Difference between revisions of "Aufgaben:Exercise 3.4Z: Various All-Pass Filters"
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− | {{quiz-Header|Buchseite= | + | {{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function |
}} | }} | ||
− | [[File: | + | [[File:EN_LZI_Z_3_4.png|right|frame|All-pass filter in two different variants<br> $Z_{\rm I}$: Internal resistance of the source <br> $Z_{\rm A}$: Terminating resistor]] |
− | + | We first assume a two-port network with the following transfer function: | |
:$$H_{\rm L}(p)= \frac {1-{p}/{A}} {1+{p}/{A}}.$$ | :$$H_{\rm L}(p)= \frac {1-{p}/{A}} {1+{p}/{A}}.$$ | ||
− | + | From this is to be determined the conventional Fourier frequency response | |
− | :$$H(f) = {\rm e}^{-a(f)\hspace{0.05cm}}\cdot {\rm e}^{- \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}b(f)}$$ | + | :$$H(f) = {\rm e}^{-a(f)\hspace{0.05cm}}\cdot {\rm e}^{- \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}b(f)},$$ |
− | + | which is representable | |
+ | *by the attenuation function $a(f)$ | ||
+ | *and the phase function $b(f)$. | ||
− | + | ||
+ | The upper diagram shows a so-called "all-pass circuit" where the complex resistance $Z_1$ denotes an inductance and the complex resistance $Z_2$ denotes a capacitance: | ||
:$$Z_1 = p \cdot L\hspace{0.05cm},\hspace{0.2cm}Z_2 = \frac{1}{p \cdot C}\hspace{0.05cm}.$$ | :$$Z_1 = p \cdot L\hspace{0.05cm},\hspace{0.2cm}Z_2 = \frac{1}{p \cdot C}\hspace{0.05cm}.$$ | ||
− | + | For reflection-free adaptation at the input and output with | |
− | :$$Z_{\rm I}=Z_{\rm A} = \sqrt{Z_1 \cdot Z_2} = \sqrt{{L}/{C}}$$ | + | :$$Z_{\rm I}=Z_{\rm A} = \sqrt{Z_1 \cdot Z_2} = \sqrt{{L}/{C}},$$ |
− | + | the following holds for the $p$–transfer function of circuit $\rm A$ (see upper diagram): | |
:$$H_{\rm L}(p)= \frac {Y_{\rm L}(p)} {X_{\rm L}(p)}= \frac {Z_2-Z_1} {Z_1+2 \cdot \sqrt{Z_1 \cdot Z_2}+Z_2}\hspace{0.05cm}.$$ | :$$H_{\rm L}(p)= \frac {Y_{\rm L}(p)} {X_{\rm L}(p)}= \frac {Z_2-Z_1} {Z_1+2 \cdot \sqrt{Z_1 \cdot Z_2}+Z_2}\hspace{0.05cm}.$$ | ||
− | |||
− | : | + | Circuit $\rm B$ is defined by the pole–zero diagram of the $p$–transfer function. It is characterized by the fact that |
+ | *all poles (in the left $p$–half-plane) | ||
+ | *are located in a mirror-imaged manner with respect to the zeros (in the right half-plane). | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | Please note: | ||
+ | *The exercise belongs to the chapter [[Linear_and_Time_Invariant_Systems/Laplace_Transform_and_p-Transfer_Function|Laplace Transform and p-Transfer Function]]. | ||
+ | |||
− | === | + | ===Questions=== |
<quiz display=simple> | <quiz display=simple> | ||
− | { | + | {Specify the zero $p_{\rm o}$ and the pole $p_{\rm x}$ of $H_{\rm L}(p)= (1 -p/A)/(1 +p/A)$ . What is the constant factor $K$? |
|type="{}"} | |type="{}"} | ||
− | $K$ | + | $K \ = \ $ { -1.03--0.97 } |
− | $ | + | $p_{\rm o} \ = \ $ { 1 3% } $\ \cdot A$ |
− | $ | + | $p_{\rm x} \ = \ $ { -1.03--0.97 } $\ \cdot A$ |
− | { | + | {Compute the attenuation function $a(f)$. Which statements are true? |
|type="[]"} | |type="[]"} | ||
− | - | + | - The attenuation function $a(f)$ exhibits low-pass filter behavior. |
− | + | + | + The attenuation function $a(f)$ is constant. |
− | + | + | + The above result is generally valid for $p_{\rm x} = - p_{\rm o}$. |
− | { | + | {Compute the phase response $b(f)$. What are the phase values for the given frequencies? |
|type="{}"} | |type="{}"} | ||
− | $b(f = A/2 \pi)$ | + | $b(f = A/2 \pi) \ = \ $ { 90 3% } $\ \rm deg$ |
− | $b(f = A/ \pi)$ | + | $b(f = A/ \pi)\ = \ $ { 126.8 3% } $ \rm deg$ |
− | $b(f → ∞)$ | + | $b(f → ∞) \ = \ $ { 180 3% } $ \rm deg$ |
− | { | + | {Compute the $p$–transfer function of circuit $\rm A$. What statements can be derived from this? |
|type="[]"} | |type="[]"} | ||
− | + | + | + The attenuation $a(f)$ is constantly equal to $0$ (Np). |
− | - | + | - The phase $b(f)$ increases linearly with frequency $f$ . |
− | - | + | - $b(f)$ is the Hilbert transform of $a(f)$. |
− | { | + | {What statements can be derived from the pole–zero diagram of circuit $\rm B$ ? |
|type="[]"} | |type="[]"} | ||
− | + | + | + The attenuation $a(f)$ is constant. |
− | + | + | + The following holds for the phase function: $b(f =0) =0$. |
Line 61: | Line 73: | ||
</quiz> | </quiz> | ||
− | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | + | '''(1)''' Transforming the given $p$–transfer function yields | |
:$$H_{\rm L}(p)= \frac {1-{p}/{A}} {1+{p}/{A}}= -1 \cdot \frac {p-A} {p+A}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\hspace{0.15cm} \underline{K =- 1}, \hspace{0,2cm} \hspace{0.15cm} \underline{p_{\rm o}/A = 1} ,\hspace{0,2cm} | :$$H_{\rm L}(p)= \frac {1-{p}/{A}} {1+{p}/{A}}= -1 \cdot \frac {p-A} {p+A}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\hspace{0.15cm} \underline{K =- 1}, \hspace{0,2cm} \hspace{0.15cm} \underline{p_{\rm o}/A = 1} ,\hspace{0,2cm} | ||
\hspace{0.15cm} \underline{ p_{\rm x}/A = -1} \hspace{0.05cm} .$$ | \hspace{0.15cm} \underline{ p_{\rm x}/A = -1} \hspace{0.05cm} .$$ | ||
− | + | ||
+ | |||
+ | |||
+ | '''(2)''' <u>Statements 2 and 3</u> are correct: | ||
+ | *If $p = {\rm j} \cdot 2 \pi f$ is set, the following is obtained: | ||
:$$H(f)= \frac {1-{\rm j \cdot 2\pi \it | :$$H(f)= \frac {1-{\rm j \cdot 2\pi \it | ||
f}/A} {1+{\rm j \cdot 2\pi \it | f}/A} {1+{\rm j \cdot 2\pi \it | ||
f}/A}\hspace{0.05cm} .$$ | f}/A}\hspace{0.05cm} .$$ | ||
− | + | *The magnitude of a quotient is equal to the quotient of the magnitudes: | |
:$$|H(f)|= \frac {|1-{{\rm j} \cdot 2\pi | :$$|H(f)|= \frac {|1-{{\rm j} \cdot 2\pi | ||
f}/A|} {|1+{\rm j \cdot 2\pi \it | f}/A|} {|1+{\rm j \cdot 2\pi \it | ||
f}/A|}= \frac {\sqrt{1+(2\pi | f}/A|}= \frac {\sqrt{1+(2\pi | ||
f/A)^2}} {\sqrt{1+(2\pi | f/A)^2}} {\sqrt{1+(2\pi | ||
− | f/A)^2}}= 1 | + | f/A)^2}}= 1\hspace{0.3cm} |
− | + | \Rightarrow \hspace{0.3cm} a(f)= -{\rm ln} \hspace{0.1cm} |H(f)|= 0\hspace{0.2cm}({\rm Np \hspace{0.2cm}or \hspace{0.2cm}dB})\hspace{0.05cm} .$$ | |
− | + | *However, $\text{statement 3}$ is also correct as can be seen from the theory page "Graphical determination of attenuation". | |
+ | |||
+ | |||
+ | |||
− | + | '''(3)''' The phase function $b(f)$ can be computed as follows: | |
− | :$$b(f)= -{\rm arc} \hspace{0.1cm} H(f) \hspace{0.25cm} = \hspace{0.2cm} {\rm arctan } \hspace{0.1cm} | + | :$$b(f)= -{\rm arc} \hspace{0.1cm} H(f) \hspace{0.25cm} = \hspace{0.2cm} {\rm arctan } \hspace{0.1cm} ({2\pi |
− | f}{A} - {\rm arctan } \hspace{0.1cm} | + | f}/{A}) - {\rm arctan } \hspace{0.1cm} ({-2\pi |
− | f}{A} = 2 \cdot {\rm arctan } \hspace{0.1cm} | + | f}/{A}) = 2 \cdot {\rm arctan } \hspace{0.1cm} ({2\pi |
− | f}{A}: | + | f}/{A}),$$ |
− | + | :$$b(f= {A}/{2\pi})= 2 \cdot {\rm arctan } | |
− | \hspace{0.1cm}(1) = 2 \cdot 45^\circ\hspace{0.15cm} \underline{ = 90^\circ}\hspace{0.05cm} | + | \hspace{0.1cm}(1) = 2 \cdot 45^\circ \hspace{0.15cm} \underline{ = 90^\circ}\hspace{0.05cm},$$ |
− | + | :$$ b(f= {A}/{\pi})=2 \cdot {\rm arctan } | |
− | |||
\hspace{0.1cm}(2) = 2 \cdot 63.4^\circ \hspace{0.15cm} \underline{= 126.8^\circ}\hspace{0.05cm} | \hspace{0.1cm}(2) = 2 \cdot 63.4^\circ \hspace{0.15cm} \underline{= 126.8^\circ}\hspace{0.05cm} | ||
− | , | + | ,$$ |
− | + | :$$ b(f \rightarrow \infty)=2 \cdot {\rm arctan } | |
\hspace{0.1cm}(\infty) = 2 \cdot 90^\circ \hspace{0.15cm} \underline{= 180^\circ}\hspace{0.05cm} | \hspace{0.1cm}(\infty) = 2 \cdot 90^\circ \hspace{0.15cm} \underline{= 180^\circ}\hspace{0.05cm} | ||
.$$ | .$$ | ||
− | + | *The same results are obtained following the approach according to the page "Graphical determination of phase" in the theory part. | |
+ | |||
+ | |||
+ | |||
− | + | '''(4)''' Only <u>statement 1</u> is correct: | |
− | :$$H_{\rm L}(p) | + | *The given $p$–transfer function can be expressed as follows: |
− | + | :$$H_{\rm L}(p)= \frac {Z_2-Z_1} {Z_1+2 \cdot \sqrt{Z_1 \cdot Z_2}+Z_2}= | |
+ | \frac {(\sqrt{Z_2}-\sqrt{Z_1})(\sqrt{Z_2}+\sqrt{Z_1})} {(\sqrt{Z_2}+\sqrt{Z_1})^2}= | ||
\frac {\sqrt{Z_2}-\sqrt{Z_1}} {\sqrt{Z_2}+\sqrt{Z_1}}\hspace{0.05cm}.$$ | \frac {\sqrt{Z_2}-\sqrt{Z_1}} {\sqrt{Z_2}+\sqrt{Z_1}}\hspace{0.05cm}.$$ | ||
− | + | *Furthermore, considering $Z_1 = p \cdot L$ and $Z_2 = 1/(p \cdot C)$ the following is obtained: | |
:$$H_{\rm L}(p)= \frac {\sqrt{{1}/(pC)}-\sqrt{pL}} {\sqrt{{1}/(pC)}+\sqrt{pL}} | :$$H_{\rm L}(p)= \frac {\sqrt{{1}/(pC)}-\sqrt{pL}} {\sqrt{{1}/(pC)}+\sqrt{pL}} | ||
= \frac {1- p \cdot \sqrt{LC}} {1+ p \cdot \sqrt{LC}} | = \frac {1- p \cdot \sqrt{LC}} {1+ p \cdot \sqrt{LC}} | ||
− | = -1 \cdot \frac {p-\sqrt{{1}/(LC)}} {p+\sqrt{{1}/(LC)}} | + | = -1 \cdot \frac {p-\sqrt{{1}/(LC)}} {p+\sqrt{{1}/(LC)}}\hspace{0.3cm} |
− | + | \Rightarrow \hspace{0.3cm}A = \sqrt{{1}/(LC)}: \hspace{0.2cm}H_{\rm L}(p)= -1 \cdot \frac {p-A} {p+A}\hspace{0.05cm}.$$ | |
− | |||
− | |||
− | |||
− | + | *The same transfer function as computed in subtask '''(1)''' is obtained. | |
− | |||
− | : | + | From this it follows that only <u>statement 1</u> is correct: |
+ | *The attenuation curve is $a(f) = 0\ \rm (Np)$. No frequency is attenuated or amplified. Therefore, it is called an "all-pass filter". | ||
+ | *The second statement is false. The phase response $b(f)$ is not linear but rather curved as computed in subtask '''(3)''' . | ||
+ | *The Hilbert transform of the constant $a(f) = 0$ should result in the phase function $b(f) = 0$ as shown in the theory part. That is: statement 3 is false. | ||
+ | *The attenuation function $a(f)$ and phase function $b(f)$ are related to each other via the Hilbert transformation only for minimum-phase systems. | ||
+ | *However, in such a minimum–phase system, all poles and zeros lie in the left $p$–half-plane which is not true here <br>⇒ '''an all-pass filter is not a minimum–phase system'''. | ||
− | |||
− | |||
− | + | '''(5)''' <u>Both statements</u> are correct: | |
+ | *As already determined in subtask '''(2)''': A constant attenuation arises as a result whenever there is a corresponding zero in the right half-plane for each pole in the left $p$–half-plane ⇒ circuit $\rm B$ also exhibits all-pass filter charakteristics. | ||
+ | *Since $b(f)$ is always an asymmetric function, $b(f= 0) = 0$ holds in general. That is, for any spectral function $H(f)$ whose inverse Fourier transform ("impulse response") is real. | ||
{{ML-Fuß}} | {{ML-Fuß}} | ||
− | [[Category: | + | [[Category:Linear and Time-Invariant Systems: Exercises|^3.2 Laplace Transform and p-Transfer Function^]] |
Latest revision as of 15:54, 15 October 2021
We first assume a two-port network with the following transfer function:
- $$H_{\rm L}(p)= \frac {1-{p}/{A}} {1+{p}/{A}}.$$
From this is to be determined the conventional Fourier frequency response
- $$H(f) = {\rm e}^{-a(f)\hspace{0.05cm}}\cdot {\rm e}^{- \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}b(f)},$$
which is representable
- by the attenuation function $a(f)$
- and the phase function $b(f)$.
The upper diagram shows a so-called "all-pass circuit" where the complex resistance $Z_1$ denotes an inductance and the complex resistance $Z_2$ denotes a capacitance:
- $$Z_1 = p \cdot L\hspace{0.05cm},\hspace{0.2cm}Z_2 = \frac{1}{p \cdot C}\hspace{0.05cm}.$$
For reflection-free adaptation at the input and output with
- $$Z_{\rm I}=Z_{\rm A} = \sqrt{Z_1 \cdot Z_2} = \sqrt{{L}/{C}},$$
the following holds for the $p$–transfer function of circuit $\rm A$ (see upper diagram):
- $$H_{\rm L}(p)= \frac {Y_{\rm L}(p)} {X_{\rm L}(p)}= \frac {Z_2-Z_1} {Z_1+2 \cdot \sqrt{Z_1 \cdot Z_2}+Z_2}\hspace{0.05cm}.$$
Circuit $\rm B$ is defined by the pole–zero diagram of the $p$–transfer function. It is characterized by the fact that
- all poles (in the left $p$–half-plane)
- are located in a mirror-imaged manner with respect to the zeros (in the right half-plane).
Please note:
- The exercise belongs to the chapter Laplace Transform and p-Transfer Function.
Questions
Solution
- $$H_{\rm L}(p)= \frac {1-{p}/{A}} {1+{p}/{A}}= -1 \cdot \frac {p-A} {p+A}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\hspace{0.15cm} \underline{K =- 1}, \hspace{0,2cm} \hspace{0.15cm} \underline{p_{\rm o}/A = 1} ,\hspace{0,2cm} \hspace{0.15cm} \underline{ p_{\rm x}/A = -1} \hspace{0.05cm} .$$
(2) Statements 2 and 3 are correct:
- If $p = {\rm j} \cdot 2 \pi f$ is set, the following is obtained:
- $$H(f)= \frac {1-{\rm j \cdot 2\pi \it f}/A} {1+{\rm j \cdot 2\pi \it f}/A}\hspace{0.05cm} .$$
- The magnitude of a quotient is equal to the quotient of the magnitudes:
- $$|H(f)|= \frac {|1-{{\rm j} \cdot 2\pi f}/A|} {|1+{\rm j \cdot 2\pi \it f}/A|}= \frac {\sqrt{1+(2\pi f/A)^2}} {\sqrt{1+(2\pi f/A)^2}}= 1\hspace{0.3cm} \Rightarrow \hspace{0.3cm} a(f)= -{\rm ln} \hspace{0.1cm} |H(f)|= 0\hspace{0.2cm}({\rm Np \hspace{0.2cm}or \hspace{0.2cm}dB})\hspace{0.05cm} .$$
- However, $\text{statement 3}$ is also correct as can be seen from the theory page "Graphical determination of attenuation".
(3) The phase function $b(f)$ can be computed as follows:
- $$b(f)= -{\rm arc} \hspace{0.1cm} H(f) \hspace{0.25cm} = \hspace{0.2cm} {\rm arctan } \hspace{0.1cm} ({2\pi f}/{A}) - {\rm arctan } \hspace{0.1cm} ({-2\pi f}/{A}) = 2 \cdot {\rm arctan } \hspace{0.1cm} ({2\pi f}/{A}),$$
- $$b(f= {A}/{2\pi})= 2 \cdot {\rm arctan } \hspace{0.1cm}(1) = 2 \cdot 45^\circ \hspace{0.15cm} \underline{ = 90^\circ}\hspace{0.05cm},$$
- $$ b(f= {A}/{\pi})=2 \cdot {\rm arctan } \hspace{0.1cm}(2) = 2 \cdot 63.4^\circ \hspace{0.15cm} \underline{= 126.8^\circ}\hspace{0.05cm} ,$$
- $$ b(f \rightarrow \infty)=2 \cdot {\rm arctan } \hspace{0.1cm}(\infty) = 2 \cdot 90^\circ \hspace{0.15cm} \underline{= 180^\circ}\hspace{0.05cm} .$$
- The same results are obtained following the approach according to the page "Graphical determination of phase" in the theory part.
(4) Only statement 1 is correct:
- The given $p$–transfer function can be expressed as follows:
- $$H_{\rm L}(p)= \frac {Z_2-Z_1} {Z_1+2 \cdot \sqrt{Z_1 \cdot Z_2}+Z_2}= \frac {(\sqrt{Z_2}-\sqrt{Z_1})(\sqrt{Z_2}+\sqrt{Z_1})} {(\sqrt{Z_2}+\sqrt{Z_1})^2}= \frac {\sqrt{Z_2}-\sqrt{Z_1}} {\sqrt{Z_2}+\sqrt{Z_1}}\hspace{0.05cm}.$$
- Furthermore, considering $Z_1 = p \cdot L$ and $Z_2 = 1/(p \cdot C)$ the following is obtained:
- $$H_{\rm L}(p)= \frac {\sqrt{{1}/(pC)}-\sqrt{pL}} {\sqrt{{1}/(pC)}+\sqrt{pL}} = \frac {1- p \cdot \sqrt{LC}} {1+ p \cdot \sqrt{LC}} = -1 \cdot \frac {p-\sqrt{{1}/(LC)}} {p+\sqrt{{1}/(LC)}}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}A = \sqrt{{1}/(LC)}: \hspace{0.2cm}H_{\rm L}(p)= -1 \cdot \frac {p-A} {p+A}\hspace{0.05cm}.$$
- The same transfer function as computed in subtask (1) is obtained.
From this it follows that only statement 1 is correct:
- The attenuation curve is $a(f) = 0\ \rm (Np)$. No frequency is attenuated or amplified. Therefore, it is called an "all-pass filter".
- The second statement is false. The phase response $b(f)$ is not linear but rather curved as computed in subtask (3) .
- The Hilbert transform of the constant $a(f) = 0$ should result in the phase function $b(f) = 0$ as shown in the theory part. That is: statement 3 is false.
- The attenuation function $a(f)$ and phase function $b(f)$ are related to each other via the Hilbert transformation only for minimum-phase systems.
- However, in such a minimum–phase system, all poles and zeros lie in the left $p$–half-plane which is not true here
⇒ an all-pass filter is not a minimum–phase system.
(5) Both statements are correct:
- As already determined in subtask (2): A constant attenuation arises as a result whenever there is a corresponding zero in the right half-plane for each pole in the left $p$–half-plane ⇒ circuit $\rm B$ also exhibits all-pass filter charakteristics.
- Since $b(f)$ is always an asymmetric function, $b(f= 0) = 0$ holds in general. That is, for any spectral function $H(f)$ whose inverse Fourier transform ("impulse response") is real.