Aufgaben:Exercise 1.6Z: Interpretation of the Frequency Response - Revision history

Oezer at 17:33, 7 September 2021

2021-09-07T17:33:49Z

Oezer at 17:32, 7 September 2021

2021-09-07T17:32:58Z

Oezer at 17:31, 7 September 2021

2021-09-07T17:31:40Z

Oezer at 17:29, 7 September 2021

2021-09-07T17:29:32Z

Oezer at 17:29, 7 September 2021

2021-09-07T17:29:07Z

Oezer at 17:27, 7 September 2021

2021-09-07T17:27:54Z

Oezer at 17:26, 7 September 2021

2021-09-07T17:26:18Z

Oezer at 17:25, 7 September 2021

2021-09-07T17:25:25Z

Oezer at 17:23, 7 September 2021

2021-09-07T17:23:49Z

Oezer at 17:08, 7 September 2021

2021-09-07T17:08:00Z

@@ Line 102: / Line 102: @@
   d}\tau =  \frac{2A_x}{2\pi   f_1 \cdot \Delta t} \cdot {\rm sin}(2\pi   f_1  \frac{\Delta t}{2}
   )= A_x \cdot {\rm si}(\pi   f_1  \Delta t ).$$
-*Taking &nbsp; $f_1 · Δt = 0.5$&nbsp;into account thus the result is:
+*Taking &nbsp; $f_1 · Δt = 0.5$&nbsp;into account the result is:
 :$$A_1 = A_x \cdot {\rm si}(\frac{\pi}{2} ) = \frac{2A_x}{\pi} \hspace{0.15cm}\underline{= 0.637\,{\rm V}}.$$
 *Correspondingly, the following is obtained using&nbsp; $f_3 · Δt = 1.5$:

@@ Line 93: / Line 93: @@
 *For the direct (DC) signal  &nbsp;$x_0(t) = A_x$&nbsp;, set &nbsp;$f_i = 0$&nbsp; and one obtains &nbsp;$A_0 = A_x \ \rm \underline{ = \ 1 \hspace{0.05cm} V}$.
 *In contrast to this, for the cosine frequencies &nbsp;$f_2 = 2 \ \rm kHz$&nbsp; and &nbsp;$f_4 = 4 \ \rm kHz$&nbsp; the integral vanishes in each case because then it is integrated over one and two periods, respectively: &nbsp; $A_2 \ \rm \underline{ = \hspace{0.05cm} 0}$&nbsp; und&nbsp; $A_4 \hspace{0.05cm} \rm \underline{ = \ 0}$.
-*In the frequency domain, the case which are dealt with here correspond to:
+*In the frequency domain, the cases which are dealt with here correspond to:
 :$$H(f=0) = 1, \hspace{0.3cm}H(f=\Delta f) = 0, \hspace{0.3cm}H(f=2\Delta f) = 0.$$

@@ Line 91: / Line 91: @@
 '''(4)'''&nbsp; <u>Approaches 2, 3 and 5</u> are correct:
-*For the direct (DC) signal  &nbsp;$x_0(t) = A_x$&nbsp; set &nbsp;$f_i = 0$&nbsp; and one obtains &nbsp;$A_0 = A_x \ \rm \underline{ = \ 1 \hspace{0.05cm} V}$.
+*For the direct (DC) signal  &nbsp;$x_0(t) = A_x$&nbsp;, set &nbsp;$f_i = 0$&nbsp; and one obtains &nbsp;$A_0 = A_x \ \rm \underline{ = \ 1 \hspace{0.05cm} V}$.
 *In contrast to this, for the cosine frequencies &nbsp;$f_2 = 2 \ \rm kHz$&nbsp; and &nbsp;$f_4 = 4 \ \rm kHz$&nbsp; the integral vanishes in each case because then it is integrated over one and two periods, respectively: &nbsp; $A_2 \ \rm \underline{ = \hspace{0.05cm} 0}$&nbsp; und&nbsp; $A_4 \hspace{0.05cm} \rm \underline{ = \ 0}$.
 *In the frequency domain, the case which are dealt with here correspond to:

@@ Line 27: / Line 27: @@
 ''Please note:''
 *The exercise belongs to the chapter&nbsp;   [[Linear_and_Time_Invariant_Systems/Some_Low-Pass_Functions_in_Systems_Theory|Some Low-Pass Functions in Systems Theory]].
-*Contrary to the usual definition of the amplitude, the "$A_i$" may well be negative. Then, this corresponds to the function „minus-cosine”.
+*Contrary to the usual definition of the amplitude, the "$A_i$" may well be negative. Then, this corresponds to the function "minus-cosine".

@@ Line 27: / Line 27: @@
 ''Please note:''
 *The exercise belongs to the chapter&nbsp;   [[Linear_and_Time_Invariant_Systems/Some_Low-Pass_Functions_in_Systems_Theory|Some Low-Pass Functions in Systems Theory]].
-*Contrary to the usual definition of the amplitude, the "$A_i$" may well be negative. This corresponds then to the function „minus-cosine”.
+*Contrary to the usual definition of the amplitude, the "$A_i$" may well be negative. Then, this corresponds to the function „minus-cosine”.

@@ Line 5: / Line 5: @@
 The task is meant to investigate the influence of a low-pass filter&nbsp; $H(f)$&nbsp; on cosinusoidal signals of the form
 :$$x_i(t) = A_x \cdot {\rm cos}(2\pi   f_i  t ).$$
-In the graph you can see the signals &nbsp;$x_i(t)$, where the index&nbsp;$i$&nbsp; indicates the frequency in&nbsp;$\rm kHz$&nbsp;. So, &nbsp;$x_2(t)$&nbsp; describes a &nbsp;$2 \hspace{0.09cm} \rm  kHz$–signal.
+In the graph you can see the signals &nbsp;$x_i(t)$ where the index&nbsp;$i$&nbsp; indicates the frequency in&nbsp;$\rm kHz$&nbsp;. So, &nbsp;$x_2(t)$&nbsp; describes a &nbsp;$2 \hspace{0.09cm} \rm  kHz$–signal.
 The signal amplitude in each case is&nbsp;$A_x = 1 \hspace{0.05cm} \rm  V$. The direct (DC) signal&nbsp; $x_0(t)$&nbsp; is to be interpreted as a limiting case of a cosine signal with frequeny&nbsp; $f_0 =0$.

@@ Line 98: / Line 98: @@
-'''(5)'''&nbsp; The result of the subtask&nbsp;  '''(3)'''&nbsp; - considering the symmetry - is for&nbsp; $f_i = f_1$:
+'''(5)'''&nbsp; The result of subtask&nbsp; '''(3)'''&nbsp; – considering the symmetry – is for&nbsp; $f_i = f_1$:
 :$$A_1= \frac{2A_x}{\Delta t} \cdot \int\limits_{ 0 }^{  \Delta t /2 } {\rm cos}(2\pi   f_1  \tau )\hspace{0.1cm}{\rm
   d}\tau =  \frac{2A_x}{2\pi   f_1 \cdot \Delta t} \cdot {\rm sin}(2\pi   f_1  \frac{\Delta t}{2}
   )= A_x \cdot {\rm si}(\pi   f_1  \Delta t ).$$
-*Mit&nbsp; $f_1 · Δt = 0.5$&nbsp; lautet somit das Ergebnis:
+*Taking &nbsp; $f_1 · Δt = 0.5$&nbsp;into account thus the result is:
 :$$A_1 = A_x \cdot {\rm si}(\frac{\pi}{2} ) = \frac{2A_x}{\pi} \hspace{0.15cm}\underline{= 0.637\,{\rm V}}.$$
-*Entsprechend erhält man mit&nbsp; $f_3 · Δt = 1.5$:
+*Correspondingly, the following is obtained using&nbsp; $f_3 · Δt = 1.5$:
 :$$A_3 = A_x \cdot {\rm si}({3\pi}/{2} ) = -\frac{2A_x}{3\pi} = -{A_1}/{3}\hspace{0.15cm}\underline{= -0.212\,{\rm V}}.$$
-*Genau zu den gleichen Ergebnissen – aber deutlich schneller – kommt man durch die Anwendung der Gleichung:
+*The exact same results are obtained – but much faster – by applying the equation:
 :$$A_i = A_x · H(f = f_i).$$
-*Bereits aus den Grafiken auf der Angabenseite erkennt man, dass das Integral über&nbsp; $x_1(t)$&nbsp; im markierten Bereich positiv und das Integral über&nbsp; $x_3(t)$&nbsp; negativ ist.
+*From the graphs on the information page it is already obvious that the integral over&nbsp; $x_1(t)$&nbsp; is positive in the marked area and the integral over&nbsp; $x_3(t)$&nbsp; is negative.
-*Es ist allerdings anzumerken, dass man im Allgemeinen als Amplitude meist den Betrag bezeichnet (siehe Hinweis auf der Angabenseite).
+*However, it should be noted that in general amplitude is usually referred to as the magnitude (See notice on the information page).
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