Difference between revisions of "Aufgaben:Exercise 3.3Z: High- and Low-Pass Filters in p-Form"
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− | { | + | {At what frequency $f_{\rm G}$ has the power transfer function $|H(f)|^2$ decreased to half with respect to the maximum value? |
|type="{}"} | |type="{}"} | ||
$f_{\rm G} \ = \ $ { 1.59 3% } $\ \rm MHz$ | $f_{\rm G} \ = \ $ { 1.59 3% } $\ \rm MHz$ | ||
− | { | + | {Which of the two RC–two-port networks results in the same transfer function as the two-port network $(1)$ if the capacitance $C$ is chosen correctly? |
|type="()"} | |type="()"} | ||
− | + | + | + Two-port network $(3)$, |
− | - | + | - Two-port network $(4)$. |
− | { | + | {Let $R = 100 \ \rm \Omega$ hold. How must $C$ be chosen so that the pole $p_{\rm x}$ coincides with that of the two-port network $(1)$? |
|type="{}"} | |type="{}"} | ||
$C \ = \ $ { 1 3% } $\ \rm nF$ | $C \ = \ $ { 1 3% } $\ \rm nF$ | ||
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===Solution=== | ===Solution=== | ||
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''(1)''' <u> | + | '''(1)''' <u>Both statements</u> are true: |
− | * | + | *The following limits hold for the two two-port networks: |
:$$\lim_{p \hspace{0.05cm}\rightarrow \hspace{0.05cm}0} H_{\rm | :$$\lim_{p \hspace{0.05cm}\rightarrow \hspace{0.05cm}0} H_{\rm | ||
TP}(p)\hspace{0.2cm} = \hspace{0.1cm}\lim_{p \hspace{0.05cm}\rightarrow \hspace{0.05cm}0}\frac{K}{p + p_{\rm x}} \hspace{0.15cm} { | TP}(p)\hspace{0.2cm} = \hspace{0.1cm}\lim_{p \hspace{0.05cm}\rightarrow \hspace{0.05cm}0}\frac{K}{p + p_{\rm x}} \hspace{0.15cm} { | ||
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\lim_{p \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} H_{\rm | \lim_{p \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} H_{\rm | ||
HP}(p)= \lim_{p \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty}\frac{K\cdot p}{p + p_{\rm x}} = K \hspace{0.05cm}.$$ | HP}(p)= \lim_{p \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty}\frac{K\cdot p}{p + p_{\rm x}} = K \hspace{0.05cm}.$$ | ||
− | * | + | *It can be seen that $H_{\rm TP}(p)$ yields zero for very high frequencies and $H_{\rm HP}(p)$ for very low frequencies. |
− | '''(2)''' | + | '''(2)''' We consider two-port network $(1)$. |
*Der Spannungsteiler liefert das Ergebnis | *Der Spannungsteiler liefert das Ergebnis | ||
:$$H_{\rm L}(p)= \frac { p L} | :$$H_{\rm L}(p)= \frac { p L} |
Revision as of 10:51, 12 October 2021
The diagram shows four simple filter configurations with low-pass and high-pass characteristics, respectively, which are composed of discrete components.
The following holds for the components of the circuits $(1)$ and $(2)$ :
- $$R = 100\,{\rm \Omega}\hspace{0.05cm},\hspace{0.2cm} L = 10\,{\rm µ H}\hspace{0.05cm}.$$
- The four-terminal networks $(1)$, ... , $(4)$ should be characterized by their $p$–transfer functions $H_{\rm L}(p)$ .
- From this (in this task, not in general), the frequency response is obtained according to the equation
- $$H(f) = H_{\rm L}(p)\Bigg |_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm}2\pi \it f}} \hspace{0.05cm}.$$
Please note:
- The exercise belongs to the chapter Laplace Transform and p-Transfer Function.
Questions
Solution
(1) Both statements are true:
- The following limits hold for the two two-port networks:
- $$\lim_{p \hspace{0.05cm}\rightarrow \hspace{0.05cm}0} H_{\rm TP}(p)\hspace{0.2cm} = \hspace{0.1cm}\lim_{p \hspace{0.05cm}\rightarrow \hspace{0.05cm}0}\frac{K}{p + p_{\rm x}} \hspace{0.15cm} { =K /{p_{\rm x}}}, \hspace{1.2cm} \lim_{p \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} H_{\rm TP}(p)= 0\hspace{0.05cm},$$
- $$ \lim_{p \hspace{0.05cm}\rightarrow \hspace{0.05cm}0}H_{\rm HP}(p) \hspace{0.2cm} = \hspace{0.1cm}0, \hspace{1.4cm} \lim_{p \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty} H_{\rm HP}(p)= \lim_{p \hspace{0.05cm}\rightarrow \hspace{0.05cm}\infty}\frac{K\cdot p}{p + p_{\rm x}} = K \hspace{0.05cm}.$$
- It can be seen that $H_{\rm TP}(p)$ yields zero for very high frequencies and $H_{\rm HP}(p)$ for very low frequencies.
(2) We consider two-port network $(1)$.
- Der Spannungsteiler liefert das Ergebnis
- $$H_{\rm L}(p)= \frac { p L} {R + pL}= \frac { p } {p +{R}/{L}} \hspace{0.05cm} .$$
- Es handelt sich um einen $\rm Hochpass$ mit dem Kennparameter $\underline {K = 1}$ und der Nullstelle bei
- $$p_{\rm x}= -\frac{R}{L}= -\frac{100\,{\rm \Omega}}{10^{-5 }\,{\rm \Omega s}}\hspace{0.15cm}\underline{= -0.1} \cdot10^{-6 }\,{1}/{\rm s} \hspace{0.05cm} .$$
(3) Zur Übertragungsfunktion kommt man mit der Substitution $p = {\rm j} \cdot 2 \pi f$:
- $$H(f)= \frac { {\rm j} \cdot 2\pi \hspace{-0.05cm}f } {{\rm j} \cdot 2\pi \hspace{-0.05cm}f +p_{\rm o}}\Rightarrow \hspace{0.3cm}\hspace{0.3cm} |H(f)|^2 = \frac { (2\pi \hspace{-0.05cm}f)^2 } {(2\pi \hspace{-0.05cm}f)^2 +p_{\rm o}^2}\hspace{0.05cm} .$$
- Aus der Bedingung $|H(f_{\rm G})|^2 = 0.5 $ erhält man folgende Bestimmungsgleichung:
- $$(2\pi \hspace{-0.05cm}f_{\rm G})^2 = p_{\rm o}^2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}\hspace{0.3cm} f_{\rm G} = -\frac { p_{\rm o}} {2 \pi}= \frac { 10^{-7 }\, 1/s} {2 \pi}\hspace{0.15cm}\underline{\approx 1.59\,{\rm MHz}}\hspace{0.05cm} .$$
(4) Richtig ist die erste Aussage:
- Für ein Gleichsignal ist eine Kapazität $C$ ein unendlich großer Widerstand, für hohe Frequenzen wirkt $C$ wie ein Kurzschluss.
- Daraus folgt: Der Vierpol $(3)$ beschreibt ebenfalls einen Hochpass. Dagegen zeigen die Schaltungen $(2)$ und $(4)$ Tiefpassverhalten.
(5) Die $p$–Übertragungsfunktion von Vierpol $(3)$ lautet:
- $$H_{\rm L}(p)= \frac { R } {{1}/{(pC)} + R}= \frac { p } {p +{1}/{(RC)}}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}p_{\rm x}= -{1}/(RC)\hspace{0.3cm}\Rightarrow \hspace{0.3cm} C = -\frac{1}{p_{\rm x} \cdot R}= \frac{-1}{-10^{-7 }\, 1/s \cdot 100\,{\rm \Omega}}\hspace{0.15cm}\underline{ = 1\,{\rm nF}} \hspace{0.05cm} .$$