Aufgaben:Exercise 3.3: p-Transfer Function - Revision history

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← Older revision		Revision as of 15:30, 16 March 2026
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	[[File:P_ID1765__LZI_A_3_3.png\|right\|frame\|Considered two-port network]]		[[File:P_ID1765__LZI_A_3_3.png\|right\|frame\|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		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
	:$$H_{\rm L}(p)= \frac {A_Z \cdot p^Z +\text{ ...} + A_1 \cdot p + A_0}		:$$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} .$$
	{B_N \cdot p^N + \text{ ...} + B_1 \cdot p + B_0}= \frac {Z(p)}{N(p)}
	\hspace{0.05cm} .$$

	*All coefficients  $A_Z$, ... ,  $A_0$,  $B_N$, ... ,  $B_0$  are real.		*All coefficients  $A_Z$, ... ,  $A_0$,  $B_N$, ... ,  $B_0$  are real.
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	An equivalent representation form of the above equation is:		An equivalent representation form of the above equation is:
	:$$H_{\rm L}(p)= K \cdot \frac {\prod\limits_{i=1}^Z p - p_{\rm o i}}		:$$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} .$$
	{\prod\limits_{i=1}^N p - p_{\rm x i}}= K \cdot \frac {(p - p_{\rm o 1})(p - p_{\rm o 2})\cdot \text{ ...} \cdot (p - p_{{\rm o} \hspace{-0.03cm} Z})}
	{(p - p_{\rm x 1})(p - p_{\rm x 2})\cdot \text{ ...} \cdot (p - p_{{\rm x} \hspace{-0.03cm} N})}
	\hspace{0.05cm} .$$

	The  $Z + N + 1$  parameters mean:		The  $Z + N + 1$  parameters mean:
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	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  According to the voltage divider principle,  the following can be written for the $p$–transfer function:		'''(1)'''  According to the voltage divider principle,  the following can be written for the $p$–transfer function:
	:$$H_{\rm L}(p)= \frac {pL +{1}/{(pC)}}		:$$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} .$$
	{R + pL + {1}/{(pC)}}= \frac { p^2 \cdot{LC}+1}
	{p^2 \cdot{LC} + p \cdot{RC}+ 1}
	\hspace{0.05cm} .$$

	*The two desired limit processes yield:		*The two desired limit processes yield:
	:$$\underline {H_{\rm L}(p \rightarrow 0)= 1, \hspace{0.2cm}H_{\rm L}(p \rightarrow \infty)= 1}		:$$\underline {H_{\rm L}(p \rightarrow 0)= 1, \hspace{0.2cm}H_{\rm L}(p \rightarrow \infty)= 1}\hspace{0.05cm} .$$
	\hspace{0.05cm} .$$

	:#From this it follows that it can be neither a low-pass filter nor a high-pass filter.		:#From this it follows that it can be neither a low-pass filter nor a high-pass filter.
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	'''(2)'''  <u>Suggested solution 2</u>  is correct:		'''(2)'''  <u>Suggested solution 2</u>  is correct:
	*Replacing  $p$  by  ${\rm j } \cdot 2\pi f$ the following is obtained:		*Replacing  $p$  by  ${\rm j } \cdot 2\pi f$ the following is obtained:
	:$$H(f)= \frac {1 - (2\pi f)^2 \cdot LC}		:$$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} .$$
	{1 - (2\pi f)^2 \cdot LC + {\rm j} \cdot 2\pi f \cdot RC}
	\hspace{0.05cm} .$$

	*So,  there is always a frequency at which the numerator is zero, namely the resonance frequency of  $L$  and  $C$.		*So,  there is always a frequency at which the numerator is zero, namely the resonance frequency of  $L$  and  $C$.
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	'''(3)'''  The following holds according to the information sheet:		'''(3)'''  The following holds according to the information sheet:
	:$$A = \frac{R}{2L}= \frac{50\,{\rm \Omega}}{2 \cdot 10\,{\rm \mu H}} = \frac{50\,{\rm \Omega}}{2 \cdot 10^{-5 }\,{\rm \Omega s}}\hspace{0.15cm} \underline {=		:$$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},$$
	2.5} \cdot 10^6 \, \,{1}/{\rm s}\hspace{0.05cm},$$		:$$ B = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{10^{-5 }\,{\rm \Omega s} \cdot 25 \cdot 10^{-9 }\,{\rm s/\Omega }}}\hspace{0.15cm} \underline {=2.0} \cdot 10^6 \, \,{1}/{\rm s}\hspace{0.05cm} .$$
	:$$ B = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{10^{-5 }\,{\rm \Omega s} \cdot 25 \cdot 10^{-9 }\,{\rm s/\Omega }}}\hspace{0.15cm} \underline {=
	2.0} \cdot 10^6 \, \,{1}/{\rm s}\hspace{0.05cm} .$$



	'''(4)'''  Using  $A=R/(2L)$  and  $B^2 = 1/(LC)$  the following is obtained from the  $p$–transfer function determined in subtask  '''(1)''' :		'''(4)'''  Using  $A=R/(2L)$  and  $B^2 = 1/(LC)$  the following is obtained from the  $p$–transfer function determined in subtask  '''(1)''' :
	:$$H_{\rm L}(p)= \frac { p^2 + {1}/(LC)}		:$$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} .$$
	{p^2 + p \cdot{R}/{L} +{1}/(LC)} = \frac { p^2 + B^2}
	{p^2 + 2A \cdot p + B^2}
	\hspace{0.05cm} .$$

	*The numerator polynomial  $Z(p)$  and the denominator polynomial  $N(p)$  are each quadratic   ⇒   $\underline {Z = N = 2}$.		*The numerator polynomial  $Z(p)$  and the denominator polynomial  $N(p)$  are each quadratic   ⇒   $\underline {Z = N = 2}$.
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	'''(5)'''  Solving the equation  $p^2 + B^2 = 0$  leads to the result  $p = \pm {\rm j} \cdot B$  and thus to the zeros		'''(5)'''  Solving the equation  $p^2 + B^2 = 0$  leads to the result  $p = \pm {\rm j} \cdot B$  and thus to the zeros
	:$${\rm Re}\{ p_{\rm o1}\} \underline {= 0}\hspace{-0.3cm} \hspace{1cm}{\rm Im}\{ p_{\rm o1}\} \underline {=+2.5} \cdot 10^6 \, {1}/{{\rm		:$${\rm Re}\{ p_{\rm o1}\} \underline {= 0}\hspace{-0.3cm} \hspace{1cm}{\rm Im}\{ p_{\rm o1}\} \underline {=+2.5} \cdot 10^6 \, {1}/{{\rms}} \hspace{0.05cm},$$
	s}} \hspace{0.05cm},$$		:$$ {\rm Re}\{ p_{\rm o2}\}\hspace{0.15cm} \underline { = 0}\hspace{-0.3cm} \hspace{1cm}{\rm Im}\{ p_{\rm o2}\} \underline {=-2.5} \cdot 10^6 \, {1}/{{\rms}} \hspace{0.05cm}.$$
	:$$ {\rm Re}\{ p_{\rm o2}\}\hspace{0.15cm} \underline { = 0}\hspace{-0.3cm} \hspace{1cm}{\rm Im}\{ p_{\rm o2}\} \underline {=-2.5} \cdot 10^6 \, {1}/{{\rm
	s}} \hspace{0.05cm}.$$

	*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.		*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.
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	'''(6)'''  If the denominator polynomial is set  $N(p) = 0$,  the following conditional equation arises as a result:		'''(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^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},$$
	p_{\rm x1,\hspace{0.05cm}2}= -A \pm \sqrt{A^2 - B^2}
	\hspace{0.05cm},$$

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

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

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

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	[[Category:Linear and Time-Invariant Systems: Exercises\|^3.2 Laplace Transform and p-Transfer Function^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^3.2 Laplace Transform and p-Transfer Function^]]
			[[de:Aufgaben:3.3_p-Übertragungsfunktion]]

Guenter at 13:16, 13 October 2021

2021-10-13T13:16:49Z

← Older revision		Revision as of 15:16, 13 October 2021
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	'''(2)'''  <u>Suggested solution 2</u> is correct:		'''(2)'''  <u>Suggested solution 2</u>  is correct:
	*Replacing  $p$  by  ${\rm j } \cdot 2\pi f$ the following is obtained:		*Replacing  $p$  by  ${\rm j } \cdot 2\pi f$ the following is obtained:
	:$$H(f)= \frac {1 - (2\pi f)^2 \cdot LC}		:$$H(f)= \frac {1 - (2\pi f)^2 \cdot LC}
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	\hspace{0.05cm} .$$		\hspace{0.05cm} .$$

	*So, there is always a frequency at which the numerator is zero, namely the resonance frequency of  $L$  and  $C$.		*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.		*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  $\rm band–stop \:filter$.		*From this it follows:  Regardless of the values of  $R$,  $L$  and  $C$ it is a  $\rm band–stop \:filter$.


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	s}} \hspace{0.05cm}.$$		s}} \hspace{0.05cm}.$$

	*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.		*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.		*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:		'''(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^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}		p_{\rm x1,\hspace{0.05cm}2}= -A \pm \sqrt{A^2 - B^2}
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	'''(7)'''  <u>Suggested solution 1</u> is correct:		'''(7)'''  <u>Suggested solution 1</u>  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.		*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.		* The resistance value  $R$  must be changed.
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	'''(8)'''  According to the result in subtask  '''(7)'''  there is a double pole for  $\underline {A = B = 2 \cdot 10^{-6} \cdot \rm 1/s}$.		'''(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  $40 \ \rm \Omega$ .		*To do this,  the ohmic resistance must be reduced from  $50 \ \rm \Omega$  to  $40 \ \rm \Omega$ .
	*Then, the double pole is at  ${-2 \cdot 10^{6} \cdot \rm 1/s}$.		*Then,  the double pole is at  ${-2 \cdot 10^{6} \cdot \rm 1/s}$.
	*Or with another normalization at  ${-2 \cdot \rm (1/µ s)}$.		*Or with another normalization at  ${-2 \cdot \rm (1/µ s)}$.
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 13:06, 13 October 2021

2021-10-13T13:06:17Z

← Older revision		Revision as of 15:06, 13 October 2021
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	[[File:P_ID1765__LZI_A_3_3.png\|right\|frame\|Considered two-port network]]		[[File:P_ID1765__LZI_A_3_3.png\|right\|frame\|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 ~~also~~ has a fractional–rational  $p$–transfer function of the form		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
	:$$H_{\rm L}(p)= \frac {A_Z \cdot p^Z +\text{ ...} + A_1 \cdot p + A_0}		:$$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)}		{B_N \cdot p^N + \text{ ...} + B_1 \cdot p + B_0}= \frac {Z(p)}{N(p)}
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	The  $Z + N + 1$  parameters mean:		The  $Z + N + 1$  parameters mean:
	* $K = A_Z/B_n$  is a constant factor. If  $Z = N$ applies, then this is dimensionless.		* $K = A_Z/B_n$  is a constant factor.  If  $Z = N$ applies, then this is dimensionless.
	* The solutions of the equation  $Z(p) = 0$  yield the  $Z$ zeros  $p_{{\rm o}1}$, ... , $p_{{\rm o}N}$  of  $H_{\rm L}(p)$.		* 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.		* The zeros of the denominator polynomial  $N(p)$  yield the  $N$  poles  $p_{{\rm x}1}$, ... , $p_{{\rm x}N}$  of the transfer function.

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	''Please note:''		Please note:
	*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function\|Laplace Transform and p-Transfer Function]].		*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function\|Laplace Transform and p-Transfer Function]].

Oezer at 22:46, 11 October 2021

2021-10-11T22:46:33Z

← Older revision		Revision as of 00:46, 12 October 2021
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	s}} \hspace{0.05cm}.$$		s}} \hspace{0.05cm}.$$

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

Oezer at 22:45, 11 October 2021

2021-10-11T22:45:33Z

← Older revision		Revision as of 00:45, 12 October 2021
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	*So, there is always a frequency at which the numerator is zero, namely the resonance frequency of  $L$  and  $C$.		*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.		*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  $\rm band-stop \:filter$.		*From this it follows:  Regardless of the values of  $R$,  $L$ and  $C$ it is a  $\rm band–stop \:filter$.

Oezer at 22:44, 11 October 2021

2021-10-11T22:44:11Z

← Older revision		Revision as of 00:44, 12 October 2021
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	*So, there is always a frequency at which the numerator is zero, namely the resonance frequency of  $L$  and  $C$.		*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.		*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  $\rm band-stop ~~ ~~filter$.		*From this it follows:  Regardless of the values of  $R$,  $L$ and  $C$ it is a  $\rm band-stop \:filter$.

Oezer at 22:43, 11 October 2021

2021-10-11T22:43:08Z

← Older revision		Revision as of 00:43, 12 October 2021
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	*So, there is always a frequency at which the numerator is zero, namely the resonance frequency of  $L$  and  $C$.		*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.		*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		*From this it follows:  Regardless of the values of  $R$,  $L$ and  $C$ it is a  $\rm band-stop  filter$.
	filter$.

Oezer at 22:42, 11 October 2021

2021-10-11T22:42:31Z

← Older revision		Revision as of 00:42, 12 October 2021
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	*So, there is always a frequency at which the numerator is zero, namely the resonance frequency of  $L$  and  $C$.		*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.		*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  $\rm band-stop filter$.		*From this it follows:  Regardless of the values of  $R$,  $L$ and  $C$ it is a  $\rm band-stop
			filter$.

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	'''(5)'''  Solving the equation  $p^2 + B^2 = 0$  leads to the result  $p = \pm {\rm j} \cdot B$  and thus to the zeros		'''(5)'''  Solving the equation  $p^2 + B^2 = 0$  leads to the result  $p = \pm {\rm j} \cdot B$  and thus to the zeros
	:$${\rm Re}\{ p_{\rm o1}\} \underline {= 0}\hspace{-0.3cm} \hspace{1cm}{\rm Im}\{ p_{\rm o1}\} \underline {=+2.5} \cdot 10^6 \, {1}/{{\~~rms~~}} \hspace{0.05cm},$$		:$${\rm Re}\{ p_{\rm o1}\} \underline {= 0}\hspace{-0.3cm} \hspace{1cm}{\rm Im}\{ p_{\rm o1}\} \underline {=+2.5} \cdot 10^6 \, {1}/{{\rm s}} \hspace{0.05cm},$$
	:$$ {\rm Re}\{ p_{\rm o2}\}\hspace{0.15cm} \underline { = 0}\hspace{-0.3cm} \hspace{1cm}{\rm Im}\{ p_{\rm o2}\} \underline {=-2.5} \cdot 10^6 \, {1}/{{\~~rms~~}} \hspace{0.05cm}.$$		:$$ {\rm Re}\{ p_{\rm o2}\}\hspace{0.15cm} \underline { = 0}\hspace{-0.3cm} \hspace{1cm}{\rm Im}\{ p_{\rm o2}\} \underline {=-2.5} \cdot 10^6 \, {1}/{{\rm s}} \hspace{0.05cm}.$$

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

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

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

	This result is only unique considering the specification  $\|p_\text{x2}\| > \|p_\text{x1}\|$.		This result is only unique considering the specification  $\|p_\text{x2}\| > \|p_\text{x1}\|$.
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	[[Category:Linear and Time-Invariant Systems: Exercises\|^3.2 Laplace Transform and p-Transfer Function^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^3.2 Laplace Transform and p-Transfer Function^]]
	[[de:Aufgaben:3.~~3_p~~-Übertragungsfunktion]]		[[de:Aufgaben:Aufgabe 3.3: p-Übertragungsfunktion]]