Difference between revisions of "Aufgaben:Exercise 1.1Z: Low-Pass Filter of 1st and 2nd Order"
| Line 6: | Line 6: | ||
This is then referred to as a first-order low-pass filter. The diagram shows the following for this filter | This is then referred to as a first-order low-pass filter. The diagram shows the following for this filter | ||
* above the ''damping curve'' $a_1(f)$, | * above the ''damping curve'' $a_1(f)$, | ||
| − | * below the ''phase | + | * below the ''phase response'' $b_1(f)$. |
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In this task, | In this task, | ||
*based on the functions $a_1(f)$ and $b_1(f)$ of the low-pass filter of first order | *based on the functions $a_1(f)$ and $b_1(f)$ of the low-pass filter of first order | ||
| − | *the | + | *the damping curve and phase response of a low-pass filter of higher order is to be analysed. |
| Line 46: | Line 46: | ||
| − | {Compute the phase response $b_1(f)$. <br> | + | {Compute the phase response $b_1(f)$. <br>What values in radian (rad) are obtained for $f = f_0$ and $f = 2f_0$? |
|type="{}"} | |type="{}"} | ||
$b_1(f = f_0)\ = \ $ { 0.786 5% } $\text{rad}$ | $b_1(f = f_0)\ = \ $ { 0.786 5% } $\text{rad}$ | ||
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| − | { | + | {What is the damping curve $a_n(f)$ of a low-pass filter of $n$–th order? <br>What $\rm dB$–values are obtained for $n = 2$ for $f = f_0$ or $f = \: –2f_0$? |
|type="{}"} | |type="{}"} | ||
$a_2(f = f_0)\ = \ $ { 6.02 5% } $\text{dB}$ | $a_2(f = f_0)\ = \ $ { 6.02 5% } $\text{dB}$ | ||
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| − | { | + | {Compute the phase function $b_2(f)$ of a low-pass filter of second order. <br>What values (in Radian) are obtained for $f = f_0$ and $f = \: –2f_0$? |
|type="{}"} | |type="{}"} | ||
$b_2(f = f_0)\ = \ $ { 1.571 5% } $\text{rad}$ | $b_2(f = f_0)\ = \ $ { 1.571 5% } $\text{rad}$ | ||
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===Sample solution=== | ===Sample solution=== | ||
{{ML-Kopf}} | {{ML-Kopf}} | ||
| − | '''(1)''' | + | '''(1)''' The amplitude response of the low-pass filter of first order is: |
:$$|H_{\rm 1}(f)| = \frac{1}{\sqrt{1+ (f/f_0)^2}}.$$ | :$$|H_{\rm 1}(f)| = \frac{1}{\sqrt{1+ (f/f_0)^2}}.$$ | ||
*Damit erhält man den Dämpfungsverlauf in Neper (Np): | *Damit erhält man den Dämpfungsverlauf in Neper (Np): | ||
Revision as of 16:52, 28 June 2021
The simplest form of a low-pass filter – for example, realisable as an RC low-pass filter according to exercise 1.1 – has the following frequency response:
- $$H_{\rm 1}(f) = \frac{1}{1+{\rm j}\cdot f/f_0}.$$
This is then referred to as a first-order low-pass filter. The diagram shows the following for this filter
- above the damping curve $a_1(f)$,
- below the phase response $b_1(f)$.
Correspondingly, for a low-pass filter of $n$–th order, the following defining equation applies:
- $$H_n(f) = H_{\rm 1}(f)^n.$$
In this task,
- based on the functions $a_1(f)$ and $b_1(f)$ of the low-pass filter of first order
- the damping curve and phase response of a low-pass filter of higher order is to be analysed.
In general, the following holds:
- $$H(f) = {\rm e}^{-a(f) - {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}b(f)}.$$
Please note:
- The task belongs to the chapter System Description in Frequency Domain.
- There is the following relationship between the Np- and dB-values of an amplitude value $|H| = 1/x$ :
- $$a_{\rm Np} = \ln (x) = \ln (10) \cdot \lg (x) = \frac{\ln (10)}{20} \cdot a_{\rm dB} \approx 0.11513 \cdot a_{\rm dB}.$$
- Further consider that for two complex quantities $z_1$ and $z_2$ the following equations hold:
- $$|z_1 \cdot z_2| = |z_1| \cdot |z_2|, \hspace{0.5 cm}{\rm arc}\hspace{0.05 cm}(z_1 \cdot z_2) = {\rm arc}\hspace{0.05 cm}(z_1) + {\rm arc}\hspace{0.05 cm}(z_2).$$
Questions
Sample solution
- $$|H_{\rm 1}(f)| = \frac{1}{\sqrt{1+ (f/f_0)^2}}.$$
- Damit erhält man den Dämpfungsverlauf in Neper (Np):
- $$a_1(f) = \ln \frac{1}{|H_1(f)|} = {1}/{2} \cdot \ln \left[1 + ({f}/{f_0})^2 \right] \Rightarrow a_1(f = f_0) = 0.3466 \hspace{0.1 cm}{\rm Np},\hspace{0.5 cm}a_1(f = 2 f_0) = 0.8047 \hspace{0.1 cm}{\rm Np}.$$
Die entsprechenden dB–Werte erhält man durch Multiplikation mit $1/0.11513 = 8.68589$ und führt zu den Ergebnissen
- $ \underline{3.01 \: {\rm dB} ≈ 3 \: {\rm dB}}$ für $ f = f_0$,
- $ \underline{6.99 \: {\rm dB}}$ für $ f = 2f_0$.
Beim Tiefpass erster Ordnung beträgt somit die 3dB–Grenzfrequenz $f_{\rm G} = f_0$.
(2) Der Frequenzgang $H_1(f)$ kann auch nach Real– und Imaginärteil getrennt dargestellt werden:
- $$H_{\rm 1}(f) = \frac{1}{ {1+ (f/f_0)^2} } - {\rm j} \cdot \frac{f/f_0}{ {1+ (f/f_0)^2} }.$$
- Damit ergibt sich für den Phasengang:
- $$b_1(f) = - \arctan \hspace{0.1cm} ( {\rm Im} /{\rm Re} ) = \arctan \hspace{0.1cm} ({f}/{f_0}).$$
- Für $f = f_0$ erhält man $\arctan(1) = π/4 \rm \underline{\: = 0.786 \: rad}$, und für $f = 2f_0$ den Wert $\arctan(2) \rm \underline{\: = 1.108 \: rad}$.
(3) Für den Amplitudengang eines Tiefpasses $n$–ter Ordnung gilt:
- $$|H_n(f)| = |H_{\rm 1}(f)|^n.$$
Bezüglich der (logarithmischen) Dämpfungsfunktion wird aus der $n$–fachen Multiplikation die $n$–fache Summe:
- $$a_n(f) = n \cdot a_1(f)= {n}/{2} \cdot \ln \left[ 1 + ({f}/{f_0})^2 \right].$$
Für den Tiefpass zweiter Ordnung ergibt sich daraus als Sonderfall:
- $$a_2(f) = \ln \left[ 1 + ({f}/{f_0})^2 \right]= 2 \cdot a_1(f).$$
Die dB–Werte lauten nun:
- $ \underline{6.02 \: {\rm dB} ≈ 6 \: {\rm dB}}$ für $f = ±f_0$,
- $\rm \underline{13.98 \: {\rm dB} ≈ 14 \: {\rm dB}}$ für $f = ±2f_0$.
Damit ist offensichtlich, dass für $n > 1$ der Parameter $f_0$ nicht mehr die 3 dB–Grenzfrequenz $f_{\rm G}$ angibt.
Für $n = 2$ ⇒ "Tiefpass zweiter Ordnung" gilt vielmehr der Zusammenhang:
- $${f_{\rm G} } = {f_0}/\sqrt{2}.$$
(4) Auch bezüglich der Phasenfunktion gilt:
- $$b_n(f) = n \cdot b_1(f), \hspace{0.3 cm} b_2(f) = 2 \cdot b_1(f).$$
Beim Tiefpass zweiter Ordnung sind somit alle Phasenwerte zwischen $±π$ möglich. Insbesondere ist
- $b_2(f = f_0) = π/2 \rm \underline{\: = 1.571 \: rad}$,
- $b_2(f = 2f_0) = \rm 2.216 \: rad$.
Da die Phase eine ungerade Funktion ist, gilt hier: $b_2(f = \: –2f_0) = \rm \underline{–2.216 \: rad}$.
