Aufgaben:Exercise 3.7Z: Rectangular Signal with Echo - Revision history

Javier: Text replacement - "Dirac impulse" to "Dirac delta"

2021-05-26T07:05:07Z

Text replacement - "Dirac impulse" to "Dirac delta"

Guenter at 12:24, 28 April 2021

2021-04-28T12:24:59Z

Guenter at 12:15, 28 April 2021

2021-04-28T12:15:51Z

Guenter: Guenter moved page Aufgaben:Exercise 3.7Z: Rectangular signal with Echo to Aufgaben:Exercise 3.7Z: Rectangular Signal with Echo

2021-04-28T12:14:02Z

Guenter moved page Exercise 3.7Z: Rectangular signal with Echo to Exercise 3.7Z: Rectangular Signal with Echo

Guenter at 12:13, 28 April 2021

2021-04-28T12:13:45Z

Guenter: Guenter moved page Aufgaben:Exercise 3.7Z: Square Wave With Echo to Aufgaben:Exercise 3.7Z: Rectangular signal with Echo

2021-04-28T12:03:19Z

Guenter moved page Exercise 3.7Z: Square Wave With Echo to Exercise 3.7Z: Rectangular signal with Echo

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

2021-03-23T12:39:08Z

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

Noah at 21:17, 28 January 2021

2021-01-28T21:17:57Z

Noah at 21:15, 28 January 2021

2021-01-28T21:15:24Z

Noah at 21:12, 28 January 2021

2021-01-28T21:12:55Z

@@ Line 62: / Line 62: @@
 {{ML-Kopf}}
 '''(1)'''&nbsp;  The <u>second suggested solution</u> is correct:
-*The impulse response is equal to the received signal&nbsp; $r(t)$, if a single Dirac impulse is present at the input at time&nbsp; $t = 0$&nbsp;:
+*The impulse response is equal to the received signal&nbsp; $r(t)$, if a single Dirac delta is present at the input at time&nbsp; $t = 0$&nbsp;:
 :$$h(t) = \delta (t) + \alpha  \cdot \delta( {t - \tau } ).$$

@@ Line 67: / Line 67: @@
 [[File:P_ID532__Sig_Z_3_7_b_neu.png|right|frame|Convolution of square wave signal&nbsp; $s(t)$&nbsp; and impulse response&nbsp; $h(t)$]]
-'''(2)'''&nbsp; It holds that&nbsp; $r(t) = s(t) ∗ h(t)$. This convolution operation is most easily performed graphically:
+'''(2)'''&nbsp; It holds&nbsp; $r(t) = s(t) ∗ h(t)$.&nbsp; This convolution operation is most easily performed graphically:
 The values of the received signal are generally:
@@ Line 82: / Line 82: @@
-'''(3)'''&nbsp;  Using a similar procedure as in '''(2)''' , a DC signal of $2\hspace{0.02cm}\text{ V}$ is obtained for $r(t)$ :
+'''(3)'''&nbsp;  Using a similar procedure as in&nbsp; '''(2)''',&nbsp; a direct signal&nbsp; $\rm (DC)$&nbsp; of&nbsp; $2\hspace{0.02cm}\text{ V}$&nbsp; is obtained for&nbsp; $r(t)$ :
-*The gaps in the signal $s(t)$ are completely filled by the echo $s(t - T/2)$.
+*The gaps in the signal&nbsp; $s(t)$&nbsp; are completely filled by the echo&nbsp; $s(t - T/2)$.
-*This result can also be derived in the frequency domain.
+*This result can also be derived in the frequency domain.&nbsp; The channel frequency response is with&nbsp; $\alpha = 1$&nbsp; and&nbsp; $\tau = T/2$:
 :$$H( f ) = 1 + 1 \cdot {\rm{e}}^{ - {\rm{j\pi }}fT}  = 1 + \cos ( {{\rm{\pi }}fT} ) - {\rm{j}} \cdot {\rm{sin}}( {{\rm{\pi }}fT} ).$$
-*Apart from the DC component, the input signal ${s(t)}$ only has components at $f = f_0 = 1/T$, $f = 3 \cdot f_0$, $f = 5 \cdot f_0$ etc..
+*Apart from the DC component, the input signal&nbsp; ${s(t)}$&nbsp; only has components at&nbsp; $f = f_0 = 1/T$,&nbsp; $f = 3 \cdot f_0$,&nbsp; $f = 5 \cdot f_0$,&nbsp; etc..
-*At these frequencies, however, both the real&ndash; and the imaginary part of ${H(f)}$ are equal to zero.
+*At these frequencies, however, both the real&ndash; and the imaginary part of&nbsp; ${H(f)}$&nbsp; are equal to zero.
-*Thus, for the output spectrum with $A_0 = 1 \text{ V}$ and $H(f = 0) = 2$ we obtain:
+*Thus, for the output spectrum with&nbsp; $A_0 = 1 \text{ V}$&nbsp; and&nbsp; $H(f = 0) = 2$&nbsp; we obtain:
 :$$R(f) = A_0  \cdot H(f = 0) \cdot \delta (f) = 2\;{\rm{V}} \cdot \delta (f).$$
-The inverse Fourier transformation thus also yields $r(t) \underline{= 2 \text{ V= const}}$.
+The inverse Fourier transformation thus also yields&nbsp; $r(t) \underline{= 2 \text{ V= const}}$.
 {{ML-Fuß}}

@@ Line 40: / Line 40: @@
 {Which statements are true regarding the impulse response&nbsp; $h(t)$?
 |type="[]"}
-- For&nbsp; $0 ≤ t < \tau$:&nbsp; &nbsp;$h(t) = 1$ is true, for&nbsp; $t  > \tau$:&nbsp; &nbsp; $h(t) = 1 + \alpha$.
+- For&nbsp; $0 ≤ t < \tau$&nbsp; holds &nbsp;$h(t) = 1$, and for&nbsp; $t  > \tau$&nbsp; holds &nbsp; $h(t) = 1 + \alpha$.
 + It holds that &nbsp;$h(t) = \delta (t) + \alpha \cdot \delta(t - \tau)$.
 - $h(t)$&nbsp; has a Gaussian shape.

@@ Line 3: / Line 3: @@
 }}
-[[File:EN_Sig_Z_3_7.png|right|frame|Transmit signal $s(t)$ & signal $r(t)$ with echo]]
+[[File:EN_Sig_Z_3_7.png|right|frame|Transmission signal&nbsp; $s(t)$, and&nbsp; <br>reception signal&nbsp; $r(t)$&nbsp; with echo]]
-We consider a periodic square wave signal&nbsp; $s(t)$&nbsp; with the possible amplitude values&nbsp; $0\text{ V}$&nbsp; and&nbsp; $2\text{ V}$&nbsp; and the period duration&nbsp; $T_0 = T = 1 \text{ ms}$. At the jump points, for example at&nbsp; $t = T/4$, the signal value is&nbsp; $1\text{ V}$. The DC component $($also i.e. the Fourier coefficient&nbsp; $A_0)$&nbsp; of the signal is also&nbsp; $1\text{ V}$.
+We consider a periodic rectangular signal&nbsp; $s(t)$&nbsp; with the possible amplitude values&nbsp; $0\text{ V}$&nbsp; and&nbsp; $2\text{ V}$&nbsp; and the period duration&nbsp; $T_0 = T = 1 \text{ ms}$.&nbsp; At the jump points, e.g. at&nbsp; $t = T/4$, the signal value are&nbsp; $1\text{ V}$.&nbsp; The DC component&nbsp; $($i.e. the Fourier coefficient&nbsp; $A_0)$&nbsp; of the signal is&nbsp; $1\text{ V}$, too.
 Further applies:
@@ Line 12: / Line 12: @@
 * The coefficients&nbsp; $A_n$&nbsp; with even&nbsp; $n$&nbsp; are also zero.
-* For odd values of&nbsp; $n$&nbsp; on the other hand, the following applies:
+* For odd values of&nbsp; $n$,&nbsp; the following applies:
 :$$A_n  = ( { - 1} )^{\left( {n - 1} \right)/2}  \cdot \frac{{4\;{\rm{V}}}}{{n \cdot {\rm{\pi }}}}.$$
@@ Line 19: / Line 19: @@
 *The latter is characterised by the attenuation factor&nbsp; $\alpha$&nbsp; and the transit time&nbsp; $\tau$&nbsp;.
 *Therefore, the following applies to the received signal:
-$$r(t) = s(t) + \alpha  \cdot s( {t - \tau } ).$$
+:$$r(t) = s(t) + \alpha  \cdot s( {t - \tau } ).$$
-The frequency response of the channel is&nbsp; $H(f) = R(f)/S(f)$, the impulse response is denoted by&nbsp; $h(t)$&nbsp;.
+The frequency response of the channel is&nbsp; $H(f) = R(f)/S(f)$,&nbsp; the impulse response is denoted by&nbsp; $h(t)$&nbsp;.
@@ Line 30: / Line 30: @@
 ''Hints:''
-*This exercise belongs to the 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 in particular 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 38: / Line 38: @@
 <quiz display=simple>
-{Which statements are true regarding the impulse response&nbsp; $h(t)$&nbsp;?
+{Which statements are true regarding the impulse response&nbsp; $h(t)$?
 |type="[]"}
-- For&nbsp; $0 ≤ t < \tau$&nbsp; &nbsp;$h(t) = 1$ is true, for&nbsp; $t  > \tau$&nbsp; &nbsp; $h(t) = 1 + \alpha$.
+- For&nbsp; $0 ≤ t < \tau$:&nbsp; &nbsp;$h(t) = 1$ is true, for&nbsp; $t  > \tau$:&nbsp; &nbsp; $h(t) = 1 + \alpha$.
 + It holds that &nbsp;$h(t) = \delta (t) + \alpha \cdot \delta(t - \tau)$.
-- $h(t)$&nbsp; has a gaussian shape.
+- $h(t)$&nbsp; has a Gaussian shape.
-{Calculate the signal&nbsp; $r(t)$&nbsp; for the channel parameters&nbsp; $\alpha = -0.5$&nbsp; and&nbsp; $\tau = T/4$. <br>What values result at the given times?
+{Calculate the reception signal&nbsp; $r(t)$&nbsp; for the channel parameters&nbsp; $\alpha = -0.5$&nbsp; and&nbsp; $\tau = T/4$. <br>What values result at the given times?
 |type="{}"}
 $r(t = 0.2 \cdot T)\ = \ $  { 1 3% } &nbsp;$\text{V}$
@@ Line 51: / Line 51: @@
-{Calculate the signal&nbsp; $r(t)$&nbsp; with&nbsp; $\alpha = 1$&nbsp; and&nbsp; $\tau = T/2$. Interpret the result in the frequency domain. <br>What value results for&nbsp; $t = T/2$?
+{Calculate the reception signal&nbsp; $r(t)$&nbsp; with&nbsp; $\alpha = 1$&nbsp; and&nbsp; $\tau = T/2$.&nbsp; Interpret the result in the frequency domain. <br>What value results for&nbsp; $t = T/2$?
 |type="{}"}
 $r(t = T/2)\ = \ $ { 2 3% } &nbsp;$\text{V}$

← Older revision		Revision as of 12:39, 23 March 2021
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	__NOEDITSECTION__		__NOEDITSECTION__
−	[[Category:~~Exercises for~~ Signal Representation\|^3.4 The Convolution Theorem^]]	+	[[Category:Signal Representation: Exercises\|^3.4 The Convolution Theorem^]]

@@ Line 59: / Line 59: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp;  Richtig ist der <u>zweite Lösungsvorschlag</u>:
+'''(1)'''&nbsp;  The <u>second suggested solution</u> is correct:
-*Die Impulsantwort ist gleich dem Empfangssignal&nbsp; $r(t)$, wenn am Eingang ein einzelner Diracimpuls zum Zeitpunkt&nbsp; $t = 0$&nbsp; anliegt:
+*The impulse response is equal to the received signal&nbsp; $r(t)$, if a single Dirac impulse is present at the input at time&nbsp; $t = 0$&nbsp;:
 :$$h(t) = \delta (t) + \alpha  \cdot \delta( {t - \tau } ).$$
 [[File:P_ID532__Sig_Z_3_7_b_neu.png|right|frame|Faltung von Rechtecksignal&nbsp; $s(t)$&nbsp; und Impulsantwort&nbsp; $h(t)$]]
-'''(2)'''&nbsp;   Es gilt&nbsp; $r(t) = s(t) ∗ h(t)$. Diese Faltungsoperation lässt sich am einfachsten grafisch ausführen:
+'''(2)'''&nbsp; It holds that&nbsp; $r(t) = s(t) ∗ h(t)$. This convolution operation is most easily performed graphically:
-Die Werte des Empfangssignals lauten allgemein:
+The values of the received signal are generally:
 * $0.00 < t/T < 0.25\text{:}\hspace{0.4cm}   r(t) = +1\hspace{0.02cm}\text{ V}$,
@@ Line 77: / Line 77: @@
-Die gesuchten Werte sind somit
+The values we are looking for are thus
 :$$r(t = 0.2 \cdot T) \hspace{0.15cm}\underline{= +1 \hspace{0.02cm}\text{ V}},$$
 :$$r(t = 0.3 · T) \hspace{0.15cm}\underline{= -1 \hspace{0.02cm}\text{ V}}.$$
-'''(3)'''&nbsp;  Bei ähnlicher Vorgehensweise wie unter '''(2)''' erhält man für $r(t)$ ein Gleichsignal von $2 \hspace{0.02cm}\text{ V}$:
+'''(3)'''&nbsp;  Using a similar procedure as in '''(2)''' , a DC signal of $2\hspace{0.02cm}\text{ V}$ is obtained for $r(t)$ :
-*Die Lücken im Signal $s(t)$ werden durch das Echo $s(t - T/2)$ vollständig aufgefüllt.
+*The gaps in the signal $s(t)$ are completely filled by the echo $s(t - T/2)$.
-*Dieses Ergebnis lässt sich auch im Frequenzbereich ableiten.
+*This result can also be derived in the frequency domain.
-*Der Kanalfrequenzgang lautet mit $\alpha = 1$ und $\tau = T/2$:
+*The channel frequency response is with $\alpha = 1$ and $\tau = T/2$:
 :$$H( f ) = 1 + 1 \cdot {\rm{e}}^{ - {\rm{j\pi }}fT}  = 1 + \cos ( {{\rm{\pi }}fT} ) - {\rm{j}} \cdot {\rm{sin}}( {{\rm{\pi }}fT} ).$$
-*Das Eingangssignal ${s(t)}$ hat außer dem Gleichanteil nur Anteile bei $f = f_0 = 1/T$, $f = 3 \cdot f_0$, $f = 5 \cdot f_0$ usw..
+*Apart from the DC component, the input signal ${s(t)}$ only has components at $f = f_0 = 1/T$, $f = 3 \cdot f_0$, $f = 5 \cdot f_0$ etc..
-*Bei diesen Frequenzen sind aber sowohl der Real&ndash; als auch der Imaginärteil von ${H(f)}$ gleich Null.
+*At these frequencies, however, both the real&ndash; and the imaginary part of ${H(f)}$ are equal to zero.
-*Damit erhält man für das Ausgangsspektrum mit $A_0 = 1 \text{ V}$ und $H(f = 0) = 2$:
+*Thus, for the output spectrum with $A_0 = 1 \text{ V}$ and $H(f = 0) = 2$ we obtain:
 :$$R(f) = A_0  \cdot H(f = 0) \cdot \delta (f) = 2\;{\rm{V}} \cdot \delta (f).$$
-Die Fourierrücktransformation liefert damit ebenfalls $r(t) \underline{= 2 \text{ V= const}}$.
+The Fourier inverse transformation thus also yields $r(t) \underline{= 2 \text{ V= const}}$.
 {{ML-Fuß}}