Difference between revisions of "Linear and Time Invariant Systems/Conclusions from the Allocation Theorem"

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{{BlaueBox|TEXT=   
 
{{BlaueBox|TEXT=   
 
$\text{Definition:}$   
 
$\text{Definition:}$   
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  [[Signal_Representation/Special_Cases_of_Impulse_Signals#Dirac_delta_or_impulse|Dirac delta]].}}
+
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  [[Signal_Representation/Special_Cases_of_Impulse_Signals#Dirac_delta_or_impulse|Dirac delta impulse]].}}
  
  
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<br>
 
<br>
 
A very pragmatic way is followed to derive Hilbert correspondences, namely as follows:
 
A very pragmatic way is followed to derive Hilbert correspondences, namely as follows:
[[File:EN_LZI_T_3_1_S4.png|right|frame|Table with Hilbert correspondences|class=fit]]
 
 
   
 
   
*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 &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 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;${\rm j \cdot 2}πf.$  
+
*$Y_{\rm L}(p)$&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 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,
 
**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#Representation with Hilbert transform|Hilbert transformation]]&nbsp; is applicable to various problems.
+
[[File:EN_LZI_T_3_1_S4.png|right|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;${\cal H} \{g(x)\}$.
 +
 +
 +
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.  
 
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; For example, if &nbsp;${\cal H} \{g(x)\} = f(x)$ holds, then from this it also follows that &nbsp;${\cal H} \{f(x)\} = \, –g(x)$. In particular, it also holds:
+
$\text{Example 4:}$&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:
 
:$${\cal H}\left \{ \cos(x) \right\} = \sin(x)\hspace{0.3cm}\Rightarrow \hspace{0.3cm}  {\cal H}\left \{ \sin(x) \right\} = -\cos(x)\hspace{0.05cm}.$$}}
 
:$${\cal H}\left \{ \cos(x) \right\} = \sin(x)\hspace{0.3cm}\Rightarrow \hspace{0.3cm}  {\cal H}\left \{ \sin(x) \right\} = -\cos(x)\hspace{0.05cm}.$$}}
  
 
==Attenuation and phase of minimum-phase systems==
 
==Attenuation and phase of minimum-phase systems==
 
<br>
 
<br>
An important application of the Hilbert transformation is the relationship between attenuation and phase in the so-called ''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.  
+
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.  
  
In general, 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;$b(f)$:
+
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;$b(f)$:
 
:$$H(f) = {\rm e}^{-g(f)} = {\rm e}^{-a(f)\hspace{0.05cm}- \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}b(f)}  \hspace{0.3cm}\Rightarrow \hspace{0.3cm}  g(f) = a(f)+ {\rm j} \cdot b(f)\hspace{0.05cm}.$$
 
:$$H(f) = {\rm e}^{-g(f)} = {\rm e}^{-a(f)\hspace{0.05cm}- \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}b(f)}  \hspace{0.3cm}\Rightarrow \hspace{0.3cm}  g(f) = a(f)+ {\rm j} \cdot b(f)\hspace{0.05cm}.$$
Now in the case of minimum-phase systems, not only does the Hilbert transformation hold  
+
Now in the case of minimum-phase systems,&nbsp; not only does the Hilbert transformation hold  
 
*regarding imaginary and real part as it does for all realizable systems &nbsp; &rArr; &nbsp; ${\rm Im} \left\{ H(f) \right \}  \, \bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow \, {\rm Re} \left\{ H(f) \right \}\hspace{0.01cm},$  
 
*regarding imaginary and real part as it does for all realizable systems &nbsp; &rArr; &nbsp; ${\rm Im} \left\{ H(f) \right \}  \, \bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow \, {\rm Re} \left\{ H(f) \right \}\hspace{0.01cm},$  
 
*but additionally also the Hilbert correspondence between the phase and attenuation functions is valid &nbsp; &rArr; &nbsp; $b(f)  \, \bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow \, a(f)\hspace{0.05cm}.$
 
*but additionally also the Hilbert correspondence between the phase and attenuation functions is valid &nbsp; &rArr; &nbsp; $b(f)  \, \bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow \, a(f)\hspace{0.05cm}.$
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{{GraueBox|TEXT=   
 
{{GraueBox|TEXT=   
$\text{Example 5:}$&nbsp;  
+
$\text{Example 5:}$ &nbsp;  
A low-pass filter has the frequency response &nbsp;$H(f) = 1$ &nbsp; ⇒  &nbsp; $a(f) =0 \ {\rm Np}$ in the pass band &ndash; thus for &nbsp;$\vert f \vert < f_{\rm G}$ &ndash; while for higher frequencies the attenuation function &nbsp;$a(f)$&nbsp; has the constant value &nbsp;$a_{\rm S}$ (in Neper).  
+
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 this stop band, &nbsp;$H(f) = {\rm e}^{–a_{\rm S} }$&nbsp; is very small but not zero.  
+
*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.  
 
 
[[File:P_ID1753__LZI_T_3_1_S5_neu.png|center|frame|Attenuation and phase functions of an exemplary minimum-phase low-pass filter|class=fit]]
 
 
 
 
*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;.  
 
*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;.  
 
*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.  
 
*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:
 
 
This function shown dashed in the graph is rectangular between &nbsp;$±f_{\rm G}$&nbsp; (negative). 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:
 
 
:$$b(f)  = {a_{\rm S} }/{\pi} \cdot {\rm ln}\hspace{0.1cm}\left\vert \frac{f+f_{\rm G} }{f-f_{\rm  G} }\right \vert \hspace{0.05cm}.$$
 
:$$b(f)  = {a_{\rm S} }/{\pi} \cdot {\rm ln}\hspace{0.1cm}\left\vert \frac{f+f_{\rm G} }{f-f_{\rm  G} }\right \vert \hspace{0.05cm}.$$
 
In contrast, any other phase response would result in a non-causal impulse response.}}
 
In contrast, any other phase response would result in a non-causal impulse response.}}
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==List of sources==
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==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.