Signal Representation/The Fourier Transform and its Inverse - Revision history

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	:$$Y(f) = \mathop {\lim }\limits_{\varepsilon \to 0 } Y_\varepsilon (f) = \mathop {\lim }\limits_{\varepsilon \hspace{0.05cm} \to \hspace{0.05cm}0 }\frac{1}{ {\varepsilon + {\rm{j} }2{\rm{\pi } }f} } = \frac{1}{ { {\rm{j2\pi } }f} }.$$		:$$Y(f) = \mathop {\lim }\limits_{\varepsilon \to 0 } Y_\varepsilon (f) = \mathop {\lim }\limits_{\varepsilon \hspace{0.05cm} \to \hspace{0.05cm}0 }\frac{1}{ {\varepsilon + {\rm{j} }2{\rm{\pi } }f} } = \frac{1}{ { {\rm{j2\pi } }f} }.$$

	*In addition there is a Dirac delta function at  $f = 0$  with weight  $1/2$, due to the equal part.  This is explained in the example in section  [[Signal_Representation/~~Fourier_Transform_Theorems~~#Assignment_Theorem\|»Assignment Theorem«]].  }}		*In addition there is a Dirac delta function at  $f = 0$  with weight  $1/2$, due to the equal part.  This is explained in the example in section  [[Signal_Representation/The_Fourier_Transform_Theorems#Assignment_Theorem\|»Assignment Theorem«]].  }}


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	{{Display}}		{{Display}}
			[[de:Signaldarstellung/Fouriertransformation_und_-rücktransformation]]

Guenter at 11:38, 14 June 2023

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	==Closer examination of the Fourier coefficients==		==Closer examination of the Fourier coefficients==
	<br>		<br>
	We assume a periodic signal  $x_{\rm P}(t)$  with period duration  $T_0$  which corresponds to the explanations in section  [[Signal_Representation/Fourier_Series#Complex_Fourier_series\|"Complex Fourier series"]].		We assume a periodic signal  $x_{\rm P}(t)$  with period duration  $T_0$  which corresponds to the explanations in section  [[Signal_Representation/Fourier_Series#Complex_Fourier_series\|»Complex Fourier series«]].
	[[File:P_ID538__Sig_T_3_1_S2b_rah.png\|right\|frame\|Periodic signal  $x_{\rm P}(t)$  and  $x_{\rm P}\hspace{0.01cm}'(t)$  and its line spectra]]		[[File:P_ID538__Sig_T_3_1_S2b_rah.png\|right\|frame\|Periodic signal  $x_{\rm P}(t)$  and  $x_{\rm P}\hspace{0.01cm}'(t)$  and its line spectra]]
	*This signal can be described as follows:		*This signal can be described as follows:

Guenter at 11:36, 14 June 2023

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Guenter at 09:04, 14 June 2023

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	Thus,  a fundamental definition is obtained,  which allows the calculation of the spectral function of an aperiodic time function.  The name of this spectral transformation goes back to the French physicist  [https://en.wikipedia.org/wiki/Joseph_Fourier $\text{Jean-Baptiste Joseph Fourier}$].		Thus,  a fundamental definition is obtained,  which allows the calculation of the spectral function of an aperiodic time function.  The name of this spectral transformation goes back to the French physicist  [https://en.wikipedia.org/wiki/Joseph_Fourier »$\text{Jean-Baptiste Joseph Fourier}$«].

	{{BlaueBox\|TEXT=		{{BlaueBox\|TEXT=
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	{{BlaueBox\|TEXT=		{{BlaueBox\|TEXT=
	$\text{Definition:}$		$\text{Definition:}$
	The generally valid functional relation  $X(f) = F\big [x(t) \big ]$  is called  »'''Fourier Transform}'''«.  For the short notation we use  $($with the  "white dot"  for the time domain and the  "filled dot"  for the spectral domain$)$:		The generally valid functional relation  $X(f) = F\big [x(t) \big ]$  is called  »'''Fourier Transform'''«.  For the short notation we use  $($with the  "white dot"  for the time domain and the  "filled dot"  for the spectral domain$)$:

	:$$X(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ x(t).$$		:$$X(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ x(t).$$

Guenter at 16:01, 13 June 2023

2023-06-13T16:01:36Z

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Guenter at 15:14, 13 June 2023

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Guenter at 14:22, 13 June 2023

2023-06-13T14:22:34Z

← Older revision		Revision as of 16:22, 13 June 2023
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	== # OVERVIEW OF THE THIRD MAIN CHAPTER # ==		== # OVERVIEW OF THE THIRD MAIN CHAPTER # ==
	<br>		<br>
	In the second chapter periodic signals were described by different harmonic oscillations  ("Fourier series").		In the second chapter periodic signals were described by different harmonic oscillations  (»Fourier series«).

	If one reduces – at least mentally – the repetition frequency of a periodic signal more and more,  i.e.,  the period duration becomes longer and longer,  then one comes from the periodic signal to the unique  ~~$\text{~~aperiodic signal}$  – often also called  ~~"$\text{~~pulse~~}$"~~.		If one reduces – at least mentally – the repetition frequency of a periodic signal more and more,  i.e.,  the period duration becomes longer and longer,  then one comes from the periodic signal to the unique  »aperiodic signal«  – often also called  »pulse«.

	In the following,  such aperiodic and pulse–shaped signals are considered and mathematically described in the time and frequency domain.		In the following,  such aperiodic and pulse–shaped signals are considered and mathematically described in the time and frequency domain.
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	# The derivation of the two  »Fourier integrals«  from the Fourier series,		# The derivation of the two  »Fourier integrals«  from the Fourier series,
	# the extension of the Fourier integral to the  »Fourier transform«  by means of distributions,		# the extension of the Fourier integral to the  »Fourier transform«  by means of distributions,
	# »some  special cases«  of pulse&~~ndash~~;~~like signals like rectangular~~ and Gaussian ~~pulses and the Dirac delta~~,		# »some  special cases«  of pulses:  »rectangular pulse«  and  »Gaussian pulse«,
	# the  »laws  of ~~the~~ Fourier transform«,  and finally		# the  »laws  of Fourier transform«,  and finally
	# the meaning of the  »convolution operation«  and its various applications.		# the meaning of the  »convolution operation«  and its various applications.


	Laplace transform  and  Hilbert transform,  which are only applicable to causal signals or systems,  will be treated in the ~~next~~ book  "Linear Time-invariant Systems".		Laplace transform  and  Hilbert transform,  which are only applicable to causal signals or systems,  will be treated in the second book  »Linear Time-invariant Systems«.

Guenter: Guenter moved page Signal Representation/Fourier Transform and its Inverse to Signal Representation/The Fourier Transform and its Inverse

2023-02-09T17:20:02Z

Guenter moved page Signal Representation/Fourier Transform and its Inverse to Signal Representation/The Fourier Transform and its Inverse

← Older revision		Revision as of 19:20, 9 February 2023
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Guenter at 16:24, 12 December 2022

2022-12-12T16:24:07Z

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	Up to now, it has only been shown how to calculate the corresponding spectral function  $x(t)$  $X(f)$  for an aperiodic, pulse-like signal  $x(t)$.  Now we turn to the reverse task, namely:   How to determine the time function  $x(t)$  from the spectral function  $X(f)$?		Up to now, it has only been shown how to calculate the corresponding spectral function  $x(t)$  $X(f)$  for an aperiodic, pulse-like signal  $x(t)$.  Now we turn to the reverse task, namely:   How to determine the time function  $x(t)$  from the spectral function  $X(f)$?

	[[File:~~P_ID399__Sig_T_3_1_S7_rah~~.png\|right\|frame\|On the second Fourier integral ~~'''KORREKTUR: Fourier integral'''~~]]		[[File:EN_Sig_T_3_1_S7.png\|right\|frame\|On the second Fourier integral]]

	With the same designations as in the first sections of this chapter, you can write the signal  $x(t)$  as a Fourier series, where now the boundary  ${f_0}' \to 0$  is to be considered:		With the same designations as in the first sections of this chapter, you can write the signal  $x(t)$  as a Fourier series, where now the boundary  ${f_0}' \to 0$  is to be considered:

Guenter at 12:05, 9 December 2022

2022-12-09T12:05:12Z

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	This signal is shown in blue color in the left sketch.		This signal is shown in blue color in the left sketch.

	Since the signal  $x(t)$  extends to infinity on both sides, we must add a suitable convergence factor  $\text{e}^{-\varepsilon \hspace{0.05cm} \cdot \hspace{0.05cm}\vert \hspace{0.05cm} t \hspace{0.05cm} \vert}$  ~~is valid~~  $($  $\varepsilon > 0)$  in order to calculate the Fourier transform for both sections.  The resulting time function is then		Since the signal  $x(t)$  extends to infinity on both sides, we must add a suitable convergence factor  $\text{e}^{-\varepsilon \hspace{0.05cm} \cdot \hspace{0.05cm}\vert \hspace{0.05cm} t \hspace{0.05cm} \vert}$  with  $($  $\varepsilon > 0)$  in order to calculate the Fourier transform for both sections.  The resulting time function is then

	:$$x_\varepsilon (t) = \left\{ {\begin{array}{*{20}c} { {\rm{e} }^{ - \varepsilon \hspace{0.05cm} \cdot \hspace{0.05cm}t} } & { {\rm{for} }\quad t > 0,} \\ { {\rm{ - e} }^{\hspace{0.05cm}\varepsilon\hspace{0.05cm} \cdot \hspace{0.05cm} t} } & { {\rm{for} }\quad t < 0.} \\\end{array} } \right.$$		:$$x_\varepsilon (t) = \left\{ {\begin{array}{*{20}c} { {\rm{e} }^{ - \varepsilon \hspace{0.05cm} \cdot \hspace{0.05cm}t} } & { {\rm{for} }\quad t > 0,} \\ { {\rm{ - e} }^{\hspace{0.05cm}\varepsilon\hspace{0.05cm} \cdot \hspace{0.05cm} t} } & { {\rm{for} }\quad t < 0.} \\\end{array} } \right.$$

	Following a similar procedure as in ~~the~~ section  [[Signal_Representation/Direct_Current_Signal_-_Limit_Case_of_a_Periodic_Signal#Dirac_.28delta.29_function_in_frequency_domain\|~~$\text{~~Dirac delta function in the frequency domain}$]]  results for the corresponding spectral function:		Following a similar procedure as in section  [[Signal_Representation/Direct_Current_Signal_-_Limit_Case_of_a_Periodic_Signal#Dirac_.28delta.29_function_in_frequency_domain\|»Dirac delta function in the frequency domain«]]  results for the corresponding spectral function:

	:$$X_\varepsilon (f) = \frac{1}{ {\varepsilon + {\rm{j} }2{\rm{\pi } }f} } - \frac{1}{ {\varepsilon - {\rm{j} }2{\rm{\pi } }f} } = \frac{ { - {\rm{j4\pi } }f} }{ {\varepsilon ^2 + \left( {2{\rm{\pi } }f} \right)^2 } }.$$		:$$X_\varepsilon (f) = \frac{1}{ {\varepsilon + {\rm{j} }2{\rm{\pi } }f} } - \frac{1}{ {\varepsilon - {\rm{j} }2{\rm{\pi } }f} } = \frac{ { - {\rm{j4\pi } }f} }{ {\varepsilon ^2 + \left( {2{\rm{\pi } }f} \right)^2 } }.$$

	But actually we are interested in the spectrum of the ~~actual~~ jump function		But actually we are interested in the spectrum of the  »jump function«

	:$$x(t) = \mathop {\lim }\limits_{\varepsilon \hspace{0.05cm}\to \hspace{0.05cm}0 } x_\varepsilon (t).$$		:$$x(t) = \mathop {\lim }\limits_{\varepsilon \hspace{0.05cm}\to \hspace{0.05cm}0 } x_\varepsilon (t).$$

	Therefore, the spectral function  $X(f) =\text{F}\big[x(t)\big]$  has to be determined as limit value of  $X_\varepsilon(f)$  for  $\varepsilon \to 0$~~ ~~:		Therefore,  the spectral function  $X(f) =\text{F}\big[x(t)\big]$  has to be determined as limit value of  $X_\varepsilon(f)$  for  $\varepsilon \to 0$:

	:$$X(f) = \mathop {\lim }\limits_{\varepsilon \hspace{0.05cm} \to \hspace{0.05cm}0 } X_\varepsilon (f) = \frac{ { - {\rm{j} } } }{ { {\rm{\pi } }f} } = \frac{1}{ { {\rm{j\pi } }f} }.$$		:$$X(f) = \mathop {\lim }\limits_{\varepsilon \hspace{0.05cm} \to \hspace{0.05cm}0 } X_\varepsilon (f) = \frac{ { - {\rm{j} } } }{ { {\rm{\pi } }f} } = \frac{1}{ { {\rm{j\pi } }f} }.$$
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	In the right graph the imaginary spectral function  $X(f)$  is shown as a blue curve.  You can see that  $\vert X(f) \vert$  decreases continuously with increasing frequency.		In the right graph the imaginary spectral function  $X(f)$  is shown as a blue curve.  You can see that  $\vert X(f) \vert$  decreases continuously with increasing frequency.

	The green curve in the left graph shows the signal  $y(t)$, which differs from  $x(t)$  only in the negative time section.		⇒   The green curve in the left graph shows the signal  $y(t)$,  which differs from  $x(t)$  only in the negative time section.

	*In this area  $y(t) = 0$.  The corresponding spectral function  $Y(f)$  is only half as large as  $X(f)$  in the entire range.  This is shown in the following calculation:		*In this area  $y(t) = 0$.  The corresponding spectral function  $Y(f)$  is only half as large as  $X(f)$  in the entire range.  This is shown in the following calculation:
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	:$$Y(f) = \mathop {\lim }\limits_{\varepsilon \to 0 } Y_\varepsilon (f) = \mathop {\lim }\limits_{\varepsilon \hspace{0.05cm} \to \hspace{0.05cm}0 }\frac{1}{ {\varepsilon + {\rm{j} }2{\rm{\pi } }f} } = \frac{1}{ { {\rm{j2\pi } }f} }.$$		:$$Y(f) = \mathop {\lim }\limits_{\varepsilon \to 0 } Y_\varepsilon (f) = \mathop {\lim }\limits_{\varepsilon \hspace{0.05cm} \to \hspace{0.05cm}0 }\frac{1}{ {\varepsilon + {\rm{j} }2{\rm{\pi } }f} } = \frac{1}{ { {\rm{j2\pi } }f} }.$$

	*In addition there is ~~now~~ a Dirac delta function at  $f = 0$  with weight  $1/2$, due to the equal part.  This is explained ~~in detail~~ in the example in section  [[Signal_Representation/Fourier_Transform_Theorems#Assignment_Theorem\|"Assignment Theorem"]].  }}		*In addition there is a Dirac delta function at  $f = 0$  with weight  $1/2$, due to the equal part.  This is explained in the example in section  [[Signal_Representation/Fourier_Transform_Theorems#Assignment_Theorem\|»Assignment Theorem«]].  }}


	==The second Fourier integral==		==The second Fourier integral==
	<br>		<br>
	Up to now, it has only been shown how to calculate the corresponding spectral function~~  $x(t)$~~  $X(f)$  for an aperiodic, pulse-like signal  $x(t)$.  ~~Now we turn to the reverse task, namely:   How to determine the time function  $x(t)$  from the spectral function  $X(f)$?~~		Up to now,  it has only been shown how to calculate the corresponding spectral function  $X(f)$  for an aperiodic,  pulse-like signal  $x(t)$.

	[[File:EN_Sig_T_3_1_S7.png\|right\|frame\|On the second Fourier integral]]		[[File:EN_Sig_T_3_1_S7.png\|right\|frame\|On the second Fourier integral]]
			Now we turn to the reverse task,  namely:   How to determine the time function  $x(t)$  from the spectral function  $X(f)$?

	With the same designations as in the first sections of this chapter, you can write the signal  $x(t)$  as a Fourier series, where now the ~~boundary~~  ${f_0}' \to 0$  is to be considered:		With the same designations as in the first sections of this chapter,  you can write the signal  $x(t)$  as Fourier series,  where now the limit  ${f_0}' \to 0$  is to be considered:

	:$$x(t)=\lim_{{f_{\rm 0}}' \hspace{0.05cm}\to \hspace{0.05cm}0} \sum^{+\infty}_{\nu = -\infty}{D_{\it \nu}}' \cdot \rm e^{j\hspace{0.03cm} 2 \hspace{0.03cm}\pi \hspace{0.03cm}\it\nu \hspace{0.03cm} {f_{\rm 0}}' t}.$$		:$$x(t)=\lim_{{f_{\rm 0}}' \hspace{0.05cm}\to \hspace{0.05cm}0} \sum^{+\infty}_{\nu = -\infty}{D_{\it \nu}}' \cdot \rm e^{j\hspace{0.03cm} 2 \hspace{0.03cm}\pi \hspace{0.03cm}\it\nu \hspace{0.03cm} {f_{\rm 0}}' t}.$$
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	The limit crossing  ${f_0}' \to 0$  has the following effects:		The limit crossing  ${f_0}' \to 0$  has the following effects:
	*The  (infinite)  sum becomes an integral, where  ${f_0}'$  has to be formally replaced by the differential quantity  $\text{d}f$  (integration variable).		#The  $($infinite$)$  sum becomes an integral,  where  ${f_0}'$  has to be formally replaced by the differential quantity  $\text{d}f$  $($integration variable$)$.
			#The quantity   $\nu \cdot{f_0}'$  in the exponent describes the physical frequency  $f$.
	*The quantity   $\nu \cdot{f_0}'$  in the exponent describes the physical frequency  $f$.		#The quotient  ${D_{\nu}}'/{f_0}'$  yields the spectral function  $X(f)$  at the frequency  $f$.

	*The quotient  ${D_{\nu}}'/{f_0}'$  yields the spectral function  $X(f)$  at the frequency  $f$.


	Taking these properties into account, the  "second Fourier integral"  is obtained.		Taking these properties into account, the  »second Fourier integral«  is obtained.

	{{BlaueBox\|TEXT=		{{BlaueBox\|TEXT=
	$\text{Second Fourier Integral:}$		$\text{Second Fourier Integral:}$  If the spectral function  $X(f)$  of an aperiodic and energy-limited signal is given,  then the corresponding  »'''time signal'''«  is:
	If the spectral function  $X(f)$  of an aperiodic and energy-limited signal is given, then the corresponding  ~~$\text{~~time ~~function}$~~  is:

	:$$x(t) = \hspace{0.01cm}\int_{-\infty} ^{ {+}\infty} X(f) \, \cdot \, { \rm e}^{\rm j 2\pi \it ft} \,{\rm d}f.$$}}		:$$x(t) = \hspace{0.01cm}\int_{-\infty} ^{ {+}\infty} X(f) \, \cdot \, { \rm e}^{\rm j 2\pi \it ft} \,{\rm d}f.$$}}