Signal Representation/Analytical Signal and its Spectral Function

1. Signal Representation
2. Bandpass Signals
3. Analytical Signal and its Spectral Function

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Definition in the Frequency Domain

We consider a real band-pass signal  $x(t)$  with the corresponding band-pass spectrum  $X(f)$, which has an even real and an odd imaginary part with respect to the frequency zero point.  It is assumed that the carrier frequency  $f_{\rm T}$  is much larger than the bandwidth of the band-pass signal  $x(t)$.

$\text{Definition:}$  The  $\text{analytical signal}$  $x_+(t)$  belonging to the physical signal  $x(t)$  is that time function, whose spectrum fulfills the following property:

Analytical signal in the frequency domain

$$X_+(f)=\big[1+{\rm sign}(f)\big] \cdot X(f) = \left\{ {2 \cdot X(f) \; \hspace{0.2cm}\rm for\hspace{0.2cm} {\it f} > 0, \atop {\,\,\,\, \rm 0 \; \hspace{0.9cm}\rm for\hspace{0.2cm} {\it f} < 0.} }\right.$$

The  $\text{signum function}$  is for positive $f$–values equal to  $+1$  and for negative  $f$-values equal to  $-1$.

• The (double sided) limit value returns  $\sign(0) = 0$.
• The index "+" should make clear that  $X_+(f)$  has only parts at positive frequencies.

From the graphic you can see the calculation rule for  $X_+(f)$:

The actual band-pass spectrum  $X(f)$  will

• be doubled at the positive frequencies, and
• set to zero at the negative frequencies.

Spectrum  $X(f)$  and Spectrum  $X_{+}(f)$  of the analytical signal

$\text{Example 1:}$ 

The graphic

• on the left shows the (discrete and complex) spectrum  $X(f)$  of the  "physical band-pass signal"

$$x(t) = 4\hspace{0.05cm}{\rm V} \cdot {\cos} ( 2 \pi f_{\rm u} \hspace{0.03cm}t) + 6\hspace{0.05cm}{\rm V} \cdot {\sin} ( 2 \pi f_{\rm o} \hspace{0.03cm}t),$$

• on the right the (also discrete and complex) spectrum  $X_{+}(f)$  of the corresponding  "analytical signal"  $x_{+}(t)$.

General calculation rule in the time domain

Now we will take a closer look at the spectrum  $X_+(f)$  of the analytical signal and divide it with respect to  $f = 0$  into

For a clear explanation of the analytical signal

• an even  (German:  "gerade"   ⇒   $\rm g$)  part  $X_{\rm +g}(f)$,  and
• an odd   (German:  "ungerade"   ⇒   $\rm u$)  part  $X_{\rm +u}(f)$:

$$X_+(f) = X_{\rm +g}(f) + X_{\rm +u}(f).$$

All these spectra are generally complex.

If one considers the  Assignment Theorem  of the Fourier transform, then the following statements are possible on basis of the graphic:

• The even part  $X_{\rm +g}(f)$  of  $X_{+}(f)$  leads after the Fourier transform to a real time signal, and the odd part  $X_{\rm +u}(f)$  to an imaginary one.

• It is obvious that  $X_{\rm +g}(f)$  is equal to the physical Fourier spectrum  $X(f)$  and thus the real part of  $x_{\rm +g}(t)$  is equal to the given physical signal  $x(t)$  with band-pass properties.

• If we denote the imaginary part with  $y(t)$, the analytical signal is:

$$x_+(t)= x(t) + {\rm j} \cdot y(t) .$$

• According to the generally valid laws of Fourier transform corresponding to the  Assignment Theorem,  the following applies to the spectral function of the imaginary part:

$${\rm j} \cdot Y(f) = X_{\rm +u}(f)= {\rm sign}(f) \cdot X(f) \hspace{0.3cm}\Rightarrow\hspace{0.3cm}Y(f) = \frac{{\rm sign}(f)}{ {\rm j}}\cdot X(f).$$

• After transforming this equation into the time domain, the multiplication becomes the  "convolution",  and one gets:

$$y(t) = \frac{1}{ {\rm \pi} t} \hspace{0.05cm}\star \hspace{0.05cm}x(t) = \frac{1}{ {\rm \pi}} \cdot \hspace{0.03cm}\int_{-\infty}^{+\infty}\frac{x(\tau)}{ {t - \tau}}\hspace{0.15cm} {\rm d}\tau.$$

Representation with Hilbert transform

At this point it is necessary to briefly discuss a further spectral transformation, which is dealt thoroughly in the book Linear and Time Invariant Systems Systeme .

$\text{Definition:}$  FFor the  Hilbert transformed   $ {\rm H}\left\{x(t)\right\}$  a time function  $x(t)$  applies:

$$y(t) = {\rm H}\left\{x(t)\right\} = \frac{1}{ {\rm \pi} } \cdot \hspace{0.03cm}\int_{-\infty}^{+\infty}\frac{x(\tau)}{ {t - \tau} }\hspace{0.15cm} {\rm d}\tau.$$

• This particular integral cannot be solved in a simple, conventional way, but must be evaluated using the  principal value of Cauchy .

• Correspondingly valid in the frequency domain:

$$Y(f) = - {\rm j} \cdot {\rm sign}(f) \cdot X(f) \hspace{0.05cm} .$$

The result of the last page can be summarized with this definition as follows:

• You get from the real, physical band-pass signal  $x(t)$  the analytic signal  $x_+(t)$ by adding to  $x(t)$  an imaginary part according to the Hilbert transform:

$$x_+(t) = x(t)+{\rm j} \cdot {\rm H}\left\{x(t)\right\} .$$

• The Hilbert transformed  $\text{H}\{x(t)\}$  disappears only in the case of  $x(t) = \rm const.$   ⇒   DC signal With all other signal forms the analytic signal  $x_+(t)$  is therefore always complex.
• From the analytical signal  $x_+(t)$  the real band-pass signal can be easily determined by real part formation:

$$x(t) = {\rm Re}\left\{x_+(t)\right\} .$$

{{GraueBox|TEXT= $\text{Example 2:}$  The principle of the Hilbert transformation is illustrated again by the following diagram:

• • According to the left representation  $\rm (A)$  ,one gets an analytical signal  $x_+(t)$ from the physical signal  $x(t)$  by adding an imaginary part   ${\rm j} \cdot y(t)$ .
• Here   $y(t) = {\rm H}\left\{x(t)\right\}$  is a real time function, which can be calculated easily in the spectral range by multiplying the spectrum  $X(f)$  with  $- {\rm j} \cdot \sign(f)$ .

On the Illustration of the Hilbert Transformed

The right representation  $\rm (B)$  is equivalent to  $\rm (A)$:

• Now applies  $x_+(t) = x(t) + z(t)$  with the purely imaginary function  $z(t)$.
• A comparison of the two images shows that actually  $z(t) = {\rm j} \cdot y(t)$  is valid.}

Vector diagram representation of the harmonic oscillation

The spectral function  $X(f)$  of a harmonic oscillation  $x(t) = A \cdot \text{cos}(2\pi f_{\rm T}t - \varphi)$  consists of two Dirac functions at the frequencies

• $+f_{\rm T}$  with the complex weight   $A/2 \cdot \text{e}^{-\text{j}\hspace{0.05cm}\varphi}$,
• $-f_{\rm T}$  with the complex weight   $A/2 \cdot \text{e}^{+\text{j}\hspace{0.05cm}\varphi}$.

Thus, the spectrum of the analytical signal is  $($without the Dirac function at the frequency  $f =-f_{\rm T})$:

$$X_+(f) = A \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\varphi}\cdot\delta (f - f_{\rm T}) .$$

The corresponding time function is obtained by applying the  Shifting Theorem:

$$x_+(t) = A \cdot {\rm e}^{\hspace{0.05cm} {\rm j}\hspace{0.05cm}\cdot\hspace{0.05cm}( 2 \pi f_{\rm T} t \hspace{0.05cm}-\hspace{0.05cm} \varphi)}.$$

This equation describes a rotating pointer with constant angular velocity  $\omega_{\rm T} = 2\pi f_{\rm T}$ .

$\text{Example 3:}$  For illustrative reasons the coordinate system in the following figure is rotated to the left (real part up, imaginary part to the left), contrary to the usual representation by  $90^\circ$&nbsp.

Vector Diagram of a Harmonic Oscillation

Anhand dieser Grafik sind folgende Aussagen möglich: On the basis of this diagram the following statements are possible:

• At the start time  $t = 0$  the pointer of length  $A$  (signal amplitude) lies with angle  $-\varphi$  in the complex plane. In the drawn example,  $\varphi = 45^\circ$.
• For the times  $t > 0$  the pointer rotates with constant angular velocity (angular frequency)  $\omega_{\rm T}$  in mathematically positive direction, i.e. counterclockwise.
• The top of the pointer thus always lies on a circle with radius  $A$  and requires exactly the time  $T_0$, i.e. the period of the harmonic oscillation  $x(t)$ for one rotation.
• The projection of the analytical signal  $x_+(t)$  onto the real axis, marked by red dots, provides the instantaneous values of  $x(t)$.

Vector diagram of a sum of harmonic oscillations

For further description we assume the following spectrum for the analytical signal:

$$X_+(f) = \sum_{i=1}^{I}A_i \cdot {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot\hspace{0.05cm} \varphi_i}\cdot\delta (f - f_{i}) .$$

The left image shows such a spectrum for the example  $I = 3$. If one chooses  $I$  relatively large and the distance between adjacent spectral lines correspondingly small, then (frequency–) continuous spectral functions  $X_+(f)$  can also be approximated with the above equation.

Vector Diagram of a Sum of 3 Oscillations

In the right picture the corresponding time function is indicated. This is in general:

$$x_+(t) = \sum_{i=1}^{I}A_i \cdot {\rm e}^{ {\rm j}\hspace{0.05cm}\cdot\hspace{0.05cm}(\omega_i \hspace{0.05cm}\cdot\hspace{0.05cm} t \hspace{0.05cm}-\hspace{0.05cm} \varphi_i)}.$$

To note about this graphic:

• The sketch shows the initial position of the pointers at the start time  $t = 0$  corresponding to the amplitudes  $A_i$  and the phase positions  $\varphi_i$.
• The tip of the resulting pointer compound is marked by the violet cross. One obtains by vectorial addition of the three individual pointers for the time  $t = 0$:

$$x_+(t= 0) = \big [1 \cdot \cos(60^\circ) - 1 \cdot {\rm j} \cdot \sin(60^\circ) \big ]+ 2 \cdot \cos(0^\circ)+1 \cdot \cos(180^\circ) = 1.500 - {\rm j} \cdot 0.866.$$

• For times  $t > 0$  the three pointers rotate at different angular speeds  $\omega_i = 2\pi f_i$. The red hand rotates faster than the green hand, but slower than the blue hand.
• Since all hands rotate counterclockwise, the resulting hand  $x_+(t)$  will also tend to move in this direction. At time  $t = 1\,µ\text {s}$  the peak of the resulting pointer for the given parameter values is

$$ \begin{align*}x_+(t = 1 {\rm \hspace{0.05cm}µ s}) & = 1 \cdot {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}60^\circ}\cdot {\rm e}^{{\rm j}\hspace{0.05cm}2 \pi \hspace{0.05cm}\cdot \hspace{0.1cm}40 \hspace{0.05cm} \cdot \hspace{0.1cm} 0.001} + 2\cdot {\rm e}^{{\rm j}\hspace{0.05cm}2 \pi \hspace{0.05cm}\cdot \hspace{0.1cm}50 \hspace{0.05cm} \cdot \hspace{0.1cm} 0.001}- 1\cdot {\rm e}^{{\rm j}\hspace{0.05cm}2 \pi \hspace{0.05cm}\cdot \hspace{0.1cm}60 \hspace{0.05cm} \cdot \hspace{0.1cm} 0.001} = \\ & = 1 \cdot {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}45.6^\circ} + 2\cdot {\rm e}^{{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}18^\circ}- 1\cdot {\rm e}^{{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}21.6^\circ} \approx 1.673- {\rm j} \cdot 0.464.\end{align*}$$

• The resulting pointer tip does not lie on a circle like a single oscillation, but a complicated geometric figure is created.

The interactive applet  Physical Signal & Analytical Signal  illustrates  $x_+(t)$  for the sum of three harmonic oscillations.

Exercises for the chapter

Exercise 4.3: Vector Diagram Representation

Exercise 4.3Z: Hilbert Transformator

Exercise 4.4: Vector Diagram for DSB-AM

Exercise 4.4Z: Vector Diagram for DSB-AM