Difference between revisions of "Aufgaben:Exercise 1.1: Simple Filter Functions"
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− | {{quiz-Header|Buchseite= | + | {{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/System_Description_in_Frequency_Domain}} |
− | [[File: | + | [[File:EN_LZI_A_1_1.png|Considered two-port networks|right|frame]] |
A filter with the frequency response | A filter with the frequency response | ||
− | :$$H_{\rm | + | :$$H_{\rm LP}(f) = \frac{1}{1+ {\rm j}\cdot f/f_0}$$ |
− | is described as a low-pass filter of first order. Out of it, a first order high-pass filter can be designed according to the following rule: | + | is described as a low-pass filter of first order. Out of it, a first order high-pass filter can be designed according to the following rule: |
− | :$$H_{\rm HP}(f) = 1- H_{\rm | + | :$$H_{\rm HP}(f) = 1- H_{\rm LP}(f) .$$ |
In both cases $f_0$ represents the so-called $\text{3 dB}$–cutoff frequency. | In both cases $f_0$ represents the so-called $\text{3 dB}$–cutoff frequency. | ||
− | The figure shows two | + | The figure shows two two-port networks $\rm A$ and $\rm B$. The task is to clarify which of the two networks has a low-pass characteristic and which has a high-pass characteristic. |
The components of circuit $\rm A$ are given as follows: | The components of circuit $\rm A$ are given as follows: | ||
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''Please note:'' | ''Please note:'' | ||
− | *The task belongs to the chapter [[Linear_and_Time_Invariant_Systems/ | + | *The task belongs to the chapter [[Linear_and_Time_Invariant_Systems/System_Description_in_Frequency_Domain |System Description in Frequency Domain]]. |
*For the subtask '''(4)''' cosine-shaped input signals are assumed. The frequency $f_x$ is variable, the power is $P_x = 10\,{\rm mW}.$ | *For the subtask '''(4)''' cosine-shaped input signals are assumed. The frequency $f_x$ is variable, the power is $P_x = 10\,{\rm mW}.$ | ||
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<quiz display=simple> | <quiz display=simple> | ||
− | {Compute the frequency response $H_{\rm A}(f)$ of the | + | {Compute the frequency response $H_{\rm A}(f)$ of the two-port network $\rm A$ and check the following statements. |
|type="()"} | |type="()"} | ||
− | + | + | + Two-port network $\rm A$ is a low-pass filter. |
− | - | + | - Two-port network $\rm A$ is a high-pass filter. |
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− | {Compute the amplitude response $|H_{\rm B}(f)|$ of the | + | {Compute the amplitude response $|H_{\rm B}(f)|$ of the two-port network $\rm B$ with the components $R$ and $L$ using the reference frequency $f_0 = R/(2πL)$. <br>What are the values for $f = 0$, $f = f_0$, $f = 2f_0$ and for $f → ∞$? |
|type="{}"} | |type="{}"} | ||
$|H_{\rm B}(f = 0)|\ = \ $ { 0 1% } | $|H_{\rm B}(f = 0)|\ = \ $ { 0 1% } | ||
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{{ML-Kopf}} | {{ML-Kopf}} | ||
'''(1)''' <u>Approach 1</u> is correct: | '''(1)''' <u>Approach 1</u> is correct: | ||
− | * | + | *The complex resistance of the capacitance $C$ is equal to $1/({\rm j}ωC)$, where $ω = 2πf$ represents the so-called angular frequency. |
− | * | + | *The frequency response can be calculated according to the voltage divider principle: |
:$$H_{\rm A}(f) = \frac{Y_{\rm A}(f)}{X_{\rm A}(f)} = \frac{1/({\rm j}\omega C)}{R+1/({\rm j}\omega C)}=\frac{1}{1+{\rm j \cdot 2\pi}\cdot f \cdot R\cdot C}.$$ | :$$H_{\rm A}(f) = \frac{Y_{\rm A}(f)}{X_{\rm A}(f)} = \frac{1/({\rm j}\omega C)}{R+1/({\rm j}\omega C)}=\frac{1}{1+{\rm j \cdot 2\pi}\cdot f \cdot R\cdot C}.$$ | ||
− | * | + | *Because of $H_{\rm A}(f = 0) = 1$ this cannot be a high-pass filter; rather it is a low-pass filter. |
− | * | + | *At low frequencies, the reactance of the capacitance is very large and $y_{\rm A}(t) ≈ x_{\rm A}(t)$ applies. |
− | * | + | *In contrast to this, at very high frequencies, the capacitor acts like a short circuit and $y_{\rm A}(t) ≈ 0$ holds. |
− | '''(2)''' | + | '''(2)''' By comparing the coefficients between $H_{\rm LP}(f)$ given in the statement of the task and $H_{\rm A}(f)$ according to subtask '''(1)''' the following is obtained: |
:$$f_0 = \frac{1}{2\pi \cdot R \cdot C} = \frac{1}{2\pi \cdot{\rm | :$$f_0 = \frac{1}{2\pi \cdot R \cdot C} = \frac{1}{2\pi \cdot{\rm | ||
50\hspace{0.05cm} \Omega}\cdot {\rm 637 \cdot 10^{-9}\hspace{0.05cm} s/\Omega}}\hspace{0.15cm}\underline{\approx 5 \, {\rm kHz}}.$$ | 50\hspace{0.05cm} \Omega}\cdot {\rm 637 \cdot 10^{-9}\hspace{0.05cm} s/\Omega}}\hspace{0.15cm}\underline{\approx 5 \, {\rm kHz}}.$$ | ||
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− | '''(3)''' | + | '''(3)''' The amplitude response is: |
:$$|H_{\rm A}(f)| = \frac{1}{\sqrt{1+ (f/f_0)^2}}.$$ | :$$|H_{\rm A}(f)| = \frac{1}{\sqrt{1+ (f/f_0)^2}}.$$ | ||
− | * | + | *For $f = f_0$ the numerical value $1/\sqrt{2}\hspace{0.1cm} \underline{≈ 0.707}$ is obtained, and |
− | * | + | *for $f = 2f_0$ approximately the value $1/\sqrt{5}\hspace{0.1cm} \underline{≈ 0.447}$. |
− | '''(4)''' | + | '''(4)''' The output power can be calculated using the following equation: |
:$$P_y = P_x \cdot |H_{\rm A}(f = f_x)|^2.$$ | :$$P_y = P_x \cdot |H_{\rm A}(f = f_x)|^2.$$ | ||
− | * | + | *For $f_x = f_0$ ⇒ $P_y = P_x/2 \hspace{0.1cm} \underline{ = 5\hspace{0.1cm} {\rm mW}}$, so this results in half of the power at the output. |
− | *In | + | *In logarithmic representation, this relationship is: |
:$$10 \cdot {\rm lg}\hspace{0.2cm} \frac{P_x(f_0)}{P_y(f_0)} = 3\,{\rm dB}.$$ | :$$10 \cdot {\rm lg}\hspace{0.2cm} \frac{P_x(f_0)}{P_y(f_0)} = 3\,{\rm dB}.$$ | ||
− | : | + | :Accordingly, for $f_0$ the term "3dB cut-off frequency" is also common. |
− | * | + | *For $f_x = 2f_0$, on the other hand, a smaller value is obtained: $P_y = P_x/5 \hspace{0.1cm}\underline{= 2\hspace{0.1cm} {\rm mW}}$. |
− | '''(5)''' | + | '''(5)''' Analogous to subtask '''(1)''' the following holds: |
:$$H_{\rm B}(f) = \frac{Y_{\rm B}(f)}{X_{\rm B}(f)} = \frac{{\rm j}\omega L}{R+{\rm j}\omega L}=\frac{{\rm j2\pi}\cdot f \cdot L/R}{1+{\rm j2\pi}\cdot f \cdot L/R}.$$ | :$$H_{\rm B}(f) = \frac{Y_{\rm B}(f)}{X_{\rm B}(f)} = \frac{{\rm j}\omega L}{R+{\rm j}\omega L}=\frac{{\rm j2\pi}\cdot f \cdot L/R}{1+{\rm j2\pi}\cdot f \cdot L/R}.$$ | ||
− | * | + | *Using the reference frequency $f_0 = R/(2πL)$ this can also be written as in the following: |
:$$H_{\rm B}(f) = \frac{{\rm j}\cdot f/f_0}{1+{\rm j}\cdot f/f_0}\hspace{0.5cm}\Rightarrow \hspace{0.5cm}|H_{\rm B}(f)| = \frac{|f/f_0|}{\sqrt{1+ (f/f_0)^2}}.$$ | :$$H_{\rm B}(f) = \frac{{\rm j}\cdot f/f_0}{1+{\rm j}\cdot f/f_0}\hspace{0.5cm}\Rightarrow \hspace{0.5cm}|H_{\rm B}(f)| = \frac{|f/f_0|}{\sqrt{1+ (f/f_0)^2}}.$$ | ||
− | * | + | *Hence, these numerical values are obtained: |
:$$|H_{\rm B}(f = 0)| \hspace{0.15cm}\underline{= 0}, \hspace{0.5cm} |H_{\rm B}( f_0)| \hspace{0.15cm}\underline{=0.707}, \hspace{0.5cm}|H_{\rm B}(2f_0)| \hspace{0.15cm}\underline{= 0.894}, | :$$|H_{\rm B}(f = 0)| \hspace{0.15cm}\underline{= 0}, \hspace{0.5cm} |H_{\rm B}( f_0)| \hspace{0.15cm}\underline{=0.707}, \hspace{0.5cm}|H_{\rm B}(2f_0)| \hspace{0.15cm}\underline{= 0.894}, | ||
\hspace{0.5cm}|H_{\rm B}(f \rightarrow \infty)|\hspace{0.15cm}\underline{ = 1}.$$ | \hspace{0.5cm}|H_{\rm B}(f \rightarrow \infty)|\hspace{0.15cm}\underline{ = 1}.$$ | ||
− | * | + | *Therefore, the circuit $\rm B$ is a high-pass filter. |
− | '''(6)''' | + | '''(6)''' According to the above definition of the reference frequency it follows that: |
:$$L = \frac{R}{2\pi \cdot f_0} = \frac{{\rm 50\hspace{0.05cm} | :$$L = \frac{R}{2\pi \cdot f_0} = \frac{{\rm 50\hspace{0.05cm} | ||
\Omega}}{2\pi \cdot{\rm 5000 \hspace{0.05cm} Hz}}= {\rm 1.59 \cdot | \Omega}}{2\pi \cdot{\rm 5000 \hspace{0.05cm} Hz}}= {\rm 1.59 \cdot |
Latest revision as of 16:11, 9 July 2021
A filter with the frequency response
- $$H_{\rm LP}(f) = \frac{1}{1+ {\rm j}\cdot f/f_0}$$
is described as a low-pass filter of first order. Out of it, a first order high-pass filter can be designed according to the following rule:
- $$H_{\rm HP}(f) = 1- H_{\rm LP}(f) .$$
In both cases $f_0$ represents the so-called $\text{3 dB}$–cutoff frequency.
The figure shows two two-port networks $\rm A$ and $\rm B$. The task is to clarify which of the two networks has a low-pass characteristic and which has a high-pass characteristic.
The components of circuit $\rm A$ are given as follows:
- $$R = 50 \,\, {\rm \Omega}, \hspace{0.2cm} C = 637 \,\, {\rm nF} .$$
The inductivity $L$ of circuit $\rm B$ is to be computed in subtask (6) .
Please note:
- The task belongs to the chapter System Description in Frequency Domain.
- For the subtask (4) cosine-shaped input signals are assumed. The frequency $f_x$ is variable, the power is $P_x = 10\,{\rm mW}.$
Questions
Sample solution
- The complex resistance of the capacitance $C$ is equal to $1/({\rm j}ωC)$, where $ω = 2πf$ represents the so-called angular frequency.
- The frequency response can be calculated according to the voltage divider principle:
- $$H_{\rm A}(f) = \frac{Y_{\rm A}(f)}{X_{\rm A}(f)} = \frac{1/({\rm j}\omega C)}{R+1/({\rm j}\omega C)}=\frac{1}{1+{\rm j \cdot 2\pi}\cdot f \cdot R\cdot C}.$$
- Because of $H_{\rm A}(f = 0) = 1$ this cannot be a high-pass filter; rather it is a low-pass filter.
- At low frequencies, the reactance of the capacitance is very large and $y_{\rm A}(t) ≈ x_{\rm A}(t)$ applies.
- In contrast to this, at very high frequencies, the capacitor acts like a short circuit and $y_{\rm A}(t) ≈ 0$ holds.
(2) By comparing the coefficients between $H_{\rm LP}(f)$ given in the statement of the task and $H_{\rm A}(f)$ according to subtask (1) the following is obtained:
- $$f_0 = \frac{1}{2\pi \cdot R \cdot C} = \frac{1}{2\pi \cdot{\rm 50\hspace{0.05cm} \Omega}\cdot {\rm 637 \cdot 10^{-9}\hspace{0.05cm} s/\Omega}}\hspace{0.15cm}\underline{\approx 5 \, {\rm kHz}}.$$
(3) The amplitude response is:
- $$|H_{\rm A}(f)| = \frac{1}{\sqrt{1+ (f/f_0)^2}}.$$
- For $f = f_0$ the numerical value $1/\sqrt{2}\hspace{0.1cm} \underline{≈ 0.707}$ is obtained, and
- for $f = 2f_0$ approximately the value $1/\sqrt{5}\hspace{0.1cm} \underline{≈ 0.447}$.
(4) The output power can be calculated using the following equation:
- $$P_y = P_x \cdot |H_{\rm A}(f = f_x)|^2.$$
- For $f_x = f_0$ ⇒ $P_y = P_x/2 \hspace{0.1cm} \underline{ = 5\hspace{0.1cm} {\rm mW}}$, so this results in half of the power at the output.
- In logarithmic representation, this relationship is:
- $$10 \cdot {\rm lg}\hspace{0.2cm} \frac{P_x(f_0)}{P_y(f_0)} = 3\,{\rm dB}.$$
- Accordingly, for $f_0$ the term "3dB cut-off frequency" is also common.
- For $f_x = 2f_0$, on the other hand, a smaller value is obtained: $P_y = P_x/5 \hspace{0.1cm}\underline{= 2\hspace{0.1cm} {\rm mW}}$.
(5) Analogous to subtask (1) the following holds:
- $$H_{\rm B}(f) = \frac{Y_{\rm B}(f)}{X_{\rm B}(f)} = \frac{{\rm j}\omega L}{R+{\rm j}\omega L}=\frac{{\rm j2\pi}\cdot f \cdot L/R}{1+{\rm j2\pi}\cdot f \cdot L/R}.$$
- Using the reference frequency $f_0 = R/(2πL)$ this can also be written as in the following:
- $$H_{\rm B}(f) = \frac{{\rm j}\cdot f/f_0}{1+{\rm j}\cdot f/f_0}\hspace{0.5cm}\Rightarrow \hspace{0.5cm}|H_{\rm B}(f)| = \frac{|f/f_0|}{\sqrt{1+ (f/f_0)^2}}.$$
- Hence, these numerical values are obtained:
- $$|H_{\rm B}(f = 0)| \hspace{0.15cm}\underline{= 0}, \hspace{0.5cm} |H_{\rm B}( f_0)| \hspace{0.15cm}\underline{=0.707}, \hspace{0.5cm}|H_{\rm B}(2f_0)| \hspace{0.15cm}\underline{= 0.894}, \hspace{0.5cm}|H_{\rm B}(f \rightarrow \infty)|\hspace{0.15cm}\underline{ = 1}.$$
- Therefore, the circuit $\rm B$ is a high-pass filter.
(6) According to the above definition of the reference frequency it follows that:
- $$L = \frac{R}{2\pi \cdot f_0} = \frac{{\rm 50\hspace{0.05cm} \Omega}}{2\pi \cdot{\rm 5000 \hspace{0.05cm} Hz}}= {\rm 1.59 \cdot 10^{-3}\hspace{0.05cm} \Omega s}\hspace{0.15cm}\underline{= {\rm 1.59 \hspace{0.1cm} mH}} .$$