Difference between revisions of "Aufgaben:Exercise 1.4Z: Representation of Oscillations"

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{{quiz-Header|Buchseite=Modulationsverfahren/ Allgemeines Modell der Modulation
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{{quiz-Header|Buchseite=Modulation_Methods/General_Model_of_Modulation
 
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[[File:P_ID969__Mod_Z_1_4.png|right|]]
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[[File:P_ID969__Mod_Z_1_4.png|right|frame|Two representations of a harmonic oscillation]]
Betrachtet wird eine harmonische Schwingung $z(t)$, die zusammen mit dem zugehörigen Signal $z_+(t)$ in der Grafik dargestellt ist.
+
Here,  we consider a harmonic oscillation  $z(t)$,  which is shown in the graph together with the corresponding analytical signal   $z_+(t)$.  These signals can be described mathematically as follows:
 +
:$$z(t)  =  A_{\rm T} \cdot \cos(2 \pi f_{\rm T} t + \phi_{\rm T})=  A_{\rm T} \cdot \cos(2 \pi f_{\rm T}( t - \tau)) \hspace{0.05cm},$$
 +
:$$ z_+(t)  =  A_{\rm 0} \cdot {\rm e}^{{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm T}\hspace{0.05cm}\cdot \hspace{0.05cm}t+ϕ_{\rm T}}.$$
 +
The two amplitude parameters  $A_{\rm T} $  and  $A_0$  are each dimensionless,  the phase value  $ϕ_{\rm T} $  is supposed to lie between  $\text{±π}$,  and the duration  $τ$   is non-negative.
  
Diese Signale können mathematisch wie folgt beschrieben werden:
+
Subtask  '''(4)'''  refers to the equivalent low-pass signal  $z_{\rm TP}(t)$,  which is related to  $z_+(t)$  as follows:
$$z(t) =  A_{\rm T} \cdot \cos(2 \pi f_{\rm T} t + \phi_{\rm T})$
+
:$$z_{\rm TP}(t) = z_+(t) \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm T}\hspace{0.05cm}\cdot \hspace{0.05cm}t}.$$
$$=  A_{\rm T} \cdot \cos(2 \pi f_{\rm T}( t - \tau)) \hspace{0.05cm},$$
 
$$ z_+(t) =  A_{\rm 0} \cdot {\rm e}^{{\rm j} \cdot \hspace{0.03cm}\omega_{\rm T}\cdot \hspace{0.05cm}t}$$
 
Die zwei Amplitudenparameter $A_T$ und $A_0$ sind jeweils dimensionslos, der Phasenwert $ϕ_T$ soll zwischen $\text{±π}$ liegen und die Laufzeit τ ist nicht negativ.
 
  
Beachten Sie weiter, dass $ϕ_T$ in obiger Gleichung mit positivem Vorzeichen erscheint. Unter Anmerkungen zur Nomenklatur finden Sie eine Begründung für die unterschiedliche Verwendung von $φ_T$ und $ϕ_T = – φ_T$.
+
Further note that  $ϕ_{\rm T}$  appears in the above equation with a positive sign.  See "Notes on Nomenclature" below for reasons for the differential usage of  $φ_{\rm T}$  and  $ϕ_{\rm T} = – φ_{\rm T}$.
  
Die Teilaufgabe (d) bezieht sich auf das äquivalente TP–Signal $z_{TP}(t)$, das mit $z_+(t)$ in folgendem Zusammenhang steht:
 
$$z_{\rm TP}(t) = z_+(t) \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.03cm}\omega_{\rm T}\cdot \hspace{0.05cm}t} \hspace{0.05cm}.$$
 
'''Hinweis:''' Die Aufgabe bezieht sich auf das [http://en.lntwww.de/Modulationsverfahren/Allgemeines_Modell_der_Modulation Kapitel 1.3] dieses Buches. Weitere Informationen zu dieser Thematik finden Sie in [http://en.lntwww.de/Signaldarstellung/Harmonische_Schwingung Kapitel 2.3] – [http://en.lntwww.de/Signaldarstellung/Analytisches_Signal_und_zugeh%C3%B6rige_Spektralfunktion Kapitel 4.2] – [http://en.lntwww.de/Signaldarstellung/%C3%84quivalentes_Tiefpass-Signal_und_zugeh%C3%B6rige_Spektralfunktion Kapitel 4.3] des Buches „Signaldarstellung” sowie bei den folgenden Interaktionsmodulen:
 
  
 +
Notes on Nomenclature:
 +
*In this tutorial,  as is common in other literature,  the phase enters the equations with a negative sign when describing harmonic oscillations,  Fourier series,  and Fourier integrals.  In the context of modulation methods,  the phase is always given a plus sign.
 +
*To distinguish these two variants,  we use  $\phi_{\rm T}$  and  $\varphi_{\rm T} = - \phi_{\rm T}$.  Both symbols denote the lowercase Greek  "phi",  with the notation  $\phi$  used predominantly in Anglo-American contexts,  and  $\varphi$  in German.
 +
*The phase values  $\varphi_{\rm T} = 90^\circ$  and  $\phi_{\rm T} = -90^\circ$  are thus equivalent and both represent the sine function:
 +
:$$\cos(2 \pi f_0 t - 90^{\circ}) = \cos(2 \pi f_0 t - \varphi_{\rm T})  = \cos(2 \pi f_0 t + \phi_{\rm T}) = \sin(2 \pi f_0 t ).$$
  
Zeigerdiagramm – Darstellung des analytischen Signals
 
  
Ortskurve – Verlauf des äquivalenten Tiefpass-Signals
+
Further hints:
 +
*This exercise belongs to the chapter   [[Modulation_Methods/Allgemeines_Modell_der_Modulation|General Model of Modulation]].
 +
*Particular reference is made to the page   [[Modulation_Methods/General_Model_of_Modulation#Describing_the_physical_signal_using_the_equivalent_low-pass_signal|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)   [[ Signal_Representation/Harmonic_Oscillation|Harmonic Oscillation]],
 +
::(2)  [[Signal_Representation/Analytical_Signal_and_Its_Spectral_Function|Analytical Signal and its Spectral Function]], 
 +
::(3)  [[Signal_Representation/Equivalent_Low-Pass_Signal_and_its_Spectral_Function| 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)  [[Applets:Physical_Signal_%26_Analytic_Signal|Physical and analytic signal]],
 +
::(2)  [[Applets:Physical_Signal_%26_Equivalent_Lowpass_Signal|Physical signal and equivalent low-pass signal]].
  
  
'''Zu Kapitel 2.3: Anmerkung zur Nomenklatur der Phase'''
 
Anzumerken ist, dass in diesem Tutorial – wie auch in anderer Literatur üblich – bei der Beschreibung von harmonischer Schwingung, Fourierreihe und Fourierintegral die Phase mit negativem Vorzeichen in die Gleichungen eingeht, während in Zusammenhang mit Modulationsverfahren die Phase stets mit einem Pluszeichen angesetzt wird. Zur Unterscheidung dieser beiden Varianten benutzen wir $φ$ und $ϕ$. Beide Symbole kennzeichnen das kleine griechische „phi”, wobei die Schreibweise $φ$ vorwiegend im deutschen und $ϕ$ im anglo-amerikanischen Sprachraum angewandt wird.
 
  
Die Phasenwerte $φ = 90°$ und $ϕ = –90°$ sind somit äquivalent und stehen beide für die Sinusfunktion:
+
 
$$cos(2\Pi f_0t - 90°) = cos(2\Pi f_0t- \varphi) = cos(2\Pi f_0t + \phi) = sin(2\Pi f_0 t)$$
+
===Questions===
===Fragebogen===
 
  
 
<quiz display=simple>
 
<quiz display=simple>
  
{Berechnen Sie die Signalparameter $A_T$, $f_T$ und $ω_T$.
+
{Calculate the signal parameters &nbsp;$A_{\rm T}$, &nbsp;$f_{\rm T}$&nbsp; and &nbsp;$ω_{\rm T}$.
 
|type="{}"}
 
|type="{}"}
$A_T$ = { 2 3% }
+
$A_{\rm T} \ = \ $ { 2 3% }
$f_T$= { 500 3% } $\text{Hz}$
+
$f_{\rm T} \ = \ $ { 500 3% } $\ \text{Hz}$
$\omega_T$ = { 3141.5 3% } $\text{Hz}$
+
$\omega_{\rm T} \ = \ $ { 3141.5 3% } $\ \text{1/s}$
  
{Bestimmen Sie die Phase $ϕ_T$ (zwischen ±180°) und die Laufzeit $τ$.
+
{Determine the phase&nbsp; $\phi_{\rm T}$&nbsp; $($between&nbsp; $±180^\circ)$&nbsp; and the duration &nbsp;$τ$.
 
|type="{}"}
 
|type="{}"}
$ϕ_T$ = { -135 3% } $\text{Grad}$  
+
$\phi_{\rm T}  \ = \ $ { -139--131 } $\ \text{deg}$  
$τ$ = { 0.75 3% } $\text{ms}$
+
\ = \ $ { 0.75 3% } $\ \text{ms}$
  
{Zu welcher Zeit $t_1 > 0$ ist das analytische Signal $z_+(t)$ erstmalig imaginär?
+
{At what time &nbsp;$t_1 > 0$&nbsp; does the analytical signal &nbsp;$z_+(t)$&nbsp; first become imaginary?
 
|type="{}"}
 
|type="{}"}
$t_1$ = { 0.25 3% } $\text{ms}$
+
$t_1 \ = \ $ { 0.25 3% } $\ \text{ms}$
  
{Wie lautet das äquivalente Tiefpass–Signal $z_{TP}(t)$? Geben Sie zur Kontrolle den Wert bei t = 1 ms ein.
+
{What is the equivalent low-pass signal &nbsp;$z_{\rm TP}(t)$?&nbsp; Enter the value at &nbsp;$t = 1 \text{ ms}$&nbsp; to check.
 
|type="{}"}  
 
|type="{}"}  
$Re[z_{TP}(t = 1 ms)]$ = { -1.414 3% }  
+
${\rm Re}\big[z_{\rm TP}(t = 1\ \rm ms)\big] \ = \ $ { -1.454--1.374 }  
$Im[z_{TP}(t = 1 ms)]$ = { -1.414 3% }  
+
${\rm Im}\big[z_{\rm TP}(t = 1\ \rm ms)\big] \ = \ $ { -1.454--1.374 }
  
{Welche der Aussagen gelten für alle harmonischen Schwingungen?
+
{Which of these statements are valid for all harmonic oscillations?
 
|type="[]"}
 
|type="[]"}
+ Das Spektrum $Z(f)$ besteht aus zwei Diracfunktionen bei $±f_T$.
+
+ The spectrum &nbsp;$Z(f)$&nbsp; consists of two Dirac delta functions at &nbsp;$±f_{\rm T}$.
- Das Spektrum $Z_+(f)$ weist eine Diracfunktion bei $–f_T$ auf.
+
- The spectrum &nbsp;$Z_+(f)$&nbsp; has one delta Dirac function at &nbsp;$–f_{\rm T}$.
+ Das Spektrum $Z_{TP}(f)$ beinhaltet eine Diracfunktion bei f = 0.
+
+ The spectrum &nbsp;$Z_{\rm TP}(f)$&nbsp; contains a Dirac delta function at &nbsp;$f = 0$.
+ Das analytische Signal $z_+(t)$ ist stets komplex.
+
+ The analytical signal &nbsp;$z_+(t)$&nbsp; is always complex.
- Das äquivalente TP–Signal $z_{TP}(t)$ ist stets komplex.
+
- The equivalent low-pass signal &nbsp;$z_{\rm TP}(t)$&nbsp; is always complex.
  
 
</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''1.'''Aus der grafischen Darstellung der Zeitfunktion $z(t)$ erkennt man die (normierte) Amplitude $A_T = 2$ und die Periodendauer $T_0 = 2$ Millisekunden. Deshalb ist die Signalfrequenz $f_T = 1/T_0 = 500 Hz$ und die Kreisfrequenz beträgt $ω_T = 2πf_T = 3141.5 1/s$.
+
'''(1)'''&nbsp; In the graphical representation of the time function&nbsp; $z(t)$,&nbsp; one can identify
 +
*the (normalized) amplitude&nbsp; $A_{\rm T}\hspace{0.15cm}\underline{ = 2}$&nbsp; and the period&nbsp; $T_0=2$&nbsp; milliseconds.  
 +
*Therefore,&nbsp; the signal frequency is &nbsp; $f_{\rm T} = 1/T_0\hspace{0.15cm}\underline{ = 500}$&nbsp; Hz and the angular frequency is&nbsp; $ω_{\rm T}= 2πf_{\rm T} \hspace{0.15cm}\underline{ = 3141.5}$&nbsp; 1/s.
 +
 
 +
 
 +
'''(2)'''&nbsp; The analytical signal is generally:
 +
:$$z_+(t) = A_{\rm T} \cdot {\rm e}^{{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}(\omega_{\rm T}\cdot \hspace{0.05cm}t + \phi_{\rm T})} = A_{\rm T} \cdot {\rm e}^{{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm} \phi_{\rm T}} \cdot {\rm e}^{{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm} \hspace{0.03cm}\omega_{\rm T}\cdot \hspace{0.05cm}t }\hspace{0.05cm}.$$
 +
*At the same time the relationship:
 +
:$$A_0 = z_+(t = 0) = A_{\rm T} \cdot {\rm e}^{{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm} \phi_{\rm T}} \hspace{0.05cm}.$$
 +
*The complex amplitude&nbsp; $A_0$&nbsp; can be read from the upper plot.
 +
:$$A_0 = - \sqrt{2} - {\rm j} \cdot \sqrt{2} = A_{\rm 0} \cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm} 0.75 \pi} \hspace{0.05cm}.$$
 +
*A comparison of both equations leads to the result:
 +
:$$ \phi_{\rm T} = - 0.75 \pi \hspace{0.15cm}\underline {= - 135^{\circ}} \hspace{0.05cm}.$$
 +
*Thereby,&nbsp; the following relationship exists with the duration&nbsp; $τ$:
 +
:$$\phi_{\rm T} = - 2 \pi \cdot f_{\rm T} \cdot \tau \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \tau = \frac{-\phi_{\rm T}}{2 \pi \cdot f_{\rm T}} = \frac{0.75 \pi}{2 \pi \cdot 0.5\,{\rm kHz}} \hspace{0.15cm}\underline {= 0.75 \,{\rm ms}} \hspace{0.05cm}.$$
 +
 
 +
 
  
 +
'''(3)'''&nbsp; The analytical signal covers exactly one revolution in the time&nbsp; $T_0$&nbsp;.
 +
*Thus,&nbsp; starting from&nbsp; $A_0$&nbsp; after &nbsp; $t_1 = T_0/8\hspace{0.15cm}\underline{ = 0.25}$&nbsp; ms,&nbsp; we reach the first moment that the analytical signal is imaginary:
 +
:$$z_+(t_1) = - 2 {\rm j}.$$
 +
*Because of the relation&nbsp; $z(t) = {\rm Re}[z_+(t)]$,&nbsp; the first zero crossing of the signal&nbsp; $z(t)$&nbsp; also occurs at time&nbsp; $t_1$.
  
'''2.'''Das analytische Signal lautet allgemein:
 
$$z_+(t) = A_{\rm T} \cdot {\rm e}^{{\rm j} \cdot \hspace{0.03cm}(\omega_{\rm T}\cdot \hspace{0.05cm}t + \phi_{\rm T})} = A_{\rm T} \cdot {\rm e}^{{\rm j} \cdot \phi_{\rm T}} \cdot {\rm e}^{{\rm j} \cdot \hspace{0.03cm}\omega_{\rm T}\cdot \hspace{0.05cm}t }\hspace{0.05cm}.$$
 
Gleichzeitig gilt der Zusammenhang:
 
$$A_0 = z_+(t = 0) = A_{\rm T} \cdot {\rm e}^{{\rm j} \cdot \phi_{\rm T}} \hspace{0.05cm}.$$
 
Die komplexe Amplitude $A_0$ kann aus der oberen Grafik abgelesen werden.
 
$$A_0 = - \sqrt{2} - {\rm j} \cdot \sqrt{2} = A_{\rm 0} \cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm} 0.75 \pi} \hspace{0.05cm}.$$
 
Ein Vergleich beider Gleichungen führt zum Ergebnis:
 
$$ \phi_{\rm T} = - 0.75 \pi \hspace{0.15cm}\underline {= - 135^{\circ}} \hspace{0.05cm}.$$
 
Dabei besteht folgender Zusammenhang mit der Laufzeit $τ$:
 
$$\phi_{\rm T} = - 2 \pi \cdot f_{\rm T} \cdot \tau \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \tau = \frac{-\phi_{\rm T}}{2 \pi \cdot f_{\rm T}} = \frac{0.75 \pi}{2 \pi \cdot 0.5\,{\rm kHz}} \hspace{0.15cm}\underline {= 0.75 \,{\rm ms}} \hspace{0.05cm}.$$
 
  
'''3.'''Das analytische Signal legt in der Zeit $T_0$ genau eine Umdrehung zurück. Ausgehend von $A_0$ erreicht man somit nach $T_0/8 = 0.25 ms$ zum ersten Mal, dass das analytische Signal imaginär ist: $z_+(t1) = – 2 j$. Wegen der Beziehung $z(t) = Re[z_+(t)]$ tritt zu diesem Zeitpunkt $t_1$ auch der erste Nulldurchgang des Signals $z(t)$ auf.
 
  
 +
'''(4)'''&nbsp; Using the result from subtask &nbsp; '''(2)''',&nbsp; we obtain:
 +
:$$ z_{\rm TP}(t) = z_+(t) \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm T}\cdot \hspace{0.05cm}t} = A_0 = A_{\rm T} \cdot {\rm e}^{{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \phi_{\rm T}} = {\rm const.}$$
 +
*Thus,&nbsp; for all times&nbsp; $t$&nbsp; and hence also&nbsp; $t = 1$ ms:
 +
:$${\rm Re}[z_{\rm TP}(t)]  =  - \sqrt{2} \hspace{0.15cm}\underline {= -1.414} \hspace{0.05cm},$$
 +
:$$ {\rm Im}[z_{\rm TP}(t)]  =  - \sqrt{2}\hspace{0.15cm}\underline {= -1.414} \hspace{0.05cm}.$$
  
'''4.'''Mit dem Ergebnis aus b) erhält man:
 
$$ z_{\rm TP}(t) = z_+(t) \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.03cm}\omega_{\rm T}\cdot \hspace{0.05cm}t} = A_0 = A_{\rm T} \cdot {\rm e}^{{\rm j} \cdot \phi_{\rm T}} = {\rm const.}$$
 
Somit gilt für alle Zeiten t und damit auch für t = 1 ms:
 
$${\rm Re}[z_{\rm TP}(t)]  =  - \sqrt{2} \hspace{0.15cm}\underline {= -1.414} \hspace{0.05cm},\\ {\rm Im}[z_{\rm TP}(t)]  =  - \sqrt{2}\hspace{0.15cm}\underline {= -1.414} \hspace{0.05cm}.$$
 
  
  
'''5.'''Richtig sind die Aussagen 1, 3 und 4:
+
'''(5)'''&nbsp; <u>Statements 1, 3 and 4</u>&nbsp; are correct:
:*Die einzige Diracfunktion von $Z_+(f)$ liegt bei $f = f_T$ und nicht bei $–f_T$.
+
*The only Dirac delta function of&nbsp; $Z_+(f)$&nbsp; is at&nbsp; $f = f_{\rm T}$&nbsp; and not at&nbsp; $–f_{\rm T}$.
:*Das analytische Signal einer harmonischen Schwingung ist immer komplex.
+
*The analytical signal of a harmonic oscillation is always complex.
:* Das äquivalente TP–Signal einer harmonischen Schwingung ist meistens komplex.
+
* The equivalent low-pass signal of a harmonic oscillation is usually complex.&nbsp; The exception: 
Ausnahme:   $z(t) = ±A_T · cos(ω_T · t).$
+
:$$z(t) = ±A_{\rm T} · \cos(ω_{\rm T} · t) \ \Rightarrow \  z_{\rm TP}(t) = ±A_{\rm T}.$$  
  
 
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[[Category:Aufgaben zu  Modulationsverfahren|^1.3 Allgemeines Modell der Modulation ^]]
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[[Category:Modulation Methods: Exercises|^1.3 General Model of Modulation ^]]

Latest revision as of 18:07, 16 November 2021

Two representations of a harmonic oscillation

Here,  we consider a harmonic oscillation  $z(t)$,  which is shown in the graph together with the corresponding analytical signal  $z_+(t)$.  These signals can be described mathematically as follows:

$$z(t) = A_{\rm T} \cdot \cos(2 \pi f_{\rm T} t + \phi_{\rm T})= A_{\rm T} \cdot \cos(2 \pi f_{\rm T}( t - \tau)) \hspace{0.05cm},$$
$$ z_+(t) = A_{\rm 0} \cdot {\rm e}^{{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm T}\hspace{0.05cm}\cdot \hspace{0.05cm}t+ϕ_{\rm T}}.$$

The two amplitude parameters  $A_{\rm T} $  and  $A_0$  are each dimensionless,  the phase value  $ϕ_{\rm T} $  is supposed to lie between  $\text{±π}$,  and the duration  $τ$   is non-negative.

Subtask  (4)  refers to the equivalent low-pass signal  $z_{\rm TP}(t)$,  which is related to  $z_+(t)$  as follows:

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

Further note that  $ϕ_{\rm T}$  appears in the above equation with a positive sign.  See "Notes on Nomenclature" below for reasons for the differential usage of  $φ_{\rm T}$  and  $ϕ_{\rm T} = – φ_{\rm T}$.


Notes on Nomenclature:

  • In this tutorial,  as is common in other literature,  the phase enters the equations with a negative sign when describing harmonic oscillations,  Fourier series,  and Fourier integrals.  In the context of modulation methods,  the phase is always given a plus sign.
  • To distinguish these two variants,  we use  $\phi_{\rm T}$  and  $\varphi_{\rm T} = - \phi_{\rm T}$.  Both symbols denote the lowercase Greek  "phi",  with the notation  $\phi$  used predominantly in Anglo-American contexts,  and  $\varphi$  in German.
  • The phase values  $\varphi_{\rm T} = 90^\circ$  and  $\phi_{\rm T} = -90^\circ$  are thus equivalent and both represent the sine function:
$$\cos(2 \pi f_0 t - 90^{\circ}) = \cos(2 \pi f_0 t - \varphi_{\rm T}) = \cos(2 \pi f_0 t + \phi_{\rm T}) = \sin(2 \pi f_0 t ).$$


Further hints:

(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

Calculate the signal parameters  $A_{\rm T}$,  $f_{\rm T}$  and  $ω_{\rm T}$.

$A_{\rm T} \ = \ $

$f_{\rm T} \ = \ $

$\ \text{Hz}$
$\omega_{\rm T} \ = \ $

$\ \text{1/s}$

2

Determine the phase  $\phi_{\rm T}$  $($between  $±180^\circ)$  and the duration  $τ$.

$\phi_{\rm T} \ = \ $

$\ \text{deg}$
$τ \ = \ $

$\ \text{ms}$

3

At what time  $t_1 > 0$  does the analytical signal  $z_+(t)$  first become imaginary?

$t_1 \ = \ $

$\ \text{ms}$

4

What is the equivalent low-pass signal  $z_{\rm TP}(t)$?  Enter the value at  $t = 1 \text{ ms}$  to check.

${\rm Re}\big[z_{\rm TP}(t = 1\ \rm ms)\big] \ = \ $

${\rm Im}\big[z_{\rm TP}(t = 1\ \rm ms)\big] \ = \ $

5

Which of these statements are valid for all harmonic oscillations?

The spectrum  $Z(f)$  consists of two Dirac delta functions at  $±f_{\rm T}$.
The spectrum  $Z_+(f)$  has one delta Dirac function at  $–f_{\rm T}$.
The spectrum  $Z_{\rm TP}(f)$  contains a Dirac delta function at  $f = 0$.
The analytical signal  $z_+(t)$  is always complex.
The equivalent low-pass signal  $z_{\rm TP}(t)$  is always complex.


Solution

(1)  In the graphical representation of the time function  $z(t)$,  one can identify

  • the (normalized) amplitude  $A_{\rm T}\hspace{0.15cm}\underline{ = 2}$  and the period  $T_0=2$  milliseconds.
  • Therefore,  the signal frequency is   $f_{\rm T} = 1/T_0\hspace{0.15cm}\underline{ = 500}$  Hz and the angular frequency is  $ω_{\rm T}= 2πf_{\rm T} \hspace{0.15cm}\underline{ = 3141.5}$  1/s.


(2)  The analytical signal is generally:

$$z_+(t) = A_{\rm T} \cdot {\rm e}^{{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}(\omega_{\rm T}\cdot \hspace{0.05cm}t + \phi_{\rm T})} = A_{\rm T} \cdot {\rm e}^{{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm} \phi_{\rm T}} \cdot {\rm e}^{{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm} \hspace{0.03cm}\omega_{\rm T}\cdot \hspace{0.05cm}t }\hspace{0.05cm}.$$
  • At the same time the relationship:
$$A_0 = z_+(t = 0) = A_{\rm T} \cdot {\rm e}^{{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm} \phi_{\rm T}} \hspace{0.05cm}.$$
  • The complex amplitude  $A_0$  can be read from the upper plot.
$$A_0 = - \sqrt{2} - {\rm j} \cdot \sqrt{2} = A_{\rm 0} \cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm} 0.75 \pi} \hspace{0.05cm}.$$
  • A comparison of both equations leads to the result:
$$ \phi_{\rm T} = - 0.75 \pi \hspace{0.15cm}\underline {= - 135^{\circ}} \hspace{0.05cm}.$$
  • Thereby,  the following relationship exists with the duration  $τ$:
$$\phi_{\rm T} = - 2 \pi \cdot f_{\rm T} \cdot \tau \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \tau = \frac{-\phi_{\rm T}}{2 \pi \cdot f_{\rm T}} = \frac{0.75 \pi}{2 \pi \cdot 0.5\,{\rm kHz}} \hspace{0.15cm}\underline {= 0.75 \,{\rm ms}} \hspace{0.05cm}.$$


(3)  The analytical signal covers exactly one revolution in the time  $T_0$ .

  • Thus,  starting from  $A_0$  after   $t_1 = T_0/8\hspace{0.15cm}\underline{ = 0.25}$  ms,  we reach the first moment that the analytical signal is imaginary:
$$z_+(t_1) = - 2 {\rm j}.$$
  • Because of the relation  $z(t) = {\rm Re}[z_+(t)]$,  the first zero crossing of the signal  $z(t)$  also occurs at time  $t_1$.


(4)  Using the result from subtask   (2),  we obtain:

$$ z_{\rm TP}(t) = z_+(t) \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm T}\cdot \hspace{0.05cm}t} = A_0 = A_{\rm T} \cdot {\rm e}^{{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \phi_{\rm T}} = {\rm const.}$$
  • Thus,  for all times  $t$  and hence also  $t = 1$ ms:
$${\rm Re}[z_{\rm TP}(t)] = - \sqrt{2} \hspace{0.15cm}\underline {= -1.414} \hspace{0.05cm},$$
$$ {\rm Im}[z_{\rm TP}(t)] = - \sqrt{2}\hspace{0.15cm}\underline {= -1.414} \hspace{0.05cm}.$$


(5)  Statements 1, 3 and 4  are correct:

  • The only Dirac delta function of  $Z_+(f)$  is at  $f = f_{\rm T}$  and not at  $–f_{\rm T}$.
  • The analytical signal of a harmonic oscillation is always complex.
  • The equivalent low-pass signal of a harmonic oscillation is usually complex.  The exception:
$$z(t) = ±A_{\rm T} · \cos(ω_{\rm T} · t) \ \Rightarrow \ z_{\rm TP}(t) = ±A_{\rm T}.$$