Signal Representation/Harmonic Oscillation - Revision history

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2026-03-16T12:28:34Z

Add German interlanguage link

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	:$$x(t)=A \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ X(f)=A \cdot \rm \delta (\it f),$$		:$$x(t)=A \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ X(f)=A \cdot \rm \delta (\it f),$$

	*the  [[Signal_Representation/~~Fourier_Transform_Theorems~~#Shifting_Theorem\|»shifting theorem«]]  $($for the frequency range$)$  in anticipation of a later chapter:		*the  [[Signal_Representation/The_Fourier_Transform_Theorems#Shifting_Theorem\|»shifting theorem«]]  $($for the frequency range$)$  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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	{{Display}}		{{Display}}
			[[de:Signaldarstellung/Harmonische_Schwingung]]

Guenter at 15:39, 9 June 2023

2023-06-09T15:39:34Z

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

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

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

Guenter at 15:35, 9 June 2023

2023-06-09T15:35:43Z

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	==General spectral representation==		==General spectral representation==
	<br>		<br>
	For a sinusoidal signal the  [https://en.wikipedia.org/wiki/Leonhard_Euler $\text{Euler}$] theorem applies in a similar way:		For a sinusoidal signal the  »[https://en.wikipedia.org/wiki/Leonhard_Euler $\text{Euler}$] '''theorem'''«  applies in a similar way:

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

	~~With the~~ amplitude  $C$  and ~~the~~ phase  $\varphi$  ~~this Fourier correlation is~~		This Fourier correlation is with amplitude  $C$  and phase  $\varphi$

	:$$x(t)=C\cdot \cos(2\pi f_0 t-\varphi)\ \, \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,\ X(f)={C}/{2}\cdot {\rm e}^{{\rm j} \varphi} \cdot \delta(f+f_0) + {C}/{2} \cdot {\rm e}^{\rm{-j} \varphi} \cdot \delta (f-f_0).$$		:$$x(t)=C\cdot \cos(2\pi f_0 t-\varphi)\ \, \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,\ X(f)={C}/{2}\cdot {\rm e}^{{\rm j} \varphi} \cdot \delta(f+f_0) + {C}/{2} \cdot {\rm e}^{\rm{-j} \varphi} \cdot \delta (f-f_0).$$
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	The following (German language) learning video illustrates the properties of harmonic oscillations using scales: <br>		The following  (German-language)  learning video illustrates the properties of harmonic oscillations using scales: <br>
	[[Harmonische_Schwingungen_(Lernvideo)\|"Harmonische Schwingungen"]]   ⇒   "Harmonic Oscillations".		[[Harmonische_Schwingungen_(Lernvideo)\|»Harmonische Schwingungen»]]   ⇒   "Harmonic Oscillations".

Guenter at 15:27, 9 June 2023

2023-06-09T15:27:15Z

Guenter at 14:15, 9 June 2023

2023-06-09T14:15:40Z

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Guenter at 16:33, 8 February 2023

2023-02-08T16:33:35Z

← Older revision		Revision as of 18:33, 8 February 2023
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	:$$x(t)=B\cdot \sin(2\pi f_0 t)\;\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\; \ X(f)={\rm j} \cdot \big [ {B}/{2} \cdot \delta (f+f_0)- {B}/{2} \cdot \delta (f-f_0) \big ].$$		:$$x(t)=B\cdot \sin(2\pi f_0 t)\;\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\; \ X(f)={\rm j} \cdot \big [ {B}/{2} \cdot \delta (f+f_0)- {B}/{2} \cdot \delta (f-f_0) \big ].$$

	[[File: P_ID2719__Sig_T_2_3_S5.png \|right\|frame\|Spectrum of a sinusoidal signal]]
	~~{{GraueBox\|TEXT=~~		{{GraueBox\|TEXT=
			[[File: P_ID2719__Sig_T_2_3_S5.png \|right\|frame\|Spectrum of a sinusoidal signal]]
	$\text{Example 4:}$		$\text{Example 4:}$
	The figure shows the imaginary spectrum of a sine wave  $x(t)$  with		The figure shows the imaginary spectrum of a sine wave  $x(t)$  with
	*amplitude  $B = 3 \;{\rm V},$		*amplitude  $B = 3 \;{\rm V},$
	*frequency  $f_0 = 5\;{\rm kHz},$
			*frequency  $f_0 = 5\;{\rm kHz},$

	*zero phase angle  $\varphi=90^{\circ}$   ⇒   $\phi = -90^{\circ}$.		*zero phase angle  $\varphi=90^{\circ}$   ⇒   $\phi = -90^{\circ}$.
	<br>$\text{Please note}$:		<br>$\text{Please note}$:

	*For the positive frequency  $(f = +f_0)$  the imaginary part is negative.		#For the positive frequency  $(f = +f_0)$  the imaginary part is negative.<br>
			#The negative frequency  $(f = -f_0)$  results in a positive imaginary part.}}
	*The negative frequency  $(f = -f_0)$  results in a positive imaginary part.}}


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	$\text{One recognizes:}$  The spectral function  $X(f)$		$\text{One recognizes:}$  The spectral function  $X(f)$
	*is not only defined for positive and negative frequencies,		*is not only defined for positive and negative frequencies,

	*but it is also complex-valued.}}		*but it is also complex-valued.}}


	[[File:Sig_T_2_3_S6_version2.png\|right\|frame\|Spectrum of the  $\varphi=30^{\circ}$  oscillation]]		{{GraueBox\|TEXT=
	~~{{GraueBox\|TEXT=~~		[[File:Sig_T_2_3_S6_version2.png\|right\|frame\|Spectrum of the  $\varphi=30^{\circ}$  oscillation]]
	$\text{Example 5:}$		$\text{Example 5:}$
	With the parameters  $C = 5 \;{\rm V}$,  $f_0 = 5\;{\rm kHz}$  and  $\varphi=30^{\circ}$  $($in radian measure $\pi/6)$  results		With the parameters  $C = 5 \;{\rm V}$,  $f_0 = 5\;{\rm kHz}$  and  $\varphi=30^{\circ}$  $($in radian measure $\pi/6)$  results
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	Because of:		Because of:

	:$$2.5 · \cos(30^{\circ}) = 2.165, ~~\hspace{0.3cm}~~ 2.5 · \sin(30^{\circ}) = 1.25.$$}}		:$$2.5 · \cos(30^{\circ}) = 2.165,$$
			:$$2.5 · \sin(30^{\circ}) = 1.25.$$}}

Guenter at 16:28, 8 February 2023

2023-02-08T16:28:31Z

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Guenter at 17:10, 2 December 2022

2022-12-02T17:10:56Z

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	{{BlaueBox\|TEXT=		{{BlaueBox\|TEXT=
	$\text{Please note:}$		$\text{Please note:}$
	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.		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:

	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:

	*For Fourier series and Fourier integral, the  $\varphi$  representation is common in literature.		*For Fourier series and Fourier integral, the  $\varphi$  representation is common in literature.

Guenter at 17:08, 2 December 2022

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	{{BlaueBox\|TEXT=		{{BlaueBox\|TEXT=
	$\text{Please ~~consider~~:}$		$\text{Please note:}$
	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 a negative sign.		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:		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:

Guenter at 17:06, 2 December 2022

2022-12-02T17:06:45Z

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	Then the following relation to the above equation applies:		Then the following relation to the above equation applies:
	*The Dirac ~~'''KORREKTUR:~~ delta~~?'''~~ function at  $-f_0$  belongs to the first term <br>$($derivable from the condition  $f + f_0 = 0)$.		*The Dirac delta function at  $-f_0$  belongs to the first term <br>$($derivable from the condition  $f + f_0 = 0)$.
	*The Dirac ~~'''KORREKTUR:~~ delta~~?'''~~ function for  $+f_0$  belongs to the second term <br>$($derivable from the condition $f - f_0 = 0)$.		*The Dirac delta function for  $+f_0$  belongs to the second term <br>$($derivable from the condition $f - f_0 = 0)$.
	*Both impulse weights are  $A/2 = 2 \;{\rm V}$.}}		*Both impulse weights are  $A/2 = 2 \;{\rm V}$.}}

← Older revision		Revision as of 17:27, 9 June 2023
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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 also can be written with the  complex exponential function  and the  [[Signal_Representation/Calculating_with_Complex_Numbers#Representation_by_magnitude_and_phase\|"Euler theorem"]]   in the following manner:		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  [[Signal_Representation/Calculating_with_Complex_Numbers#Representation_by_magnitude_and_phase\|»Euler theorem«]]   in the following manner:

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

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

	{{BlaueBox\|TEXT=		{{BlaueBox\|TEXT=
	$\text{Proof:}$  For the mathematical derivation of the spectral function we use the following relations:		$\text{Proof:}$  For the mathematical derivation of the spectral function we use the following relations:
	*~~the~~ equation from the section  [[Signal_Representation/Direct_Current_Signal_-_Limit_Case_of_a_Periodic_Signal#Dirac_.28delta.29_function_in_frequency_domain\|"Dirac delta function in frequency domain"]]:		*The equation from the section  [[Signal_Representation/Direct_Current_Signal_-_Limit_Case_of_a_Periodic_Signal#Dirac_.28delta.29_function_in_frequency_domain\|»Dirac delta function in frequency domain«]]:
	:$$x(t)=A \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ X(f)=A \cdot \rm \delta (\it f).$$		:$$x(t)=A \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ X(f)=A \cdot \rm \delta (\it f),$$

	*the  [[Signal_Representation/Fourier_Transform_Theorems#Shifting_Theorem\|"shifting theorem"]]  (for the frequency range)  in anticipation of a later chapter:		*the  [[Signal_Representation/Fourier_Transform_Theorems#Shifting_Theorem\|»shifting theorem«]]  $($for the frequency range$)$  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  $X(f)$  of a cosine signal with ~~the~~ frequency $f_0$  is composed of two Dirac delta functions at  $\pm f_0$.		*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.}}		* The impulse weights are each equal to half the signal amplitude.}}
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	[[File:P_ID1365__Sig_T_2_3_S4_neu.png\|right\|frame\|Spectrum of a cosine signal]]		[[File:P_ID1365__Sig_T_2_3_S4_neu.png\|right\|frame\|Spectrum of a cosine signal]]
	$\text{Example 3:}$		$\text{Example 3:}$
	The diagram shows the spectrum of a cosine oscillation		The diagram shows the spectrum of a cosine oscillation with
	*~~with~~ amplitude  $A = 4 \;{\rm V}$  and		*amplitude  $A = 4 \;{\rm V}$  and

	*~~the~~ frequency  $f_0 = 5 \;{\rm kHz}$   ⇒   $T_0 = 200 \;{\rm µ s}$.		*frequency  $f_0 = 5 \;{\rm kHz}$   ⇒   $T_0 = 200 \;{\rm µ s}$.


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	$\text{Note:}$		$\text{Note:}$
	The spectral function of any real time function except the DC signal has components at both positive and negative frequencies.		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  [[Signal_Representation/Calculating_with_Complex_Numbers#Representation_by_magnitude_and_phase\|"Euler theorem"]].		*This fact,  which often causes problems for first-year students,  is formally derived from the  [[Signal_Representation/Calculating_with_Complex_Numbers#Representation_by_magnitude_and_phase\|»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".		*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  [[Signal_Representation/Harmonic_Oscillation#Definition_and_properties\|$\text{definition}$]]  is not applicable anymore:   You cannot interpret  $\rm -5 kHz$  as  ~~<br>~~"minus 5000 oscillations per second".		*However,  for a negative frequency the  [[Signal_Representation/Harmonic_Oscillation#Definition_and_properties\|$\text{definition}$]]  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'''.}}		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'''.}}