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}} |
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− | + | [[File:EN_LZI_A_1_1.png|Considered two-port networks|right|frame]] | |
+ | 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)''' . | |
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− | === | + | |
+ | |||
+ | |||
+ | ''Please note:'' | ||
+ | *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}.$ | ||
+ | |||
+ | |||
+ | |||
+ | ===Questions=== | ||
<quiz display=simple> | <quiz display=simple> | ||
− | { | + | {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. |
− | { | + | {Compute the reference frequency $f_0$ using the components $R$ and $C$. |
|type="{}"} | |type="{}"} | ||
− | $f_0$ | + | $f_0 \ = \ $ { 5 1% } $\text{kHz}$ |
− | { | + | {Compute the amplitude response $|H_{\rm A}(f)|$. What are the numerical values for $f = f_0$ and $f = 2f_0$? |
|type="{}"} | |type="{}"} | ||
− | $|H_{\rm A}(f = f_0)|$ | + | $|H_{\rm A}(f = f_0)|\ = \ $ { 0.707 1% } |
− | $|H_{\rm A}(f = 2f_0)|$ | + | $|H_{\rm A}(f = 2f_0)|\ = \ $ { 0.447 1% } |
− | { | + | {What is the power $P_y$ of the output signal $y(t)$ if a cosine signal of frequency $f_x = 5\,{\rm kHz}$ or $f_x = 10\,{\rm kHz}$ is applied to the input? |
|type="{}"} | |type="{}"} | ||
− | $P_y(f_x = 5 \rm kHz)$ | + | $P_y(f_x = 5 \ \rm kHz)\ = \ $ { 5 1% } $\text{mW}$ |
− | $P_y(f_x = 10 \rm kHz)$ | + | $P_y(f_x = 10 \ \rm kHz)\ = \ $ { 2 1% } $\text{mW}$ |
− | { | + | {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)|$ | + | $|H_{\rm B}(f = 0)|\ = \ $ { 0 1% } |
− | $|H_{\rm B}(f = f_0)|$ | + | $|H_{\rm B}(f = f_0)|\ = \ $ { 0.707 1% } |
− | $|H_{\rm B}(f = 2f_0)|$ | + | $|H_{\rm B}(f = 2f_0)|\ = \ ${ 0.894 1% } |
− | $|H_{\rm B}(f → ∞)|$ | + | $|H_{\rm B}(f → ∞)|\ = \ $ { 1 1% } |
− | { | + | {What inductivity $L$ results in the reference frequency $f_0 = 5 \,\text{kHz}$? |
|type="{}"} | |type="{}"} | ||
− | $L$ | + | $L\ = \ $ { 1.59 1% } $\text{mH}$ |
</quiz> | </quiz> | ||
− | === | + | ===Sample solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''1 | + | '''(1)''' <u>Approach 1</u> is correct: |
− | $$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}.$$ | + | *The complex resistance of the capacitance $C$ is equal to $1/({\rm j}ωC)$, where $ω = 2πf$ represents the so-called angular frequency. |
− | + | *The frequency response can be calculated according to the voltage divider principle: | |
+ | :$$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: $P_y = P_x/5 \hspace{0.1cm}\underline{= 2\hspace{0.1cm} {\rm mW}}$. | |
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+ | '''(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. | ||
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− | '''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 | ||
− | 10^{-3}\hspace{0.05cm} \Omega s}\hspace{0.15cm}\underline{= {\rm 1.59 \hspace{0. | + | 10^{-3}\hspace{0.05cm} \Omega s}\hspace{0.15cm}\underline{= {\rm 1.59 \hspace{0.1cm} |
mH}} .$$ | mH}} .$$ | ||
{{ML-Fuß}} | {{ML-Fuß}} | ||
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− | [[Category: | + | [[Category:Linear and Time-Invariant Systems: Exercises|^1.1 System Description in Frequency Domain^]] |
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:
- $$L = \frac{R}{2\pi \cdot f_0} = \frac{{\rm 50\hspace{0.05cm} \Omega}}{2\pi \cdot{\rm 5000 \hspace{0.05cm} Hz}}= {\rm 1.59 \cdot 10^{-3}\hspace{0.05cm} \Omega s}\hspace{0.15cm}\underline{= {\rm 1.59 \hspace{0.1cm} mH}} .$$