Difference between revisions of "Signal Representation/Harmonic Oscillation"

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==Definition and Properties==
+
==Definition and properties==
 
<br>
 
<br>
Harmonic oscillations are of particular importance for communications engineering as well as in many natural sciences. The following diagram shows an exemplary signal waveform.
+
[[File:Sig_T_2_3_S1_Version3.png|right|frame|Example of a harmonic oscillation]]
  
[[File:Sig_T_2_3_S1_Version3.png|right|frame|Example of a Harmonic Oscillation]]
+
Harmonic oscillations are of particular importance for Communications Engineering as well as in many natural sciences.&nbsp; The diagram shows an exemplary signal waveform.
  
Its importance is also related to the fact that the harmonic oscillation represents the solution of a&nbsp; ''Differential equation''&nbsp; which is found in many disciplines and reads as follows
+
Its importance is also related to the fact that the harmonic oscillation represents the solution of a&nbsp; differential equation&nbsp; which is found in many disciplines and reads as follows:
 
   
 
   
 
:$$ x(t) + k \cdot\ddot{x} (t) = 0.$$
 
:$$ x(t) + k \cdot\ddot{x} (t) = 0.$$
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{{BlaueBox|TEXT=   
 
{{BlaueBox|TEXT=   
 
$\text{Definition:}$&nbsp;
 
$\text{Definition:}$&nbsp;
Any&nbsp; '''harmonic oscillation''' &nbsp; can be represented in most general form as follows:
+
Any&nbsp; &raquo;'''harmonic oscillation'''&laquo;&nbsp; can be represented in most general form as follows:
  
 
:$$x(t)= C \cdot \cos(2\pi f_0 t - \varphi).$$
 
:$$x(t)= C \cdot \cos(2\pi f_0 t - \varphi).$$
  
 
The following signal parameters are used:
 
The following signal parameters are used:
*the&nbsp; '''amplitude'''&nbsp; $C$&nbsp; &ndash; simultaneously the maximum value of the signal,
+
*the&nbsp; &raquo;'''amplitude'''&laquo;&nbsp; $C$&nbsp; &ndash; simultaneously the maximum value of the signal,
*the&nbsp; '''signal frequency'''&nbsp; $f_{0}$ &nbsp; &rArr; &nbsp; the reciprocal of the period duration&nbsp; $T_{0}$, and
 
*the&nbsp; '''zero phase angle'''&nbsp; (or briefly the ''phase'')&nbsp; $\varphi$&nbsp; of the oscillation.}}
 
  
 +
*the&nbsp; &raquo;'''signal frequency'''&laquo;&nbsp; $f_{0}$ &nbsp; &rArr; &nbsp; the reciprocal of the period duration&nbsp; $T_{0}$, and
  
The german learning video&nbsp; [[Harmonische_Schwingungen_(Lernvideo)|Harmonic Oscillations]]&nbsp; illustrates the properties of harmonic oscillations using scales.
+
*the&nbsp; &raquo;'''zero phase angle'''&laquo;&nbsp; $($or briefly the&nbsp; &raquo;phase&laquo;$)$ &nbsp; $\varphi$&nbsp; of the oscillation.}}
 +
 
 +
 
 +
The following&nbsp; $($German-language$)$&nbsp; learning video illustrates the properties of harmonic oscillations using scales: <br>
 +
&nbsp; &nbsp; &nbsp; [[Harmonische_Schwingungen_(Lernvideo)|&raquo;Harmonische Schwingungen&laquo;]] &nbsp; &rArr; &nbsp; "Harmonic Oscillations".  
  
 
{{BlaueBox|TEXT=   
 
{{BlaueBox|TEXT=   
 
$\text{Comments on nomenclature:}$&nbsp;
 
$\text{Comments on nomenclature:}$&nbsp;
*In this tutorial - as usual in other literature - when describing harmonic oscillations, Fourier series and Fourier integral, the phase is entered into the equations with a negative sign, whereas in connection with all modulation methods the phase is always entered with a plus sign.
+
 
*To distinguish between the two variants we use in&nbsp; $\rm LNTwww$&nbsp; $\varphi$&nbsp; and&nbsp; $\phi$. Both symbols denote the small Greek "phi". The spelling&nbsp; $\varphi$&nbsp; is mainly used in the German and&nbsp; $\phi$&nbsp; in the anglo&ndash;american language area.
+
In this tutorial &ndash; as usual in other literature when describing harmonic oscillations,&nbsp; Fourier series and Fourier integral &ndash;  the phase is entered into the equations with negative sign,&nbsp; whereas in connection with all modulation methods the phase is entered with a plus sign.
 +
*To distinguish between the two variants we use in our tutorial &nbsp; $\varphi$&nbsp; and&nbsp; $\phi$.&nbsp; Both symbols denote the small Greek "phi".&nbsp; The spelling&nbsp; $\varphi$&nbsp; is mainly used in the German and&nbsp; $\phi$&nbsp; in the anglo&ndash;american language area.
 +
 
 
*The indications&nbsp; $\varphi = 90^{\circ}$&nbsp; and&nbsp; $\phi = -90^{\circ}$&nbsp; are thus equivalent and both stand for the sine function:
 
*The indications&nbsp; $\varphi = 90^{\circ}$&nbsp; and&nbsp; $\phi = -90^{\circ}$&nbsp; are thus equivalent and both stand for the sine function:
 
   
 
   
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==Time Signal Representation==
+
==Time domain representation==
 
<br>
 
<br>
[[File:P_ID129__Sig_T_2_3_S2a.png|right|frame|Signal Parameters of a Harmonic Oscillation]]
+
The amplitude&nbsp; $C$&nbsp; can be read directly from the adjacent graph.&nbsp; The signal frequency $f_0$&nbsp; is the reciprocal of the period duration&nbsp; $T_0$.&nbsp;
The amplitude&nbsp; $C$&nbsp; can be read directly from the adjacent graphic. The signal frequency&nbsp; $f_0$&nbsp; is equal to the reciprocal of the period duration&nbsp; $T_0$. If the above equation is written in the form
+
[[File:P_ID129__Sig_T_2_3_S2a.png|right|frame|Signal parameters of a harmonic oscillation]]
 +
 
 +
If the above equation is written in the form
  
 
:$$x(t)  =  C \cdot \cos(2\pi f_0 t - \varphi) =  C \cdot \cos \big(2\pi f_0  (t - \tau) \big), $$
 
:$$x(t)  =  C \cdot \cos(2\pi f_0 t - \varphi) =  C \cdot \cos \big(2\pi f_0  (t - \tau) \big), $$
 
   
 
   
it becomes clear that the zero phase angle&nbsp; $\varphi$&nbsp; and the displacement&nbsp; $\tau$&nbsp; relative to a cosine signal are related as follows
+
it becomes clear that the zero phase angle&nbsp; $\varphi$&nbsp; and the shift&nbsp; $\tau$&nbsp; relative to a cosine signal are related as follows
  
 
:$$\varphi  =  \frac{\tau}{T_0} \cdot 2{\pi}. $$  
 
:$$\varphi  =  \frac{\tau}{T_0} \cdot 2{\pi}. $$  
  
*For a ''cosine signal''&nbsp; the parameters&nbsp; $\tau$&nbsp; and&nbsp; $\varphi$&nbsp; are zero.  
+
*For a cosine signal both the parameters&nbsp; $\tau$&nbsp; and&nbsp; $\varphi$&nbsp; are zero.
*In contrast, a ''sinusoidal signal''&nbsp; is shifted by&nbsp; $\tau = T_0/4$&nbsp; and accordingly applies to the zero phase angle&nbsp; $\varphi = \pi/2$&nbsp; (in radians) or &nbsp; $90^{\circ}$.
+
 +
*In contrast,&nbsp; a sinusoidal signal&nbsp; is shifted by&nbsp; $\tau = T_0/4$&nbsp; and accordingly applies to the zero phase angle&nbsp; $\varphi = \pi/2$&nbsp; $($in radians$)$&nbsp; or &nbsp; $90^{\circ}$.
 +
 
 +
*So it can be stated that &ndash; as assumed for the above sketch &ndash; at a positive value of&nbsp; $\tau$&nbsp; resp. &nbsp; $\varphi$&nbsp; the $($referring&nbsp; $t = 0)$&nbsp; nearest signal maximum comes later than at the cosine signal and at negative values earlier.  
  
 +
*If a cosine signal is present at the system input and the output signal is delayed by a value&nbsp; $\tau$&nbsp; then&nbsp; $\tau$&nbsp; is also called the&nbsp; &raquo;runtime&laquo;&nbsp; of the system.
  
So it can be stated, that - as assumed for the above sketch - at a positive value of&nbsp; $\tau$&nbsp; resp. &nbsp; $\varphi$&nbsp; the $($referring&nbsp; $t = 0)$&nbsp; nearest signal maximum comes later than at the cosine signal and at negative values earlier. If a cosine signal is present at the system input and the output signal is delayed by a value&nbsp; $\tau$&nbsp; then the system is called&nbsp; $\tau$&nbsp; also called the&nbsp; '''runtime'''&nbsp;.
+
*Since a harmonic oscillation is clearly defined by three parameters,&nbsp; the entire time course from&nbsp; $-\infty$&nbsp; to&nbsp; $+\infty$&nbsp; can be calculated  analytically from only three signal values&nbsp; $x_1=x(t_1)$,&nbsp; $x_2=x(t_2)$,&nbsp; $x_3=x(t_3)$&nbsp; if the times&nbsp; $t_1$,&nbsp; $t_2$&nbsp; and&nbsp; $t_3$&nbsp; have been determined appropriately.
  
Since a harmonic oscillation is clearly defined by only three signal parameters, the entire time course from&nbsp; $-\infty$&nbsp; to&nbsp; $+\infty$&nbsp; can be calculated from only three signal values&nbsp; $x_1=x(t_1)$,&nbsp; $x_2=x(t_2)$,&nbsp; $x_3=x(t_3)$&nbsp; can be described analytically if the times&nbsp; $t_1$,&nbsp; $t_2$&nbsp; and&nbsp; $t_3$&nbsp; have been determined appropriately.
 
  
[[File:P_ID2717__Sig_T_2_3_S2b_neu.png|right|frame|Harmonic Oscillation, defined by <br>Three Samples]]
+
{{GraueBox|TEXT=
{{GraueBox|TEXT= 
+
[[File:P_ID2717__Sig_T_2_3_S2b_neu.png|right|frame|Harmonic oscillation, defined by only&nbsp; $3$&nbsp; samples]]  
 
$\text{Example 1:}$&nbsp;
 
$\text{Example 1:}$&nbsp;
 
From the three sample values,
 
From the three sample values,
 
:$$x_1 = x(t_1 = 3.808 \;{\rm ms}) = +1.609,$$
 
:$$x_1 = x(t_1 = 3.808 \;{\rm ms}) = +1.609,$$
 
:$$x_2 = x(t_2 = 16.696 \;{\rm ms})=\hspace{0.05 cm} -0.469,$$
 
:$$x_2 = x(t_2 = 16.696 \;{\rm ms})=\hspace{0.05 cm} -0.469,$$
:$$x_3 = x(t_3 = 33.84 \;{\rm ms}) = +1.227$$
+
:$$x_3 = x(t_3 = 33.84 \;{\rm ms}) = +1.227,$$
  
 
you get the following system of equations:
 
you get the following system of equations:
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After solving this nonlinear system of equations the following signal parameters are obtained:
 
After solving this nonlinear system of equations the following signal parameters are obtained:
*Signal amplitude&nbsp; $C = 2$,
+
*Signal amplitude:&nbsp; $C = 2$,
*Period duration&nbsp; $T_0 = 8 \;{\rm ms}$ &nbsp; ⇒ &nbsp; Signal frequency&nbsp; $f_0 = 125 \;{\rm Hz}$,
+
 
*displacement with respect to a cosine&nbsp; $\tau = 3 \;{\rm ms}$ &nbsp; ⇒ &nbsp; zero phase angle&nbsp; $\varphi = 3\pi /4 = 135^\circ$.
+
*period duration:&nbsp; $T_0 = 8 \;{\rm ms}$ &nbsp; ⇒ &nbsp; signal frequency $f_0 = 125 \;{\rm Hz}$,
 +
 
 +
*shift with respect to a cosine:&nbsp; $\tau = 3 \;{\rm ms}$ &nbsp; ⇒ &nbsp; zero phase angle&nbsp; $\varphi = 3\pi /4 = 135^\circ$.
  
  
''Note'': &nbsp; If you set all sampling times&nbsp; $t_1$,&nbsp; $t_2$,&nbsp; $t_3$&nbsp; in maxima, minima and/or zeros, there is no unique solution for the nonlinear system of equations.}}
+
If you set all sampling times&nbsp; $t_1$,&nbsp; $t_2$,&nbsp; $t_3$&nbsp; in maxima, minima and/or zeros, there is no unique solution for the nonlinear equation system.}}
  
  
==Representation with Cosine and Sine Components==
+
==Representation with cosine and sine components==
 
<br>
 
<br>
 
Another representation of the harmonic oscillation is as follows:
 
Another representation of the harmonic oscillation is as follows:
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:$$x(t)=A\cdot\cos(2\pi  f_0 t)+ B\cdot\sin(2\pi f_0 t).$$
 
:$$x(t)=A\cdot\cos(2\pi  f_0 t)+ B\cdot\sin(2\pi f_0 t).$$
  
*The terms&nbsp; $A$&nbsp; and&nbsp; $B$&nbsp; for the amplitudes of the cosine and sine components are chosen to match the nomenclature of the following chapter[Signal_Representation/Fourier_Series| Fourier Series]]&nbsp;.
+
*The terms&nbsp; $A$&nbsp; and&nbsp; $B$&nbsp; for the amplitudes of the cosine and sine components are chosen to match the nomenclature of the following chapter&nbsp; [[Signal_Representation/Fourier_Series|&raquo;Fourier Series&laquo;]].
  
*By applying trigonometric transformations we obtain from the illustration on the last page
+
*By applying trigonometric transformations we obtain from the illustration in the last section:
 
   
 
   
 
:$$x(t)=C\cdot \cos(2\pi f_0 t-\varphi)=C\cdot\cos(\varphi)\cdot\cos(2\pi f_0  t)+C\cdot\sin(\varphi)\cdot\sin(2\pi f_0  t).$$
 
:$$x(t)=C\cdot \cos(2\pi f_0 t-\varphi)=C\cdot\cos(\varphi)\cdot\cos(2\pi f_0  t)+C\cdot\sin(\varphi)\cdot\sin(2\pi f_0  t).$$
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*From this follows directly by equating the coefficients:
 
*From this follows directly by equating the coefficients:
 
   
 
   
:$$A=C\cdot\cos(\varphi),\hspace{0.5cm}B=C\cdot\sin(\varphi).$$
+
:$$A=C\cdot\cos(\varphi),$$
 +
:$$B=C\cdot\sin(\varphi).$$
  
*The magnitude and the zero phase angle of the harmonic oscillation can be calculated from the parameters&nbsp; $A$&nbsp; and&nbsp; $B$&nbsp; also according to simple trigonometric considerations
+
*The magnitude and the zero phase angle of the harmonic oscillation can be calculated from the parameters&nbsp; $A$&nbsp; and&nbsp; $B$&nbsp; also according to simple trigonometric considerations:
 
   
 
   
:$$C=\sqrt{A^2+B^2}, \hspace{0.5 cm}\varphi = \arctan\left({-B}/{A}\right).$$
+
:$$C=\sqrt{A^2+B^2},$$
 +
:$$\varphi = \arctan\left({-B}/{A}\right).$$
  
 
{{BlaueBox|TEXT=   
 
{{BlaueBox|TEXT=   
$\text{Please consider:}$&nbsp;
+
$\text{Please note:}$&nbsp;
The minus-sign at the calculation of the zero-phase-angle&nbsp; $\varphi$&nbsp; is related to the fact, that&nbsp; $\varphi$&nbsp; enters the argument of the cosine function with a negative sign.  
+
The minus&ndash;sign at the calculation of the zero phase angle&nbsp; $\varphi$&nbsp; is related to the fact&nbsp; that&nbsp; $\varphi$&nbsp; enters the argument of the cosine function with negative sign.&nbsp; If one would use the notation&nbsp; $\cos(2\pi f_0 t  +\phi)$&nbsp; instead of&nbsp; $\cos(2\pi f_0 t - \varphi)$,&nbsp; then&nbsp; $\phi= \arctan(B/A)$.&nbsp; Note the following here:
  
If one would use the notation&nbsp; "$\cos(2\pi f_0 t - +\phi)$"&nbsp; instead of&nbsp; "$\cos(2\pi f_0 t - \varphi)$"&nbsp;, then&nbsp; $\phi= \arctan(B/A)$. Note the following here:
+
#For Fourier series and Fourier integral, the&nbsp; $\varphi$&nbsp; representation is common in literature.
 +
#For description of the modulation methods, however, the&nbsp; $\phi$&nbsp; representation is almost always used.}}
  
*For Fourier series and Fourier integral, the&nbsp; $\varphi$ representation is common in literature.
 
*For the description of the modulation methods, however, the&nbsp; $\phi$ representation is almost always used.}}
 
  
 
[[File:Sig_T_2_3_S3_version2.png|right|frame|Harmonic oscillation, represented in the complex plane]]
 
 
{{GraueBox|TEXT=   
 
{{GraueBox|TEXT=   
$\text{Example 2:}$&nbsp;
+
$\text{Example 2:}$&nbsp; The oscillation shown in the left graphic as time course is characterized by the parameters[[File:Sig_T_2_3_S3_version2.png|right|frame|Harmonic oscillation. On the right: Represented in the complex plane]]
The oscillation shown in the left graphic as time course is characterized by the parameters  
+
*$C=2,$
 
*$f_0 = 125 \;{\rm Hz}$,   
 
*$f_0 = 125 \;{\rm Hz}$,   
*$\varphi = +135^{\circ}$  &nbsp; ⇒  &nbsp; $\phi = -135^{\circ}$ .  
+
*$\varphi = +135^{\circ}$  &nbsp; ⇒  &nbsp; $\phi = -135^{\circ}$.  
  
  
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:$$x(t) \hspace{-0.05cm}=\hspace{-0.05cm} -\sqrt{2}\cdot \cos(2\pi f_0 t) \hspace{-0.05cm}+\hspace{-0.05cm}\sqrt{2}\cdot \sin(2\pi f_0 t)\hspace{0.05cm}.$$
 
:$$x(t) \hspace{-0.05cm}=\hspace{-0.05cm} -\sqrt{2}\cdot \cos(2\pi f_0 t) \hspace{-0.05cm}+\hspace{-0.05cm}\sqrt{2}\cdot \sin(2\pi f_0 t)\hspace{0.05cm}.$$
  
The right sketch illustrates the trigonometric transformation:
+
The sketch on the right illustrates the trigonometrical calculation:
  
 
:$$A = 2\cdot \cos(-135^{\circ}) = -\sqrt{2}\hspace{0.05cm},$$
 
:$$A = 2\cdot \cos(-135^{\circ}) = -\sqrt{2}\hspace{0.05cm},$$
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==Spectral Representation of a Cosine Signal==
+
==Spectral representation of a cosine signal==
 
<br>
 
<br>
In order to derive the spectral function, we first restrict ourselves to a cosine signal, which can be written with the ''complex exponential function''&nbsp; and the&nbsp;[[Signal_Representation/Calculating_With_Complex_Numbers#Darstellung_nach_Betrag_und_Phase|Euler Theorem]]&nbsp; also in the following manner:
+
In order to derive the spectral function,&nbsp; we first restrict ourselves to a cosine signal,&nbsp; which also can be written with the&nbsp; complex exponential function&nbsp; and the&nbsp; [[Signal_Representation/Calculating_with_Complex_Numbers#Representation_by_magnitude_and_phase|&raquo;Euler theorem&laquo;]] &nbsp; in the following manner:
 
   
 
   
 
:$$x(t)=A \cdot \cos(2\pi f_0 t)={A}/{2}\cdot \big [{\rm e}^{\rm -j2 \pi \it f_{\rm 0} t} + {\rm e}^{\rm j2\pi \it f_{\rm 0} t} \big ].$$
 
:$$x(t)=A \cdot \cos(2\pi f_0 t)={A}/{2}\cdot \big [{\rm e}^{\rm -j2 \pi \it f_{\rm 0} t} + {\rm e}^{\rm j2\pi \it f_{\rm 0} t} \big ].$$
  
From this time domain representation it can already be seen that the cosine signal - spectrally seen - contains only one single (physical) frequency, namely the frequency&nbsp; $f_0$.
+
From this time domain representation it can be seen that the cosine signal contains  &ndash; spectrally seen &ndash; only one single&nbsp; $($physical$)$&nbsp; frequency,&nbsp; namely the frequency&nbsp; $f_0$.
  
 
{{BlaueBox|TEXT=   
 
{{BlaueBox|TEXT=   
 
$\text{Proof:}$&nbsp; For the mathematical derivation of the spectral function we use the following relations:
 
$\text{Proof:}$&nbsp; For the mathematical derivation of the spectral function we use the following relations:
*the one on the page&nbsp; [[Signal_Representation/Direct_Current_Signal_-_Limit_Case_of_a_Periodic_Signal#Diracfunktion_im_Frequenzbereich|Dirac Function in Frequency Domain]]&nbsp; derived functinal relation
+
*The equation from the section&nbsp; [[Signal_Representation/Direct_Current_Signal_-_Limit_Case_of_a_Periodic_Signal#Dirac_.28delta.29_function_in_frequency_domain|&raquo;Dirac delta function in frequency domain&laquo;]]:&nbsp;
:$$x(t)=A \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ X(f)=A \cdot \rm \delta (\it f).$$
+
:$$x(t)=A \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ X(f)=A \cdot \rm \delta (\it f),$$
  
*the&nbsp; [[Signal_Representation/Fourier_Transform_Laws#Verschiebungssatz|Displacement set (for the frequency range)]]&nbsp; in anticipation of a later chapter:
+
*the&nbsp; [[Signal_Representation/Fourier_Transform_Theorems#Shifting_Theorem|&raquo;shifting theorem&laquo;]]&nbsp; $($for the frequency range$)$&nbsp; in anticipation of a later chapter:
 
:$$x(t) \cdot {\rm e}^{\rm j2\pi\it f_{\rm 0} t}\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ X(f-f_0). $$
 
:$$x(t) \cdot {\rm e}^{\rm j2\pi\it f_{\rm 0} t}\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ X(f-f_0). $$
  
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This means:  
 
This means:  
*The spectral function&nbsp; $X(f)$&nbsp; of a cosine signal with the frequency&nbsp; $f_0$&nbsp; is composed of two Dirac functions at&nbsp; $\pm f_0$&nbsp;.
+
*The spectral function&nbsp; $X(f)$&nbsp; of a cosine signal with frequency $f_0$&nbsp; is composed of two Dirac delta functions at&nbsp; $\pm f_0$.
* The pulse weights are each equal to half the signal amplitude.}}
 
  
 +
* The impulse weights are each equal to half the signal amplitude.}}
  
[[File:P_ID1365__Sig_T_2_3_S4_neu.png|right|frame|Spectrum of a cosine signal]]
+
 
{{GraueBox|TEXT=  
+
{{GraueBox|TEXT=
 +
[[File:P_ID1365__Sig_T_2_3_S4_neu.png|right|frame|Spectrum of a cosine signal]]   
 
$\text{Example 3:}$&nbsp;
 
$\text{Example 3:}$&nbsp;
The diagram shows the spectrum of a cosine oscillation  
+
The diagram shows the spectrum of a cosine oscillation with
*with amplitude&nbsp; $A = 4 \;{\rm V}$&nbsp; and  
+
*amplitude&nbsp; $A = 4 \;{\rm V}$&nbsp; and
*the frequency&nbsp; $f_0 = 5 \;{\rm kHz}$  &nbsp; ⇒ &nbsp;  $T_0 = 200 \;{\rm &micro; s}$.  
+
 +
*frequency&nbsp; $f_0 = 5 \;{\rm kHz}$  &nbsp; ⇒ &nbsp;  $T_0 = 200 \;{\rm &micro; s}$.  
  
  
Dann gilt folgender Zusammenhang mit obiger Gleichung:
+
Then the following relation to the above equation applies:
*Die Diracfunktion bei&nbsp; $-f_0$&nbsp; gehört zum ersten Term $($ableitbar aus der Bedingung&nbsp; $f + f_0 = 0)$.
+
#The Dirac delta function at&nbsp; $-f_0$&nbsp; belongs to the first term <br>$($derivable from the condition&nbsp; $f + f_0 = 0)$.
*Die Diracfunktion bei&nbsp; $+f_0$&nbsp; gehört zum zweiten Term $($ableitbar aus der Bedingung $f - f_0 = 0)$.  
+
#The Dirac delta function for&nbsp; $+f_0$&nbsp; belongs to the second term <br>$($derivable from the condition $f - f_0 = 0)$.  
*Die Impulsgewichte sind jeweils&nbsp; $A/2 = 2 \;{\rm V}$.}}
+
#Both impulse weights are&nbsp; $A/2 = 2 \;{\rm V}$.}}
  
  
 
{{BlaueBox|TEXT=   
 
{{BlaueBox|TEXT=   
$\text{Please consider:}$&nbsp;  
+
$\text{Note:}$&nbsp;
Die Spektralfunktion einer jeden reellen Zeitfunktion mit Ausnahme des Gleichsignals weist sowohl Anteile bei positiven als auch bei negativen Frequenzen auf.  
+
The spectral function of any real time function except the DC signal has components at both positive and negative frequencies.  
*Diese Tatsache, die Studienanfängern oft Probleme bereitet, ergibt sich ganz formal aus dem&nbsp; [[Signal_Representation/Calculating_With_Complex_Numbers#Darstellung_nach_Betrag_und_Phase|Satz von Euler]] .  
+
*This fact,&nbsp; which often causes problems for first-year students,&nbsp; is formally derived from the&nbsp; [[Signal_Representation/Calculating_with_Complex_Numbers#Representation_by_magnitude_and_phase|&raquo;Euler theorem&laquo;]].
*Durch die Erweiterung des Frequenzwertebereichs von&nbsp; $f \ge 0$&nbsp; auf die Menge der reellen Zahlen kommt man von der physikalischen zur ''mathematischen Frequenz''.  
+
*Allerdings ist für eine negative Frequenz die vorne angegebene&nbsp; [[ Signal_Representation/Harmonic_Oscillation#Definition_und_Eigenschaften|Definition]]&nbsp; nicht mehr anwendbar: &nbsp; Man kann &bdquo;–5 kHz&rdquo; nicht als „minus 5000 Schwingungen pro Sekunde” interpretieren.
+
*By extending the frequency range from&nbsp; $f \ge 0$&nbsp; to the set of real numbers you get from the physical to the&nbsp; &raquo;mathematical frequency&laquo;.
 +
 +
*However,&nbsp; for a negative frequency the&nbsp; [[Signal_Representation/Harmonic_Oscillation#Definition_and_properties|$\text{definition}$]]&nbsp; is not applicable anymore: &nbsp; You cannot interpret&nbsp; $\rm -5\ kHz$&nbsp; as&nbsp; "minus 5000 oscillations per second".
  
  
Im Verlauf dieses Kurses werden Sie feststellen, dass durch die Verkomplizierung dieses einfachen Sachverhaltes später kompliziertere Sachverhalte sehr elegant und einfach beschrieben werden können.}}
+
In the process of this course you will find that&nbsp; '''by complicating a simple subject,&nbsp; more complex matters can later be described very elegantly and simply'''.}}
 
   
 
   
 
 
 
 
==Allgemeine Spektraldarstellung==
+
==General spectral representation==
 
<br>
 
<br>
Für ein sinusförmiges Signal gilt mit dem Satz von&nbsp; [https://de.wikipedia.org/wiki/Leonhard_Euler Euler]&nbsp; in ähnlicher Weise:
+
For a sinusoidal signal the&nbsp; &raquo;[https://en.wikipedia.org/wiki/Leonhard_Euler $\text{Euler}$] '''theorem'''&laquo;&nbsp; applies in a similar way:
  
 
:$$x(t)=B\cdot \sin(2 \pi f_0 t)= \frac{\it B}{2  \rm j} \cdot \big [{\rm e}^{+{\rm j} 2 \pi \it f_{\rm 0} t}-{\rm e}^{-\rm j2 \pi \it f_{\rm 0} t}\big ]=\rm j\cdot  {\it B}/{2} \cdot \big  [{\rm e}^{-j2 \pi \it f_{\rm 0} t}-{\rm e}^{+\rm j2 \pi \it f_{\rm 0} t} \big ] .$$  
 
:$$x(t)=B\cdot \sin(2 \pi f_0 t)= \frac{\it B}{2  \rm j} \cdot \big [{\rm e}^{+{\rm j} 2 \pi \it f_{\rm 0} t}-{\rm e}^{-\rm j2 \pi \it f_{\rm 0} t}\big ]=\rm j\cdot  {\it B}/{2} \cdot \big  [{\rm e}^{-j2 \pi \it f_{\rm 0} t}-{\rm e}^{+\rm j2 \pi \it f_{\rm 0} t} \big ] .$$  
  
Daraus folgt für die Spektralfunktion, die jetzt rein imaginär ist:
+
From this follows for the spectral function, which is now purely imaginary:
 +
:$$x(t)=B\cdot \sin(2\pi f_0 t)\;\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\; \ X(f)={\rm j} \cdot \big  [ {B}/{2} \cdot \delta (f+f_0)- {B}/{2} \cdot \delta (f-f_0) \big  ].$$
  
:$$x(t)=B\cdot \sin(2\pi f_0 t)\;\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\; \ X(f)={\rm j} \cdot \big  [ {B}/{2} \cdot \delta (f+f_0)- {B}/{2} \cdot \delta (f-f_0) \big  ].$$
 
  
[[File: P_ID2719__Sig_T_2_3_S5.png |right|frame|Spektrum eines Sinussignals]]
+
{{GraueBox|TEXT=
{{GraueBox|TEXT= 
+
[[File: P_ID2719__Sig_T_2_3_S5.png |right|frame|Spectrum of a sinusoidal signal]]  
$\text{Beispiel 4:}$&nbsp;
+
$\text{Example 4:}$&nbsp;
Das Bild zeigt die rein imaginäre Spektralfunktion einer Sinusschwingung&nbsp; $x(t)$&nbsp; mit
+
The figure shows the imaginary spectrum of a sine wave&nbsp; $x(t)$&nbsp; with
*Amplitude&nbsp; $B = 3 \;{\rm V},$
+
*amplitude&nbsp; $B = 3 \;{\rm V},$
*Frequenz&nbsp; $f_0 = 5\;{\rm kHz},$
 
*Phase&nbsp;  $\varphi=90^{\circ}$  &nbsp; ⇒  &nbsp; $\phi = -90^{\circ}$.
 
<br>$\text{Bitte beachten Sie}$:
 
  
*Bei der positiven Frequenz $(f = +f_0)$ ist der Imaginärteil negativ.
+
*frequency&nbsp; $f_0 = 5\;{\rm kHz},$
 
   
 
   
*Bei der negativen Frequenz $(f = -f_0)$ ergibt sich ein positiver Imaginärteil.}}
+
*zero phase angle&nbsp;  $\varphi=90^{\circ}$  &nbsp; ⇒  &nbsp; $\phi = -90^{\circ}$.
 +
<br>$\text{Please note}$:
  
 +
#For the positive frequency&nbsp; $(f = +f_0)$&nbsp; the imaginary part is negative.<br>
 +
#The negative frequency&nbsp; $(f = -f_0)$&nbsp; results in a positive imaginary part.}}
  
Bei Überlagerung von Cosinus– und Sinusanteil entsprechend der Beziehung
+
 
 +
When the cosine and sine components are superimposed according to the relationship
  
 
:$$x(t)=A \cdot \cos(2\pi  f_0  t) +B \cdot \sin(2 \pi  f_0  t)$$
 
:$$x(t)=A \cdot \cos(2\pi  f_0  t) +B \cdot \sin(2 \pi  f_0  t)$$
 
   
 
   
überlagern sich auch die einzelnen Spektralfunktionen und man erhält:
+
the individual spectral functions also overlap and you get
 
   
 
   
 
:$$X(f)=\frac{A+{\rm j} \cdot B}{2}\cdot {\rm \delta} (f+f_0)+\frac{A-{\rm j} \cdot B}{2} \cdot \delta (f-f_0).$$
 
:$$X(f)=\frac{A+{\rm j} \cdot B}{2}\cdot {\rm \delta} (f+f_0)+\frac{A-{\rm j} \cdot B}{2} \cdot \delta (f-f_0).$$
  
Mit dem Betrag&nbsp; $C$&nbsp; und der Phase&nbsp; $\varphi$&nbsp; lautet diese Fourierkorrespondenz:
+
This Fourier correlation is with amplitude&nbsp; $C$&nbsp; and phase&nbsp; $\varphi$:&nbsp;  
 
   
 
   
 
:$$x(t)=C\cdot \cos(2\pi f_0 t-\varphi)\ \, \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,\ X(f)={C}/{2}\cdot {\rm e}^{{\rm j} \varphi} \cdot \delta(f+f_0) + {C}/{2} \cdot {\rm e}^{\rm{-j} \varphi} \cdot \delta (f-f_0).$$
 
:$$x(t)=C\cdot \cos(2\pi f_0 t-\varphi)\ \, \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,\ X(f)={C}/{2}\cdot {\rm e}^{{\rm j} \varphi} \cdot \delta(f+f_0) + {C}/{2} \cdot {\rm e}^{\rm{-j} \varphi} \cdot \delta (f-f_0).$$
  
 
{{BlaueBox|TEXT=   
 
{{BlaueBox|TEXT=   
$\text{Man erkennt:}$&nbsp; Die Spektralfunktion&nbsp; $X(f)$  
+
$\text{One recognizes:}$&nbsp; The spectral function&nbsp; $X(f)$  
*ist nicht nur für positive und negative Frequenzen definiert,  
+
*is not only defined for positive and negative frequencies,  
*sondern im Allgemeinen auch noch komplexwertig.}}
 
  
 +
*but it is also complex-valued.}}
  
[[File:Sig_T_2_3_S6_version2.png|right|frame|Allgemeine Spektralfunktion einer harmonischen Schwingung]]
 
{{GraueBox|TEXT= 
 
$\text{Beispiel 5:}$&nbsp;
 
Mit den Parametern&nbsp; $C = 5 \;{\rm V}$,&nbsp; $f_0 = 5\;{\rm kHz}$&nbsp; und&nbsp; $\varphi=90^{\circ}$&nbsp; $($im Bogenmaß $\pi/6)$&nbsp; ergibt sich wegen
 
  
:$$2.5 · \cos(30^{\circ}) = 2.165, \hspace{0.3cm} 2.5 · \sin(30^{\circ}) = 1.25$$  
+
{{GraueBox|TEXT=
der Real- bzw. der Imaginärteil von $X(f)$ gemäß der Grafik:
+
[[File:Sig_T_2_3_S6_version2.png|right|frame|Spectrum of the&nbsp; $\varphi=30^{\circ}$&nbsp; oscillation]] 
 +
$\text{Example 5:}$&nbsp;
 +
With the parameters&nbsp; $C = 5 \;{\rm V}$,&nbsp; $f_0 = 5\;{\rm kHz}$&nbsp; and&nbsp; $\varphi=30^{\circ}$&nbsp; $($in radian measure $\pi/6)$&nbsp; results
 +
the real and/or imaginary part of $X(f)$ according to the graph:
  
 
:$${\rm Re}\big[X(f)\big]=2.165\,{\rm V} \cdot \delta(f+f_0) + 2.165\,{\rm V}  \cdot \delta (f-f_0).$$
 
:$${\rm Re}\big[X(f)\big]=2.165\,{\rm V} \cdot \delta(f+f_0) + 2.165\,{\rm V}  \cdot \delta (f-f_0).$$
  
:$${\rm Im}\big[X(f)\big]=1.25\,{\rm V} \cdot \delta(f+f_0) -1.25\,{\rm V}  \cdot \delta (f-f_0).$$}}
+
:$${\rm Im}\big[X(f)\big]=1.25\,{\rm V} \cdot \delta(f+f_0) -1.25\,{\rm V}  \cdot \delta (f-f_0).$$
 +
 
 +
Because of:
 +
 
 +
:$$2.5 · \cos(30^{\circ}) = 2.165,$$
 +
:$$2.5 · \sin(30^{\circ}) = 1.25.$$}}  
  
  
Das Lernvideo&nbsp;
+
The following&nbsp; (German-language)&nbsp; learning video illustrates the properties of harmonic oscillations using scales: <br>
[[Harmonische_Schwingungen_(Lernvideo)|Harmonische Schwingungen]]&nbsp; verdeutlicht die Eigenschaften harmonischer Schwingungen anhand der so genannten Tonleiter.  
+
&nbsp; &nbsp; &nbsp; [[Harmonische_Schwingungen_(Lernvideo)|&raquo;Harmonische Schwingungen&raquo;]] &nbsp; &rArr; &nbsp; "Harmonic Oscillations".  
  
  
==Aufgaben zum Kapitel==
+
==Exercises for the chapter==
 
<br>
 
<br>
[[Aufgaben:2.3 cos- und sin-Anteil|Aufgabe 2.3: Cosinus- und Sinusanteil]]
+
[[Aufgaben:Exercises_2.3:_Cosine_and_Sine_Components|Exercise 2.3: Cosine and Sine Components]]
  
[[Aufgaben:Aufgabe 2.3Z:_ Schwingungsparameter|Aufgabe 2.3: Schwingungsparameter]]
+
[[Aufgaben:Exercise_2.3Z:_Oscillation_Parameters|Exercise 2.3Z: Oscillation Parameters]]
  
  
 
{{Display}}
 
{{Display}}

Latest revision as of 16:39, 9 June 2023

Definition and properties


Example of a harmonic oscillation

Harmonic oscillations are of particular importance for Communications Engineering as well as in many natural sciences.  The diagram shows an exemplary signal waveform.

Its importance is also related to the fact that the harmonic oscillation represents the solution of a  differential equation  which is found in many disciplines and reads as follows:

$$ x(t) + k \cdot\ddot{x} (t) = 0.$$

Here the two dots mark the second derivative of the function  $x(t)$  after time.

$\text{Definition:}$  Any  »harmonic oscillation«  can be represented in most general form as follows:

$$x(t)= C \cdot \cos(2\pi f_0 t - \varphi).$$

The following signal parameters are used:

  • the  »amplitude«  $C$  – simultaneously the maximum value of the signal,
  • the  »signal frequency«  $f_{0}$   ⇒   the reciprocal of the period duration  $T_{0}$, and
  • the  »zero phase angle«  $($or briefly the  »phase«$)$   $\varphi$  of the oscillation.


The following  $($German-language$)$  learning video illustrates the properties of harmonic oscillations using scales:
      »Harmonische Schwingungen«   ⇒   "Harmonic Oscillations".

$\text{Comments on nomenclature:}$ 

In this tutorial – as usual in other literature when describing harmonic oscillations,  Fourier series and Fourier integral – the phase is entered into the equations with negative sign,  whereas in connection with all modulation methods the phase is entered with a plus sign.

  • To distinguish between the two variants we use in our tutorial   $\varphi$  and  $\phi$.  Both symbols denote the small Greek "phi".  The spelling  $\varphi$  is mainly used in the German and  $\phi$  in the anglo–american language area.
  • The indications  $\varphi = 90^{\circ}$  and  $\phi = -90^{\circ}$  are thus equivalent and both stand for the sine function:
$$\cos(2 \pi f_0 t - 90^{\circ}) = \cos(2 \pi f_0 t - \varphi) = \cos(2 \pi f_0 t + \phi) = \sin(2 \pi f_0 t ).$$


Time domain representation


The amplitude  $C$  can be read directly from the adjacent graph.  The signal frequency $f_0$  is the reciprocal of the period duration  $T_0$. 

Signal parameters of a harmonic oscillation

If the above equation is written in the form

$$x(t) = C \cdot \cos(2\pi f_0 t - \varphi) = C \cdot \cos \big(2\pi f_0 (t - \tau) \big), $$

it becomes clear that the zero phase angle  $\varphi$  and the shift  $\tau$  relative to a cosine signal are related as follows

$$\varphi = \frac{\tau}{T_0} \cdot 2{\pi}. $$
  • For a cosine signal both the parameters  $\tau$  and  $\varphi$  are zero.
  • In contrast,  a sinusoidal signal  is shifted by  $\tau = T_0/4$  and accordingly applies to the zero phase angle  $\varphi = \pi/2$  $($in radians$)$  or   $90^{\circ}$.
  • So it can be stated that – as assumed for the above sketch – at a positive value of  $\tau$  resp.   $\varphi$  the $($referring  $t = 0)$  nearest signal maximum comes later than at the cosine signal and at negative values earlier.
  • If a cosine signal is present at the system input and the output signal is delayed by a value  $\tau$  then  $\tau$  is also called the  »runtime«  of the system.
  • Since a harmonic oscillation is clearly defined by three parameters,  the entire time course from  $-\infty$  to  $+\infty$  can be calculated analytically from only three signal values  $x_1=x(t_1)$,  $x_2=x(t_2)$,  $x_3=x(t_3)$  if the times  $t_1$,  $t_2$  and  $t_3$  have been determined appropriately.


Harmonic oscillation, defined by only  $3$  samples

$\text{Example 1:}$  From the three sample values,

$$x_1 = x(t_1 = 3.808 \;{\rm ms}) = +1.609,$$
$$x_2 = x(t_2 = 16.696 \;{\rm ms})=\hspace{0.05 cm} -0.469,$$
$$x_3 = x(t_3 = 33.84 \;{\rm ms}) = +1.227,$$

you get the following system of equations:

$$C \cdot \cos(2\pi \hspace{0.05 cm} f_0 \hspace{0.05 cm} t_1 - \varphi) = +1.609\hspace{0.05 cm},$$
$$C \cdot \cos(2\pi \hspace{0.05 cm} f_0 \hspace{0.05 cm} t_2 - \varphi) = \hspace{0.05 cm} -0.469\hspace{0.05 cm},$$
$$C \cdot \cos(2\pi \hspace{0.05 cm} f_0 \hspace{0.05 cm} t_3 - \varphi) = +1.227\hspace{0.05 cm}.$$

After solving this nonlinear system of equations the following signal parameters are obtained:

  • Signal amplitude:  $C = 2$,
  • period duration:  $T_0 = 8 \;{\rm ms}$   ⇒   signal frequency $f_0 = 125 \;{\rm Hz}$,
  • shift with respect to a cosine:  $\tau = 3 \;{\rm ms}$   ⇒   zero phase angle  $\varphi = 3\pi /4 = 135^\circ$.


If you set all sampling times  $t_1$,  $t_2$,  $t_3$  in maxima, minima and/or zeros, there is no unique solution for the nonlinear equation system.


Representation with cosine and sine components


Another representation of the harmonic oscillation is as follows:

$$x(t)=A\cdot\cos(2\pi f_0 t)+ B\cdot\sin(2\pi f_0 t).$$
  • The terms  $A$  and  $B$  for the amplitudes of the cosine and sine components are chosen to match the nomenclature of the following chapter  »Fourier Series«.
  • By applying trigonometric transformations we obtain from the illustration in the last section:
$$x(t)=C\cdot \cos(2\pi f_0 t-\varphi)=C\cdot\cos(\varphi)\cdot\cos(2\pi f_0 t)+C\cdot\sin(\varphi)\cdot\sin(2\pi f_0 t).$$
  • From this follows directly by equating the coefficients:
$$A=C\cdot\cos(\varphi),$$
$$B=C\cdot\sin(\varphi).$$
  • The magnitude and the zero phase angle of the harmonic oscillation can be calculated from the parameters  $A$  and  $B$  also according to simple trigonometric considerations:
$$C=\sqrt{A^2+B^2},$$
$$\varphi = \arctan\left({-B}/{A}\right).$$

$\text{Please note:}$  The minus–sign at the calculation of the zero phase angle  $\varphi$  is related to the fact  that  $\varphi$  enters the argument of the cosine function with negative sign.  If one would use the notation  $\cos(2\pi f_0 t +\phi)$  instead of  $\cos(2\pi f_0 t - \varphi)$,  then  $\phi= \arctan(B/A)$.  Note the following here:

  1. For Fourier series and Fourier integral, the  $\varphi$  representation is common in literature.
  2. For description of the modulation methods, however, the  $\phi$  representation is almost always used.


$\text{Example 2:}$  The oscillation shown in the left graphic as time course is characterized by the parameters
Harmonic oscillation. On the right: Represented in the complex plane
  • $C=2,$
  • $f_0 = 125 \;{\rm Hz}$,
  • $\varphi = +135^{\circ}$   ⇒   $\phi = -135^{\circ}$.


The oscillation is completely described by each of the two equations:

$$x(t) \hspace{-0.05cm}=\hspace{-0.05cm} 2\cdot \cos(2\pi f_0 t-135^{\circ})\hspace{0.05cm},$$
$$x(t) \hspace{-0.05cm}=\hspace{-0.05cm} -\sqrt{2}\cdot \cos(2\pi f_0 t) \hspace{-0.05cm}+\hspace{-0.05cm}\sqrt{2}\cdot \sin(2\pi f_0 t)\hspace{0.05cm}.$$

The sketch on the right illustrates the trigonometrical calculation:

$$A = 2\cdot \cos(-135^{\circ}) = -\sqrt{2}\hspace{0.05cm},$$
$$B = 2\cdot \sin(-135^{\circ}) = +\sqrt{2}\hspace{0.05cm}.$$


Spectral representation of a cosine signal


In order to derive the spectral function,  we first restrict ourselves to a cosine signal,  which also can be written with the  complex exponential function  and the  »Euler theorem«   in the following manner:

$$x(t)=A \cdot \cos(2\pi f_0 t)={A}/{2}\cdot \big [{\rm e}^{\rm -j2 \pi \it f_{\rm 0} t} + {\rm e}^{\rm j2\pi \it f_{\rm 0} t} \big ].$$

From this time domain representation it can be seen that the cosine signal contains – spectrally seen – only one single  $($physical$)$  frequency,  namely the frequency  $f_0$.

$\text{Proof:}$  For the mathematical derivation of the spectral function we use the following relations:

$$x(t)=A \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ X(f)=A \cdot \rm \delta (\it f),$$
$$x(t) \cdot {\rm e}^{\rm j2\pi\it f_{\rm 0} t}\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ X(f-f_0). $$

This results in the following Fourier correspondence:

$$x(t)=A\cdot \cos(2\pi f_0t)\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ X(f)={A}/{\rm 2}\cdot {\rm \delta} (f+f_{\rm 0})+{A}/{\rm 2}\cdot {\rm \delta}(\ f-f_{\rm 0}).$$

This means:

  • The spectral function  $X(f)$  of a cosine signal with frequency $f_0$  is composed of two Dirac delta functions at  $\pm f_0$.
  • The impulse weights are each equal to half the signal amplitude.


Spectrum of a cosine signal

$\text{Example 3:}$  The diagram shows the spectrum of a cosine oscillation with

  • amplitude  $A = 4 \;{\rm V}$  and
  • frequency  $f_0 = 5 \;{\rm kHz}$   ⇒   $T_0 = 200 \;{\rm µ s}$.


Then the following relation to the above equation applies:

  1. The Dirac delta function at  $-f_0$  belongs to the first term
    $($derivable from the condition  $f + f_0 = 0)$.
  2. The Dirac delta function for  $+f_0$  belongs to the second term
    $($derivable from the condition $f - f_0 = 0)$.
  3. Both impulse weights are  $A/2 = 2 \;{\rm V}$.


$\text{Note:}$  The spectral function of any real time function except the DC signal has components at both positive and negative frequencies.

  • This fact,  which often causes problems for first-year students,  is formally derived from the  »Euler theorem«.
  • By extending the frequency range from  $f \ge 0$  to the set of real numbers you get from the physical to the  »mathematical frequency«.
  • However,  for a negative frequency the  $\text{definition}$  is not applicable anymore:   You cannot interpret  $\rm -5\ kHz$  as  "minus 5000 oscillations per second".


In the process of this course you will find that  by complicating a simple subject,  more complex matters can later be described very elegantly and simply.


General spectral representation


For a sinusoidal signal the  »$\text{Euler}$ theorem«  applies in a similar way:

$$x(t)=B\cdot \sin(2 \pi f_0 t)= \frac{\it B}{2 \rm j} \cdot \big [{\rm e}^{+{\rm j} 2 \pi \it f_{\rm 0} t}-{\rm e}^{-\rm j2 \pi \it f_{\rm 0} t}\big ]=\rm j\cdot {\it B}/{2} \cdot \big [{\rm e}^{-j2 \pi \it f_{\rm 0} t}-{\rm e}^{+\rm j2 \pi \it f_{\rm 0} t} \big ] .$$

From this follows for the spectral function, which is now purely imaginary:

$$x(t)=B\cdot \sin(2\pi f_0 t)\;\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\; \ X(f)={\rm j} \cdot \big [ {B}/{2} \cdot \delta (f+f_0)- {B}/{2} \cdot \delta (f-f_0) \big ].$$


Spectrum of a sinusoidal signal

$\text{Example 4:}$  The figure shows the imaginary spectrum of a sine wave  $x(t)$  with

  • amplitude  $B = 3 \;{\rm V},$
  • frequency  $f_0 = 5\;{\rm kHz},$
  • zero phase angle  $\varphi=90^{\circ}$   ⇒   $\phi = -90^{\circ}$.


$\text{Please note}$:

  1. For the positive frequency  $(f = +f_0)$  the imaginary part is negative.
  2. The negative frequency  $(f = -f_0)$  results in a positive imaginary part.


When the cosine and sine components are superimposed according to the relationship

$$x(t)=A \cdot \cos(2\pi f_0 t) +B \cdot \sin(2 \pi f_0 t)$$

the individual spectral functions also overlap and you get

$$X(f)=\frac{A+{\rm j} \cdot B}{2}\cdot {\rm \delta} (f+f_0)+\frac{A-{\rm j} \cdot B}{2} \cdot \delta (f-f_0).$$

This Fourier correlation is with amplitude  $C$  and phase  $\varphi$: 

$$x(t)=C\cdot \cos(2\pi f_0 t-\varphi)\ \, \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,\ X(f)={C}/{2}\cdot {\rm e}^{{\rm j} \varphi} \cdot \delta(f+f_0) + {C}/{2} \cdot {\rm e}^{\rm{-j} \varphi} \cdot \delta (f-f_0).$$

$\text{One recognizes:}$  The spectral function  $X(f)$

  • is not only defined for positive and negative frequencies,
  • but it is also complex-valued.


Spectrum of the  $\varphi=30^{\circ}$  oscillation

$\text{Example 5:}$  With the parameters  $C = 5 \;{\rm V}$,  $f_0 = 5\;{\rm kHz}$  and  $\varphi=30^{\circ}$  $($in radian measure $\pi/6)$  results the real and/or imaginary part of $X(f)$ according to the graph:

$${\rm Re}\big[X(f)\big]=2.165\,{\rm V} \cdot \delta(f+f_0) + 2.165\,{\rm V} \cdot \delta (f-f_0).$$
$${\rm Im}\big[X(f)\big]=1.25\,{\rm V} \cdot \delta(f+f_0) -1.25\,{\rm V} \cdot \delta (f-f_0).$$

Because of:

$$2.5 · \cos(30^{\circ}) = 2.165,$$
$$2.5 · \sin(30^{\circ}) = 1.25.$$


The following  (German-language)  learning video illustrates the properties of harmonic oscillations using scales:
      »Harmonische Schwingungen»   ⇒   "Harmonic Oscillations".


Exercises for the chapter


Exercise 2.3: Cosine and Sine Components

Exercise 2.3Z: Oscillation Parameters