Linear and Time Invariant Systems/Laplace Transform and p-Transfer Function

1. Linear and Time Invariant Systems
2. Beschreibung kausaler realisierbarer Systeme
3. Laplace Transform and p-Transfer Function

• Previous page
• Next page

• Previous page
• Next page

Considered system model

We consider a linear time-invariant system with the impulse response  $h(t)$,  whose input the signal  $x(t)$  is applied to.  The output signal  $y(t)$  is then obtained as the convolution product  $x(t) ∗ h(t)$.

General (also non-causal) and causal system model

For non-causal systems and signals, the  »first Fourier integral«  must always be applied to describe the spectral behavior and the following is valid for the output spectrum:

$$Y(f) = X(f) \cdot H(f) \hspace{0.05cm}.$$

The Fourier integral also continues to be valid for causal systems and signals,  i.e. under the assumption

$$x(t) = 0 \hspace{0.2cm}{\rm{for}} \hspace{0.2cm} t<0\hspace{0.05cm},\hspace{0.2cm} h(t) = 0 \hspace{0.2cm}{\rm{for}} \hspace{0.2cm} t<0 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} y(t) = 0 \hspace{0.2cm}{\rm{for}} \hspace{0.2cm} t<0 \hspace{0.05cm}.$$

In this case,  however,  there are significant advantages in applying the  »Laplace transformation«  while taking into account certain restrictions:

1. The systems treated in this way are always realizable by a circuit.  The developer is not tempted to offer unrealistic solutions.
2. The Laplace transform  $X_{\rm L}(p)$  is always a real function of the spectral variable  $p$. 
3. The fact that this variable is derived from the multiplication of the physical angular frequency  $ω = 2πf$  by the imaginary unit  $\rm j$  according to  $p = {\rm j} · 2πf$  does not matter for the user.
4. The implicit condition  $x(t) = 0$  for  $t < 0$  specifically allows for simpler analysis of transient
   behavior after switching-on processes than with the Fourier integral.

Definition of the Laplace transformation

Starting from the  »first Fourier integral«

$$X(f) = \int_{-\infty}^{+\infty} { x(t) \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{-{\rm j}\hspace{0.05cm} 2\pi f t}}\hspace{0.1cm}{\rm d}t,$$

the Laplace transformation is obtained directly by using the formal substitution  $p = {\rm j} · 2πf$  for a causal time function  $x(t) = 0 \ \ \ \text{for} \ \ \ t < 0.$

$\text{Definition:}$  The   »Laplace transform«  of a causal time function  $x(t)$  is:

$$X_{\rm L}(p) = \int_{0}^{\infty} { x(t) \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{-p t} }\hspace{0.1cm}{\rm d}t\hspace{0.05cm}, \hspace{0.3cm}{\rm briefly}\hspace{0.3cm} X_{\rm L}(p) \quad \bullet\!\!-\!\!\!-^{\hspace{-0.25cm}\rm L}\!\!\!-\!\!\circ\quad x(t)\hspace{0.05cm}.$$

The relationship between the Laplace transform  $X_{\rm L}(p)$  and the physical spectrum  $X(f)$  is often given as follows:

$$X(f) = X_{\rm L}(p) \Bigg |_{{\hspace{0.1cm} p\hspace{0.05cm}={\rm \hspace{0.05cm} j\hspace{0.05cm}2\pi \it f}}}.$$

1. However,  if the signal  $x(t)$  has periodic components and thus the spectral function  $X(f)$  contains Dirac delta functions,  then this equation is not applicable.
2. In this case,  $p = α + {\rm j} · 2πf$  must be applied and then the limit  $α → 0$  must be formed.

$\text{Example 1:}$  We assume the unilaterally and exponentially decreasing signal corresponding to  $\text{Example 1}$  of chapter  »Conclusions from the Allocation Theorem«:

$$x(t) = \left\{ \begin{array}{c} 0 \\ 0.5 \\ {\rm e}^{-t/T} \end{array} \right.\quad \quad \begin{array}{c} {\rm{for} } \\ {\rm{for} } \\ {\rm{for} } \end{array}\begin{array}{*{20}c}{ t < 0\hspace{0.05cm},} \\ { t = 0\hspace{0.05cm},} \\{ t > 0\hspace{0.05cm}.} \end{array}$$

• Thus,  the Laplace transform is:

$$X_{\rm L}(p) = \int_{0}^{\infty} {\rm e}^{-t/T} \cdot {\rm e}^{-pt} \hspace{0.1cm}{\rm d}t= \frac {1}{p + 1/T} \cdot {{\rm e}^{-(p+1/T) \hspace{0.08cm}\cdot \hspace{0.08cm}t}}\hspace{0.15cm}\Bigg \vert_{t \hspace{0.05cm}=\hspace{0.05cm} 0}^{\infty}= \frac {1}{p + 1/T} \hspace{0.05cm} .$$

• Considering  $p = {\rm j} · 2πf$,  the conventional spectral function with respect to $f$  is obtained:

$$X(f) = \frac {1}{{\rm j \cdot 2\pi \it f} + 1/T} = \frac {T}{1+{\rm j \cdot 2\pi \it fT}} \hspace{0.05cm} .$$

• In contrast,  if we consider the frequency response of a low-pass filter of first-order whose impulse response  $h(t)$  differs from the above time function by the factor  $1/T$,  then the following holds for the Laplace transform and the Fourier transform,  respectively:

$$H_{\rm L}(p)= \frac {1/T}{p + 1/T}= \frac {1}{1 + p \cdot T} \hspace{0.05cm} , \hspace{0.8cm}H(f) = \frac {1}{1+{\rm j \cdot 2\pi \it fT} } = \frac {1}{1+{\rm j} \cdot f/f_{\rm G} } \hspace{0.05cm} .$$

• In this equation,  the  »3 dB cut-off frequency«  $f_{\rm G} = 1/(2πT)$   $($German:  "Grenzfrequenz"   ⇒   subscript  "G"$)$ is used instead of the parameter  $T$.

Some important Laplace correspondences

Some important Laplace correspondences are compiled subsequently.  All time signals  $x(t)$  considered here are assumed to be dimensionless.  For this reason,  $X_{\rm L}(p)$  then always has the unit  »second«  as an integral over time.

Table with some Laplace transforms

1. The Laplace transform of the  »Dirac delta function«  $δ(t)$  is  $X_{\rm L}(p) = 1$  $($diagram $\rm A)$. 
2. $X_{\rm L}(p) = 1/p$  is obtained for the  »unit jump function«  $γ(t)$  $($diagram $\rm B)$  by applying the  »integration theorem«. 
3. From this the Laplace transform of the  »linearly increasing function«  $x(t) = t/T$  for  $t > 0$  is obtained by multiplication by  $1/(pT)$  $($diagram $\rm C)$.
4. The  »rectangular function«  can be generated by subtraction of two jump functions  $γ(t)$  and  $γ(t – T)$  separated by  $T$  so that the Laplace transform  $X_{\rm L}(p) = (1 – {\rm e}^{–pT})/p$  is obtained according to the  »shifting theorem«  $($diagram $\rm D)$.
5. By integration the  »ramp function«  or after multiplication by  $1/(pT)$  its Laplace transform  is obtained from it $($diagram $\rm E)$.
6. The  »exponential function«  $($diagram $\rm F)$  has already been considered in the  »last section« .  Considering the factor  $1/T$,  this is at the same time the impulse response of a low-pass filter of first-order.
7. The Laplace transform of a second-order low-pass   ⇒   function $x(t) = t/T · {\rm e}^{–t/T}$   is obtained by squaring  $($diagram  $\rm G)$.
8. In addition to the causal  $\rm sinc$–function  $($diagram $\rm H)$,  the Laplace transforms of the  »causal cosine and sine functions«  $($diagrams  $\rm I$  and  $\rm J)$,  which result in  $p/(p^2 + ω_0^2)$  resp.  $ω_0/(p^2 + ω_0^2)$  are also given in the table.
9. Here,  $ω_0 = 2πf_0 = 2π/T$  denotes the  »angular frequency«.

Pole-zero representation of circuits

Any  »linear time-invariant system«  $\rm (LTI)$  which can be realized by a circuit of discrete time-constant components such as

• resistances  $(R)$,
• capacitances  $(C)$,
• inductances  $(L)$  and
• amplifier elements,

have a fractional-rational  »$p$–transfer function«:

$$H_{\rm L}(p)= \frac {A_Z \cdot p^Z +\text{...} + A_2 \cdot p^2 + A_1 \cdot p + A_0} {B_N \cdot p^N +\text{...} \ + B_2 \cdot p^2 + B_1 \cdot p + B_0}= \frac {Z(p)}{N(p)} \hspace{0.05cm} .$$

All coefficients of the numerator   ⇒   $A_Z, \text{...} \ , A_0$  and of the denominator   ⇒   $B_N, \text{...} , B_0$  are real.  Furthermore:

1. $Z$  denotes the degree of the numerator polynomial  $Z(p)$,
2. $N$  denotes the degree of the denominator polynomial  $N(p)$.

$\text{Equivalent pole-zero representation:}$   The following can be formulated for the  »$p$–transfer function«,  too:

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

The  $Z + N + 1$  parameters signify:

1. $K = A_Z/B_N$  is a constant factor.   If  $Z = N$,  this is dimensionless.
2. The solutions of the equation  $Z(p) = 0$  result in the  $Z$  »zeros«  $p_{\rm o1},\text{...} \ , p_{\rm oZ}$  of  $H_{\rm L}(p)$.
3. The zeros of the denominator polynomial  $N(p)$  yield the  $N$  pole locations  $($or  »poles«  for short$)$.

The transformation is unique.  This can be seen from the fact that the  $p$–transfer function is also determined only by  $Z + N + 1$  free parameters according to the first equation since one of the coefficients  $A_Z, \text{...} \ , A_0, B_N, \text{...} \ , B_0$  can be normalized to  $1$  without changing the quotient.

Considered two-port network and associated pole-zero diagram

$\text{Example 2:}$  We consider the drawn two-port network with

• an inductance  $L$  $($complex resistance  $pL)$  in the longitudinal branch as well as

• the series connection of an ohmic resistance  $R$  and a capacitance  $C$  with the complex resistance  $1/(pC)$ in the transverse branch.

Thus,  the  $p$–transfer function is:

$$H_{\rm L}(p)= \frac {Y_{\rm L}(p)} {X_{\rm L}(p)}= \frac {R + {1}/{(pC)} } {pL + R +{1}/{(pC)} }$$

$$\Rightarrow \hspace{0.3cm} H_{\rm L}(p)= \frac {1 + p \cdot{RC} } {1 + p \cdot{RC}+ p^2 \cdot{LC} } \hspace{0.05cm} .$$

If  $p = {\rm j} · 2πf$  is set,  then the Fourier transfer function  $($or the  »frequency response«$)$  is obtained.  If the numerator and denominator in the above equation are divided by  $LC$,  then the following is obtained:

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

⇒   In the right-hand side of the equation,  the transfer function  $H_{\rm L}(p)$  is given in  »pole-zero notation«. 

By comparison of coefficients the following values are obtained for  $R = 50 \ \rm Ω$,  $L = 25\ \rm µ H$  and  $C = 62.5 \ \rm nF$ :

1. the constant  $K = R/L = 2 · 10^6 \cdot 1/{\rm s}$,
2. the zero  $p_{\rm o} = -1/(RC) = -0.32 · 10^6 \cdot 1/{\rm s},$
3. the two poles  $p_{\rm x1}$  and  $p_{\rm x2}$  as the solution of the equation

$$p^2 + \frac {R} {L}\cdot p + \frac{1}{LC} = 0 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm x 1,\hspace{0.05cm}2 }= -\frac {R} {2L}\pm \sqrt{\frac {R^2} {4L^2}- \frac{1}{LC} }$$

$$\Rightarrow \hspace{0.3cm} p_{\rm x 1,\hspace{0.05cm}2 }= -10^6 \cdot {1}/{\rm s} \pm \sqrt{10^{12} \cdot {1} /{\rm s^2}-0.64 \cdot 10^{12} \cdot {1}/ {\rm s^2} }\hspace{0.6cm} \Rightarrow \hspace{0.3cm} p_{\rm x 1 }= -0.4 \cdot 10^6\cdot {1}/ {\rm s},\hspace{0.2cm}p_{\rm x 2 }= -1.6 \cdot 10^6\cdot {1}/ {\rm s} \hspace{0.05cm} .$$

In the above graph,  the pole-zero diagram is given on the right-hand side.

• The two axes denote the real and imaginary parts of the variable  $p$,  each normalized to the value  $10^6 · \rm 1/s\; (= 1/µs)$.

• The zero at  $p_{\rm o} =\, –0.32$  can be seen as a circle and the poles at  $p_{\rm x1} = \,–0.4$  and  $p_{\rm x2} = \,–1.6$  as crosses.

Properties of poles and zeros

The transfer function  $H_{\rm L}(p)$  of any realizable circuit is fully described by  $Z$  »zeros«  and  $N$  »poles«  together with a constant  $K$  where the following restrictions apply:

• $Z ≤ N$  always holds.  The  $p$–transfer function would also be  »infinitely large«  with  $Z > N$  in the limiting case for  $p → ∞$  $($i.e. for very high frequencies$)$.

• The zeros  $p_{\rm oi}$  and the poles  $p_{ {\rm x}i}$  are generally complex,  and have like  $p$  the unit  $\rm 1/s$.  If  $Z < N$  holds,  then the constant  $K$  has also a unit.

• The poles and zeros can be real as shown in the last example.  If they are complex,  then two poles resp. two zeros always occur as complex conjugates,  since  $H_{\rm L}(p)$  always represents a real fractional-rational function.

• All poles lie in the left half-plane or on the imaginary axis  $($limiting case$)$.  This property follows from the required and assumed causality together with the  »fundamental function theory«,  which will be stated in the next chapter.

• Zeros can occur in both the left and right  $p$–half-planes or also on the imaginary axis.  An example of zeros in the right half-plane can be found in  $\text{Exercise 3.4Z}$,  which deals with all-pass filters.

• In so-called  »minimum-phase systems«,  not only poles are forbidden in the right  $p$–half-plane but also zeros.  The real part of all singularities here is never positive.

These properties are now illustrated by three examples.

$\text{Example 3:}$  Starting from the  »two-port network«  $(L$  in the longitudinal branch,  $R$  and  $C$  in the transverse branch$)$  the characteristic quantities of the transfer function can be given as follows:

Location of the zero and the poles for  $Z = 1$  and  $N = 2$

$$K = 2A, \hspace{0.2cm}p_{\rm x 1,\hspace{0.05cm}2 }= -A \pm \sqrt{A^2-B^2},$$

$$p_{\rm o }= - \frac{B^2}{2A} \hspace{0.05cm} \hspace{0.2cm} {\rm with } \hspace{0.2cm} A = \frac {R} {2L}, \hspace{0.2cm}B = \frac{1}{\sqrt{LC} } \hspace{0.05cm}.$$

The graph shows three different diagrams with different capacitance values  $C$. 

• $R = 50 \ \rm Ω$  and  $L = 25 \ \rm µ H$  always hold.

• The axes are normalized to  $A = R/(2L) = 10^6 · \rm 1/s$.

• The constant factor is  $K = 2A = 2 · 10^6 · \rm 1/s.$

$\text{Example 3.1:}$ 
For  $B < A$,  »two real poles«  and  »one zero«  to the right of  $-A/2$  are obtained. The following arises as a result for  $C = 62.5 \ \rm nF$   ⇒   $ {B}/ {A}= 0.8 $  $($left diagram$)$:

$$ p_{\rm x 1}/A = -0.4 , \hspace{0.2cm}p_{\rm x 2}/A= -1.6 , \hspace{0.2cm}p_{\rm o}/A= -0.32 \hspace{0.05cm} .$$

$\text{Example 3.2:}$ 
For  $B > A$,  »two conjugate-complex poles«  and  »one zero«  to the left of   $-A/2$  are obtained. For  $C = 8 \ \rm nF$   ⇒   $ {B}/ {A}= \sqrt{5} $  $($right diagram$)$:

$$p_{\rm x 1,\hspace{0.05cm}2 }/A= -1\pm {\rm j}\cdot 2,\hspace{0.2cm}p_{\rm o}/A\approx -2.5 \hspace{0.05cm} .$$

$\text{Example 3.3:}$ 
The case  $A = B$  leads to  »a real double pole«  and  »one zero«  at  $– A/2$.  For  $C = 400 \ \rm nF$   ⇒   $ {B}/ {A}= 1 $  $($middle diagram$)$:

$$ p_{\rm x 1}/A= p_{\rm x 2}/A= -1, \hspace{0.2cm}p_{\rm o}/A= -0.5 \hspace{0.05cm} .$$

The impulse responses  $h(t)$  are obtained according to the following chapter  »Inverse Laplace Transform«  as follows:

1. For the left constellation,  $h(t)$  is  »aperiodically decaying«.
2. For the right constellation,  $h(t)$  is  »attenuated-oscillatory«.
3. The middle constellation is called the  »critically attenuated case«.

Graphical determination of attenuation and phase

For the computation of attenuation and phase

The  $p$–transfer function is given in pole-zero notation:

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

• The conventional frequency response  $H(f)$  is obtained by replacing the argument  $p$  of  $H_{\rm L}(p)$  by  ${\rm j} \cdot 2πf$ :

$$H(f)= K \cdot \frac {({\rm j} \cdot 2\pi \hspace{-0.05cm}f - p_{\rm o 1})({\rm j} \cdot 2\pi \hspace{-0.05cm}f - p_{\rm o 2})\cdot \text{...} \cdot ({\rm j} \cdot 2\pi \hspace{-0.05cm}f - p_{ {\rm o} \hspace{-0.03cm} Z})} {({\rm j} \cdot 2\pi \hspace{-0.05cm}f - p_{\rm x 1})({\rm j} \cdot 2\pi \hspace{-0.05cm}f - p_{\rm x 2})\cdot \text{...}\cdot ({\rm j} \cdot 2\pi \hspace{-0.05cm}f - p_{ {\rm x} \hspace{-0.03cm} N})} \hspace{0.05cm} .$$

• We now consider a fixed frequency $f$  and describe the distances and angles of all  »zeros«  $($circles$)$  by vectors:

$$R_{ {\rm o} i} = {\rm j} \cdot 2\pi \hspace{-0.05cm}f - p_{ {\rm o} i}= |R_{{\rm o} i}| \cdot {\rm e}^{\hspace{0.03cm}{\rm j}\hspace{0.03cm}\cdot\hspace{0.03cm}\phi_{ {\rm o} i} }, \hspace{0.3cm}i= 1, \text{...}\ , Z \hspace{0.05cm} .$$

• We proceed in the same way for the  »poles«  $($crosses$)$:

$$R_{ {\rm x} i} = {\rm j} \cdot 2\pi \hspace{-0.05cm}f - p_{ {\rm x} i}= |R_{ {\rm x} i}| \cdot {\rm e}^{\hspace{0.03cm}{\rm j}\hspace{0.03cm}\cdot\hspace{0.03cm}\phi_{ {\rm x} i} }, \hspace{0.3cm}i= 1, \text{...}\ , N \hspace{0.05cm} .$$

• The graph shows the magnitudes and phase angles for a system with

1. $Z = 2$  zeros  (circles)  in the right half-plane
2. $N = 2$  poles  (crosses)  in the left half-plane.
3. constant  $K$.

• The following can be written for the frequency response with this vector representation:

$$H(f)= K \cdot \frac {|R_{ {\rm o} 1}| \cdot |R_{ {\rm o} 2}|\cdot ... \cdot |R_{ {\rm o} \hspace{-0.03cm} Z}|} {|R_{ {\rm x} 1}| \cdot |R_{ {\rm x} 2}|\cdot \text{...} \cdot |R_{ {\rm x} \hspace{-0.03cm} N}|} \cdot {\rm e^{\hspace{0.03cm}{\rm j} \hspace{0.05cm}\cdot [ \phi_{ {\rm o} 1}\hspace{0.1cm}+ \hspace{0.1cm}\phi_{ {\rm o} 2} \hspace{0.1cm}+ \hspace{0.1cm}\hspace{0.1cm}\text{...}. \hspace{0.1cm} + \hspace{0.1cm}\phi_{ {\rm o} \hspace{-0.03cm}{\it Z}}\hspace{0.1cm}- \hspace{0.1cm}\phi_{ {\rm x} 1}\hspace{0.1cm}- \hspace{0.1cm}\phi_{ {\rm x} 2} \hspace{0.1cm}- \hspace{0.1cm}... \hspace{0.1cm} - \hspace{0.1cm} \phi_{ {\rm x} \hspace{-0.03cm}{\it N} }]} } \hspace{0.05cm} .$$

If  $H(f)$  is expressed by the attenuation function  $a(f)$  and the phase function  $b(f)$  according to the generally valid relation  $H(f) = {\rm e}^{-a(f)\hspace{0.05cm}- \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}b(f)}$,  then the following result is obtained by comparing with the above equation:

• With a suitable normalization of all dimensional quantities the following holds for the  »attenuation function«  in Neper   $(1 \ \rm Np$ corresponds to  $8.686 \ \rm dB)$:

$$a(f) = -{\rm ln} \hspace{0.1cm} K + \sum \limits_{i=1}^N {\rm ln} \hspace{0.1cm} |R_{ {\rm x} i}|- \sum \limits_{i=1}^Z {\rm ln} \hspace{0.1cm} |R_{ {\rm o} i}| \hspace{0.05cm} .$$

• The  »phase function«  in radian  $\rm (rad)$  arises as a result according to the upper sketch:

$$b(f) = \phi_K + \sum \limits_{i=1}^N \phi_{ {\rm x} i}- \sum \limits_{i=1}^Z \phi_{ {\rm o} i}\hspace{0.2cm}{\rm with} \hspace{0.2cm} \phi_K = \left\{ \begin{array}{c} 0 \\ \pi \end{array} \right. \begin{array}{c} {\rm{for} } \\ {\rm{for} } \end{array}\begin{array}{*{20}c} { K > 0\hspace{0.05cm},} \\ { K <0\hspace{0.05cm}.} \end{array}$$

$\text{Example 4:}$  The graphic illustrates the calculation of

For the computation of the attenuation and phase functions  $($screen capture of the program »Causal Systems & Laplace Transformation» in an earlier German version$)$

1.   the attenuation function  $a(f)$   ⇒   red curve,  and
2.   the phase function  $b(f)$   ⇒   green curve

of a two-port network which is defined by

• the factor  $K = 1.5$, 
• one zero at  $-3$  and
• two poles at  $–1 \pm {\rm j} · 4$.

The given numerical values are valid for the frequency  $2πf = 3$:

$$a \big [f = {3}/({2\pi}) \big ] = 0.453\,\,{\rm Np}= 3.953\,\,{\rm dB}$$

$$\Rightarrow \hspace{0.4cm}\big \vert H \big [f = {3}/({2\pi}) \big ]\big \vert = 0.636,$$

$$ b\big [f = {3}/({2\pi}) \big ] = -8.1^\circ \hspace{0.05cm} .$$

The derivation of these numerical values is illustrated in the framed block.  For the magnitude frequency response   $\vert H(f)\vert$   ⇒   blue curve,  a band-pass-like curve shape is obtained with

$$\vert H(f = 0)\vert \approx 0.25\hspace{0.05cm},$$

$$\vert H(f = {3}/(2\pi)\vert \approx 0636\hspace{0.05cm},$$

$$\vert H(f \rightarrow \infty)\vert= 0 \hspace{0.05cm}.$$

Exercises for the chapter

Exercise 3.2: Laplace Transform

Exercise 3.2Z: Laplace and Fourier

Exercise 3.3: p-Transfer Function

Exercise 3.3Z: High- and Low-Pass Filters in p-Form

Exercise 3.4: Attenuation and Phase Response

Exercise 3.4Z: Various All-Pass Filters