Aufgaben:Exercise 2.1Z: Distortion and Equalisation - Revision history

Javier: Text replacement - "equalisation" to "equalization"

2021-09-21T12:50:08Z

Text replacement - "equalisation" to "equalization"

Oezer at 22:01, 9 September 2021

2021-09-09T22:01:17Z

Oezer at 21:57, 9 September 2021

2021-09-09T21:57:27Z

Oezer at 21:57, 9 September 2021

2021-09-09T21:57:07Z

Oezer at 21:54, 9 September 2021

2021-09-09T21:54:17Z

Oezer at 21:53, 9 September 2021

2021-09-09T21:53:16Z

Oezer at 21:52, 9 September 2021

2021-09-09T21:52:38Z

Oezer at 21:50, 9 September 2021

2021-09-09T21:50:51Z

Oezer at 21:49, 9 September 2021

2021-09-09T21:49:37Z

Oezer at 21:46, 9 September 2021

2021-09-09T21:46:14Z

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@@ Line 18: / Line 18: @@
 Furthermore, we consider
 *a linearly distorting system&nbsp; $S_{\rm V}$&nbsp; with&nbsp; $X(f)$&nbsp; at the input and&nbsp; $Y(f)$&nbsp; at the output, and
-*the equalisation system&nbsp; $S_{\rm E}$&nbsp; with the input spectrum&nbsp; $Y(f)$&nbsp; and output spectrum&nbsp; $Z(f)$.
+*the equalization system&nbsp; $S_{\rm E}$&nbsp; with the input spectrum&nbsp; $Y(f)$&nbsp; and output spectrum&nbsp; $Z(f)$.
@@ Line 32: / Line 32: @@
 ''Please note:''
 *The exercise belongs to the chapter&nbsp;  [[Linear_and_Time_Invariant_Systems/Classification_of_the_Distortions|Classification of the Distortions]].
-*A complete equalisation means that&nbsp; $Z(f) = X(f)$&nbsp; holds.
+*A complete equalization means that&nbsp; $Z(f) = X(f)$&nbsp; holds.
@@ Line 45: / Line 45: @@
-{$X(f) = A(f)$&nbsp; and &nbsp;$Y(f) = B(f)$ still hold true. Is complete equalisation possible with a linear filter&nbsp;$H_{\rm E}(f)$&nbsp;? <br>If YES, please specify  &nbsp;$H_{\rm E}(f)$.
+{$X(f) = A(f)$&nbsp; and &nbsp;$Y(f) = B(f)$ still hold true. Is complete equalization possible with a linear filter&nbsp;$H_{\rm E}(f)$&nbsp;? <br>If YES, please specify  &nbsp;$H_{\rm E}(f)$.
 |type="()"}
 + Yes.
@@ Line 57: / Line 57: @@
-{$X(f) = C(f)$&nbsp; und &nbsp;$Y(f) = B(f)$ still hold true. Is complete equalisation possible with a linear filter&nbsp;$H_{\rm E}(f)$&nbsp;? <br>If YES, please specify &nbsp;$H_{\rm E}(f)$&nbsp;.
+{$X(f) = C(f)$&nbsp; und &nbsp;$Y(f) = B(f)$ still hold true. Is complete equalization possible with a linear filter&nbsp;$H_{\rm E}(f)$&nbsp;? <br>If YES, please specify &nbsp;$H_{\rm E}(f)$&nbsp;.
 |type="()"}
 - Yes.
@@ Line 80: / Line 80: @@
 '''(2)'''&nbsp; <u>Yes</u> is correct:
-*For this constellation, a complete linear equalisation is also possible with
+*For this constellation, a complete linear equalization is also possible with
 :$$H_{\rm E}(f)  = \frac{Z(f)}{Y(f)} = \frac{A(f)}{B(f)} = \frac{1}{H_{\rm V}(f)}$$
@@ Line 93: / Line 93: @@
 '''(4)'''&nbsp; <u>No</u> is correct:
-*A complete equalisation is not possible here:
+*A complete equalization is not possible here:
 *The parts of the Gaussian spectrum, which are completely eliminated by&nbsp;  $H_{\rm V}(f)$&nbsp;, cannot be recovered by the linear system.

@@ Line 99: / Line 99: @@
 '''(5)'''&nbsp; <u>No</u> is correct:
-*This constellation is not possible with a linear system since there cannot be any spectral components in the spectrum &nbsp;$C(f) = A(f) \cdot H_{\rm V}(f)$&nbsp; that do not exist in&nbsp;$A(f)$&nbsp;.
+*This constellation is not possible with a linear system since there cannot be any spectral components included in the spectrum &nbsp;$C(f) = A(f) \cdot H_{\rm V}(f)$&nbsp; that do not exist in&nbsp;$A(f)$&nbsp;.
 *The question whether there is a non-linear system which forms a Gaussian spectrum out of the&nbsp; $\cos^2$-spectrum is not asked and therefore does not need to be answered: &nbsp; The authors rather believe "no".
 {{ML-Fuß}}

@@ Line 80: / Line 80: @@
 '''(2)'''&nbsp; <u>Yes</u> is correct:
-*For this constellation a complete linear equalisation is also possible with
+*For this constellation, a complete linear equalisation is also possible with
 :$$H_{\rm E}(f)  = \frac{Z(f)}{Y(f)} = \frac{A(f)}{B(f)} = \frac{1}{H_{\rm V}(f)}$$

@@ Line 80: / Line 80: @@
 '''(2)'''&nbsp; <u>Yes</u> is correct:
-*A complete linear equalisation is also possible for this constellation with
+*For this constellation a complete linear equalisation is also possible with
 :$$H_{\rm E}(f)  = \frac{Z(f)}{Y(f)} = \frac{A(f)}{B(f)} = \frac{1}{H_{\rm V}(f)}$$

@@ Line 80: / Line 80: @@
 '''(2)'''&nbsp; <u>Yes</u> is correct:
-*A complete linear equalisation is also possible with this constellation with
+*A complete linear equalisation is also possible for this constellation with
 :$$H_{\rm E}(f)  = \frac{Z(f)}{Y(f)} = \frac{A(f)}{B(f)} = \frac{1}{H_{\rm V}(f)}$$

@@ Line 74: / Line 74: @@
 '''(1)'''&nbsp; <u>Yes</u> is correct:
 *This constellation is possible because&nbsp; $X(f)$&nbsp; is also always different from zero for all&nbsp; $Y(f) \ne 0$&nbsp;.
-*For all frequencies less than&nbsp; $0.5 \ \rm kHz$&nbsp; causes&nbsp; $H_{\rm V} = B(f)/A(f) < 1$&nbsp; an attenuation.
+*For all frequencies less than&nbsp; $0.5 \ \rm kHz$&nbsp;, &nbsp; $H_{\rm V} = B(f)/A(f) < 1$&nbsp; causes an attenuation.
 *In contrast, the frequencies between&nbsp; $0.5 \ \rm kHz$&nbsp; and&nbsp; $1 \ \rm kHz$&nbsp; are amplified by the system.