Aufgaben:Exercise 1.4Z: Representation of Oscillations - Revision history

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Reed: /* Solution */

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Solution

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 {Determine the phase&nbsp; $\phi_{\rm T}$&nbsp; $($between&nbsp; $±180^\circ)$&nbsp; and the duration &nbsp;$τ$.
 |type="{}"}
-$\phi_{\rm T}  \ = \ $ { -139--131 } $\ \text{Grad}$
+$\phi_{\rm T}  \ = \ $ { -139--131 } $\ \text{deg}$
 $τ \ = \ $ { 0.75 3% } $\ \text{ms}$

@@ Line 73: / Line 73: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; In the graphical representation of the time function &nbsp; $z(t)$&nbsp;, one can identify
+'''(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, 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.
+*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.
@@ Line 86: / Line 86: @@
 *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&nbsp; $τ$:
+*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, starting from&nbsp; $A_0$&nbsp; after &nbsp; $t_1 = T_0/8\hspace{0.15cm}\underline{ = 0.25}$&nbsp; ms, we reach the first moment that the analytical signal is imaginary:
+'''(3)'''&nbsp; The analytical signal covers exactly one revolution in the time&nbsp; $T_0$&nbsp;.
 :$$z_+(t_1) = - 2 {\rm j}.$$
-*Because of the relation&nbsp; $z(t) = {\rm Re}[z_+(t)]$&nbsp; , the first zero crossing of the signal $z(t)$&nbsp; also occurs at time&nbsp; $t_1$&nbsp;.
+*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; $t_1$.
-'''(4)'''&nbsp; Using the result from subtask &nbsp; '''(2)''', we obtain:
+'''(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, for all times &nbsp; $t$&nbsp; and hence also &nbsp; $t = 1$ ms:
+*Thus,&nbsp; for all times&nbsp; $t$&nbsp; and hence also&nbsp; $t = 1$ ms:
 :$${\rm Re}[z_{\rm TP}(t)]  =  - \sqrt{2} \hspace{0.15cm}\underline {= -1.414} \hspace{0.05cm},$$
 :$$ {\rm Im}[z_{\rm TP}(t)]  =  - \sqrt{2}\hspace{0.15cm}\underline {= -1.414} \hspace{0.05cm}.$$
@@ Line 105: / Line 106: @@
-'''(5)'''&nbsp; <u>Statements 1, 3 and 4</u> are correct:
+'''(5)'''&nbsp; <u>Statements 1, 3 and 4</u>&nbsp; are correct:
-*The only Dirac function of $Z_+(f)$&nbsp; is at $f = f_{\rm T}$&nbsp; and not at&nbsp; $–f_{\rm T}$.
+*The only Dirac delta function of&nbsp; $Z_+(f)$&nbsp; is at&nbsp; $f = f_{\rm T}$&nbsp; and not at&nbsp; $–f_{\rm T}$.
 *The analytical signal of a harmonic oscillation is always complex.
-* The equivalent lowpass signal of a harmonic oscillation is usually complex. The exception:
+* The equivalent low-pass signal of a harmonic oscillation is usually complex.&nbsp; The exception:
 :$$z(t) = ±A_{\rm T} · \cos(ω_{\rm T} · t) \ \Rightarrow \  z_{\rm TP}(t) = ±A_{\rm T}.$$

← Older revision		Revision as of 16:31, 16 November 2021
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	*The phase values  $\varphi_{\rm T} = 90^\circ$  and  $\phi_{\rm T} = -90^\circ$  are thus equivalent and both represent the sine function:		*The phase values  $\varphi_{\rm T} = 90^\circ$  and  $\phi_{\rm T} = -90^\circ$  are thus equivalent and both represent the sine function:
	:$$\cos(2 \pi f_0 t - 90^{\circ}) = \cos(2 \pi f_0 t - \varphi_{\rm T}) = \cos(2 \pi f_0 t + \phi_{\rm T}) = \sin(2 \pi f_0 t ).$$		:$$\cos(2 \pi f_0 t - 90^{\circ}) = \cos(2 \pi f_0 t - \varphi_{\rm T}) = \cos(2 \pi f_0 t + \phi_{\rm T}) = \sin(2 \pi f_0 t ).$$
−
−

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 Further hints:
 *This exercise belongs to the chapter &nbsp; [[Modulation_Methods/Allgemeines_Modell_der_Modulation|General Model of Modulation]].
-*Particular reference is made to the page&nbsp;  [[Modulation_Methods/General_Model_of_Modulation#Describing_the_physical_signal_using_the_equivalent_low-pass_signal|Describing the physical signal using the equivalent lowpass signal]].
+*Particular reference is made to the page&nbsp;  [[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) &nbsp; [[ Signal_Representation/Harmonic_Oscillation|Harmonic Oscillation]],
@@ Line 36: / Line 36: @@
 ::(2) &nbsp;[[Applets:Physical_Signal_%26_Equivalent_Lowpass_Signal|Physical signal and equivalent low-pass signal]].

@@ Line 24: / Line 24: @@
 Further hints:
 *This exercise belongs to the chapter &nbsp; [[Modulation_Methods/Allgemeines_Modell_der_Modulation|General Model of Modulation]].
 *Particular reference is made to the page&nbsp;  [[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]].

@@ Line 4: / Line 4: @@
 [[File:P_ID969__Mod_Z_1_4.png|right|frame|Two representations of a harmonic oscillation]]
-Here, we consider a harmonic oscillation &nbsp;$z(t)$, which is shown in the graph together with the corresponding analytical signal  &nbsp;$z_+(t)$&nbsp;. These signals can be described mathematically as follows:
+Here,&nbsp; we consider a harmonic oscillation &nbsp;$z(t)$,&nbsp; which is shown in the graph together with the corresponding analytical signal  &nbsp;$z_+(t)$.&nbsp; These signals can be described mathematically as follows:
 :$$z(t)  =  A_{\rm T} \cdot \cos(2 \pi f_{\rm T} t + \phi_{\rm T})=  A_{\rm T} \cdot \cos(2 \pi f_{\rm T}( t - \tau)) \hspace{0.05cm},$$
-:$$ z_+(t)  =  A_{\rm 0} \cdot {\rm e}^{{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm T}\hspace{0.05cm}\cdot \hspace{0.05cm}t}$$
+:$$ z_+(t)  =  A_{\rm 0} \cdot {\rm e}^{{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm T}\hspace{0.05cm}\cdot \hspace{0.05cm}t+ϕ_{\rm T}}.$$
-The two amplitude parameters &nbsp;$A_{\rm T} $&nbsp; and &nbsp;$A_0$&nbsp; are each dimensionless, the phase value &nbsp;$ϕ_{\rm T} $&nbsp; is supposed to lie between &nbsp;$\text{±π}$&nbsp;, and the duration τ&nbsp;$τ$&nbsp; is non-negative.
+The two amplitude parameters &nbsp;$A_{\rm T} $&nbsp; and &nbsp;$A_0$&nbsp; are each dimensionless,&nbsp; the phase value &nbsp;$ϕ_{\rm T} $&nbsp; is supposed to lie between &nbsp;$\text{±π}$,&nbsp; and the duration &nbsp;$τ$ &nbsp; is non-negative.
-Subtask &nbsp; '''(4)'''&nbsp; refers to the equivalent lowpass signal &nbsp;$z_{\rm TP}(t)$, which is related to &nbsp;$z_+(t)$&nbsp; as follows:
+Subtask&nbsp; '''(4)'''&nbsp; refers to the equivalent low-pass signal &nbsp;$z_{\rm TP}(t)$,&nbsp; which is related to &nbsp;$z_+(t)$&nbsp; 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 &nbsp;$ϕ_{\rm T}$&nbsp; ppears in the above equation with a positive sign.&nbsp; See "Notes on Nomenclature" below for reasons for the differential usage of &nbsp;$φ_{\rm T}$&nbsp; and &nbsp;$ϕ_{\rm T} = – φ_{\rm T}$.
+Further note that &nbsp;$ϕ_{\rm T}$&nbsp; appears in the above equation with a positive sign.&nbsp; See "Notes on Nomenclature" below for reasons for the differential usage of &nbsp;$φ_{\rm T}$&nbsp; and &nbsp;$ϕ_{\rm T} = – φ_{\rm T}$.
-''Notes on Nomenclature:''
+Notes on Nomenclature:
-*In this tutorial, as is common in other literature, the phase enters the equations with a negative sign when describing harmonic oscillation, Fourier series, and Fourier integrals, whereas in the context of modulation methods, the phase is always given a plus sign.
+*In this tutorial,&nbsp; as is common in other literature,&nbsp; the phase enters the equations with a negative sign when describing harmonic oscillations,&nbsp; Fourier series,&nbsp; and Fourier integrals.&nbsp; In the context of modulation methods,&nbsp; the phase is always given a plus sign.
-*To distinguish these two variants, we use &nbsp;$\phi_{\rm T}$ and $\varphi_{\rm T} = - \phi_{\rm T}$.&nbsp; Both symbols denote the lowercase Greek "phi", with the notation &nbsp;$\phi$&nbsp; used predominantly in Anglo-American contexts, and $\varphi$ in German.
+*To distinguish these two variants,&nbsp; we use &nbsp;$\phi_{\rm T}$&nbsp; and&nbsp; $\varphi_{\rm T} = - \phi_{\rm T}$.&nbsp; Both symbols denote the lowercase Greek&nbsp; "phi",&nbsp; with the notation &nbsp;$\phi$&nbsp; used predominantly in Anglo-American contexts,&nbsp; and&nbsp; $\varphi$&nbsp; in German.
-*The phase values &nbsp;$\varphi_{\rm T} = 90^\circ$ and $\phi_{\rm T} = -90^\circ$&nbsp; are thus equivalent and both represent the sine function:
+*The phase values &nbsp;$\varphi_{\rm T} = 90^\circ$&nbsp; and&nbsp; $\phi_{\rm T} = -90^\circ$&nbsp; are thus equivalent and both represent the sine function:
 :$$\cos(2 \pi f_0 t - 90^{\circ}) = \cos(2 \pi f_0 t - \varphi_{\rm T})  = \cos(2 \pi f_0 t + \phi_{\rm T}) = \sin(2 \pi f_0 t ).$$
@@ Line 24: / Line 24: @@
+Further hints:
 *This exercise belongs to the chapter &nbsp; [[Modulation_Methods/Allgemeines_Modell_der_Modulation|General Model of Modulation]].
 *Particular reference is made to the page&nbsp;  [[Modulation_Methods/Allgemeines_Modell_der_Modulation#Beschreibung_des_physikalischen_Signals_mit_Hilfe_des_.C3.A4quivalenten_TP-Signals|Describing the physical signal using the equivalent lowpass signal]].
-*Further information on this topic can be found in the following chapters of the book "Signal Representation":
+*You will find further information on these topics in these chapters of the book „Signal Representation”:
 ::(1) &nbsp; [[ Signal_Representation/Harmonic_Oscillation|Harmonic Oscillation]],
-::(2)&nbsp; [[Signal_Representation/Analytical_Signal_and_Its_Spectral_Function|Analytical Signal and its Spectral Function]]&nbsp; and
+::(2)&nbsp; [[Signal_Representation/Analytical_Signal_and_Its_Spectral_Function|Analytical Signal and its Spectral Function]],&nbsp;
 ::(3)&nbsp; [[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 &nbsp;$s_+(t)$&nbsp;  in the complex plane is sometimes referred to as the "pointer diagram", while the "locus curve" gives the time course of the equivalent lowpass signal &nbsp;$s_{\rm TP}(t)$&nbsp;. We refer you to the corresponding interactive Applets
@@ Line 51: / Line 52: @@
 $\omega_{\rm T} \ = \ $ { 3141.5 3% } $\ \text{1/s}$
-{Determine the phase &nbsp;$\phi_{\rm T}$&nbsp; $($zwischen $±180^\circ)$ and the duration &nbsp;$τ$.
+{Determine the phase&nbsp; $\phi_{\rm T}$&nbsp; $($between&nbsp; $±180^\circ)$&nbsp; and the duration &nbsp;$τ$.
 |type="{}"}
 $\phi_{\rm T}  \ = \ $ { -139--131 } $\ \text{Grad}$
@@ Line 60: / Line 61: @@
 $t_1 \ = \ $ { 0.25 3% } $\ \text{ms}$
-{What is the equivalent low-pass signal &nbsp;$z_{\rm TP}(t)$?&nbsp; Enter the value at &nbsp;$t = 1 \text{ ms}$ ms to check.
+{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="{}"}
 ${\rm Re}\big[z_{\rm TP}(t = 1\ \rm ms)\big] \ = \ $ { -1.454--1.374 }
@@ Line 67: / Line 68: @@
 {Which of these statements are valid for all harmonic oscillations?
 |type="[]"}
-+ The spectrum &nbsp;$Z(f)$&nbsp; consists of two Dirac functions at &nbsp;$±f_{\rm T}$.
++ The spectrum &nbsp;$Z(f)$&nbsp; consists of two Dirac delta functions at &nbsp;$±f_{\rm T}$.
-- The spectrum &nbsp;$Z_+(f)$&nbsp; has one Dirac function at &nbsp;$–f_{\rm T}$.
+- The spectrum &nbsp;$Z_+(f)$&nbsp; has one delta Dirac function at &nbsp;$–f_{\rm T}$.
-+ The spectrum &nbsp;$Z_{\rm TP}(f)$&nbsp; contains a Dirac function at &nbsp;$f = 0$.
++ The spectrum &nbsp;$Z_{\rm TP}(f)$&nbsp; contains a Dirac delta function at &nbsp;$f = 0$.
 + The analytical signal &nbsp;$z_+(t)$&nbsp; is always complex.
-- The equivalent lowpass signal &nbsp;$z_{\rm TP}(t)$&nbsp; is always complex.
+- The equivalent low-pass signal &nbsp;$z_{\rm TP}(t)$&nbsp; is always complex.
 </quiz>

@@ Line 94: / Line 94: @@
-'''(3)'''&nbsp; Das analytische Signal legt in der Zeit&nbsp; $T_0$&nbsp; genau eine Umdrehung zurück.
+'''(3)'''&nbsp; The analytical signal covers exactly one revolution in the time &nbsp; $T_0$&nbsp;.
-*Ausgehend von&nbsp; $A_0$&nbsp; erreicht man somit nach&nbsp; $t_1 = T_0/8\hspace{0.15cm}\underline{ = 0.25}$&nbsp; ms zum ersten Mal, dass das analytische Signal imaginär ist:
+*Thus, starting from&nbsp; $A_0$&nbsp; after &nbsp; $t_1 = T_0/8\hspace{0.15cm}\underline{ = 0.25}$&nbsp; ms, we reach the first moment that the analytical signal is imaginary:
 :$$z_+(t_1) = - 2 {\rm j}.$$
-*Wegen der Beziehung&nbsp; $z(t) = {\rm Re}[z_+(t)]$&nbsp; tritt zu diesem Zeitpunkt&nbsp; $t_1$&nbsp; auch der erste Nulldurchgang des Signals&nbsp; $z(t)$&nbsp; auf.
+*Because of the relation&nbsp; $z(t) = {\rm Re}[z_+(t)]$&nbsp; , the first zero crossing of the signal $z(t)$&nbsp; also occurs at time&nbsp; $t_1$&nbsp;.
-'''(4)'''&nbsp; Mit dem Ergebnis der Teilaufgabe&nbsp; '''(2)'''&nbsp; erhält man:
+'''(4)'''&nbsp; Using the result from subtask &nbsp; '''(2)''', we obtain:
 :$$ z_{\rm TP}(t) = z_+(t) \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm T}\cdot \hspace{0.05cm}t} = A_0 = A_{\rm T} \cdot {\rm e}^{{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \phi_{\rm T}} = {\rm const.}$$
-*Somit gilt für alle Zeiten&nbsp; $t$&nbsp; und damit auch für&nbsp; $t = 1$ ms:
+*Thus, for all times &nbsp; $t$&nbsp; and hence also &nbsp; $t = 1$ ms:
 :$${\rm Re}[z_{\rm TP}(t)]  =  - \sqrt{2} \hspace{0.15cm}\underline {= -1.414} \hspace{0.05cm},$$
 :$$ {\rm Im}[z_{\rm TP}(t)]  =  - \sqrt{2}\hspace{0.15cm}\underline {= -1.414} \hspace{0.05cm}.$$
@@ Line 109: / Line 109: @@
-'''(5)'''&nbsp; Richtig sind die <u>Aussagen 1, 3 und 4</u>:
+'''(5)'''&nbsp; <u>Statements 1, 3 and 4</u> are correct:
-*Die einzige Diracfunktion von&nbsp; $Z_+(f)$&nbsp; liegt bei&nbsp; $f = f_{\rm T}$&nbsp; und nicht bei&nbsp; $–f_{\rm T}$.
+*The only Dirac function of&nbsp; $Z_+(f)$&nbsp; is atnbsp; $f = f_{\rm T}$&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.&nbsp; Ausnahme:
+* The equivalent lowpass signal of a harmonic oscillation is usually complex. The exception::
 :$$z(t) = ±A_{\rm T} · \cos(ω_{\rm T} · t) \ \Rightarrow \  z_{\rm TP}(t) = ±A_{\rm T}.$$