Aufgaben:Exercise 2.5: Half-Wave Rectification - Revision history

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	[[Category:Signal Representation: Exercises\|^2.4 Fourier Series^]]		[[Category:Signal Representation: Exercises\|^2.4 Fourier Series^]]
	[[de:Aufgaben:2.~~5_Einweggleichrichtung~~]]		[[de:Aufgaben:Aufgabe 2.5: Einweggleichrichtung]]

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	*The Fourier coefficients sought are thus:		*The Fourier coefficients sought are thus:
	:$$A_0 \hspace{0.1cm}\underline{=0.5},\hspace{1cm}		:$$A_0 \hspace{0.1cm}\underline{=0.5},\hspace{1cm}B_1 = \pi /4 \hspace{0.1cm}\underline{= 0.785},\hspace{1cm}A_2\hspace{0.1cm}\underline{ = -0.333}.$$
	B_1 = \pi /4 \hspace{0.1cm}\underline{= 0.785},\hspace{1cm}
	A_2\hspace{0.1cm}\underline{ = -0.333}.$$
	{{ML-Fuß}}		{{ML-Fuß}}

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	[[Category:Signal Representation: Exercises\|^2.4 Fourier Series^]]		[[Category:Signal Representation: Exercises\|^2.4 Fourier Series^]]
			[[de:Aufgaben:2.5_Einweggleichrichtung]]

Guenter at 13:14, 15 April 2021

2021-04-15T13:14:26Z

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	'''(1)'''  The shifted signal  $v(t)$  is also even and all sine coefficients are accordingly zero.		'''(1)'''  The shifted signal  $v(t)$  is also even and all sine coefficients are accordingly zero.
	*Nothing changes in the DC signal coefficient ~~either~~:   $A_0 = 1$.		*Nothing changes in the DC signal coefficient:   $A_0 = 1$.

	*From the signal curves it can be seen that  $v(t) = u(t - T/4)$  applies:		*From the signal curves it can be seen that  $v(t) = u(t - T/4)$  applies:
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	'''(2)'''  ~~BecaUSE OF~~  $w(t) = \pi /2 \cdot \sin(\omega_0 t)$  all Fourier coefficients except  $B_1 = \pi /2 \hspace{0.1cm}\underline{=1.571}$  are zero.		'''(2)'''  Because of  $w(t) = \pi /2 \cdot \sin(\omega_0 t)$  all Fourier coefficients except  $B_1 = \pi /2 \hspace{0.1cm}\underline{=1.571}$  are zero.

Guenter at 13:09, 15 April 2021

2021-04-15T13:09:38Z

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	<quiz display=simple>		<quiz display=simple>
	{Calculate the Fourier coefficients of the signal  $v(t)$. What is ~~the value of~~ the coefficient  $A_2$?		{Calculate the Fourier coefficients of the signal  $v(t)$.  What is the coefficient  $A_2$?
	\|type="{}"}		\|type="{}"}
	$v(t)$:   $A_2\ = \ $ { -0.67--0.66 }		$v(t)$:   $A_2\ = \ $ { -0.67--0.66 }

	{Calculate the Fourier coefficients of the signal  $w(t)$. What is ~~the value of~~ the coefficient  $B_1$?		{Calculate the Fourier coefficients of the signal  $w(t)$.  What is the coefficient  $B_1$?
	\|type="{}"}		\|type="{}"}
	$w(t)$:  $B_1\ = \ $ { 1.571 3% }		$w(t)$:  $B_1\ = \ $ { 1.571 3% }

	{How can  $x(t)$  be composed of  $v(t)$  and  $w(t)$ ? Give the corresponding Fourier coefficients of the signal  $x(t)$  in particular		{How can  $x(t)$  be composed of  $v(t)$  and  $w(t)$?  Give the corresponding Fourier coefficients of the signal  $x(t)$.  In particular:
	\|type="{}"}		\|type="{}"}
	$x(t)$:  $A_0\ = \ $ { 0.5 3% }		$x(t)$:  $A_0\ = \ $ { 0.5 3% }

Guenter at 13:06, 15 April 2021

2021-04-15T13:06:47Z

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	It should be noted:		It should be noted:
	*The ~~fundamental angular~~ frequency is denoted by  $\omega_1$ . But since the period of the signals  $u(t)$  and  $v(t)$  is  $T/2$~~ ~~,   $\omega_1 = 2\pi /(T/2) = 4 \pi /T$.		*The basic circular frequency is denoted by  $\omega_1$ .  But since the period of the signals  $u(t)$  and  $v(t)$  is  $T/2$,   $\omega_1 = 2\pi /(T/2) = 4 \pi /T$.
	*Because in this task the signals  $u(t)$,  $w(t)$  and  $x(t)$  are to be related to each other, the signal  $u(t)$  must also be represented with the period duration  $T$  of the signal  $x(t)$ .		*Because in this task the signals  $u(t)$,  $w(t)$  and  $x(t)$  are to be related to each other, the signal  $u(t)$  must also be represented with the period duration  $T$  of the signal  $x(t)$.
	*With  $\omega_0 = 2\pi /T = \omega_1/2$  the same applies:		*With  $\omega_0 = 2\pi /T = \omega_1/2$  the same applies:

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	''Hints:''		''Hints:''
	*This exercise belongs to the chapter  [[Signal_Representation/Fourier_Series\|~~Fourierreihe~~]].		*This exercise belongs to the chapter  [[Signal_Representation/Fourier_Series\|Fourier Series]].
	*You can find a compact summary of the topic in the two learning videos		*You can find a compact summary of the topic in the two learning videos
	::[[Zur_Berechnung_der_Fourierkoeffizienten_(Lernvideo)\|Zur Berechnung der Fourierkoeffizienten]],		::[[Zur_Berechnung_der_Fourierkoeffizienten_(Lernvideo)\|Zur Berechnung der Fourierkoeffizienten]]  ⇒   "To calculate the Fourier coefficients",
	::[[Eigenschaften_der_Fourierreihendarstellung_(Lernvideo)\|Eigenschaften der Fourierreihendarstellung]].		:: [[Eigenschaften_der_Fourierreihendarstellung_(Lernvideo)\|Eigenschaften der Fourierreihendarstellung]]   ⇒   "Properties of the Fourier series representation".

Guenter at 13:01, 15 April 2021

2021-04-15T13:01:39Z

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	[[File:P_ID311__Sig_A_2_5.png\|right\|frame\|Rectified cosine functions]]		[[File:P_ID311__Sig_A_2_5.png\|right\|frame\|Rectified cosine functions]]

	We are looking for the Fourier coefficients of the signal  $x(t)$, sketched below, which results from the one-way rectification of the sinusoidal signal  $w(t)$  with amplitude  $\pi /2$~~ ~~.		We are looking for the Fourier coefficients of the signal  $x(t)$, sketched below, which results from the one-way rectification of the sinusoidal signal  $w(t)$  with amplitude  $\pi /2$.

	The Fourier series representation of the signal  $u(t)$ sketched above is assumed to be known. This was already determined in  [[Aufgaben:Aufgabe_2.4:_Gleichgerichteter_Cosinus\|~~Aufgabe~~ 2.4]] . Taking into account the amplitude  $\pi /2$  the following applies:		The Fourier series representation of the signal  $u(t)$ sketched above is assumed to be known.  This was already determined in  [[Aufgaben:Aufgabe_2.4:_Gleichgerichteter_Cosinus\|Exercise 2.4]].  Taking into account the amplitude  $\pi /2$  the following applies:
	:$$u(t)=1+\frac{2}{3} \cdot \cos(\omega_1t)-\frac{2}{15}\cdot \cos(2\omega_1t)+\frac{2}{35}\cdot \cos(3\omega_1t)-\dots$$		:$$u(t)=1+\frac{2}{3} \cdot \cos(\omega_1t)-\frac{2}{15}\cdot \cos(2\omega_1t)+\frac{2}{35}\cdot \cos(3\omega_1t)-\dots$$

Noah at 09:25, 26 March 2021

2021-03-26T09:25:35Z

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	[[File:P_ID311__Sig_A_2_5.png\|right\|frame\|~~Gleichgerichtete Cosinusfunktionen~~]]		[[File:P_ID311__Sig_A_2_5.png\|right\|frame\|Rectified cosine functions]]

	We are looking for the Fourier coefficients of the signal  $x(t)$, sketched below, which results from the one-way rectification of the sinusoidal signal  $w(t)$  with amplitude  $\pi /2$ .		We are looking for the Fourier coefficients of the signal  $x(t)$, sketched below, which results from the one-way rectification of the sinusoidal signal  $w(t)$  with amplitude  $\pi /2$ .

Javier: Text replacement - "Category:Exercises for Signal Representation" to "Category:Signal Representation: Exercises"

2021-03-23T12:39:29Z

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	[[Category:~~Exercises for~~ Signal Representation\|^2.4 Fourier Series^]]		[[Category:Signal Representation: Exercises\|^2.4 Fourier Series^]]

Noah at 18:51, 16 January 2021

2021-01-16T18:51:34Z

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  ~~Auch das verschobene Signal~~  $v(t)$  ~~ist gerade und alle Sinuskoeffizienten sind dementsprechend Null~~.		'''(1)'''  The shifted signal  $v(t)$  is also even and all sine coefficients are accordingly zero.
	*~~Am Gleichsignalkoeffizienten ändert sich ebenfalls nichts~~:  $A_0 = 1$.		*Nothing changes in the DC signal coefficient either:   $A_0 = 1$.

	*~~Aus den Signalverläufen ist zu erkennen, dass~~  $v(t) = u(t - T/4)$  ~~gilt~~:		*From the signal curves it can be seen that  $v(t) = u(t - T/4)$  applies:

	:$$v(t)=1+\frac{2}{3}\cdot \cos(2\omega_0(t-\frac{T}{4}))-\frac{2}{15}\cdot \cos(4\omega_0(t-\frac{T}{4}))+\frac{2}{35}\cdot \cos(6\omega_0(t-\frac{T}{4}))-\dots$$		:$$v(t)=1+\frac{2}{3}\cdot \cos(2\omega_0(t-\frac{T}{4}))-\frac{2}{15}\cdot \cos(4\omega_0(t-\frac{T}{4}))+\frac{2}{35}\cdot \cos(6\omega_0(t-\frac{T}{4}))-\dots$$

	*~~Die Cosinusterme können nun mit~~  $\omega_0 \cdot T = 2 \pi$  ~~umgeformt werden~~:		*The cosine terms can now be transformed with  $\omega_0 \cdot T = 2 \pi$  :

	:$$\cos(2\omega_0(t-\frac{T}{4}))=\cos(2\omega_0t-\pi)=-\cos(2\omega_0t),$$		:$$\cos(2\omega_0(t-\frac{T}{4}))=\cos(2\omega_0t-\pi)=-\cos(2\omega_0t),$$
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	:$$\cos(6\omega_0(t-\frac{T}{4}))=\cos(6\omega_0t-3\pi)=-\cos(6\omega_0t).$$		:$$\cos(6\omega_0(t-\frac{T}{4}))=\cos(6\omega_0t-3\pi)=-\cos(6\omega_0t).$$

	*~~Damit erhält man für die Fourierreihe~~:		*This gives us for the Fourier series:

	:$$v(t)=1-{2}/{3}\cdot \cos(2\omega_0t)-{2}/{15}\cdot \cos(4\omega_0t)-{2}/{35}\cdot \cos(6\omega_0t)-\dots$$		:$$v(t)=1-{2}/{3}\cdot \cos(2\omega_0t)-{2}/{15}\cdot \cos(4\omega_0t)-{2}/{35}\cdot \cos(6\omega_0t)-\dots$$

	:~~bzw. für die Cosinuskoeffizienten mit geradzahligem~~  $n$:		:or for the cosine coefficients with even-numbered  $n$:

	:$$A_n=\frac{-2}{n^2-1}\hspace{0.5cm}\Rightarrow\hspace{0.5cm}A_2=-\hspace{-0.05cm}2/3 \hspace{0.1cm}\underline{= -\hspace{-0.05cm}0.667}.$$		:$$A_n=\frac{-2}{n^2-1}\hspace{0.5cm}\Rightarrow\hspace{0.5cm}A_2=-\hspace{-0.05cm}2/3 \hspace{0.1cm}\underline{= -\hspace{-0.05cm}0.667}.$$
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	'''(2)'''  ~~Wegen~~  $w(t) = \pi /2 \cdot \sin(\omega_0 t)$  ~~sind alle Fourierkoeffizienten außer~~  $B_1 = \pi /2 \hspace{0.1cm}\underline{=1.571}$  ~~gleich Null~~.		'''(2)'''  BecaUSE OF  $w(t) = \pi /2 \cdot \sin(\omega_0 t)$  all Fourier coefficients except  $B_1 = \pi /2 \hspace{0.1cm}\underline{=1.571}$  are zero.


	'''(3)'''  ~~Aus der grafischen Darstellung erkennt man den Zusammenhang~~  $x(t)={1}/{2} \cdot \big [v(t)+w(t) \big].$ ~~Das bedeutet~~:		'''(3)'''  From the graphical representation one can see the relationship  $x(t)={1}/{2} \cdot \big [v(t)+w(t) \big].$ This means:

	:$$x(t)=\frac{1}{2}+\frac{\pi}{4}\cdot \sin(\omega_0 t)-\frac{1}{3}\cdot \cos(2\omega_0 t)-\frac{1}{15}\cdot \cos(4\omega_0 t)-\frac{1}{35}\cdot \cos(6\omega_0 t)-\ldots$$		:$$x(t)=\frac{1}{2}+\frac{\pi}{4}\cdot \sin(\omega_0 t)-\frac{1}{3}\cdot \cos(2\omega_0 t)-\frac{1}{15}\cdot \cos(4\omega_0 t)-\frac{1}{35}\cdot \cos(6\omega_0 t)-\ldots$$

	*~~Die gesuchten Fourierkoeffizienten sind somit~~:		*The Fourier coefficients sought are thus:
	:$$A_0 \hspace{0.1cm}\underline{=0.5},\hspace{1cm}		:$$A_0 \hspace{0.1cm}\underline{=0.5},\hspace{1cm}
	B_1 = \pi /4 \hspace{0.1cm}\underline{= 0.785},\hspace{1cm}		B_1 = \pi /4 \hspace{0.1cm}\underline{= 0.785},\hspace{1cm}