# Inverse Laplace Transform

## Problem formulation and prerequisites

$\text{Task:}$  This chapter deals with the following problem:

• The  $p$–spectral function  $Y_{\rm L}(p)$  is given in  "pole-zero"  notation.
• The  inverse Laplace transform, i.e. the associated time function  $y(t)$  is searched-for,  where the following notation should hold:
$$y(t) = {\rm L}^{-1}\{Y_{\rm L}(p)\}\hspace{0.05cm} , \hspace{0.3cm}{\rm briefly}\hspace{0.3cm} y(t) \quad \circ\!\!-\!\!\!-^{\hspace{-0.25cm}\rm L}\!\!\!-\!\!\bullet\quad Y_{\rm L}(p)\hspace{0.05cm} .$$

Prerequisites for the chapter "Inverse Laplace Transform"

The graph summarizes the prerequisites for this task.

• $H_{\rm L}(p)$  describes the transfer function of the causal system and  $Y_{\rm L}(p)$  specifies the Laplace transform of the output signal  $y(t)$  considering the input signal  $x(t)$ .  $Y_{\rm L}(p)$  is characterized by  $N$  poles,  by  $Z ≤ N$  zeros and by the constant  $K.$
• Poles and zeros exhibit the properties mentioned in the  last chapter:  Poles are only allowed in the left  $p$–half plane or on the imaginary axis;  zeros are also allowed in the right  $p$–half plane.
• All singularities – this is the generic term for poles and zeros – are either real or exist as pairs of conjugate-complex singularities.  Multiple poles and zeros are also allowed.
• With the input  $x(t) = δ(t)$   ⇒   $X_{\rm L}(p) = 1$   ⇒   $Y_{\rm L}(p) = H_{\rm L}(p)$, the output signal  $y(t)$  then describes the impulse response  $h(t)$  of the transmission system.  For this purpose, only the singularities drawn in green in the graph may be used for computation.
• A step function  $x(t) = γ(t)$   ⇒   $X_{\rm L} = 1/p$  at the input causes the output signal  $y(t)$  to be equal to the  step response   $σ(t)$ of $H_{\rm L}(p)$ .  In addition to the singularities of  $H_{\rm L}(p)$,  the pole (shown in red in the graph) at  $p = 0$  must now also be taken into account for computation.
• Only signals for which  $X_{ \rm L}(p)$  can be expressed in pole-zero notation are possible as input  $x(t)$  (see the  table  in the chapter "Laplace Transform and $p$–Transfer Function"), for example a cosine or sine signal switched on at time  $t = 0$ .
• So, a rectangle as input signal  $x(t)\ \ ⇒ \ \ X_{\rm L}(p) = (1 - {\rm e}^{\hspace{0.05cm}p\hspace{0.05cm}\cdot \hspace{0.05cm} T})/p$  is not possible in the approach described here.  However, the rectangular response  $y(t)$  can be computed indirectly as the difference of two step responses.

## Some results of the theory of functions

In contrast to the  Fourier integrals,  which differ only slightly in the two directions of transformation,  for  "Laplace"  the computation of  $y(t)$  from  $Y_{\rm L}(p)$ – that is the inverse transformation – is

• much more difficult than computing  $Y_{\rm L}(p)$  from  $y(t)$,
• unresolvable or solvable only very laboriously by elementary means.

$\text{Definition:}$  In general, the following holds for the  inverse Laplace trans:

$$y(t) = {\rm L}^{-1}\{Y_{\rm L}(p)\}= \lim_{\beta \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} \hspace{0.15cm} \frac{1}{ {\rm j} \cdot 2 \pi}\cdot \int_{ \alpha - {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2 \pi \beta } ^{\alpha+{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2 \pi \beta} Y_{\rm L}(p) \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm} t}\hspace{0.1cm}{\rm d}p \hspace{0.05cm} .$$
• The integration is parallel to the imaginary axis.
• The real part  $α$  is to be chosen such that all poles are located to the left of the integration path.

Line integral together with left and right circular integral

The left graph illustrates this line integral along the red dotted vertical  ${\rm Re}\{p\}= α$.  This integral is solvable using  Jordan's lemma of complex analysis.  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.
• According to the theory of functions, the right circular integral yields the time function  $y(t)$  for negative times.  Due to causality,  $y(t < 0)$  must be identical to zero,  but according to the fundamental theorem of the theory of functions this is only true if there are no poles in the right  $p$–half-plane.
• In contrast,  the integral over the left semicircle yields the time function for  $t ≥ 0$.  This encloses all poles and can be computed using the  "residue theorem"  in a (relatively) simple way,  as it will be shown on the next pages.

## Formulation of the residue theorem

It is further assumed that the transfer function  $Y_{\rm L}(p)$  can be expressed in pole-zero notation by

• the constant factor  $K$,
• the  $Z$  "zeros"  $p_{{\rm o}i}$  $(i = 1$, ... , $Z)$  and
• the  $N$  "poles"  $p_{{\rm x}i}$  $(i = 1$, ... , $N$).

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  sketch  in the last section considering the double pole:   $N = 5$  and  $I = 4$.

$\text{Residue Theorem:}$  Considering the above conditions, the  inverse Laplace transform  of  $Y_{\rm L}(p)$  for times  $t ≥ 0$  is obtained as the sum of  $I$  natural oscillations of the poles,  which are called the  "residuals"  – abbreviated as „Res”:

$$y(t) = \sum_{i=1}^{I}{\rm Res} \bigg \vert _{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}_i}} \hspace{-0.7cm}\{Y_{\rm L}(p)\cdot {\rm e}^{p \hspace{0.05cm}t}\} \hspace{0.05cm} .$$

Since  $Y_{\rm L}(p)$  is only specifiable for causal signals,  $y(t < 0) = 0$  always holds for negative times.

• In general,  the following holds for a pole of multiplicity  $l$ :
$${\rm Res} \bigg \vert _{p \hspace{0.05cm}= \hspace{0.05cm}p_{ {\rm x}_i} } \hspace{-0.7cm}\{Y_{\rm L}(p)\cdot {\rm e}^{p t}\}= \frac{1}{(l-1)!}\cdot \frac{ {\rm d}^{\hspace{0.05cm}l-1} }{ {\rm d}p^{\hspace{0.05cm}l-1} }\hspace{0.15cm} \left \{Y_{\rm L}(p)\cdot (p - p_{ {\rm x}_i})^{\hspace{0.05cm}l}\cdot {\rm e}^{p \hspace{0.05cm}t}\right\} \bigg \vert_{p \hspace{0.05cm}= \hspace{0.05cm}p_{ {\rm x}_i} } \hspace{0.05cm} .$$
• The following is obtained out of it with  $l = 1$  for a simple pole as a special case:
$${\rm Res} \bigg\vert_{p \hspace{0.05cm}= \hspace{0.05cm}p_{ {\rm x}_i} } \hspace{-0.7cm}\{Y_{\rm L}(p)\cdot {\rm e}^{p t}\}= Y_{\rm L}(p)\cdot (p - p_{ {\rm x}_i} )\cdot {\rm e}^{p \hspace{0.05cm}t} \bigg \vert _{p \hspace{0.05cm}= \hspace{0.05cm}p_{ {\rm x}_i} } \hspace{0.05cm} .$$

On the next pages,  the  "residue theorem"  is illustrated by three detailed examples corresponding to the three constellations in  $\text{Example 3}$  in the chapter  "Laplace Transform and p-Transfer Function":

• So, we consider again the two-port network with an inductance  $L = 25 \ \rm µH$  in the longitudinal branch as well as the the series connection of an ohmic resistance  $R = 50 \ \rm Ω$  and a capacitance  $C$  in the transverse branch.
• For the latter,  we again consider three different values,  namely  $C = 62.5 \ \rm nF$,  $C = 8 \ \rm nF$  and  $C = 40 \ \rm nF$.
• The following is always assumed:  $x(t) = δ(t) \; ⇒ \; X_{\rm L}(p) = 1$   ⇒   $Y_{\rm L}(p) = H_{\rm L}(p)$   ⇒   the output signal  $y(t)$  is equal to the impulse response  $h(t)$.

## Aperiodically decaying impulse response

The following is obtained for the  $p$–transfer function computed on the page  pole-zero representation of circuits  with the capacitance  $C = 62.5 \ \rm nF$  and the other numerical values given in the graph below:

$$H_{\rm L}(p)= K \cdot \frac {p - p_{\rm o }} {(p - p_{\rm x 1})(p - p_{\rm x 2})}= 2 \cdot \frac {p + 0.32 } {(p +0.4)(p +1.6 )} \hspace{0.05cm} .$$

Note the normalization of  $p$,  $K$ and also of all poles and zeros by the factor  ${\rm 10^6} · 1/\rm s$.

The impulse response is composed of  $I = N = 2$  natural oscillations. For $t < 0$,  these are equal to zero.

• The residual of the pole at  $p_{{\rm x}1} =\ –0.4$  yields the following time function:
$$h_1(t) = {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}} \hspace{-0.7cm}\{H_{\rm L}(p)\cdot {\rm e}^{p t}\}= H_{\rm L}(p)\cdot (p - p_{{\rm x}1})\cdot {\rm e}^{p \hspace{0.05cm}t} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}=2 \cdot \frac {p + 0.32 } {p +0.4}\cdot {\rm e}^{p \hspace{0.05cm}t} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}-0.4}= - \frac {2 } {15}\cdot {\rm e}^{-0.4 \hspace{0.05cm} t} \hspace{0.05cm}.$$
• In the same way, the residual of the second pole at  $p_{{\rm x}2} = \ –1.6$  can be computed:
$$h_2(t) = {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}} \hspace{-0.7cm}\{H_{\rm L}(p)\cdot {\rm e}^{p t}\}= H_{\rm L}(p)\cdot (p - p_{{\rm x}2})\cdot {\rm e}^{p \hspace{0.05cm}t} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}=2 \cdot \frac {p + 0.32 } {p +1.6}\cdot {\rm e}^{p \hspace{0.05cm}t} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}-1.6}= \frac {32 } {15}\cdot {\rm e}^{-1.6 \hspace{0.05cm} t} \hspace{0.05cm}.$$
Decaying impulse response

The graph shows  $h_1(t)$  and  $h_2(t)$  as well as the sum signal  $h(t)$.

• Also, the normalization factor  $1/T = 10^6 · \rm 1/s$ is taken into account here so that the time is normalized to  $T = 1 \ \rm µ s$ .
• For  $t =0$,  $T \cdot h(t=0) = {32 }/ {15} -{2 }/ {15}= 2 \hspace{0.05cm}$  is obtained as a result.
• For times  $t > 2 \ \rm µ s$,  the impulse response is negative
(although only slightly and difficult to see in the graph).

## Attenuated-oscillatory impulse response

The component values  $R = 50 \ \rm Ω$,  $L = 25 \ \rm µ H$ and  $C = 8 \ \rm nF$ result in two conjugate complex poles at  $p_{{\rm x}1} = \ –1 + {\rm j} · 2$  and  $p_{{\rm x}2} = \ –1 - {\rm j} · 2$.  The zero is located at  $p_{\rm o} = \ –2.5$.  $K = 2$  holds and all numerical values are to be multiplied again by the factor  $1/T$  $(T = 1\ \rm µ s$).

When applying the residue theorem to this configuration the following is obtained:

$$h_1(t) = \text{ ...} = K \cdot \frac {p_{\rm x 1} - p_{\rm o }} {p_{\rm x 1} - p_{\rm x 2}}\cdot {\rm e}^{\hspace{0.05cm}p_{\rm x 1} \cdot \hspace{0.05cm}t}= 2 \cdot \frac {-1 + {\rm j}\cdot 2 +2.5} {(-1 + {\rm j}\cdot 2) - (-1 - {\rm j}\cdot 2)}\cdot {\rm e}^{\hspace{0.05cm}p_{\rm x 1} \cdot\hspace{0.05cm}t}= 2 \cdot \frac {1.5 + {\rm j}\cdot 2} {{\rm j}\cdot 4}\cdot {\rm e}^{\hspace{0.05cm}p_{\rm x 1} \cdot\hspace{0.05cm}t}$$
$$\Rightarrow \hspace{0.3cm}h_1(t) = 2 \cdot \frac {1.5 + {\rm j}\cdot 2} {{\rm j}\cdot 4}\cdot {\rm e}^{\hspace{0.05cm}p_{\rm x 1} \cdot\hspace{0.05cm}t}= (1 - {\rm j}\cdot 0.75)\cdot {\rm e}^{-t}\cdot {\rm e}^{\hspace{0.03cm}{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} 2t} \hspace{0.05cm} ,$$
$$h_2(t) = \text{ ...} = K \cdot \frac {p_{\rm x 2} - p_{\rm o }} {p_{\rm x 2} - p_{\rm x 1}}\cdot {\rm e}^{\hspace{0.05cm}p_{\rm x 2} \hspace{0.03cm}\cdot \hspace{0.05cm}t}= 2 \cdot \frac {-1 - {\rm j}\cdot 2 +2.5} {(-1 - {\rm j}\cdot 2) - (-1 + {\rm j}\cdot 2)}\cdot {\rm e}^{\hspace{0.05cm}p_{\rm x 2} \hspace{0.03cm}\cdot\hspace{0.05cm}t}=2 \cdot \frac {1.5 - {\rm j}\cdot 2} {-{\rm j}\cdot 4}\cdot {\rm e}^{\hspace{0.05cm}p_{\rm x 2} \hspace{0.03cm}\cdot\hspace{0.05cm}t}$$
Gedämpft oszillierende Impulsantwort
$$\Rightarrow \hspace{0.3cm}h_2(t) = (1 + {\rm j}\cdot 0.75)\cdot {\rm e}^{-t}\cdot {\rm e}^{\hspace{0.03cm}-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} 2t} \hspace{0.05cm} .$$

Using  Euler's theorem  the following is thus obtained for the sum signal:

$$h(t) = h_1(t) + h_2(t)$$
$$\Rightarrow \hspace{0.3cm}h(t) = {\rm e}^{-t}\cdot \big [ (1 - {\rm j}\cdot 0.75)\cdot (\cos() + {\rm j}\cdot \sin())+$$
$$\hspace{3.1cm}+ (1 + {\rm j}\cdot 0.75)\cdot (\cos() - {\rm j}\cdot \sin())\big ]$$
$$\Rightarrow \hspace{0.3cm}h(t) ={\rm e}^{-t}\cdot \big [ 2\cdot \cos(2t) + 1.5 \cdot \sin(2t)\big ]\hspace{0.05cm} .$$

The graph shows the attenuated-oscillatory impulse response  $h(t)$  attenuated by  ${\rm e}^{–t}$  for this pole–zero configuration.

## Critically-damped case

Mit  $R = 50 \ \rm Ω$,  $L = 25 \ \rm µ H$ und  $C = 40 \ \rm nF$ ergibt sich der so genannte aperiodische Grenzfall:

$$H_{\rm L}(p)= K \cdot \frac {p - p_{\rm o }} {(p - p_{\rm x })^2}= 2 \cdot \frac {p + 0.5 } {(p +1)^2} \hspace{0.05cm} .$$

Der Kapazitätswert  $C = 40 \ \rm nF$  ist der kleinstmögliche Wert, für den sich gerade noch reelle Polstellen ergeben.  Diese fallen zusammen, das heißt  $p_{\rm x} = \ –1$  ist eine doppelte Polstelle.  Die Zeitfunktion lautet somit entsprechend dem Residuensatz mit  $l = 2$:

$$h(t) = {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{ {\rm x} } } \hspace{-0.7cm}\{H_{\rm L}(p)\cdot {\rm e}^{p t}\}= \frac{ {\rm d} }{ {\rm d}p}\hspace{0.15cm} \left \{H_{\rm L}(p)\cdot (p - p_{ {\rm x} })^2\cdot {\rm e}^{p \hspace{0.05cm}t}\right\} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{ {\rm x} } } = K \cdot \frac{ {\rm d} }{ {\rm d}p}\hspace{0.15cm}\left \{ (p - p_{ {\rm o} })\cdot {\rm e}^{p \hspace{0.05cm}t}\right\} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{ {\rm x} } } \hspace{0.05cm} .$$

Mit der Produktregel  der Differentialrechnung ergibt sich daraus:

Impulsantwort und Sprungantwort des aperiodischen Grenzfalls
$$h(t) = K \cdot \left [ {\rm e}^{p \hspace{0.05cm}t} + (p - p_{ {\rm o} })\cdot t \cdot {\rm e}^{p \hspace{0.05cm}t} \right ] \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}-1} = {\rm e}^{-t}\cdot \left ( 2 - t \right) \hspace{0.05cm} .$$

Die Grafik zeigt diese Impulsantwort (grüne Kurve) in normierter Darstellung.  Sie unterscheidet sich von derjenigen mit den beiden unterschiedlichen Polen bei  $-0.4$  und  $-1.6$  nur geringfügig.

Das rot gezeichnete Signal  $y(t) = 1 - {\rm e}^{-t} + t \cdot {\rm e}^{-t}$  ergibt sich, wenn man am Eingang zusätzlich eine Sprungfunktion berücksichtigt   ⇒   Sprungantwort.

Zur Berechnung der Sprungantwort  $\sigma(t) = y(t)$  kann man alternativ

• bei der Residuenberechnung einen zusätzlichen Pol bei  $p = 0$   (rot markiert) berücksichtigen,
• oder das Integral über die Impulsantwort  $h(t)$  bilden.

## Partial fraction decomposition

Voraussetzung für die Anwendung des Residuensatzes ist, dass es weniger Nullstellen als Pole gibt   ⇒   $Z$  muss stets kleiner als  $N$  sein.

Gilt dagegen wie bei einem Hochpass  $Z = N$, so

• ist der Grenzwert der Spektralfunktion für großes  $p$  ungleich Null,
• beinhaltet das zugehörige Zeitsignal  $y(t)$  auch einen  Diracimpuls,
• versagt der Residuensatz und es ist eine Partialbruchzerlegung  vorzunehmen.

Die Vorgehensweise soll beispielhaft für einen Hochpass erster Ordnung verdeutlicht werden.

Impulsantwort von Tiefpass (blau) und Hochpass (rot)

$\text{Example 1:}$  Die  $p$–Übertragungsfunktion eines  $RC$–Hochpasses erster Ordnung kann durch Abspaltung einer Konstanten wie folgt umgewandelt werden:

$$\frac{p}{p + RC} = 1- \frac{RC}{p + RC}\hspace{0.05cm} .$$

Damit lautet die Hochpass–Impulsantwort:

$$h_{\rm HP}(t) = \delta(t) - h_{\rm TP}(t) \hspace{0.05cm} .$$

Die Grafik zeigt

• als rote Kurve die Hochpass–Impulsantwort  $h_{\rm HP}(t)$,
• als blaue Kurve die Impulsantwort  $h_{\rm TP}(t)$  des äquivalenten Tiefpasses.

Die Diracfunktion ist die Laplace–Transformierte des konstanten Wertes $1$, während die zweite Funktion die Impulsantwort des äquivalenten Tiefpasses angibt, die mit  $Z = 0$,  $N =1$ und  $K = RC$  durch den Residuensatz angebbar ist.