Aufgaben:Exercise 3.7: Synchronous Demodulator - Revision history

Hwang at 13:19, 18 January 2023

2023-01-18T13:19:00Z

Guenter at 11:45, 28 April 2021

2021-04-28T11:45:14Z

Guenter at 11:35, 28 April 2021

2021-04-28T11:35:49Z

Guenter at 11:34, 28 April 2021

2021-04-28T11:34:52Z

Guenter: Guenter moved page Aufgaben:Exercise 3.7: Carrier Recovery to Aufgaben:Exercise 3.7: Synchronous Demodulator

2021-04-28T11:09:26Z

Guenter moved page Exercise 3.7: Carrier Recovery to Exercise 3.7: Synchronous Demodulator

Noah at 13:45, 27 March 2021

2021-03-27T13:45:38Z

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

2021-03-23T12:38:57Z

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

Noah at 18:52, 27 January 2021

2021-01-27T18:52:40Z

Show changes

Oezdemir: Oezdemir moved page Aufgaben:Aufgabe 3.7: Synchrondemodulator to Aufgaben:Exercise 3.7: Carrier Recovery

2020-11-18T22:05:02Z

Oezdemir moved page Aufgabe 3.7: Synchrondemodulator to Exercise 3.7: Carrier Recovery

Oezdemir at 22:02, 18 November 2020

2020-11-18T22:02:49Z

@@ Line 12: / Line 12: @@
 The spectrum&nbsp; $R(f)$&nbsp; of the received signal&nbsp; $r(t)$&nbsp; sketched above is produced by&nbsp; "Two-sideband Amplitude Modulation"&nbsp; of a sinusoidal source signal&nbsp; $q(t)$&nbsp; with the frequency&nbsp; $5\,\text{kHz}$&nbsp; and the amplitude&nbsp; $8\,\text{V}$.&nbsp; A cosine signal&nbsp; $z_{\rm S}(t)$&nbsp; with the frequency&nbsp; $30\,\text{kHz}$&nbsp; was used as the transmission-side carrier signal.
-The spectrum of the carrier signal&nbsp; $z_{\rm E}(t)$&nbsp; on the receiver side consists of two Dirac lines according to the sketch below, each with the weight&nbsp; $A/2$.&nbsp; Since&nbsp; $z_{\rm E}(t)$&nbsp; is not to contain a unit, the weights of the Dirac functions are also dimensionless.
+The spectrum of the carrier signal&nbsp; $z_{\rm E}(t)$&nbsp; on the receiver side consists of two Dirac deltalines according to the sketch below, each with the weight&nbsp; $A/2$.&nbsp; Since&nbsp; $z_{\rm E}(t)$&nbsp; is not to contain a unit, the weights of the Dirac functions are also dimensionless.
@@ Line 50: / Line 50: @@
 {{ML-Kopf}}
 '''(1)'''&nbsp; If we name the signal after the multiplier with&nbsp; $m(t) = r(t) \cdot z_{\rm E}(t)$, the corresponding spectrum&nbsp; $M(f)$&nbsp; is the convolution product of&nbsp; $R(f)$&nbsp; and&nbsp; $Z_{\rm E}(f)$.
-*Convolution of the spectrum&nbsp; $R(f)$&nbsp; with the right Dirac line at&nbsp; $+30 \text{ kHz}$&nbsp; leads to discrete spectral lines at&nbsp;  $-\hspace{-0.08cm}5\, \text{kHz}$,&nbsp; $+5 \,\text{kHz}$,&nbsp; $+55 \,\text{kHz}$&nbsp; and&nbsp; $+65 \,\text{kHz}$.&nbsp; These are all imaginary and smaller than the impulse weights of&nbsp; $R(f)$&nbsp; by a factor of&nbsp; $A/2 = 0.5$.
+*Convolution of the spectrum&nbsp; $R(f)$&nbsp; with the right Dirac delta line at&nbsp; $+30 \text{ kHz}$&nbsp; leads to discrete spectral lines at&nbsp;  $-\hspace{-0.08cm}5\, \text{kHz}$,&nbsp; $+5 \,\text{kHz}$,&nbsp; $+55 \,\text{kHz}$&nbsp; and&nbsp; $+65 \,\text{kHz}$.&nbsp; These are all imaginary and smaller than the impulse weights of&nbsp; $R(f)$&nbsp; by a factor of&nbsp; $A/2 = 0.5$.
 *Convolution of&nbsp; $R(f)$&nbsp; with the Dirac at&nbsp; $-\hspace{-0.08cm}30 \,\text{kHz}$&nbsp; yields lines at&nbsp; $-\hspace{-0.08cm}65 \,\text{kHz}$,&nbsp; $-55 \,\text{kHz}$, $-5 \,\text{kHz}$&nbsp; and&nbsp;  $+5 \,\text{kHz}$.
@@ Line 68: / Line 68: @@
-'''(3)'''&nbsp; The Dirac lines at&nbsp; $\pm f_{\rm T}$&nbsp; each have weight&nbsp; $1$.&nbsp; All spectral lines mentioned below are imaginary and equal in magnitude to&nbsp; $2 \text{ V}$.
+'''(3)'''&nbsp; The Dirac delta lines at&nbsp; $\pm f_{\rm T}$&nbsp; each have weight&nbsp; $1$.&nbsp; All spectral lines mentioned below are imaginary and equal in magnitude to&nbsp; $2 \text{ V}$.
 *The convolution of&nbsp; $R(f)$&nbsp; with the right Diracl ine of&nbsp; $z_{\rm E}(t)$&nbsp;  yields components at&nbsp; $-\hspace{-0.08cm}4\, \text{kHz (p: positive)}$,&nbsp;  $+6 \,\text{kHz (n: negative)}$, $+56 \,\text{kHz (p)}$&nbsp; and&nbsp; $+66 \,\text{kHz (n)}$.
 *In contrast, the convolution with the left Dirac function leads to spectral lines at&nbsp; $-\hspace{-0.08cm}66 \,\text{kHz (p)}$,&nbsp; $-\hspace{-0.08cm}56 \,\text{kHz (n)}$,&nbsp; $-\hspace{-0.08cm}6 \,\text{kHz (p)}$&nbsp; und&nbsp; $+4 \,\text{kHz (n)}$, all also with the (magnitude-related) impulse weights&nbsp; $2 \text{ V}$.

@@ Line 50: / Line 50: @@
 {{ML-Kopf}}
 '''(1)'''&nbsp; If we name the signal after the multiplier with&nbsp; $m(t) = r(t) \cdot z_{\rm E}(t)$, the corresponding spectrum&nbsp; $M(f)$&nbsp; is the convolution product of&nbsp; $R(f)$&nbsp; and&nbsp; $Z_{\rm E}(f)$.
-*Convolution of the spectrum&nbsp; $R(f)$&nbsp; with the right diracline at&nbsp; $+30 \text{ kHz}$&nbsp; leads to discrete spectral lines at&nbsp;  $-\hspace{-0.08cm}5\, \text{kHz}$,&nbsp; $+5 \,\text{kHz}$,&nbsp; $+55 \,\text{kHz}$&nbsp; and&nbsp; $+65 \,\text{kHz}$. These are all imaginary and smaller than the impulse weights of&nbsp; $R(f)$&nbsp; by a factor of&nbsp; $A/2 = 0.5$&nbsp; kleiner.
+*Convolution of the spectrum&nbsp; $R(f)$&nbsp; with the right Dirac line at&nbsp; $+30 \text{ kHz}$&nbsp; leads to discrete spectral lines at&nbsp;  $-\hspace{-0.08cm}5\, \text{kHz}$,&nbsp; $+5 \,\text{kHz}$,&nbsp; $+55 \,\text{kHz}$&nbsp; and&nbsp; $+65 \,\text{kHz}$.&nbsp; These are all imaginary and smaller than the impulse weights of&nbsp; $R(f)$&nbsp; by a factor of&nbsp; $A/2 = 0.5$.
 *Convolution of&nbsp; $R(f)$&nbsp; with the Dirac at&nbsp; $-\hspace{-0.08cm}30 \,\text{kHz}$&nbsp; yields lines at&nbsp; $-\hspace{-0.08cm}65 \,\text{kHz}$,&nbsp; $-55 \,\text{kHz}$, $-5 \,\text{kHz}$&nbsp; and&nbsp;  $+5 \,\text{kHz}$.
-By superimposing the two intermediate results and taking into account the low-pass filter, which suppresses the lines at&nbsp; $\pm 55 \text{ kHz}$&nbsp; and&nbsp; $\pm 65 \text{ kHz}$&nbsp;, it thus follows for the spectrum of the descent signal:
+By superimposing the two intermediate results and taking into account the low-pass filter, which suppresses the lines at&nbsp; $\pm 55 \text{ kHz}$&nbsp; and&nbsp; $\pm 65 \text{ kHz}$&nbsp;, it thus follows for the spectrum of the sink signal:
 :$$V( f) =  - {\rm{j}} \cdot 2\;{\rm{V}} \cdot \delta ( {f - f_{\rm N} }) + {\rm{j}} \cdot 2\;{\rm{V}} \cdot \delta ( {f + f_{\rm N} } )\hspace{0.3cm}{\rm with}\hspace{0.3cm}f_{\rm N} = 5\;{\rm kHz}.$$
@@ Line 64: / Line 64: @@
-'''(2)'''&nbsp; With&nbsp; $A = 1$&nbsp; &nbsp; $v(t)$&nbsp; is only half as large as&nbsp;  $q(t)$ &nbsp; &rArr; &nbsp; With&nbsp; $\underline{A = 2}$&nbsp; both signals would be equal.
+'''(2)'''&nbsp; With&nbsp; $A = 1$&nbsp; the sink signal &nbsp; $v(t)$&nbsp; is only half as large as&nbsp;  $q(t)$ &nbsp; &rArr; &nbsp; With&nbsp; $\underline{A = 2}$&nbsp; both signals would be equal.
-'''(3)'''&nbsp; The diraclines at&nbsp; $\pm f_{\rm T}$&nbsp; each have weight&nbsp; $1$. All spectral lines mentioned below are imaginary and equal in magnitude to&nbsp; $2 \text{ V}$.
+'''(3)'''&nbsp; The Dirac lines at&nbsp; $\pm f_{\rm T}$&nbsp; each have weight&nbsp; $1$.&nbsp; All spectral lines mentioned below are imaginary and equal in magnitude to&nbsp; $2 \text{ V}$.
-*The convolution of&nbsp; $R(f)$&nbsp; with the right diracline of&nbsp; $z_{\rm E}(t)$&nbsp;  yields components at&nbsp; $-\hspace{-0.08cm}4\, \text{kHz (p: positiv)}$,&nbsp;  $+6 \,\text{kHz (n: negativ)}$, $+56 \,\text{kHz (p)}$&nbsp; and&nbsp; $+66 \,\text{kHz (n)}$.
+*The convolution of&nbsp; $R(f)$&nbsp; with the right Diracl ine of&nbsp; $z_{\rm E}(t)$&nbsp;  yields components at&nbsp; $-\hspace{-0.08cm}4\, \text{kHz (p: positive)}$,&nbsp;  $+6 \,\text{kHz (n: negative)}$, $+56 \,\text{kHz (p)}$&nbsp; and&nbsp; $+66 \,\text{kHz (n)}$.
 *In contrast, the convolution with the left Dirac function leads to spectral lines at&nbsp; $-\hspace{-0.08cm}66 \,\text{kHz (p)}$,&nbsp; $-\hspace{-0.08cm}56 \,\text{kHz (n)}$,&nbsp; $-\hspace{-0.08cm}6 \,\text{kHz (p)}$&nbsp; und&nbsp; $+4 \,\text{kHz (n)}$, all also with the (magnitude-related) impulse weights&nbsp; $2 \text{ V}$.
-*Taking the low pass into account, only the four spectral lines at&nbsp; $\pm 4 \,\text{kHz}$&nbsp; and&nbsp; $\pm 6 \,\text{kHz}$.
+*Taking the low&ndash;pass into account, there are only the four spectral lines at&nbsp; $\pm 4 \,\text{kHz}$&nbsp; and&nbsp; $\pm 6 \,\text{kHz}$.
 *The associated time signal is thus with&nbsp; $f_4 = 4 \,\text{kHz}$&nbsp; and&nbsp; $f_6 = 6 \,\text{kHz}$:

@@ Line 23: / Line 23: @@
 ''Hints:''
 *This exercise belongs to the  chapter&nbsp; [[Signal_Representation/The_Convolution_Theorem_and_Operation|The Convolution Theorem and Operation]].
-*Important information can be found on the page&nbsp; [[Signal_Representation/The_Convolution_Theorem_and_Operation#Convolution_of_a_Function_With_a_Dirac_Function|Convolution of a function with a Dirac function]].
+*Important information can be found on the page&nbsp; [[Signal_Representation/The_Convolution_Theorem_and_Operation#Convolution_of_a_function_with_a_Dirac_function|Convolution of a function with a Dirac function]].

@@ Line 5: / Line 5: @@
 [[File:Sig_A_3_7_version2.png|right|frame|The spectral functions&nbsp; $R(f)$&nbsp; and&nbsp; $Z_{\rm E}(f)$]]
-To reset an amplitude-modulated signal to the original frequency range, a&nbsp; [[Modulation_Methods/Synchrondemodulation#Blockschaltbild_und_Zeitbereichsdarstellung|synchronous demodulator]] is often used::
+To reset an amplitude-modulated signal to the original frequency range, a&nbsp; [[Modulation_Methods/Synchrondemodulation#Blockschaltbild_und_Zeitbereichsdarstellung|"Synchronous Demodulator"]]&nbsp; is often used:
-*This multiplies the AM input signal&nbsp; $r(t)$&nbsp; with a carrier signal&nbsp; $z_{\rm E}(t)$, on the receiving side, which should match the carrier signal&nbsp; $z_{\rm S}(t)$&nbsp;on the transmitting side with regard to both frequency &nbsp; $f_{\rm T}$&nbsp; and phase&nbsp; $\varphi_{\rm T}$&nbsp;.
+*This multiplies the AM input signal&nbsp; $r(t)$&nbsp; with a carrier signal&nbsp; $z_{\rm E}(t)$&nbsp; on the receiver side, which should match the carrier signal&nbsp; $z_{\rm S}(t)$&nbsp; on the transmitter side with regard to both frequency&nbsp; $f_{\rm T}$&nbsp; and phase&nbsp; $\varphi_{\rm T}$.
-*This is followed by a rectangular low-pass filter to eliminate all spectral components above the carrier frequency&nbsp; $f_{\rm T}$.  We call the output signal of the synchronous demodulator&nbsp; $v(t)$.
+*This is followed by a rectangular low-pass filter to eliminate all spectral components above the carrier frequency&nbsp; $f_{\rm T}$.&nbsp;  We call the output signal of the synchronous demodulator&nbsp; $v(t)$.
-The spectrum&nbsp; $R(f)$&nbsp; of the received signal&nbsp; $r(t)$&nbsp; sketched above is produced by two-sideband amplitude modulation of a sinusoidal source signal&nbsp; $q(t)$&nbsp; with the frequency&nbsp; $5\,\text{kHz}$&nbsp; and the amplitude&nbsp; $8\,\text{V}$&nbsp;. A cosine signal with the frequency&nbsp; $z_{\rm S}(t)$&nbsp; was used as the transmission-side carrier signal&nbsp; $30\,\text{kHz}$ verwendet.
+The spectrum&nbsp; $R(f)$&nbsp; of the received signal&nbsp; $r(t)$&nbsp; sketched above is produced by&nbsp; "Two-sideband Amplitude Modulation"&nbsp; of a sinusoidal source signal&nbsp; $q(t)$&nbsp; with the frequency&nbsp; $5\,\text{kHz}$&nbsp; and the amplitude&nbsp; $8\,\text{V}$.&nbsp; A cosine signal&nbsp; $z_{\rm S}(t)$&nbsp; with the frequency&nbsp; $30\,\text{kHz}$&nbsp; was used as the transmission-side carrier signal.
-The spectrum of the carrier signal&nbsp; $z_{\rm E}(t)$&nbsp; on the receiving side consists of two dirac lines according to the sketch below, each with the weight&nbsp; $A/2$. Since&nbsp; $z_{\rm E}(t)$&nbsp; is not to contain a unit, the weights of the dirac functions are also dimensionless.
+The spectrum of the carrier signal&nbsp; $z_{\rm E}(t)$&nbsp; on the receiver side consists of two Dirac lines according to the sketch below, each with the weight&nbsp; $A/2$.&nbsp; Since&nbsp; $z_{\rm E}(t)$&nbsp; is not to contain a unit, the weights of the Dirac functions are also dimensionless.
@@ Line 22: / Line 22: @@
 ''Hints:''
-*This exercise belongs to the&nbsp;  chapter&nbsp; [[Signal_Representation/The_Convolution_Theorem_and_Operation|The Convolution Theorem and Operation]].
+*This exercise belongs to the  chapter&nbsp; [[Signal_Representation/The_Convolution_Theorem_and_Operation|The Convolution Theorem and Operation]].
-*Important information can be found on the page&nbsp; [[Signal_Representation/The_Convolution_Theorem_and_Operation#Convolution_of_a_Function_With_a_Dirac_Function|Convolution of a Function With a Dirac Function]].
+*Important information can be found on the page&nbsp; [[Signal_Representation/The_Convolution_Theorem_and_Operation#Convolution_of_a_Function_With_a_Dirac_Function|Convolution of a function with a Dirac function]].
@@ Line 32: / Line 32: @@
 <quiz display=simple>
-{Let&nbsp; $f_{\rm T} = 30\,\text{kHz}$&nbsp; and&nbsp; $A=1$. Calculate the output signal&nbsp; $v(t)$. <br>What signal value occurs at time&nbsp; $t = 50\, {\rm  &micro;} \text{s}$?
+{Let $f_{\rm T} = 30\,\text{kHz}$&nbsp; and&nbsp; $A=1$.&nbsp; Calculate the output signal&nbsp; $v(t)$. <br>What signal value occurs at time&nbsp; $t = 50\, {\rm  &micro;} \text{s}$?
 |type="{}"}
 $v(t = 50\, µ\text{s})\ = \ $ { 4 3% } &nbsp;$\text{V}$
-{How large must the amplitude of the carrier signal&nbsp; $z_{\rm E}(t)$&nbsp; on the receiving side be chosen so that&nbsp; $v(t) = q(t)$&nbsp; is valid?
+{How large must the amplitude of the carrier signal&nbsp; $z_{\rm E}(t)$&nbsp; on the receiver side be chosen so that&nbsp; $v(t) = q(t)$&nbsp; is valid?
 |type="{}"}
 $A\ = \ $ { 2 3% }
-{Calculate the output signal&nbsp; $v(t)$&nbsp; under the conditions&nbsp; $A = 2$&nbsp; and&nbsp; $f_{\rm T} =  31\,\text{kHz}$. <br>What signal value occurs at time&nbsp; $ t = 50\, µ\text{s}$ auf?
+{Calculate the output signal&nbsp; $v(t)$&nbsp; under the conditions&nbsp; $A = 2$&nbsp; and $f_{\rm T} =  31\,\text{kHz}$. <br>What signal value occurs at time&nbsp; $ t = 50\, µ\text{s}$?
 |type="{}"}
 $v(t = 50\, µ\text{s})\ = \ $  { 7.608 3% } &nbsp;$\text{V}$

@@ Line 3: / Line 3: @@
 }}
-[[File:Sig_A_3_7_version2.png|right|frame|Die Spektralfunktionen&nbsp; $R(f)$&nbsp; und&nbsp; $Z_{\rm E}(f)$]]
+[[File:Sig_A_3_7_version2.png|right|frame|The spectral functions&nbsp; $R(f)$&nbsp; and&nbsp; $Z_{\rm E}(f)$]]
 To reset an amplitude-modulated signal to the original frequency range, a&nbsp; [[Modulation_Methods/Synchrondemodulation#Blockschaltbild_und_Zeitbereichsdarstellung|synchronous demodulator]] is often used::
@@ Line 56: / Line 56: @@
 By superimposing the two intermediate results and taking into account the low-pass filter, which suppresses the lines at&nbsp; $\pm 55 \text{ kHz}$&nbsp; and&nbsp; $\pm 65 \text{ kHz}$&nbsp;, it thus follows for the spectrum of the descent signal:
-:$$V( f) =  - {\rm{j}} \cdot 2\;{\rm{V}} \cdot \delta ( {f - f_{\rm N} }) + {\rm{j}} \cdot 2\;{\rm{V}} \cdot \delta ( {f + f_{\rm N} } )\hspace{0.3cm}{\rm mit}\hspace{0.3cm}f_{\rm N} = 5\;{\rm kHz}.$$
+:$$V( f) =  - {\rm{j}} \cdot 2\;{\rm{V}} \cdot \delta ( {f - f_{\rm N} }) + {\rm{j}} \cdot 2\;{\rm{V}} \cdot \delta ( {f + f_{\rm N} } )\hspace{0.3cm}{\rm with}\hspace{0.3cm}f_{\rm N} = 5\;{\rm kHz}.$$
 *The sink signal&nbsp; $v(t)$&nbsp; is therefore a&nbsp; $5 \text{ kHz}$–sine signal with amplitude&nbsp; $4 \text{ V}$.

← Older revision	Revision as of 11:09, 28 April 2021
(No difference)

← Older revision		Revision as of 12:38, 23 March 2021
Line 83:		Line 83:

	__NOEDITSECTION__		__NOEDITSECTION__
−	[[Category:~~Exercises for~~ Signal Representation\|^3.4 The Convolution Theorem^]]	+	[[Category:Signal Representation: Exercises\|^3.4 The Convolution Theorem^]]

← Older revision	Revision as of 22:05, 18 November 2020
(No difference)

← Older revision		Revision as of 22:02, 18 November 2020
Line 83:		Line 83:

	__NOEDITSECTION__		__NOEDITSECTION__
−	[[Category:Exercises for Signal Representation\|^3. ~~Aperiodische Signale - Impulse~~^]]	+	[[Category:Exercises for Signal Representation\|^3.4 The Convolution Theorem^]]