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== # OVERVIEW OF THE THIRD MAIN CHAPTER # ==
 
== # OVERVIEW OF THE THIRD MAIN CHAPTER # ==
 
<br>
 
<br>
In the first two chapters, filter functions with real-valued frequency responses were mostly considered for reasons of presentation so that the associated time function is symmetric about zero-time. However, the impulse response of a realisable system must always be causal, that is, &nbsp;$h(t)$&nbsp; must be identical to zero for &nbsp;$t < 0$&nbsp;. This strong asymmetry of the time function&nbsp; $h(t)$,&nbsp; though, implies at the same time that the frequency response &nbsp;$H(f)$&nbsp; of a realisable system is always complex-valued with the exception of &nbsp;$H(f) = K$&nbsp; where there is a fixed relation between its real part and imaginary part.
+
In the first two chapters,&nbsp; filter functions with real-valued frequency responses were mostly considered for reasons of presentation so that the associated time function is symmetric about zero-time.&nbsp; However,&nbsp; the impulse response of a realizable system must always be causal,&nbsp; that is, &nbsp;$h(t)$&nbsp; must be identical to zero for &nbsp;$t < 0$.&nbsp; This strong asymmetry of the time function&nbsp; $h(t)$&nbsp; implies at the same time that the frequency response &nbsp;$H(f)$&nbsp; of a realizable system is always complex-valued with the exception of &nbsp;$H(f) = K$&nbsp; where there is a fixed relation between its real part and imaginary part.
  
This third chapter provides a recapitulatory account of the description of causal realizable systems, which differ also in the mathematical methods from those commonly used with non-causal systems.  
+
This third chapter provides a recapitulatory account of the description of causal realizable systems,&nbsp; which differ also in the mathematical methods from those commonly used with non-causal systems.  
  
 
In detail, the following is dealt with:
 
In detail, the following is dealt with:
  
*the Hilbert transformation, which states how real and imaginary parts of&nbsp; $H(f)$&nbsp; are related,
+
*the Hilbert transform,&nbsp; which states how real and imaginary parts of&nbsp; $H(f)$&nbsp; are related,
*the Laplace transformation, which yields another spectral function &nbsp;$H_{\rm L}(p)$&nbsp; for causal &nbsp;$h(t)$&nbsp;,
+
*the Laplace transform,&nbsp; which yields another spectral function &nbsp;$H_{\rm L}(p)$&nbsp; for causal &nbsp;$h(t)$,
*the description of realizable systems by the pole-zero plot, as well as
+
*the description of realizable systems by the pole-zero plot,&nbsp; as well as
*the inverse Laplace transformation using the theory of functions (residue theorem).
+
*the inverse Laplace transform using the&nbsp; "theory of functions"&nbsp; ("residue theorem").
  
  
For this chapter we recommend
+
For this chapter, we recommend two of our multimedia offerings:
*the educational video &nbsp;[[Rechnen_mit_komplexen_Zahlen_(Lernvideo)|Arithmetic with complex numbers]] for preparation, as well as
+
*the&nbsp; (German language)&nbsp; learning video &nbsp;[[Rechnen_mit_komplexen_Zahlen_(Lernvideo)|"Rechnen mit komplexen Zahlen"]] &nbsp; &rArr; &nbsp; "Arithmetic operations involving complex numbers",
*the interactive applet &nbsp;[[Applets:Kausale_Systeme_-_Laplacetransformation|Causal systems – Laplace transformation]] - a coherent depiction.
+
*the&nbsp; (German language)&nbsp; interactive SWF applet &nbsp;[[Applets:Kausale_Systeme_-_Laplacetransformation|"Kausale Systeme - Laplacetransformation" ]] &nbsp; &rArr; &nbsp; "Causal systems – Laplace transform".
  
  
==Prerequisites for the entire "realizable systems" chapter==
+
==Prerequisites for the entire third main chapter==
 
<br>
 
<br>
In the first two chapters, mostly real&nbsp; [[Linear_and_Time_Invariant_Systems/System_Description_in_Frequency_Domain#Frequency_response_–_Transfer_function|transfer functions]]&nbsp; &nbsp;$H(f)$&nbsp; were considered for which the associated impulse response &nbsp;$h(t)$&nbsp; is consequently always symmetric with respect to the reference time &nbsp;$t = 0$&nbsp;. Such transfer functions  
+
In the first two chapters,&nbsp; mostly real&nbsp; [[Linear_and_Time_Invariant_Systems/System_Description_in_Frequency_Domain#Frequency_response_–_Transfer_function|transfer functions]]&nbsp; &nbsp;$H(f)$&nbsp; were considered for which the associated impulse response &nbsp;$h(t)$&nbsp; is consequently always symmetric with respect to the reference time &nbsp;$t = 0$.&nbsp; Such transfer functions  
 
*are suitable to explain basic relationships in a simple way,  
 
*are suitable to explain basic relationships in a simple way,  
 
*but unfortunately are not realizable for reasons of causality.  
 
*but unfortunately are not realizable for reasons of causality.  
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{{BlaueBox|TEXT=   
 
{{BlaueBox|TEXT=   
 
$\text{Definition:}$&nbsp;  
 
$\text{Definition:}$&nbsp;  
The&nbsp; '''impulse response''' &nbsp;$h(t)$&nbsp; is equal to the output signal &nbsp;$y(t)$&nbsp; of the system if an infinitely short impulse with an infinitely large amplitude is applied to the input at time &nbsp;$t = 0$&nbsp;: &nbsp; $x(t) = δ(t)$. Such an impulse is called a [[Signal_Representation/Special_Cases_of_Impulse_Signals#Dirac_delta_or_impulse|Dirac impulse]].}}
+
The&nbsp; '''impulse response''' &nbsp;$h(t)$&nbsp; is equal to the output signal &nbsp;$y(t)$&nbsp; of the system if an infinitely short impulse with an infinitely large amplitude is applied to the input at time &nbsp;$t = 0$&nbsp;: &nbsp; $x(t) = δ(t)$.&nbsp; Such an impulse is called a&nbsp; [[Signal_Representation/Special_Cases_of_Impulse_Signals#Dirac_delta_or_impulse|Dirac delta impulse]].}}
  
  
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{{BlaueBox|TEXT=   
 
{{BlaueBox|TEXT=   
 
$\text{Definition:}$&nbsp;  
 
$\text{Definition:}$&nbsp;  
For a&nbsp; '''causal system'''&nbsp; the impulse response $h(t)$ is identical to zero for all times&nbsp; $t < 0$&nbsp;.}}
+
For a&nbsp; '''causal system'''&nbsp; the impulse response&nbsp; $h(t)$&nbsp; is identical to zero for all times&nbsp; $t < 0$.}}
  
  
The only real transfer function that satisfies the causality condition "the output signal cannot start before the input signal" is:
+
The only real transfer function that satisfies the causality condition&nbsp; "the output signal cannot start before the input signal"&nbsp; is:
 
:$$H(f) = K \quad \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\quad h(t) = K \cdot \delta(t).$$
 
:$$H(f) = K \quad \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\quad h(t) = K \cdot \delta(t).$$
 
All other real-valued transfer functions&nbsp; $H(f)$&nbsp; describe non-causal systems and thus cannot be realized by an (electrical) circuit network.  
 
All other real-valued transfer functions&nbsp; $H(f)$&nbsp; describe non-causal systems and thus cannot be realized by an (electrical) circuit network.  
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{{BlaueBox|TEXT=   
 
{{BlaueBox|TEXT=   
 
$\text{In other words:}$ &nbsp; Except for the transfer function&nbsp; $H(f) = K,$&nbsp; '''any realistic transfer function is complex'''.  
 
$\text{In other words:}$ &nbsp; Except for the transfer function&nbsp; $H(f) = K,$&nbsp; '''any realistic transfer function is complex'''.  
*If &nbsp; $K=1$ holds additionally, the transfer function is said to be ideal.&nbsp;  
+
*If&nbsp; $K=1$&nbsp; holds additionally,&nbsp; the transfer function is said to be ideal.&nbsp;  
*The output &nbsp;$y(t)$&nbsp; is then identical to the input &nbsp;$x(t)$&nbsp; &ndash; even without attenuation or amplification.}}
+
*Then,&nbsp; the output signal &nbsp;$y(t)$&nbsp; is identical to the input signal &nbsp;$x(t)$&nbsp; &ndash; even without attenuation or amplification.}}
  
 
==Real and imaginary part of a causal transfer function==
 
==Real and imaginary part of a causal transfer function==
 
<br>
 
<br>
Any causal impulse response&nbsp; $h(t)$&nbsp; can be represented as the sum of an even part&nbsp; $h_{\rm g}(t)$&nbsp; and an odd part&nbsp; $h_{\rm u}(t)$&nbsp;:
+
Any causal impulse response&nbsp; $h(t)$&nbsp; can be represented as the sum  
 +
*of an even&nbsp; (German:&nbsp; "gerade" &nbsp; &rArr; &nbsp; "g")&nbsp; part&nbsp; $h_{\rm g}(t)$&nbsp;  
 +
*and an odd&nbsp; (German:&nbsp; "ungerade" &nbsp; &rArr; &nbsp; "u")&nbsp; part&nbsp; $h_{\rm u}(t)$:
  
 
:$$\begin{align*} h_{ {\rm g}}(t)  & =  {1}/{2}\cdot \big[  h(t) + h(-t) \big]\hspace{0.05cm},\\ h_{ {\rm u}}(t) & =  {1}/{2}\cdot \big[  h(t) - h(-t) \big] = h_{ {\rm g}}(t) \cdot {\rm sign}(t)\hspace{0.05cm} .\end{align*}$$
 
:$$\begin{align*} h_{ {\rm g}}(t)  & =  {1}/{2}\cdot \big[  h(t) + h(-t) \big]\hspace{0.05cm},\\ h_{ {\rm u}}(t) & =  {1}/{2}\cdot \big[  h(t) - h(-t) \big] = h_{ {\rm g}}(t) \cdot {\rm sign}(t)\hspace{0.05cm} .\end{align*}$$
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$\text{Example 1:}$&nbsp;  
 
$\text{Example 1:}$&nbsp;  
 
The graph shows this splitting for a causal exponentially decreasing impulse response of a low-pass filter of first-order corresponding to&nbsp; [[Aufgaben:Exercise_1.3Z:_Exponentially_Decreasing_Impulse_Response|Exercise 1.3Z]]:
 
The graph shows this splitting for a causal exponentially decreasing impulse response of a low-pass filter of first-order corresponding to&nbsp; [[Aufgaben:Exercise_1.3Z:_Exponentially_Decreasing_Impulse_Response|Exercise 1.3Z]]:
 +
[[File: P_ID1750__LZI_T_3_1_S2a_neu.png |right|frame| Splitting of the impulse response into an even part and an odd part|class=fit]]
 
:$$h(t) = \left\{ \begin{array}{c} 0 \\
 
:$$h(t) = \left\{ \begin{array}{c} 0 \\
 
  0.5/T  \\ 1/T \cdot {\rm e}^{-t/T} \end{array} \right.\quad  
 
  0.5/T  \\ 1/T \cdot {\rm e}^{-t/T} \end{array} \right.\quad  
\begin{array}{c}  {\rm{f\ddot{u}r} }  \\ {\rm{f\ddot{u}r} }
+
\begin{array}{c}  {\rm{for} }  \\ {\rm{for} }
\\  {\rm{f\ddot{u}r} }  \end{array}\begin{array}{*{20}c}
+
\\  {\rm{for} }  \end{array}\begin{array}{*{20}c}
 
{  t  < 0\hspace{0.05cm},}  \\
 
{  t  < 0\hspace{0.05cm},}  \\
 
{ t  = 0\hspace{0.05cm},}  \\{ t  > 0\hspace{0.05cm}.}
 
{ t  = 0\hspace{0.05cm},}  \\{ t  > 0\hspace{0.05cm}.}
  
 
\end{array}$$
 
\end{array}$$
 
[[File: P_ID1750__LZI_T_3_1_S2a_neu.png |center|frame| Splitting of the impulse response into an even and an odd part||class=fit]]
 
  
 
It can be seen that:  
 
It can be seen that:  
*$h_{\rm g}(t) = h_{\rm u}(t) = h(t)/2$ holds for positive times.
+
*$h_{\rm g}(t) = h_{\rm u}(t) = h(t)/2$&nbsp; holds for positive times,
*$h_{\rm g}(t)$&nbsp; and &nbsp;$h_{\rm u}(t)$&nbsp; differ only by the sign for negative times.
+
*$h_{\rm g}(t)$&nbsp; and &nbsp;$h_{\rm u}(t)$&nbsp; differ only by the sign for negative times,
*$h(t) = h_{\rm g}(t) + h_{\rm u}(t)$ holds for all times, also at time &nbsp;$t = 0$ (marked by circles). }}
+
*$h(t) = h_{\rm g}(t) + h_{\rm u}(t)$&nbsp; holds for all times, also at time &nbsp;$t = 0$&nbsp; (marked by circles). }}
  
  
Let us now consider the same issue in the spectral domain. According to the&nbsp; [[Signal_Representation/Fourier_Transform_Theorems#Assignment_Theorem|Assignment Theorem]]&nbsp; the following holds for the complex transfer function: &nbsp;
+
Let us now consider the same issue in the spectral domain.&nbsp; According to the&nbsp; [[Signal_Representation/Fourier_Transform_Theorems#Assignment_Theorem|Assignment Theorem]]&nbsp; the following holds for the complex transfer function: &nbsp;
 
:$$H(f) = {\rm Re} \left\{ H(f) \right \} + {\rm j} \cdot {\rm Im} \left\{ H(f) \right \}
 
:$$H(f) = {\rm Re} \left\{ H(f) \right \} + {\rm j} \cdot {\rm Im} \left\{ H(f) \right \}
 
  ,$$
 
  ,$$
where the following assignment holds:
+
where the following assignment is valid:
  
 
:$${\rm Re} \left\{ H(f) \right \} \quad \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\quad h_{ {\rm g}}(t)\hspace{0.05cm},$$
 
:$${\rm Re} \left\{ H(f) \right \} \quad \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\quad h_{ {\rm g}}(t)\hspace{0.05cm},$$
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:$${\rm j} \cdot {\rm Im} \left\{ H(f) \right\}  \quad  \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\quad h_{ {\rm u}}(t)\hspace{0.05cm}.$$
 
:$${\rm j} \cdot {\rm Im} \left\{ H(f) \right\}  \quad  \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\quad h_{ {\rm u}}(t)\hspace{0.05cm}.$$
  
First, this relationship between the real and imaginary parts of &nbsp;$H(f)$&nbsp; shall be worked out using another example.
+
First,&nbsp; this relationship between the real part and the imaginary part of &nbsp;$H(f)$&nbsp; shall be worked out using another example.
  
 
{{GraueBox|TEXT=   
 
{{GraueBox|TEXT=   
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A low-pass filter of first-order is assumed and the following holds for its transfer function:
 
A low-pass filter of first-order is assumed and the following holds for its transfer function:
 
:$$H(f) = \frac{1}{1+{\rm j}\cdot f/f_{\rm G} } = \frac{1}{1+(f/f_{\rm G})^2}- {\rm j} \cdot \frac{f/f_{\rm G} }{1+(f/f_{\rm G})^2} \hspace{0.05cm}.$$
 
:$$H(f) = \frac{1}{1+{\rm j}\cdot f/f_{\rm G} } = \frac{1}{1+(f/f_{\rm G})^2}- {\rm j} \cdot \frac{f/f_{\rm G} }{1+(f/f_{\rm G})^2} \hspace{0.05cm}.$$
Here,&nbsp; $f_{\rm G}$&nbsp; represents the 3dB cut-off frequency at which&nbsp; $\vert H(f)\vert^2$&nbsp; has decreased to half of its maximum&nbsp; $($at&nbsp; $f = 0)$&nbsp;. The corresponding impulse response&nbsp;  $h(t)$&nbsp; has already been shown in the above&nbsp; $\text{Example 1}$&nbsp; for&nbsp;  $f_{\rm G} = 1/(2πT)$&nbsp;.
+
[[File:P_ID1754__LZI_T_3_1_S2b_neu.png|right|frame|Frequency response of a low-pass filter of first-order|class=fit]]
 
+
Here,&nbsp; $f_{\rm G}$&nbsp; represents the&nbsp; $\rm 3dB$&nbsp; cut-off frequency at which&nbsp; $\vert H(f)\vert^2$&nbsp; has decreased to half of its maximum&nbsp; $($at&nbsp; $f = 0)$.&nbsp; The corresponding impulse response&nbsp;  $h(t)$&nbsp; has already been shown in&nbsp; $\text{Example 1}$&nbsp; for&nbsp;  $f_{\rm G} = 1/(2πT)$.
The graph shows the real part (blue) and the imaginary part (red) of&nbsp; $H(f)$. In addition, the magnitude is shown dashed in green.
 
 
 
[[File:P_ID1754__LZI_T_3_1_S2b_neu.png|right|frame|Frequency response of a low-pass filter of first-order&nbsp; (real and imaginary part)|class=fit]]
 
 
 
  
 +
The graph shows the real part (blue) and the imaginary part (red) of&nbsp; $H(f)$.&nbsp;  In addition, the magnitude is shown dashed in green.
  
 
Since the time functions &nbsp;$h_{\rm g}(t)$&nbsp; and &nbsp;$h_{\rm u}(t)$&nbsp; are related by the sign function, there also exists a fixed relationship
 
Since the time functions &nbsp;$h_{\rm g}(t)$&nbsp; and &nbsp;$h_{\rm u}(t)$&nbsp; are related by the sign function, there also exists a fixed relationship
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of the transfer function <br><br>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &rArr; &nbsp; the&nbsp;  '''Hilbert transformation'''.
+
of the transfer function&nbsp; $\{H(f)\}$  &nbsp; &rArr; &nbsp;   '''Hilbert transform'''.
 
   
 
   
 
This is described below.}}
 
This is described below.}}
  
==Hilbert transformation==
+
==Hilbert transform==
 
<br>
 
<br>
Here, two time functions&nbsp; $u(t)$&nbsp; and&nbsp; $w(t) = \sign(t) · u(t)$ are considered in the most general sense.  
+
Here,&nbsp; two time functions&nbsp; $u(t)$&nbsp; and&nbsp; $w(t) = \sign(t) · u(t)$&nbsp; are considered in the most general sense.  
*The associated spectral functions are denoted by&nbsp; $U(f)$&nbsp; and&nbsp; ${\rm j} · W(f)$&nbsp;.  
+
*The associated spectral functions are denoted by&nbsp; $U(f)$&nbsp; and&nbsp; ${\rm j} · W(f)$.  
*That is, in this section&nbsp;  ${w(t) \, \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet \, {\rm j}  \cdot W(f) }$&nbsp; is valid and not the usual Fourier correspondence&nbsp; ${w(t) \, \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet \, W(f)}.$
+
*That is: &nbsp; In this section&nbsp;  ${w(t) \, \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet \, {\rm j}  \cdot W(f) }$&nbsp; is valid and not the usual Fourier correspondence&nbsp; ${w(t) \, \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet \, W(f)}.$
  
  
 
Using the correspondence &nbsp; ${\rm sign}(t) \,  \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet \, {1}/({{\rm j} \, \pi f })$ &nbsp; the following is obtained after writing the [[Signal_Representation/The_Convolution_Theorem_and_Operation#Convolution_in_the_time domain|convolution integral]] out in full with the integration variable&nbsp; $ν$ :
 
Using the correspondence &nbsp; ${\rm sign}(t) \,  \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet \, {1}/({{\rm j} \, \pi f })$ &nbsp; the following is obtained after writing the [[Signal_Representation/The_Convolution_Theorem_and_Operation#Convolution_in_the_time domain|convolution integral]] out in full with the integration variable&nbsp; $ν$ :
 
:$${\rm j} \cdot W(f) =  \frac{1}{{\rm j} \, \pi f }\, \star \, U(f) \quad \Rightarrow \quad W(f) = -\frac{1}{\pi }\int\limits_{-\infty}^{+\infty} { \frac{U(\nu)}{f - \nu}}\hspace{0.1cm}{\rm d}\nu \hspace{0.05cm}.$$
 
:$${\rm j} \cdot W(f) =  \frac{1}{{\rm j} \, \pi f }\, \star \, U(f) \quad \Rightarrow \quad W(f) = -\frac{1}{\pi }\int\limits_{-\infty}^{+\infty} { \frac{U(\nu)}{f - \nu}}\hspace{0.1cm}{\rm d}\nu \hspace{0.05cm}.$$
However, since at the same time &nbsp; $u(t) = \sign(t) · w(t)$ &nbsp; also holds, the following is valid in the same way:
+
However,&nbsp; since at the same time &nbsp; $u(t) = \sign(t) · w(t)$ &nbsp; also holds,&nbsp; the following is valid in the same way:
 
:$$U(f) =  \frac{1}{{\rm j} \, \pi f }\, \star \, {\rm j} \cdot W(f) \quad \Rightarrow \quad U(f) = \frac{1}{\pi }\int\limits_{-\infty}^{+\infty} { \frac{W(\nu)}{f - \nu}}\hspace{0.1cm}{\rm d}\nu \hspace{0.05cm}.$$
 
:$$U(f) =  \frac{1}{{\rm j} \, \pi f }\, \star \, {\rm j} \cdot W(f) \quad \Rightarrow \quad U(f) = \frac{1}{\pi }\int\limits_{-\infty}^{+\infty} { \frac{W(\nu)}{f - \nu}}\hspace{0.1cm}{\rm d}\nu \hspace{0.05cm}.$$
  
These ''integral transformations''&nbsp; are named after their discoverer&nbsp; [https://en.wikipedia.org/wiki/David_Hilbert David Hilbert]&nbsp;.   
+
These&nbsp; "integral transformations"&nbsp; are named after their discoverer&nbsp; [https://en.wikipedia.org/wiki/David_Hilbert David Hilbert].   
  
 
{{BlaueBox|TEXT=   
 
{{BlaueBox|TEXT=   
$\text{Definition:}$&nbsp; Both variants of the&nbsp; '''Hilbert transformation'''&nbsp; will be denoted by the following abbreviations in the further course:
+
$\text{Definitions:}$&nbsp; Both variants of the&nbsp; '''Hilbert transformation'''&nbsp; will be denoted by the following abbreviations in the further course:
 
:$$W(f)  \quad \bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow\quad U(f) \hspace{0.8cm}{\rm or}\hspace{0.8cm}W(f)= {\cal H}\left\{U(f) \right \}\hspace{0.05cm}.$$
 
:$$W(f)  \quad \bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow\quad U(f) \hspace{0.8cm}{\rm or}\hspace{0.8cm}W(f)= {\cal H}\left\{U(f) \right \}\hspace{0.05cm}.$$
*To calculate the spectrum marked by the arrowhead – here &nbsp;$U(f)$&nbsp; – the equation with the positive sign is taken from the two otherwise identical upper equations:
+
*To calculate the spectrum marked by the arrowhead&nbsp; &nbsp; here &nbsp;$U(f)$ &nbsp; –&nbsp; the equation with the positive sign is taken from the two otherwise identical upper equations:
 
:$$U(f) = \frac{1}{\pi }\int\limits_{-\infty}^{+\infty} { \frac{W(\nu)}{f - \nu} }\hspace{0.1cm}{\rm d}\nu \hspace{0.05cm}.$$  
 
:$$U(f) = \frac{1}{\pi }\int\limits_{-\infty}^{+\infty} { \frac{W(\nu)}{f - \nu} }\hspace{0.1cm}{\rm d}\nu \hspace{0.05cm}.$$  
*The spectrum marked by the circle – here &nbsp;$W(f)$&nbsp; – arises as a result from the equation with the negative sign:  
+
*The spectrum marked by the circle&nbsp; &nbsp; here &nbsp;$W(f)$ &nbsp; –&nbsp; arises as a result from the equation with the negative sign:  
 
:$$
 
:$$
 
W(f) = -\frac{1}{\pi }\int\limits_{-\infty}^{+\infty} { \frac{U(\nu)}{f - \nu} }\hspace{0.1cm}{\rm d}\nu \hspace{0.05cm}.$$}}
 
W(f) = -\frac{1}{\pi }\int\limits_{-\infty}^{+\infty} { \frac{U(\nu)}{f - \nu} }\hspace{0.1cm}{\rm d}\nu \hspace{0.05cm}.$$}}
  
  
Applying the Hilbert transformation twice yields the original function with a change of sign, and applying it four times yields the original function including the correct sign:  
+
Applying the Hilbert transformation twice yields the original function with a change of sign,&nbsp; and applying it four times yields the original function including the correct sign:  
:$${\cal H}\left\{ {\cal H}\left\{ U(f) \right \} \right \} = -U(f), \hspace{0.2cm}  {\cal H}\left\{ {\cal H}\left\{ {\cal H}\left\{ {\cal H}\left\{ U(f) \right \} \right \} \right \} \right \}= U(f)\hspace{0.05cm}.$$
+
:$${\cal H}\left\{ {\cal H}\left\{ U(f) \right \} \right \} = -U(f),$$
 +
:$${\cal H}\left\{ {\cal H}\left\{ {\cal H}\left\{ {\cal H}\left\{ U(f) \right \} \right \} \right \} \right \}= U(f)\hspace{0.05cm}.$$
  
 
{{GraueBox|TEXT=   
 
{{GraueBox|TEXT=   
 
$\text{Example 3:}$&nbsp;  
 
$\text{Example 3:}$&nbsp;  
In&nbsp; [Mar94]<ref name ='Mar94'>Marko, H.: ''Methoden der Systemtheorie.'' 3. Auflage. Berlin – Heidelberg: Springer, 1994.</ref>&nbsp; the following Hilbert correspondence can be found:
+
In&nbsp; [Mar94]<ref name ='Mar94'>Marko, H.:&nbsp; Methoden der Systemtheorie.&nbsp; 3. Auflage. Berlin – Heidelberg: Springer, 1994.</ref>&nbsp; the following Hilbert correspondence can be found:
 
:$$\frac{1}{1+x^2}  \quad \bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow\quad \frac{x}{1+x^2}\hspace{0.05cm}.$$
 
:$$\frac{1}{1+x^2}  \quad \bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow\quad \frac{x}{1+x^2}\hspace{0.05cm}.$$
*Here &nbsp;$x$&nbsp; is representative of a suitably normalized time or frequency variable.  
+
*Here, &nbsp;$x$&nbsp; is representative of a suitably normalized time or frequency variable.  
*For example, if we use &nbsp;$x = f/f_{\rm G}$&nbsp; as a normalized frequency variable, then we obtain the correspondence:
+
*For example,&nbsp; if we use &nbsp;$x = f/f_{\rm G}$&nbsp; as a normalized frequency variable,&nbsp; then we obtain the correspondence:
 
:$$\frac{1}{1+(f/f_{\rm G})^2}  \quad \bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow\quad \frac{f/f_{\rm G} }{1+(f/f_{\rm G})^2}\hspace{0.05cm}.$$
 
:$$\frac{1}{1+(f/f_{\rm G})^2}  \quad \bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow\quad \frac{f/f_{\rm G} }{1+(f/f_{\rm G})^2}\hspace{0.05cm}.$$
 
Based on the equation
 
Based on the equation
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==Some pairs of Hilbert correspondences==
 
==Some pairs of Hilbert correspondences==
 
<br>
 
<br>
Zur Herleitung von Hilbert–Korrespondenzen geht man sehr pragmatisch vor, nämlich wie folgt:  
+
A very pragmatic way is followed to derive Hilbert correspondences, namely as follows:
*The &nbsp;[[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function#Definition of the Laplace transformation|Laplace transform]]&nbsp;  $Y_{\rm L}(p)$&nbsp; of the function &nbsp;$y(t)$ is computed as described below. This is already implicitly causal.  
+
*The spectral function &nbsp;$Y_{\rm L}(p)$&nbsp; is converted into the associated Fourier spectrum &nbsp;$Y(f)$&nbsp; which is split it into real and imaginary parts.&nbsp; To do this, the variable &nbsp;$p$&nbsp; is replaced by &nbsp;${\rm j \cdot 2}πf.$  
+
*The &nbsp;[[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function#Definition of the Laplace transformation|Laplace transform]]&nbsp;  $Y_{\rm L}(p)$&nbsp; of the function &nbsp;$y(t)$ is computed as described in chapter&nbsp; [[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function|Laplace Transform and p-Transfer Function]].&nbsp; This is already implicitly causal.  
*The real&ndash; and imaginary parts &ndash; so &nbsp;${\rm Re} \{Y(f)\}$&nbsp; and &nbsp;${\rm Im} \{Y(f)\}$ &ndash; are thus a pair of Hilbert transforms. Furthermore,
+
*$Y_{\rm L}(p)$&nbsp; is converted into the associated Fourier spectrum &nbsp;$Y(f)$&nbsp; which is split it into real and imaginary parts.&nbsp; To do this, the variable &nbsp;$p$&nbsp; is replaced by &nbsp;${\rm j \cdot 2}πf.$
**the frequency variable &nbsp;$f$&nbsp; is substituted by &nbsp;$x$,
 
** ${\rm Re} \{Y(f)\}$&nbsp; by &nbsp;$g(x)$, and
 
** ${\rm Im} \{Y(f)\}$&nbsp; by  &nbsp;${\cal H} \{g(x)\}$.
 
  
*The new variable &nbsp;$x$&nbsp; can describe both a (suitably) normalized frequency or a (suitably) normalized time. Hence, the&nbsp; [[Signal_Representation/Analytical_Signal_and_Its_Spectral_Function#Darstellung_mit_der_Hilberttransformation|Hilbert transformation]]&nbsp; is applicable to various problems.
+
[[File:EN_LZI_T_3_1_S4.png|right|frame|Table with Hilbert correspondences|class=fit]]  
  
  
[[File:EN_LZI_T_3_1_S4.png|center|frame|Table with Hilbert correspondences|class=fit]]
+
The real and imaginary parts &ndash; so &nbsp;${\rm Re} \{Y(f)\}$&nbsp; and &nbsp;${\rm Im} \{Y(f)\}$ &ndash; are thus a pair of Hilbert transforms. Furthermore,
 +
#&nbsp; the frequency variable &nbsp;$f$&nbsp; is substituted by &nbsp;$x$,
 +
#&nbsp;  the real part&nbsp; ${\rm Re} \{Y(f)\}$&nbsp; by &nbsp;$g(x)$,&nbsp; and
 +
#&nbsp;  the imaginary part&nbsp; ${\rm Im} \{Y(f)\}$&nbsp; by  &nbsp;${\cal H} \{g(x)\}$.  
  
Die Tabelle zeigt einige solcher Hilbertpaare. &nbsp; Auf die Vorzeichen wurde verzichtet, so dass beide Richtungen gültig sind.
 
  
 +
The new variable &nbsp;$x$&nbsp; can describe both a&nbsp; (suitably)&nbsp; normalized frequency or a&nbsp; (suitably)&nbsp; normalized time. Hence, the&nbsp; [[Signal_Representation/Analytical_Signal_and_Its_Spectral_Function#Representation with Hilbert transform|Hilbert transformation]]&nbsp; is applicable to various problems.
 +
 +
The table shows some of such Hilbert pairs. The signs have been omitted so that both directions are valid.
 +
<br clear=all>
 
{{GraueBox|TEXT=   
 
{{GraueBox|TEXT=   
$\text{Example 4:}$&nbsp; Gilt beispielsweise &nbsp;${\cal H} \{g(x)\} = f(x)$, so folgt daraus auch &nbsp;${\cal H} \{f(x)\} = \, –g(x)$. Insbesondere gilt auch:
+
$\text{Example 4:}$&nbsp; For example,&nbsp; if &nbsp;${\cal H} \{g(x)\} = f(x)$&nbsp; holds,&nbsp; then from this it also follows that &nbsp;${\cal H} \{f(x)\} = \, –g(x)$.&nbsp; In particular, it also holds:
 
:$${\cal H}\left \{ \cos(x) \right\} = \sin(x)\hspace{0.3cm}\Rightarrow \hspace{0.3cm}  {\cal H}\left \{ \sin(x) \right\} = -\cos(x)\hspace{0.05cm}.$$}}
 
:$${\cal H}\left \{ \cos(x) \right\} = \sin(x)\hspace{0.3cm}\Rightarrow \hspace{0.3cm}  {\cal H}\left \{ \sin(x) \right\} = -\cos(x)\hspace{0.05cm}.$$}}
  
==Dämpfung und Phase von Minimum–Phasen–Systemen==
+
==Attenuation and phase of minimum-phase systems==
 
<br>
 
<br>
Eine wichtige Anwendung der Hilbert–Transformation stellt der Zusammenhang zwischen Dämpfung und Phase bei den so genannten ''Minimum–Phasen–Systemen''&nbsp; dar. Im Vorgriff auf das folgende Kapitel &nbsp;[[Linear_and_Time_Invariant_Systems/Laplace–Transformation_und_p–Übertragungsfunktion|Laplace–Transformation und p–Übertragungsfunktion]]&nbsp; sei erwähnt, dass diese Systeme in der rechten&nbsp; $p$–Halbebene weder Pole noch Nullstellen aufweisen dürfen.  
+
An important application of the Hilbert transformation is the relationship between attenuation and phase in the so-called&nbsp; '''minimum-phase systems'''.&nbsp; In anticipation of the following chapter &nbsp;[[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function|Laplace Transform and p-Transfer Function]],&nbsp; it should be mentioned that these systems may have neither poles nor zeros in the right&nbsp; $p$–half plane.  
  
Allgemein gilt für die Übertragungsfunktion &nbsp;$H(f)$&nbsp; mit dem &nbsp;[[Linear_and_Time_Invariant_Systems/Einige_Ergebnisse_der_Leitungstheorie#Ersatzschaltbild_eines_kurzen_Leitungsabschnitts|komplexen Übertragungsmaß]]&nbsp; $g(f)$&nbsp; sowie der Dämpfungsfunktion &nbsp;$a(f)$&nbsp; und der Phasenfunktion &nbsp;$b(f)$:
+
In general,&nbsp; the following holds for the transfer function &nbsp;$H(f)$&nbsp; with the &nbsp;[[Linear_and_Time_Invariant_Systems/Some_Results_from_Transmission_Line_Theory#Equivalent circuit diagram of a short transmission line section|complex transmission function]]&nbsp; $g(f)$&nbsp; and the attenuation function &nbsp;$a(f)$&nbsp; as well as the phase function &nbsp;$b(f)$:
 
:$$H(f) = {\rm e}^{-g(f)} = {\rm e}^{-a(f)\hspace{0.05cm}- \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}b(f)}  \hspace{0.3cm}\Rightarrow \hspace{0.3cm}  g(f) = a(f)+ {\rm j} \cdot b(f)\hspace{0.05cm}.$$
 
:$$H(f) = {\rm e}^{-g(f)} = {\rm e}^{-a(f)\hspace{0.05cm}- \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}b(f)}  \hspace{0.3cm}\Rightarrow \hspace{0.3cm}  g(f) = a(f)+ {\rm j} \cdot b(f)\hspace{0.05cm}.$$
Bei den Minimum–Phasen–Systemen gilt nun aber nicht nur wie bei allen realisierbaren Systemen die Hilbert–Transformation
+
Now in the case of minimum-phase systems,&nbsp; not only does the Hilbert transformation hold
*bezüglich Imaginär– und Realteil &nbsp; &rArr; &nbsp; ${\rm Im} \left\{ H(f) \right \}  \, \bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow \, {\rm Re} \left\{ H(f) \right \}\hspace{0.01cm},$  
+
*regarding imaginary and real part as it does for all realizable systems &nbsp; &rArr; &nbsp; ${\rm Im} \left\{ H(f) \right \}  \, \bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow \, {\rm Re} \left\{ H(f) \right \}\hspace{0.01cm},$  
*sondern zusätzlich auch noch die Hilbert–Korrespondenz zwischen der Phasen– und der Dämpfungsfunktion &nbsp; &rArr; &nbsp; $b(f)  \, \bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow \, a(f)\hspace{0.05cm}.$
+
*but additionally also the Hilbert correspondence between the phase and attenuation functions is valid &nbsp; &rArr; &nbsp; $b(f)  \, \bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow \, a(f)\hspace{0.05cm}.$
  
  
 
{{GraueBox|TEXT=   
 
{{GraueBox|TEXT=   
$\text{Example 5:}$&nbsp;  
+
$\text{Example 5:}$ &nbsp;  
Ein Tiefpass besitze im Durchlassbereich &ndash; also für &nbsp;$\vert f \vert < f_{\rm G}$ &ndash; den Frequenzgang &nbsp;$H(f) = 1$ &nbsp; ⇒  &nbsp; $a(f) =0 \ {\rm Np}$, während für größere Frequenzen die Dämpfungsfunktion &nbsp;$a(f)$&nbsp; den konstanten Wert &nbsp;$a_{\rm S}$ (in Neper) besitzt.  
+
A low-pass filter has the frequency response &nbsp;$H(f) = 1$ &nbsp; ⇒  &nbsp; $a(f) =0$&nbsp; Np in the&nbsp; "pass band" &nbsp; &rArr; &nbsp; &nbsp;$\vert f \vert < f_{\rm G}$,&nbsp; while for higher frequencies the attenuation function &nbsp;$a(f)$&nbsp; has the constant value &nbsp;$a_{\rm S}$&nbsp; (in Neper).  
 
+
[[File:P_ID1753__LZI_T_3_1_S5_neu.png|right|frame|Attenuation and phase functions of an exemplary minimum-phase low-pass filter|class=fit]]
In diesem Sperrbereich ist &nbsp;$H(f) = {\rm e}^{–a_{\rm S} }$&nbsp; zwar sehr klein, aber nicht Null.
+
*In this&nbsp; "stop band"&nbsp; &rArr; &nbsp; &nbsp;$\vert f \vert > f_{\rm G}$,&nbsp; &nbsp;$H(f) = {\rm e}^{–a_{\rm S} }$&nbsp; is very small but not zero.  
 
+
*If the low-pass filter is to be causal and thus realizable, then the phase function &nbsp;$b(f)$&nbsp; must be equal to the Hilbert transform of the attenuation &nbsp;$a(f)$&nbsp;.  
[[File:P_ID1753__LZI_T_3_1_S5_neu.png|center|frame|Dämpfungs&ndash; und Phasenfunktion eines beispielhaften Minimum–Phasen–Tiefpasses|class=fit]]
+
*Since the Hilbert transform of a constant is equal to zero, the function &nbsp;$a(f) - a_{\rm S}$&nbsp; can be assumed in the same way.  
 
+
*This function shown dashed in the graph is&nbsp;  (negative)&nbsp; rectangular between &nbsp;$±f_{\rm G}$.&nbsp; According to the &nbsp;[[Linear_and_Time_Invariant_Systems/Conclusions_from_the_Allocation_Theorem#Some pairs of Hilbert correspondences|table]]&nbsp; on the last page the following thus holds:
*Soll der Tiefpass kausal und damit realisierbar sein, so muss die Phasenfunktion &nbsp;$b(f)$&nbsp; gleich der Hilbert–Transformierten der Dämpfung &nbsp;$a(f)$&nbsp; sein.  
 
* Da die Hilbert–Transformierte einer Konstanten gleich Null ist, kann in gleicher Weise von der Funktion &nbsp;$a(f) - a_{\rm S}$&nbsp; ausgegangen werden.  
 
 
 
 
 
Diese in der Grafik gestrichelt eingezeichnete Funktion ist zwischen &nbsp;$±f_{\rm G}$&nbsp; (negativ) rechteckförmig. Entsprechend der &nbsp;[[Linear_and_Time_Invariant_Systems/Folgerungen_aus_dem_Zuordnungssatz#Einige_Paare_von_Hilbert.E2.80.93Korrespondenzen|Tabelle]]&nbsp; auf der letzten Seite gilt deshalb:
 
 
:$$b(f)  = {a_{\rm S} }/{\pi} \cdot {\rm ln}\hspace{0.1cm}\left\vert \frac{f+f_{\rm G} }{f-f_{\rm  G} }\right \vert \hspace{0.05cm}.$$
 
:$$b(f)  = {a_{\rm S} }/{\pi} \cdot {\rm ln}\hspace{0.1cm}\left\vert \frac{f+f_{\rm G} }{f-f_{\rm  G} }\right \vert \hspace{0.05cm}.$$
Jeder andere Phasenverlauf würde dagegen zu einer akausalen Impulsantwort führen.}}
+
In contrast, any other phase response would result in a non-causal impulse response.}}
  
 
==Exercises for the chapter==
 
==Exercises for the chapter==
  
[[Aufgaben:3.1_Kausalitätsbetrachtungen|Exercise 3.1: Kausalitätsbetrachtungen]]
+
[[Aufgaben:Exercise_3.1:_Causality_Considerations|Exercise 3.1: Causality Considerations]]
  
[[Aufgaben:3.1Z_Hilbert-Transformierte|Exercise 3.1Z: Hilbert-Transformierte]]
+
[[Aufgaben:Exercise_3.1Z:_Hilbert_Transform|Exercise 3.1Z: Hilbert Transform]]
  
  
==List of sources==
+
==References==
 
<references/>
 
<references/>
  
 
{{Display}}
 
{{Display}}

Revision as of 12:39, 17 February 2022

# OVERVIEW OF THE THIRD MAIN CHAPTER #


In the first two chapters,  filter functions with real-valued frequency responses were mostly considered for reasons of presentation so that the associated time function is symmetric about zero-time.  However,  the impulse response of a realizable system must always be causal,  that is,  $h(t)$  must be identical to zero for  $t < 0$.  This strong asymmetry of the time function  $h(t)$  implies at the same time that the frequency response  $H(f)$  of a realizable system is always complex-valued with the exception of  $H(f) = K$  where there is a fixed relation between its real part and imaginary part.

This third chapter provides a recapitulatory account of the description of causal realizable systems,  which differ also in the mathematical methods from those commonly used with non-causal systems.

In detail, the following is dealt with:

  • the Hilbert transform,  which states how real and imaginary parts of  $H(f)$  are related,
  • the Laplace transform,  which yields another spectral function  $H_{\rm L}(p)$  for causal  $h(t)$,
  • the description of realizable systems by the pole-zero plot,  as well as
  • the inverse Laplace transform using the  "theory of functions"  ("residue theorem").


For this chapter, we recommend two of our multimedia offerings:


Prerequisites for the entire third main chapter


In the first two chapters,  mostly real  transfer functions   $H(f)$  were considered for which the associated impulse response  $h(t)$  is consequently always symmetric with respect to the reference time  $t = 0$.  Such transfer functions

  • are suitable to explain basic relationships in a simple way,
  • but unfortunately are not realizable for reasons of causality.


This becomes clear if the definition of the impulse response is considered:

$\text{Definition:}$  The  impulse response  $h(t)$  is equal to the output signal  $y(t)$  of the system if an infinitely short impulse with an infinitely large amplitude is applied to the input at time  $t = 0$ :   $x(t) = δ(t)$.  Such an impulse is called a  Dirac delta impulse.


It is obvious that no impulse response can be realized for which  $h(t < 0) ≠ 0$  holds.

$\text{Definition:}$  For a  causal system  the impulse response  $h(t)$  is identical to zero for all times  $t < 0$.


The only real transfer function that satisfies the causality condition  "the output signal cannot start before the input signal"  is:

$$H(f) = K \quad \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\quad h(t) = K \cdot \delta(t).$$

All other real-valued transfer functions  $H(f)$  describe non-causal systems and thus cannot be realized by an (electrical) circuit network.

$\text{In other words:}$   Except for the transfer function  $H(f) = K,$  any realistic transfer function is complex.

  • If  $K=1$  holds additionally,  the transfer function is said to be ideal. 
  • Then,  the output signal  $y(t)$  is identical to the input signal  $x(t)$  – even without attenuation or amplification.

Real and imaginary part of a causal transfer function


Any causal impulse response  $h(t)$  can be represented as the sum

  • of an even  (German:  "gerade"   ⇒   "g")  part  $h_{\rm g}(t)$ 
  • and an odd  (German:  "ungerade"   ⇒   "u")  part  $h_{\rm u}(t)$:
$$\begin{align*} h_{ {\rm g}}(t) & = {1}/{2}\cdot \big[ h(t) + h(-t) \big]\hspace{0.05cm},\\ h_{ {\rm u}}(t) & = {1}/{2}\cdot \big[ h(t) - h(-t) \big] = h_{ {\rm g}}(t) \cdot {\rm sign}(t)\hspace{0.05cm} .\end{align*}$$

Here, the so-called  sign function  is used:

$${\rm sign}(t) = \left\{ \begin{array}{c} -1 \\ +1 \\ \end{array} \right.\quad \quad \begin{array}{c} {\rm{for}} \\ {\rm{for}} \\ \end{array}\begin{array}{*{20}c} { t < 0,} \\ { t > 0.} \\ \end{array}$$


$\text{Example 1:}$  The graph shows this splitting for a causal exponentially decreasing impulse response of a low-pass filter of first-order corresponding to  Exercise 1.3Z:

Splitting of the impulse response into an even part and an odd part
$$h(t) = \left\{ \begin{array}{c} 0 \\ 0.5/T \\ 1/T \cdot {\rm e}^{-t/T} \end{array} \right.\quad \begin{array}{c} {\rm{for} } \\ {\rm{for} } \\ {\rm{for} } \end{array}\begin{array}{*{20}c} { t < 0\hspace{0.05cm},} \\ { t = 0\hspace{0.05cm},} \\{ t > 0\hspace{0.05cm}.} \end{array}$$

It can be seen that:

  • $h_{\rm g}(t) = h_{\rm u}(t) = h(t)/2$  holds for positive times,
  • $h_{\rm g}(t)$  and  $h_{\rm u}(t)$  differ only by the sign for negative times,
  • $h(t) = h_{\rm g}(t) + h_{\rm u}(t)$  holds for all times, also at time  $t = 0$  (marked by circles).


Let us now consider the same issue in the spectral domain.  According to the  Assignment Theorem  the following holds for the complex transfer function:  

$$H(f) = {\rm Re} \left\{ H(f) \right \} + {\rm j} \cdot {\rm Im} \left\{ H(f) \right \} ,$$

where the following assignment is valid:

$${\rm Re} \left\{ H(f) \right \} \quad \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\quad h_{ {\rm g}}(t)\hspace{0.05cm},$$
$${\rm j} \cdot {\rm Im} \left\{ H(f) \right\} \quad \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\quad h_{ {\rm u}}(t)\hspace{0.05cm}.$$

First,  this relationship between the real part and the imaginary part of  $H(f)$  shall be worked out using another example.

$\text{Example 2:}$  A low-pass filter of first-order is assumed and the following holds for its transfer function:

$$H(f) = \frac{1}{1+{\rm j}\cdot f/f_{\rm G} } = \frac{1}{1+(f/f_{\rm G})^2}- {\rm j} \cdot \frac{f/f_{\rm G} }{1+(f/f_{\rm G})^2} \hspace{0.05cm}.$$
Frequency response of a low-pass filter of first-order

Here,  $f_{\rm G}$  represents the  $\rm 3dB$  cut-off frequency at which  $\vert H(f)\vert^2$  has decreased to half of its maximum  $($at  $f = 0)$.  The corresponding impulse response  $h(t)$  has already been shown in  $\text{Example 1}$  for  $f_{\rm G} = 1/(2πT)$.

The graph shows the real part (blue) and the imaginary part (red) of  $H(f)$.  In addition, the magnitude is shown dashed in green.

Since the time functions  $h_{\rm g}(t)$  and  $h_{\rm u}(t)$  are related by the sign function, there also exists a fixed relationship

  • between the real part   ⇒   ${\rm Re} \{H(f)\}$ 
  • and the imaginary part   ⇒  ${\rm Im} \{H(f)\}$ 


of the transfer function  $\{H(f)\}$   ⇒   Hilbert transform.

This is described below.

Hilbert transform


Here,  two time functions  $u(t)$  and  $w(t) = \sign(t) · u(t)$  are considered in the most general sense.

  • The associated spectral functions are denoted by  $U(f)$  and  ${\rm j} · W(f)$.
  • That is:   In this section  ${w(t) \, \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet \, {\rm j} \cdot W(f) }$  is valid and not the usual Fourier correspondence  ${w(t) \, \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet \, W(f)}.$


Using the correspondence   ${\rm sign}(t) \, \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet \, {1}/({{\rm j} \, \pi f })$   the following is obtained after writing the convolution integral out in full with the integration variable  $ν$ :

$${\rm j} \cdot W(f) = \frac{1}{{\rm j} \, \pi f }\, \star \, U(f) \quad \Rightarrow \quad W(f) = -\frac{1}{\pi }\int\limits_{-\infty}^{+\infty} { \frac{U(\nu)}{f - \nu}}\hspace{0.1cm}{\rm d}\nu \hspace{0.05cm}.$$

However,  since at the same time   $u(t) = \sign(t) · w(t)$   also holds,  the following is valid in the same way:

$$U(f) = \frac{1}{{\rm j} \, \pi f }\, \star \, {\rm j} \cdot W(f) \quad \Rightarrow \quad U(f) = \frac{1}{\pi }\int\limits_{-\infty}^{+\infty} { \frac{W(\nu)}{f - \nu}}\hspace{0.1cm}{\rm d}\nu \hspace{0.05cm}.$$

These  "integral transformations"  are named after their discoverer  David Hilbert.

$\text{Definitions:}$  Both variants of the  Hilbert transformation  will be denoted by the following abbreviations in the further course:

$$W(f) \quad \bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow\quad U(f) \hspace{0.8cm}{\rm or}\hspace{0.8cm}W(f)= {\cal H}\left\{U(f) \right \}\hspace{0.05cm}.$$
  • To calculate the spectrum marked by the arrowhead  –  here  $U(f)$   –  the equation with the positive sign is taken from the two otherwise identical upper equations:
$$U(f) = \frac{1}{\pi }\int\limits_{-\infty}^{+\infty} { \frac{W(\nu)}{f - \nu} }\hspace{0.1cm}{\rm d}\nu \hspace{0.05cm}.$$
  • The spectrum marked by the circle  –  here  $W(f)$   –  arises as a result from the equation with the negative sign:
$$ W(f) = -\frac{1}{\pi }\int\limits_{-\infty}^{+\infty} { \frac{U(\nu)}{f - \nu} }\hspace{0.1cm}{\rm d}\nu \hspace{0.05cm}.$$


Applying the Hilbert transformation twice yields the original function with a change of sign,  and applying it four times yields the original function including the correct sign:

$${\cal H}\left\{ {\cal H}\left\{ U(f) \right \} \right \} = -U(f),$$
$${\cal H}\left\{ {\cal H}\left\{ {\cal H}\left\{ {\cal H}\left\{ U(f) \right \} \right \} \right \} \right \}= U(f)\hspace{0.05cm}.$$

$\text{Example 3:}$  In  [Mar94][1]  the following Hilbert correspondence can be found:

$$\frac{1}{1+x^2} \quad \bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow\quad \frac{x}{1+x^2}\hspace{0.05cm}.$$
  • Here,  $x$  is representative of a suitably normalized time or frequency variable.
  • For example,  if we use  $x = f/f_{\rm G}$  as a normalized frequency variable,  then we obtain the correspondence:
$$\frac{1}{1+(f/f_{\rm G})^2} \quad \bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow\quad \frac{f/f_{\rm G} }{1+(f/f_{\rm G})^2}\hspace{0.05cm}.$$

Based on the equation

$${\rm Im} \left\{ H(f) \right \} \quad \bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow\quad {\rm Re} \left\{ H(f) \right \}$$

the result found in  $\text{Example 2}$  is thus confirmed:

$${\rm Im} \left\{ H(f) \right \} = \frac{-f/f_{\rm G} }{1+(f/f_{\rm G})^2}\hspace{0.05cm}.$$

Some pairs of Hilbert correspondences


A very pragmatic way is followed to derive Hilbert correspondences, namely as follows:

  • The  Laplace transform  $Y_{\rm L}(p)$  of the function  $y(t)$ is computed as described in chapter  Laplace Transform and p-Transfer Function.  This is already implicitly causal.
  • $Y_{\rm L}(p)$  is converted into the associated Fourier spectrum  $Y(f)$  which is split it into real and imaginary parts.  To do this, the variable  $p$  is replaced by  ${\rm j \cdot 2}πf.$
Table with Hilbert correspondences


The real and imaginary parts – so  ${\rm Re} \{Y(f)\}$  and  ${\rm Im} \{Y(f)\}$ – are thus a pair of Hilbert transforms. Furthermore,

  1.   the frequency variable  $f$  is substituted by  $x$,
  2.   the real part  ${\rm Re} \{Y(f)\}$  by  $g(x)$,  and
  3.   the imaginary part  ${\rm Im} \{Y(f)\}$  by  ${\cal H} \{g(x)\}$.


The new variable  $x$  can describe both a  (suitably)  normalized frequency or a  (suitably)  normalized time. Hence, the  Hilbert transformation  is applicable to various problems.

The table shows some of such Hilbert pairs. The signs have been omitted so that both directions are valid.

$\text{Example 4:}$  For example,  if  ${\cal H} \{g(x)\} = f(x)$  holds,  then from this it also follows that  ${\cal H} \{f(x)\} = \, –g(x)$.  In particular, it also holds:

$${\cal H}\left \{ \cos(x) \right\} = \sin(x)\hspace{0.3cm}\Rightarrow \hspace{0.3cm} {\cal H}\left \{ \sin(x) \right\} = -\cos(x)\hspace{0.05cm}.$$

Attenuation and phase of minimum-phase systems


An important application of the Hilbert transformation is the relationship between attenuation and phase in the so-called  minimum-phase systems.  In anticipation of the following chapter  Laplace Transform and p-Transfer Function,  it should be mentioned that these systems may have neither poles nor zeros in the right  $p$–half plane.

In general,  the following holds for the transfer function  $H(f)$  with the  complex transmission function  $g(f)$  and the attenuation function  $a(f)$  as well as the phase function  $b(f)$:

$$H(f) = {\rm e}^{-g(f)} = {\rm e}^{-a(f)\hspace{0.05cm}- \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}b(f)} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} g(f) = a(f)+ {\rm j} \cdot b(f)\hspace{0.05cm}.$$

Now in the case of minimum-phase systems,  not only does the Hilbert transformation hold

  • regarding imaginary and real part as it does for all realizable systems   ⇒   ${\rm Im} \left\{ H(f) \right \} \, \bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow \, {\rm Re} \left\{ H(f) \right \}\hspace{0.01cm},$
  • but additionally also the Hilbert correspondence between the phase and attenuation functions is valid   ⇒   $b(f) \, \bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow \, a(f)\hspace{0.05cm}.$


$\text{Example 5:}$   A low-pass filter has the frequency response  $H(f) = 1$   ⇒   $a(f) =0$  Np in the  "pass band"   ⇒    $\vert f \vert < f_{\rm G}$,  while for higher frequencies the attenuation function  $a(f)$  has the constant value  $a_{\rm S}$  (in Neper).

Attenuation and phase functions of an exemplary minimum-phase low-pass filter
  • In this  "stop band"  ⇒    $\vert f \vert > f_{\rm G}$,   $H(f) = {\rm e}^{–a_{\rm S} }$  is very small but not zero.
  • If the low-pass filter is to be causal and thus realizable, then the phase function  $b(f)$  must be equal to the Hilbert transform of the attenuation  $a(f)$ .
  • Since the Hilbert transform of a constant is equal to zero, the function  $a(f) - a_{\rm S}$  can be assumed in the same way.
  • This function shown dashed in the graph is  (negative)  rectangular between  $±f_{\rm G}$.  According to the  table  on the last page the following thus holds:
$$b(f) = {a_{\rm S} }/{\pi} \cdot {\rm ln}\hspace{0.1cm}\left\vert \frac{f+f_{\rm G} }{f-f_{\rm G} }\right \vert \hspace{0.05cm}.$$

In contrast, any other phase response would result in a non-causal impulse response.

Exercises for the chapter

Exercise 3.1: Causality Considerations

Exercise 3.1Z: Hilbert Transform


References

  1. Marko, H.:  Methoden der Systemtheorie.  3. Auflage. Berlin – Heidelberg: Springer, 1994.