Aufgaben:Exercise 2.2: Distortion Power - Revision history

Oezer at 20:51, 12 September 2021

2021-09-12T20:51:49Z

Oezer at 20:49, 12 September 2021

2021-09-12T20:49:53Z

Oezer at 20:48, 12 September 2021

2021-09-12T20:48:08Z

Oezer at 20:45, 12 September 2021

2021-09-12T20:45:01Z

Oezer at 20:44, 12 September 2021

2021-09-12T20:44:05Z

Oezer at 20:43, 12 September 2021

2021-09-12T20:43:41Z

Oezer at 20:40, 12 September 2021

2021-09-12T20:40:39Z

Oezer at 20:36, 12 September 2021

2021-09-12T20:36:47Z

Oezer at 20:04, 12 September 2021

2021-09-12T20:04:01Z

Oezer at 20:02, 12 September 2021

2021-09-12T20:02:56Z

@@ Line 118: / Line 118: @@
 *Despite the same distortion power&nbsp;$10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm V2}$&nbsp; is less than &nbsp;$10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm V1}$&nbsp; by about&nbsp;$6 \ \rm dB$&nbsp;.
-*The signal&nbsp;$y_2(t)$&nbsp; is thus significantly less favorable in terms of SNR than&nbsp;$y_1(t)$.
+*The signal&nbsp;$y_2(t)$&nbsp; is thus significantly less favourable in terms of SNR than&nbsp;$y_1(t)$.
 *It is considered that now the power of the output signal is only a quarter of the input power due to &nbsp;$\alpha = 0.5$&nbsp;.
 *If this attenuation at the output was to be compensated by amplifying it by $1/\alpha$, the distortion power would indeed increase by $\alpha^2$.

← Older revision		Revision as of 20:49, 12 September 2021
Line 103:		Line 103:

	:and afterwards the mean squared error for very (infinitely) many $\alpha$–  and  $\tau$–values would have to be determined, in doing so the integration interval is to be adjusted to $\tau$  in each case.		:and afterwards the mean squared error for very (infinitely) many $\alpha$–  and  $\tau$–values would have to be determined, in doing so the integration interval is to be adjusted to $\tau$  in each case.
−	*~~Also then~~ the smallest possible result would be obtained for $\alpha \; \underline{= 0.5}$  and  $\tau \; \underline{= 1 \ \rm ms}$ . However, for this optimization of $\alpha$  and  $\tau$  the useage of a computer program should be granted.	+	*Then, the smallest possible result would also be obtained for $\alpha \; \underline{= 0.5}$  and  $\tau \; \underline{= 1 \ \rm ms}$ . However, for this optimization of $\alpha$  and  $\tau$  the useage of a computer program should be granted.

@@ Line 102: / Line 102: @@
 :$$\varepsilon_2(t) = y_2(t) - \alpha \cdot x(t - \tau)$$
-:for very (infinitely) many&nbsp;$\alpha$&ndash;&nbsp; and &nbsp;$\tau$&ndash;values and afterwards the mean squared error would have to be determined, in doing so the integration interval is to be adjusted to&nbsp;$\tau$&nbsp; in each case.
+:and afterwards the mean squared error for very (infinitely) many&nbsp;$\alpha$&ndash;&nbsp; and &nbsp;$\tau$&ndash;values would have to be determined, in doing so the integration interval is to be adjusted to&nbsp;$\tau$&nbsp; in each case.
 *Also then the smallest possible result would be obtained for&nbsp;$\alpha \; \underline{= 0.5}$&nbsp; and &nbsp;$\tau \; \underline{= 1 \ \rm ms}$&nbsp;. However, for this optimization of&nbsp;$\alpha$&nbsp; and &nbsp;$\tau$&nbsp; the useage of a computer program should be granted.

@@ Line 102: / Line 102: @@
 :$$\varepsilon_2(t) = y_2(t) - \alpha \cdot x(t - \tau)$$
-:for very (infinitely) manynbsp;$\alpha$&ndash;&nbsp; and &nbsp;$\tau$&ndash;values and afterwards the mean squared error would have to be determined, in doing so the integration interval is to be adjusted to&nbsp;$\tau$&nbsp; in each case.
+:for very (infinitely) many&nbsp;$\alpha$&ndash;&nbsp; and &nbsp;$\tau$&ndash;values and afterwards the mean squared error would have to be determined, in doing so the integration interval is to be adjusted to&nbsp;$\tau$&nbsp; in each case.
 *Also then the smallest possible result would be obtained for&nbsp;$\alpha \; \underline{= 0.5}$&nbsp; and &nbsp;$\tau \; \underline{= 1 \ \rm ms}$&nbsp;. However, for this optimization of&nbsp;$\alpha$&nbsp; and &nbsp;$\tau$&nbsp; the useage of a computer program should be granted.

@@ Line 97: / Line 97: @@
 '''(3)'''&nbsp; The sketch on the information sheet makes it clear that even without the distortions occuring &ndash; but due to attenuation and runtime alone &ndash; the signal&nbsp;$y(t)$&nbsp; would differ significantly from&nbsp;$x(t)$&nbsp;.
-*The following would arise as a result:&nbsp;$y(t) = 0.5 \cdot x(t-1\ {\rm ms}) $&nbsp;.
+*The following would arise as a result: &nbsp;$y(t) = 0.5 \cdot x(t-1\ {\rm ms}) $&nbsp;.
 *If someone does not immediately perceive these values from the graph, then first the error signal

@@ Line 19: / Line 19: @@
 *Use the interval&nbsp; $0$ ... $4 \hspace{0.08cm} \rm ms$&nbsp; for&nbsp;$y_1(t)$&nbsp; and the interval&nbsp; $1 \hspace{0.08cm} {\rm ms}$ ... $5 \hspace{0.08cm} \rm ms$ for&nbsp; &nbsp;$y_2(t)$&nbsp;.
 *Thus, the measurement time is&nbsp;$T_{\rm M} = 4 \hspace{0.08cm} \rm ms$ in both cases.
-*It is obvious that with respect to&nbsp;$y_1(t)$&nbsp; die Parameter &nbsp;$\alpha = 1$&nbsp; and &nbsp;$\tau = 0$&nbsp; respectively result in the minimum distortion power.
+*It is obvious that with respect to&nbsp;$y_1(t)$&nbsp; the parameters &nbsp;$\alpha = 1$&nbsp; and &nbsp;$\tau = 0$&nbsp; respectively result in the minimum distortion power.

@@ Line 97: / Line 97: @@
 '''(3)'''&nbsp; The sketch on the information sheet makes it clear that even without the distortions occuring &ndash; but due to attenuation and runtime alone &ndash; the signal&nbsp;$y(t)$&nbsp; would differ significantly from&nbsp;$x(t)$&nbsp;.
-*Es würde sich &nbsp;$y(t) = 0.5 \cdot x(t-1\ {\rm ms}) $&nbsp; ergeben.
+*The following would arise as a result&nbsp;$y(t) = 0.5 \cdot x(t-1\ {\rm ms}) $&nbsp;.
-*Wenn jemand diese Werte nicht sofort aus der Grafik erkennt, so müsste er für sehr (unendlich) viele &nbsp;$\alpha$&ndash;&nbsp; und &nbsp;$\tau$&ndash;Werte zunächst das Fehlersignal
+*If someone does not immediately perceive these values from the graph, then first the error signal
 :$$\varepsilon_2(t) = y_2(t) - \alpha \cdot x(t - \tau)$$
-:and anschließend den mittleren quadratischen Fehler ermitteln, wobei das Integrationsintervall jeweils an &nbsp;$\tau$&nbsp; anzupassen ist.
+:for very (infinitely) manynbsp;$\alpha$&ndash;&nbsp; and &nbsp;$\tau$&ndash;values and afterwards the mean squared error would have to be determined, in doing so the integration interval is to be adjusted to&nbsp;$\tau$&nbsp; in each case.
-*Auch dann würde man das kleinstmögliche Ergebnis für &nbsp;$\alpha \; \underline{= 0.5}$&nbsp; und &nbsp;$\tau \; \underline{= 1 \ \rm ms}$&nbsp; erhalten. Für diese Optimierung von &nbsp;$\alpha$&nbsp; und &nbsp;$\tau$&nbsp; sollte man sich allerdings schon ein Computerprogramm gönnen.
+*Also then the smallest possible result would be obtained for&nbsp;$\alpha \; \underline{= 0.5}$&nbsp; and &nbsp;$\tau \; \underline{= 1 \ \rm ms}$&nbsp;. However, for this optimization of&nbsp;$\alpha$&nbsp; and &nbsp;$\tau$&nbsp; the useage of a computer program should be granted.
-'''(4)'''&nbsp; Die obige Skizze zeigt, dass &nbsp;$\varepsilon_2(t)$&nbsp;  bis auf eine Verschiebung um &nbsp;$1 \ \rm ms$&nbsp; gleich dem Fehlersignal &nbsp;$\varepsilon_1(t)$&nbsp; ist. Mit dem Integrationsintervall &nbsp;$1 \ {\rm ms}$ ... $5 \ {\rm ms}$&nbsp; ergibt sich somit auch die gleiche Verzerrungsleistung:
+'''(4)'''&nbsp; The above sketch shows that&nbsp;$\varepsilon_2(t)$&nbsp; is equal to the error signal&nbsp;$\varepsilon_1(t)$&nbsp; except for a shift by&nbsp;$1 \ \rm ms$&nbsp;. Considering the integration interval&nbsp;$1 \ {\rm ms}$ ... $5 \ {\rm ms}$&nbsp; the same distortion power is obtained:
 :$$P_{\rm V2}  =  P_{\rm V1} \hspace{0.15cm}\underline{ =  5 \cdot 10^{-3} \, \rm   V^2}.$$
-'''(5)'''&nbsp; Entsprechend dem Angabenblatt gilt:
+'''(5)'''&nbsp; According to the information sheet the following holds:
 :$$\rho_{\rm V2} = \frac{ \alpha^2 \cdot P_{x}}{P_{\rm V2}}= \frac{ 0.5^2 \cdot {1 \, \rm
    V^2}}{0.005 \,   \rm V^2}\hspace{0.05cm}\rm = 50\hspace{0.3cm} \Rightarrow \hspace{0.3cm}
 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm V2} \hspace{0.15cm}\underline{= {16.99 \, \rm dB}}.$$
-*Trotz gleicher Verzerrungsleistung ist &nbsp;$10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm V2}$&nbsp; gegenüber &nbsp;$10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm V1}$&nbsp; um etwa &nbsp;$6 \ \rm dB$&nbsp; geringer.
+*Despite the same distortion power&nbsp;$10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm V2}$&nbsp; is less than &nbsp;$10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm V1}$&nbsp; by about&nbsp;$6 \ \rm dB$&nbsp;.
-*Das Signal &nbsp;$y_2(t)$&nbsp; ist also hinsichtlich des SNR  deutlich ungünstiger als &nbsp;$y_1(t)$.
+*The signal&nbsp;$y_2(t)$&nbsp; is thus significantly less favorable in terms of SNR than&nbsp;$y_1(t)$.
-* Es ist berücksichtigt, dass nun wegen &nbsp;$\alpha = 0.5$&nbsp; die Leistung des Ausgangssignals nur noch ein Viertel der Eingangsleistung beträgt.
+*It is considered that now the power of the output signal is only a quarter of the input power due to &nbsp;$\alpha = 0.5$&nbsp;.
-*Würde man diese Dämpfung am Ausgang durch eine Verstärkung um $1/\alpha$ kompensieren, so würde zwar die Verzerrungsleistung um $\alpha^2$ größer.
+*If this attenuation at the output was to be compensated by amplifying it by $1/\alpha$, the distortion power would indeed increase by $\alpha^2$.
-*Das Signal-zu-Verzerrungs-Leistungsverhältnis $\rho_{\rm V2}$ bliebe jedoch erhalten, weil auch das "Nutzsignal" um den gleichen Betrag angehoben wird.
+*The signal-to-distortion-power ratio $\rho_{\rm V2}$ would, however, remain the same because the "useful signal" would also be increased by the same value.
 {{ML-Fuß}}