Aufgaben:Exercise 1.4: "Pointer diagram" and "Locality Curve"

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A given analytical signal in the complex plane

The accompanying graph shows the analytical signal  $s_+(t)$  in the complex plane.

• The numbers shown in the rectangles indicate the time points in microseconds.
• For all multiples of  $5 \ \rm µ s$ ,  $s_+(t)$  is always real with the following values:

$$s_+(t = 0) =s_+(t = 50\;{\rm µ s})= 1.500\hspace{0.05cm},$$

$$s_+(t = 5\;{\rm µ s}) = s_+(t = 45\;{\rm µ s})= -1.405\hspace{0.05cm},$$

$$s_+(t = 10\;{\rm µ s}) = s_+(t = 40\;{\rm µ s})= 1.155\hspace{0.05cm},$$

$$\text{.....................................} $$

$$s_+(t = 25\;{\rm µ s}) = -0.500\hspace{0.05cm}.$$

It is assumed that the corresponding physical signal has the following form:

$$s(t) = A_{\rm T} \cdot \cos \left(\omega_{\rm T}\cdot t\right) + {A_0}/{2}\cdot \cos\left(\left(\omega_{\rm T} + \omega_{\rm 0}\right)\cdot t \right) + {A_0}/{2}\cdot \cos\left(\left(\omega_{\rm T} - \omega_{\rm 0}\right)\cdot t \right)\hspace{0.05cm}.$$

• The frequency of the carrier signal is given as  $f_{\rm T} = 100\text{ kHz}$.
• The three further parameters $f_0$,  $A_{\rm T}$  and  $A_0$ are to be determined.

Reference will also be made to the  equivalent lowpass signal  $s_{\rm TP}(t)$, with the following relationship to the analytical signal:

$$s_{\rm TP}(t) = s_+(t) \cdot {\rm e}^{-{\rm j}\hspace{0.03cm} \cdot \hspace{0.03cm}\omega_{\rm T}\cdot \hspace{0.05cm}t} \hspace{0.05cm}.$$

Hints:

• This exercise belongs to the chapter   General Model of Modulation.
• Particular reference is made to the page  Describing the physical signal using the equivalent low-pass signal.
• You will find further information on these topics in these chapters of the book "Signal Representation":

(1)   Harmonic Oscillation,

(2)  Analytical Signal and its Spectral Function, 

(3)  Equivalent Low-Pass Signal and its Spectral Function.

• In our tutorial $\rm LNTwww$, the plot of the analytical signal $s_+(t)$  n the complex plane is sometimes referred to as the "pointer diagram", while the "locus curve" gives the time course of the equivalent lowpass signal  $s_{\rm TP}(t)$ . We refer you to the corresponding interactive Applets

(1)  Physical and analytic signal,

(2)  Physical signal and equivalent low-pass signal.

Questions

1

Using $s(t)$, give the equation for  $s_+(t)$  and simplify it. Which equation is valid for the equivalent low-pass signal?

	$s_{\rm TP}(t) = A_0 · {\rm e}^{–{\rm j}ω_0t}.$
	$s_{\rm TP}(t) = A_{\rm T} + A_0 · {\rm e}^{+{\rm j}ω_0t}.$
	$s_{\rm TP}(t) = A_{\rm T} + A_0 · \cos(ω_0t).$

2

Determine the signal parameter  $f_0$.

$f_0 \ = \ $ $\ \text{kHz}$

3

Determine the additional signal parameters  $A_{\rm T}$  and  $A_0$.

$A_{\rm T} \ = \ $

$A_0 \ = \ $

4

Calculate the values of the analytical signal  $s_+(t)$  at times  $t = 15 \;{\rm µ s}$  and  $t = 20\;{\rm µ s}$.

$s_+(t = 15 \ \rm µ s) \ = \ $

$s_+(t = 20 \ \rm µ s) \ = \ $

Solution

Solution

(1)  Convert all cosine functions to corresponding complex exponential functions:

$$s_+(t) = A_{\rm T} \cdot {\rm e}^{\hspace{0.03cm}{\rm j} \cdot \hspace{0.03cm}\omega_{\rm T}\cdot \hspace{0.03cm}t} + \frac{A_0}{2}\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \cdot \hspace{0.03cm}(\omega_{\rm T} + \omega_{\rm 0})\hspace{0.03cm}\cdot \hspace{0.05cm}t} + \frac{A_0}{2}\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \cdot \hspace{0.03cm}(\omega_{\rm T} - \omega_{\rm 0})\hspace{0.03cm}\cdot \hspace{0.03cm}t} = {\rm e}^{\hspace{0.03cm}{\rm j} \cdot \hspace{0.03cm}\omega_{\rm T}\cdot \hspace{0.03cm}t} \cdot \left[ A_{\rm T}+ \frac{A_0}{2} \cdot \left( {\rm e}^{\hspace{0.03cm}{\rm j} \cdot \hspace{0.03cm}\omega_{\rm 0}\cdot \hspace{0.05cm}t} + {\rm e}^{\hspace{0.03cm}-{\rm j} \cdot \hspace{0.03cm}\omega_{\rm 0}\cdot \hspace{0.05cm}t}\right)\right]\hspace{0.05cm}.$$

• Using the equation  ${\rm e}^{{\rm j} \hspace{0.05cm}· \hspace{0.05cm}α} + {\rm e}^{-{\rm j} \hspace{0.05cm}·\hspace{0.05cm} α} = 2 · \cos(α)$  it further follows that:

$$s_+(t) = {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}\omega_{\rm T}\cdot \hspace{0.03cm}t} \cdot \left[ A_{\rm T}+ {A_0} \cdot \cos (\omega_{\rm 0}\cdot t) \right] \hspace{0.05cm}.$$

• Thus, for the equivalent lowpass signal, we obtain:

$$s_{\rm TP}(t) = s_+(t) \cdot {\rm e}^{-{\rm j}\hspace{0.03cm} \cdot \hspace{0.03cm}\omega_{\rm T}\cdot \hspace{0.05cm}t} = A_{\rm T}+ {A_0} \cdot \cos (\omega_{\rm 0}\cdot t) \hspace{0.05cm}.$$

Therefore the last answer is correct.

• In the chapter  "Envelope Demodulation"  of the present book we will see that we are dealing with  "DSB-AM of a cosine signal with a cosine carrier".

(2)  The period of the analytical signal  $s_+(t)$  is  $T_0 = 50$  μs.

• The physical signal  $s(t)$  has the same period duration.
• Assuming that  $f_{\rm T}$ is an integer multiple of  $f_0$  (which always first needs confirming, but is true for this example),  this results in

$$f_0 = 1/T_0 \hspace{0.15cm}\underline{ = 20 \ \rm kHz}.$$

(3)  At the given times  (multiples of $5$  μs),  the following applies to the complex rotating pointer of the carrier:

$${\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2 \pi \cdot \hspace{0.05cm} {100\,{\rm kHz}}\cdot \hspace{0.05cm}(k \hspace{0.05cm}\cdot \hspace{0.05cm} 5\,{\rm \mu s})} = {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}k \hspace{0.03cm} \cdot \hspace{0.05cm} \pi } = \left\{ \begin{array}{c} +1 \\ -1 \\ \end{array} \right. \begin{array}{*{10}c} {\rm{if}} \\ {\rm{if}} \\ \end{array}\begin{array}{*{20}c} k \hspace{0.1cm}{\rm even} , \\ k \hspace{0.1cm}{\rm odd} . \\ \end{array}$$

• Therefore,  it follows from the equation calculated in Question  (1):

$$k = 0 \Rightarrow \hspace{0.2cm} s_{\rm +}(t = 0) = A_{\rm T}+ {A_0} \cdot \cos (\omega_{\rm 0}\cdot 0) = A_{\rm T}+ {A_0} \hspace{0.05cm},$$

$$k = 5 \Rightarrow \hspace{0.2cm} s_{\rm +}(t = 25\;{\rm µ s}) = - \left[ A_{\rm T}+ {A_0} \cdot \cos (\omega_{\rm 0}\cdot {T_0}/{2}) \right] = -A_{\rm T}+ {A_0} \hspace{0.05cm}.$$

• A comparison with the first and last equation on the sheet shows:

$$ s_{\rm +}(t = 0) = A_{\rm T}+ {A_0}=1.5 \hspace{0.05cm}, $$

$$ s_{\rm +}(t = 25\;{\rm \mu s}) = -A_{\rm T}+ {A_0} = -0.5 \hspace{0.05cm}.$$

• This gives  $A_{\rm T} \hspace{0.15cm}\underline{ = 1}$  and  $A_0 \hspace{0.15cm}\underline{ = 0.5}$.

(4)  At time  $t = 15$ μs  $(k = 3$,  odd$)$  it holds that:

$$ s_{\rm +}(t = 15\;{\rm µ s}) = - \left[ 1+ 0.5 \cdot \cos (2 \pi \cdot 20\,{\rm kHz} \cdot 0.015\,{\rm ms}) \right] \hspace{0.05cm} = -1- 0.5 \cdot \cos (108^{\circ})\hspace{0.15cm}\underline {= -0.845} \hspace{0.05cm}.$$

• In contrast, for time   $t = 20$  μs  $(k = 4$,  even$)$:

$$ s_{\rm +}(t = 20\;{\rm µ s}) = 1 + 0.5 \cdot \cos (144^{\circ})\hspace{0.15cm}\underline {= 0.595} \hspace{0.05cm}.$$

At all these considered time points, the physical signal $s(t) = {\rm Re}[s_+(t)]$ is exactly as large.