rm x1}\}\hspace{0.15cm} \underline { = 0} \hspace{0.05cm},$$

$${\rm Re}\{ p_{\rm x2}\}\hspace{0.15cm} \underline {= -4} \cdot 10^6 \cdot {1}/{{\rm s}}\hspace{0.15cm} \underline {= -4} \cdot {1}/{{\rm \mu s}}, \hspace{0.2cm}{\rm Im}\{ p_{\rm x1}\}\hspace{0.15cm} \underline { = 0} \hspace{0.05cm}.$$

This result is only unique considering the specification  $|p_\text{x2}| > |p_\text{x1}|$.

(7)  Suggested solution 1  is correct:

• Since only one of the components is to be changed,  $L$  and  $C$  must remain the same because otherwise the zeros would also be shifted.
• The resistance value  $R$  must be changed.

(8)  According to the result in subtask  (7)  there is a double pole for  $\underline {A = B = 2 \cdot 10^{-6} \cdot \rm 1/s}$.

• To do this,  the ohmic resistance must be reduced from  $50 \ \rm \Omega$  to  $40 \ \rm \Omega$ .
• Then,  the double pole is at  ${-2 \cdot 10^{6} \cdot \rm 1/s}$.
• Or with another normalization at  ${-2 \cdot \rm (1/µ s)}$.