Linear and Time Invariant Systems/Inverse Laplace Transform - Revision history

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@@ Line 149: / Line 149: @@
 ==Critically attenuated case==
 <br>
-With &nbsp;$R = 50 \ \rm Ω$, &nbsp;$L = 25 \ \rm &micro; H$&nbsp; and&nbsp; &nbsp;$C = 40 \ \rm nF$&nbsp; we get the so-called&nbsp; "critically attenuated case":
+With &nbsp;$R = 50 \ \rm Ω$, &nbsp;$L = 25 \ \rm &micro; H$&nbsp; and&nbsp; &nbsp;$C = 40 \ \rm nF$&nbsp; we get the so-called&nbsp; &raquo;critically attenuated case&laquo;:
 :$$H_{\rm L}(p)= K \cdot \frac {p - p_{\rm o }} {(p - p_{\rm x })^2}= 2 \cdot \frac {p + 0.5 } {(p +1)^2}  \hspace{0.05cm} .$$
-The capacitance value &nbsp;$C = 40 \ \rm nF$&nbsp; is the smallest possible value for which there are just real pole places.&nbsp;
+The capacitance &nbsp;$C = 40 \ \rm nF$&nbsp; is the smallest possible value for which there are just real pole places.&nbsp; These coincide,&nbsp; that &nbsp;$p_{\rm x} = \ -1$&nbsp; is a double pole place.&nbsp; The time function is thus according to the residue theorem with &nbsp;$l = 2$:
 :$$h(t) = {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{ {\rm x} } } \hspace{-0.7cm}\{H_{\rm L}(p)\cdot {\rm e}^{p t}\}= \frac{ {\rm d} }{ {\rm d}p}\hspace{0.15cm}
   \left \{H_{\rm L}(p)\cdot (p - p_{ {\rm x} })^2\cdot {\rm e}^{p \hspace{0.05cm}t}\right\} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{ {\rm x} } } = K \cdot \frac{ {\rm d} }{ {\rm d}p}\hspace{0.15cm}\left \{ (p - p_{ {\rm o} })\cdot {\rm e}^{p \hspace{0.05cm}t}\right\}  \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{ {\rm x} } } \hspace{0.05cm} .$$
 [[File:EN_LZI_T_3_3_S3c.png |right|frame| Impulse response and step response of the critically attenuated case]]
-Using the&nbsp; "product rule"&nbsp; of differential calculus, this gives:
+Using the&nbsp; &raquo;product rule&laquo;&nbsp; of differential calculus,&nbsp; this gives:
 :$$h(t) = K \cdot \left [ {\rm e}^{p \hspace{0.05cm}t} + (p - p_{ {\rm o} })\cdot t \cdot {\rm e}^{p \hspace{0.05cm}t} \right ] \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}-1} = {\rm  e}^{-t}\cdot \left ( 2 - t \right)
   \hspace{0.05cm} .$$
-The graph shows this impulse response (green curve) in normalized representation.&nbsp; It differs only slightly from the one with the two different poles at&nbsp; $-0.4$&nbsp; and&nbsp; $-1.6$&nbsp;.
+The graph shows this impulse response&nbsp; $($green curve$)$&nbsp; in normalized representation.&nbsp; It differs only slightly from the one with two different poles at&nbsp; $-0.4$&nbsp; and&nbsp; $-1.6$&nbsp;.
-The signal drawn in red &nbsp; &rArr; &nbsp; $y(t) = 1 - {\rm e}^{-t} + t \cdot {\rm e}^{-t}$&nbsp; results when a step function is also considered at the input &nbsp; &rArr; &nbsp; "step response".
+The signal drawn in red &nbsp; &rArr; &nbsp; $y(t) = 1 - {\rm e}^{-t} + t \cdot {\rm e}^{-t}$&nbsp; results when a step function&nbsp; $\gamma(t)$&nbsp; is considered at the input &nbsp; &rArr; &nbsp; &raquo;step response&laquo;.
 To calculate the step response &nbsp;$\sigma(t) = y(t)$&nbsp; one can alternatively
-*consider an additional pole at &nbsp;$p = 0$ &nbsp; in the residual calculation (marked in red), or
+*consider the additional red pole at &nbsp;$p = 0$ &nbsp; in the residual calculation,&nbsp;
 *form the integral over the impulse response &nbsp;$h(t)$.
@@ Line 176: / Line 175: @@
 Prerequisite for the application of the residue theorem is that there are less zeros than poles &nbsp; &rArr; &nbsp; $Z$&nbsp; must always be smaller than &nbsp;$N$&nbsp;.
-If,&nbsp; on the other hand,&nbsp; as in the case of a high-pass filter &nbsp;$Z = N$,&nbsp; then
+*If,&nbsp; on the other hand,&nbsp; as in the case of a high-pass filter &nbsp;$Z = N$,&nbsp; then the limit of the p&ndash;transfer function&nbsp; $H_{\rm L}(p)$&nbsp; for large &nbsp;$p$&nbsp; is not equal to zero,
-*is the limit of the spectral function&nbsp; $H_{\rm L}(p)$&nbsp; for large &nbsp;$p$&nbsp; not equal to zero,
-*if the associated time signal &nbsp;$y(t)$&nbsp; also contains a &nbsp;[[Signal_Representation/Direct_Current_Signal_-_Limit_Case_of_a_Periodic_Signal#Dirac_.28delta.29_function_in_frequency_domain|$\text{Dirac delta functions}$]],
+*If the associated time signal &nbsp;$y(t)$&nbsp; also contains &nbsp;[[Signal_Representation/Direct_Current_Signal_-_Limit_Case_of_a_Periodic_Signal#Dirac_.28delta.29_function_in_frequency_domain|&raquo;Dirac delta functions&laquo;]],&nbsp;  the residue theorem fails and a&nbsp; [https://en.wikipedia.org/wiki/Partial_fraction_decomposition&nbsp; &raquo;'''partial fraction decomposition'''&laquo;]&nbsp; must be performed.
 The procedure is to be clarified exemplarily for a high-pass of first order.
 {{GraueBox|TEXT=
 $\text{Example 1:}$&nbsp;
-The&nbsp; $p$-transfer function of a&nbsp; "first-order RC high-pass filter"&nbsp; can be transformed by splitting off a constant as follows:
+The&nbsp; $p$-transfer function of a&nbsp; &raquo;first-order RC high-pass filter&laquo;&nbsp; can be transformed by splitting off a constant as follows:
 :$$\frac{p}{p + RC} = 1- \frac{RC}{p + RC}\hspace{0.05cm} .$$
-Thus, the high-pass impulse response is:
+Thus,&nbsp; the high-pass impulse response is:
 :$$h_{\rm HP}(t) = \delta(t) - h_{\rm TP}(t) \hspace{0.05cm} .$$
 The graph shows
-*as a red curve the high&ndash;pass impulse response &nbsp;$h_{\rm HP}(t)$,
+*as blue curve the impulse response &nbsp;$h_{\rm TP}(t)$&nbsp; of the equivalent low-pass,
-*as a blue curve the impulse response &nbsp;$h_{\rm TP}(t)$&nbsp; of the equivalent low-pass.
-The Dirac delta function is the Laplace transform of the constant value&nbsp; $1$,&nbsp; <br>while the second function to be subtracted gives the impulse response of the equivalent low-pass filter, which is given by  the residue theorem with &nbsp;
+&rArr; &nbsp; The Dirac delta function is the Laplace transform of the constant value&nbsp; $1$,&nbsp; <br>while the second function to be subtracted gives the impulse response of the equivalent low-pass filter,&nbsp; which is given by  the residue theorem with &nbsp;
 :$$Z = 0,\hspace{0.2cm} N =1,\hspace{0.2cm} K = RC.$$ }}

@@ Line 119: / Line 119: @@
 [[File:EN_LZI_T_3_3_S3b.png|right|frame| Attenuated-oscillatory impulse response]]
 *The zero is located at &nbsp;$p_{\rm o} = \ –2.5$.
-*$K = 2$&nbsp; holds and all numerical values are to be multiplied again by the factor &nbsp;$1/T$&nbsp; $(T = 1\ \rm &micro; s$).
 :$$h_1(t) =   K \cdot \frac {p_{\rm x 1} - p_{\rm o }} {p_{\rm x 1} - p_{\rm x 2}} \cdot  {\rm e}^{\hspace{0.05cm}p_{\rm x 1} \cdot \hspace{0.05cm}t} $$
 :$$\Rightarrow \hspace{0.3cm}h_1(t) =    2 \cdot \frac {-1 + {\rm j}\cdot 2 +2.5} {(-1 + {\rm j}\cdot 2) - (-1 - {\rm j}\cdot 2)}\cdot  {\rm e}^{\hspace{0.05cm}p_{\rm x 1}
@@ Line 134: / Line 138: @@
 :$$\Rightarrow \hspace{0.3cm}h_2(t) =2 \cdot \frac {1.5 - {\rm j}\cdot 2} {-{\rm j}\cdot 4}\cdot  {\rm e}^{\hspace{0.05cm}p_{\rm x 2} \hspace{0.03cm}\cdot\hspace{0.05cm}t}= (1 + {\rm j}\cdot 0.75)\cdot {\rm e}^{-t}\cdot {\rm  e}^{\hspace{0.03cm}-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} 2t} \hspace{0.05cm} . $$
-Using&nbsp; [[Signal_Representation/Calculating_With_Complex_Numbers#Representation_by_magnitude_and_phase|$\text{Euler's theorem}$]]&nbsp; the following is obtained for the sum signal:
+Using&nbsp; [[Signal_Representation/Calculating_With_Complex_Numbers#Representation_by_magnitude_and_phase|&raquo;Euler's theorem&laquo;]]&nbsp; the following is obtained for the sum signal:
 :$$h(t) =  h_1(t) + h_2(t)\hspace{0.3cm}
 \Rightarrow \hspace{0.3cm}h(t) = {\rm  e}^{-t}\cdot \big [ (1 - {\rm j}\cdot 0.75)\cdot (\cos() + {\rm j}\cdot \sin())+

@@ Line 84: / Line 84: @@
 In the next sections,&nbsp; the&nbsp; &raquo;residue theorem&laquo;&nbsp; is illustrated by three detailed examples corresponding to the three constellations in&nbsp; [[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function#Properties_of_poles_and_zeros|$\text{Example 3}$]]&nbsp; of chapter&nbsp; &raquo;Laplace transform and p-transfer function&laquo;:
-*So, we consider again the two-port network with an inductance &nbsp;$L = 25 \ \rm &micro;H$&nbsp; in the longitudinal branch as well as the the series connection of an ohmic resistance&nbsp; $R = 50 \ \rm Ω$&nbsp; and a capacitance&nbsp; $C$&nbsp; in the transverse branch.
+*So,&nbsp; we consider again the two-port network with an inductance &nbsp;$L = 25 \ \rm &micro;H$&nbsp; in the longitudinal branch as well as the the series connection of an ohmic resistance&nbsp; $R = 50 \ \rm Ω$&nbsp; and a capacitance&nbsp; $C$&nbsp; in the transverse branch.
-*For the latter,&nbsp; we again consider three different values,&nbsp; namely &nbsp;$C = 62.5 \ \rm nF$, &nbsp;$C = 8 \ \rm nF$&nbsp; and &nbsp;$C = 40 \ \rm nF$.
 *The following is always assumed: &nbsp;$x(t) = δ(t) \; ⇒  \; X_{\rm L}(p) = 1$ &nbsp; &rArr; &nbsp; $Y_{\rm L}(p) = H_{\rm L}(p)$  &nbsp; &rArr; &nbsp; the output signal&nbsp; $y(t)$&nbsp; is equal to the impulse response &nbsp;$h(t)$.
 ==Aperiodically decaying impulse response==
 <br>
-The following is obtained for the&nbsp; $p$&ndash;transfer function computed in the section &nbsp;[[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function#Pole-zero_representation_of_circuits|$\text{pole-zero representation of circuits}$]]&nbsp; with the capacitance &nbsp;$C = 62.5 \ \rm nF$&nbsp; and the other numerical values given in the graph below:
+The following is obtained for the&nbsp; $p$&ndash;transfer function computed in the section &nbsp;[[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function#Pole-zero_representation_of_circuits|&raquo;pole-zero representation of circuits&laquo;]]&nbsp; with the capacitance &nbsp;$C = 62.5 \ \rm nF$.&nbsp; The other numerical values are given in the graph below:
 [[File: EN_LZI_T_3_3_S3a.png|right|frame|Aperiodically decaying impulse response]]
@@ Line 98: / Line 100: @@
 Note the normalization of &nbsp;$p$, &nbsp;$K$ and also of all poles and zeros by the factor &nbsp;${\rm 10^6} · 1/\rm s$.
-The impulse response is composed of &nbsp;$I = N = 2$&nbsp; natural oscillations. For $t < 0$,&nbsp; these are equal to zero.
+&rArr; &nbsp; The impulse response is composed of &nbsp;$I = N = 2$&nbsp; natural oscillations. For $t < 0$,&nbsp; these are equal to zero.
 *The residual of the pole at &nbsp;$p_{{\rm x}1} =\  –0.4$&nbsp; yields the following time function:
 :$$h_1(t)  =  {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}} \hspace{-0.7cm}\{H_{\rm L}(p)\cdot {\rm e}^{p t}\}= H_{\rm L}(p)\cdot (p - p_{{\rm x}1})\cdot {\rm e}^{p \hspace{0.05cm}t} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}$$
@@ Line 107: / Line 109: @@
 The graph shows &nbsp;$h_1(t)$&nbsp; and &nbsp;$h_2(t)$&nbsp; as well as the sum signal &nbsp;$h(t)$.
-*The normalization factor &nbsp;$1/T = 10^6 · \rm 1/s$ is taken into account here so that the time is normalized to &nbsp;$T = 1 \ \rm &micro; s$&nbsp;.
+#The normalization factor &nbsp;$1/T = 10^6 · \rm 1/s$&nbsp; is taken into account here so that the time is normalized to &nbsp;$T = 1 \ \rm &micro; s$&nbsp;.
-*For &nbsp;$t =0$,&nbsp; $T \cdot h(t=0) =  {32 }/ {15} -{2 }/ {15}= 2 \hspace{0.05cm}$&nbsp; is obtained as a result.
+#For &nbsp;$t =0$,&nbsp; $T \cdot h(t=0) =  {32 }/ {15} -{2 }/ {15}= 2 \hspace{0.05cm}$&nbsp; is obtained as a result.
-*For times &nbsp;$t > 2 \ \rm &micro; s$,&nbsp; the impulse response is negative&nbsp; (although only slightly and difficult to see in the graph).
+#For times &nbsp;$t > 2 \ \rm &micro; s$,&nbsp; the impulse response is negative&nbsp; $($although only slightly and difficult to see in the graph$)$.

@@ Line 83: / Line 83: @@
-In the next sections,&nbsp; the&nbsp; &raquo;residue theorem&laquo;&nbsp; is illustrated by three detailed examples corresponding to the three constellations in&nbsp; [[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function#Properties_of_poles_and_zeros|$\text{Example 3}$]]&nbsp; of chapter&nbsp; &raquo;Laplace transform and p-transfer function":
+In the next sections,&nbsp; the&nbsp; &raquo;residue theorem&laquo;&nbsp; is illustrated by three detailed examples corresponding to the three constellations in&nbsp; [[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function#Properties_of_poles_and_zeros|$\text{Example 3}$]]&nbsp; of chapter&nbsp; &raquo;Laplace transform and p-transfer function&laquo;:
 *So, we consider again the two-port network with an inductance &nbsp;$L = 25 \ \rm &micro;H$&nbsp; in the longitudinal branch as well as the the series connection of an ohmic resistance&nbsp; $R = 50 \ \rm Ω$&nbsp; and a capacitance&nbsp; $C$&nbsp; in the transverse branch.
 *For the latter,&nbsp; we again consider three different values,&nbsp; namely &nbsp;$C = 62.5 \ \rm nF$, &nbsp;$C = 8 \ \rm nF$&nbsp; and &nbsp;$C = 40 \ \rm nF$.

← Older revision		Revision as of 14:05, 21 November 2023
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−	We also assume  $Z < N$.  The number of  »distinguishable poles«  is denoted by  $I$.  Multiple poles are counted only once to determine  $I$.  Thus,  the following holds for the  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#~~Some_results_of_the_theory_of_functions~~\|$\text{sketch}$]]  in the last section considering the double pole:	+	We also assume  $Z < N$.  The number of  »distinguishable poles«  is denoted by  $I$.  Multiple poles are counted only once to determine  $I$.  Thus,  the following holds for the  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Some_results_of_function_theory\|$\text{sketch}$]]  in the last section considering the double pole:
	:$$N = 5,\hspace{0.3cm} I = 4.$$		:$$N = 5,\hspace{0.3cm} I = 4.$$

@@ Line 33: / Line 33: @@
 *So,&nbsp; a rectangular signal &nbsp;$x(t)\ \  ⇒ \ \ X_{\rm L}(p) = (1 - {\rm e}^{\hspace{0.05cm}p\hspace{0.05cm}\cdot \hspace{0.05cm} T})/p$&nbsp; is not possible in the approach described here.&nbsp; However, the rectangular response &nbsp;$y(t)$&nbsp; can be computed indirectly as the difference of two step responses.
-==Some results of the theory of functions==
+==Some results of function theory==
 <br>
-In contrast to the&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse#The_first_Fourier_integral|$\text{Fourier integrals}$]],&nbsp; which differ only slightly in the two directions of transformation,&nbsp; for&nbsp; "Laplace"&nbsp; the computation of &nbsp;$y(t)$&nbsp; from &nbsp;$Y_{\rm L}(p)$ – that is the inverse transformation – is
+In contrast to the&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse#The_first_Fourier_integral|&raquo;Fourier integrals&laquo;]],&nbsp; which differ only slightly in the two directions of transformation,&nbsp; for&nbsp; &raquo;Laplace&laquo;&nbsp; the computation of &nbsp;$y(t)$&nbsp; from &nbsp;$Y_{\rm L}(p)$ – that is the inverse transformation – is
 *much more difficult than computing &nbsp;$Y_{\rm L}(p)$&nbsp; from &nbsp;$y(t)$,
 *unresolvable or solvable only very laboriously by elementary means.
@@ Line 45: / Line 46: @@
 :$$y(t) = {\rm L}^{-1}\{Y_{\rm L}(p)\}= \lim_{\beta \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty}  \hspace{0.15cm} \frac{1}{ {\rm j} \cdot 2 \pi}\cdot     \int_{ \alpha - {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2 \pi \beta } ^{\alpha+{\rm j}  \hspace{0.05cm}\cdot \hspace{0.05cm}2 \pi \beta}  Y_{\rm L}(p) \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm} t}\hspace{0.1cm}{\rm
   d}p \hspace{0.05cm} .$$
-*The integration is parallel to the imaginary axis.
+#The integration is parallel to the imaginary axis.
-*The real part &nbsp;$α$&nbsp; is to be chosen such that all poles are located to the left of the integration path.}}
+#The real part &nbsp;$α$&nbsp; is to be chosen such that all poles are located to the left of the integration path.}}
 [[File:EN_LZI_T_3_3_S2.png |right|frame|Line integral together with left and right circular integral]]
-−
-The left graph illustrates this line integral along the red dotted vertical &nbsp;${\rm Re}\{p\}= α$.&nbsp; This integral is solvable using &nbsp;[https://en.wikipedia.org/wiki/Jordan%27s_lemma $\text{Jordan's lemma of complex analysis}$].&nbsp; In this tutorial only a very short and simple summary of the approach is depicted:
+#The line integral can be divided into two circular integrals so that all poles are located in the left circular integral while the right circular integral may only contain zeros.
-*The line integral can be divided into two circular integrals so that all poles are located in the left circular integral while the right circular integral may only contain zeros.
+#According to the theory of functions, the right circular integral yields the time function &nbsp;$y(t)$&nbsp; for negative times.&nbsp;
-*According to the theory of functions, the right circular integral yields the time function &nbsp;$y(t)$&nbsp; for negative times.&nbsp; Due to causality, &nbsp;$y(t < 0)$&nbsp; must be identical to zero,&nbsp; but according to the fundamental theorem of the theory of functions this is only true if there are no poles in the right &nbsp;$p$–half-plane.
+#Due to causality, &nbsp;$y(t < 0)$&nbsp; must be identical to zero,&nbsp; but according to the fundamentals of function theorem this is only true if there are no poles in the right &nbsp;$p$–half-plane.
-*In contrast,&nbsp; the integral over the left semicircle yields the time function for &nbsp;$t ≥ 0$.&nbsp; This encloses all poles and can be computed using the&nbsp; "residue theorem"&nbsp; in a (relatively) simple way,&nbsp; as it will be shown in the next sections.
+#In contrast,&nbsp; the integral over the left semicircle yields the time function for &nbsp;$t ≥ 0$.&nbsp;
+#This encloses all poles and can be computed using the&nbsp; &raquo;'''residue theorem'''&laquo;&nbsp; in a&nbsp; $($relatively$)$&nbsp; simple way,&nbsp; as it will be shown in the next sections.
 ==Formulation of the residue theorem==
 <br>
 It is further assumed that the transfer function &nbsp;$Y_{\rm L}(p)$&nbsp; can be expressed in pole-zero notation  by
 *the constant factor&nbsp; $K$,
-*the &nbsp;$Z$&nbsp; "zeros" &nbsp;$p_{{\rm o}i}$&nbsp; $(i = 1$, ... , $Z)$&nbsp; and
+*the &nbsp;$Z$&nbsp; &raquo;zeros&laquo; &nbsp;$p_{{\rm o}i}$&nbsp; $(i = 1$, ... , $Z)$&nbsp; and
-*the &nbsp;$N$&nbsp; "poles" &nbsp;$p_{{\rm x}i}$&nbsp; $(i = 1$, ... , $N$).
+*the &nbsp;$N$&nbsp; &raquo;poles&laquo; &nbsp;$p_{{\rm x}i}$&nbsp; $(i = 1$, ... , $N$).
-The number of distinguishable poles is denoted by &nbsp;$I$.&nbsp; Multiple poles are counted only once to determine &nbsp;$I$.&nbsp; Thus, the following holds for the&nbsp; [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Some_results_of_the_theory_of_functions|$\text{sketch}$]]&nbsp; in the last section considering the  double pole: &nbsp;  $N = 5$&nbsp; and &nbsp;$I = 4$.
+We also assume &nbsp;$Z < N$.&nbsp; The number of&nbsp; &raquo;distinguishable poles&laquo;&nbsp; is denoted by &nbsp;$I$.&nbsp; Multiple poles are counted only once to determine &nbsp;$I$.&nbsp; Thus,&nbsp; the following holds for the&nbsp; [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Some_results_of_the_theory_of_functions|$\text{sketch}$]]&nbsp; in the last section considering the  double pole: &nbsp;
 {{BlaueBox|TEXT=
 $\text{Residue Theorem:}$&nbsp;
-Considering the above conditions, the&nbsp; &raquo;'''inverse Laplace transform'''&laquo;&nbsp; of &nbsp;$Y_{\rm L}(p)$&nbsp; for times&nbsp; $t ≥ 0$&nbsp; is obtained as the sum of&nbsp; $I$&nbsp; natural oscillations of the poles,&nbsp; which are called the&nbsp; "residuals"&nbsp; – abbreviated as "Res":
+Considering the above conditions,&nbsp; the&nbsp; &raquo;'''inverse Laplace transform'''&laquo;&nbsp; of &nbsp;$Y_{\rm L}(p)$&nbsp; for times&nbsp; $t ≥ 0$&nbsp; is obtained as the sum of&nbsp; $I$&nbsp; natural oscillations of the poles,&nbsp; which are called the&nbsp; &raquo;residuals&laquo;&nbsp; – abbreviated as&nbsp; $\rm Res$:
 :$$y(t) = \sum_{i=1}^{I}{\rm Res} \bigg \vert _{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}_i}} \hspace{-0.7cm}\{Y_{\rm L}(p)\cdot  {\rm e}^{p \hspace{0.05cm}t}\} \hspace{0.05cm} .$$
@@ Line 81: / Line 83: @@
-In the next sections,&nbsp; the&nbsp; "residue theorem"&nbsp; is illustrated by three detailed examples corresponding to the three constellations in&nbsp; [[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function#Properties_of_poles_and_zeros|$\text{Example 3}$]]&nbsp; in the chapter&nbsp; "Laplace Transform and p-Transfer Function":
+In the next sections,&nbsp; the&nbsp; &raquo;residue theorem&laquo;&nbsp; is illustrated by three detailed examples corresponding to the three constellations in&nbsp; [[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function#Properties_of_poles_and_zeros|$\text{Example 3}$]]&nbsp; of chapter&nbsp; &raquo;Laplace transform and p-transfer function":
 *So, we consider again the two-port network with an inductance &nbsp;$L = 25 \ \rm &micro;H$&nbsp; in the longitudinal branch as well as the the series connection of an ohmic resistance&nbsp; $R = 50 \ \rm Ω$&nbsp; and a capacitance&nbsp; $C$&nbsp; in the transverse branch.
 *For the latter,&nbsp; we again consider three different values,&nbsp; namely &nbsp;$C = 62.5 \ \rm nF$, &nbsp;$C = 8 \ \rm nF$&nbsp; and &nbsp;$C = 40 \ \rm nF$.

@@ Line 9: / Line 9: @@
 {{BlaueBox|TEXT=
 $\text{Task:}$&nbsp; This chapter deals with the following problem:
-*The&nbsp; $p$–spectral function&nbsp; $Y_{\rm L}(p)$&nbsp;  is given in&nbsp; "pole-zero"&nbsp; notation.
+*The&nbsp; $p$–spectral function&nbsp; $Y_{\rm L}(p)$&nbsp;  is given in&nbsp; &raquo;pole-zero notation&laquo;.
 *The&nbsp; &raquo;'''inverse Laplace transform'''&laquo;, i.e. the associated time function&nbsp; $y(t)$&nbsp; is searched-for,&nbsp; where the following notation should hold:
 :$$y(t) = {\rm L}^{-1}\{Y_{\rm L}(p)\}\hspace{0.05cm} , \hspace{0.3cm}{\rm briefly}\hspace{0.3cm}
   y(t) \quad \circ\!\!-\!\!\!-^{\hspace{-0.25cm}\rm L}\!\!\!-\!\!\bullet\quad Y_{\rm L}(p)\hspace{0.05cm} .$$}}
 The graph summarizes the prerequisites for this task.
 *$H_{\rm L}(p)$&nbsp; describes the transfer function of the causal system and &nbsp;$Y_{\rm L}(p)$&nbsp; specifies the Laplace transform of the output signal &nbsp;$y(t)$&nbsp; considering the input signal &nbsp;$x(t)$&nbsp;. &nbsp;$Y_{\rm L}(p)$&nbsp; is characterized by &nbsp;$N$&nbsp; poles,&nbsp; by &nbsp;$Z ≤ N$&nbsp; zeros and by the constant &nbsp;$K.$
-*Poles and zeros exhibit the properties mentioned in the&nbsp; [[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function#Properties_of_poles_and_zeros|"last chapter"]]:&nbsp; Poles are only allowed in the left &nbsp;$p$–half plane or on the imaginary axis;&nbsp; zeros are also allowed in the right &nbsp;$p$–half plane.
-*All singularities – this is the generic term for poles and zeros – are either real or exist as pairs of conjugate-complex singularities.&nbsp; Multiple poles and zeros are also allowed.
+*Poles and zeros exhibit the properties mentioned in the&nbsp; [[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function#Properties_of_poles_and_zeros|&raquo;last chapter&laquo;]]:&nbsp; Poles are only allowed in the left &nbsp;$p$–half plane or on the imaginary axis;&nbsp; zeros are also allowed in the right &nbsp;$p$–half plane.
-*With the input &nbsp;$x(t) = δ(t)$ &nbsp; &rArr; &nbsp;  $X_{\rm L}(p) = 1$  &nbsp; &rArr; &nbsp;  $Y_{\rm L}(p) = H_{\rm L}(p)$, the output signal &nbsp;$y(t)$&nbsp; then describes the [[Linear_and_Time_Invariant_Systems/System_Description_in_Time_Domain#Impulse_response|$\text{impulse response}$]]  &nbsp;$h(t)$&nbsp; of the transmission system.&nbsp; For this purpose, only the singularities drawn in green in the graph may be used for computation.
-*A step function &nbsp;$x(t) = γ(t)$ &nbsp; &rArr; &nbsp;  $ X_{\rm L} = 1/p$&nbsp; at the input causes the output signal &nbsp;$y(t)$&nbsp; to be equal to the &nbsp;[[Linear_and_Time_Invariant_Systems/System_Description_in_Time_Domain#Step_response|$\text{step response}$]] &nbsp; $σ(t)$ of $H_{\rm L}(p)$&nbsp;.&nbsp; In addition to the singularities of &nbsp;$H_{\rm L}(p)$,&nbsp; the pole (shown in red in the graph) at &nbsp;$p = 0$&nbsp; must now also be taken into account for computation.
+*All&nbsp; &raquo;singularities&laquo;&nbsp; – this is the generic term for poles and zeros – are either real or exist as pairs of conjugate-complex singularities.&nbsp; Multiple poles and zeros are also allowed.
-*Only signals for which &nbsp;$X_{ \rm L}(p)$&nbsp; can be expressed in pole-zero notation are possible as input &nbsp;$x(t)$&nbsp; (see the &nbsp;[[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function#Some_important_Laplace_correspondences|$\text{table}$]]&nbsp; in the chapter "Laplace Transform and $p$–Transfer Function"), for example a cosine or sine signal switched on at time &nbsp;$t = 0$&nbsp;.
-*So, a rectangle as input signal &nbsp;$x(t)\ \  ⇒ \ \ X_{\rm L}(p) = (1 - {\rm e}^{\hspace{0.05cm}p\hspace{0.05cm}\cdot \hspace{0.05cm} T})/p$&nbsp; is not possible in the approach described here.&nbsp; However, the rectangular response &nbsp;$y(t)$&nbsp; can be computed indirectly as the difference of two step responses.
+*With the input &nbsp;$x(t) = δ(t)$ &nbsp; &rArr; &nbsp;  $X_{\rm L}(p) = 1$  &nbsp; &rArr; &nbsp;  $Y_{\rm L}(p) = H_{\rm L}(p)$, the output signal &nbsp;$y(t)$&nbsp; then describes the [[Linear_and_Time_Invariant_Systems/System_Description_in_Time_Domain#Impulse_response|&raquo;impulse response&laquo;]]  &nbsp;$h(t)$&nbsp; of the transmission system.&nbsp; For this purpose,&nbsp; only the singularities drawn in green in the graph may be used for computation.
 ==Some results of the theory of functions==