Conclusions from the Allocation Theorem

# 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:

the (German language) learning video "Rechnen mit komplexen Zahlen" ⇒ "Arithmetic operations involving complex numbers",
the (German language) interactive SWF applet "Kausale Systeme - Laplacetransformation" ⇒ "Causal systems – Laplace transform".

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 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.
The output $y(t)$ is then identical to the input $x(t)$ – 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 part $h_{\rm g}(t)$ and an odd 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:

$$h(t) = \left\{ \begin{array}{c} 0 \\ 0.5/T \\ 1/T \cdot {\rm e}^{-t/T} \end{array} \right.\quad \begin{array}{c} {\rm{for} } \\ {\rm{for} } \\ {\rm{for} } \end{array}\begin{array}{*{20}c} { t < 0\hspace{0.05cm},} \\ { t = 0\hspace{0.05cm},} \\{ t > 0\hspace{0.05cm}.} \end{array}$$

Splitting of the impulse response into an even and an odd part

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 and imaginary parts 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:

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

Here, $f_{\rm G}$ represents the 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 the above $\text{Example 1}$ 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.

Frequency response of a low-pass filter of first-order (real and imaginary part)

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

⇒ the Hilbert transformation.

This is described below.

Hilbert transformation

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 .

$\text{Definition:}$ 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:

$${\cal H}\left\{ {\cal H}\left\{ U(f) \right \} \right \} = -U(f), \hspace{0.2cm} {\cal H}\left\{ {\cal H}\left\{ {\cal H}\left\{ {\cal H}\left\{ U(f) \right \} \right \} \right \} \right \}= U(f)\hspace{0.05cm}.$$

$\text{Example 3:}$ 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 below. This is already implicitly causal.
The spectral function $Y_{\rm L}(p)$ 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 ${\rm j \cdot 2}πf.$
The real and imaginary parts – so ${\rm Re} \{Y(f)\}$ and ${\rm Im} \{Y(f)\}$ – are thus a pair of Hilbert transforms. Furthermore,
- the frequency variable $f$ is substituted by $x$,
- ${\rm Re} \{Y(f)\}$ by $g(x)$, and
- ${\rm Im} \{Y(f)\}$ 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.

Table with Hilbert correspondences

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 $H(f) = 1$ ⇒ $a(f) =0 \ {\rm Np}$ in the pass band – thus for $\vert f \vert < f_{\rm G}$ – while for higher frequencies the attenuation function $a(f)$ has the constant value $a_{\rm S}$ (in Neper).

In this stop band, $H(f) = {\rm e}^{–a_{\rm S} }$ is very small but not zero.

Attenuation and phase functions of an exemplary minimum-phase low-pass filter

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 rectangular between $±f_{\rm G}$ (negative). 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

List of sources

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

[Mar94-1] Marko, H.: Methoden der Systemtheorie. 3. Auflage. Berlin – Heidelberg: Springer, 1994.

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