Difference between revisions of "Aufgaben:Exercise 2.9: Symmetrical Distortions"
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− | {{quiz-Header|Buchseite= | + | {{quiz-Header|Buchseite=Modulation_Methods/Envelope_Demodulation |
}} | }} | ||
− | [[File:P_ID1040__Mod_Z_2_8.png|right|frame| | + | [[File:P_ID1040__Mod_Z_2_8.png|right|frame|Transmitter and receiver spectrum in the equivalent low-pass region]] |
− | + | The source signal made up of two components | |
:$$q(t) = A_1 \cdot \cos(2 \pi f_1 t ) + A_2 \cdot \cos(2 \pi f_2 t )$$ | :$$q(t) = A_1 \cdot \cos(2 \pi f_1 t ) + A_2 \cdot \cos(2 \pi f_2 t )$$ | ||
− | + | is amplitude modulated and transmitted through a linearly distorting transmission channel. | |
+ | *The carrier frequency is $f_{\rm T}$ and the added DC component $A_{\rm T}$. | ||
+ | *Thus, a "double-sideband amplitude moduluation" $\rm (DSB–AM)$ with carrier" is present. | ||
− | |||
+ | The upper graph shows the spectrum $S_{\rm TP}(f)$ of the equivalent low-pass signal in schematic form. This means that the lengths of the Dirac delta lines drawn do not correspond to the actual values of $A_{\rm T}$, $A_1/2$ and $A_2/2$. | ||
− | |||
− | + | The spectral function $R(f)$ of the received signal was measured. In the lower graph we can observe the equivalent low-pass spectrum $R_{\rm TP}(f)$ calculated from this. | |
+ | |||
+ | The channel frequency response is characterized with sufficient accuracy with a few auxiliary values: | ||
:$$ H_{\rm K}(f = f_{\rm T}) = 0.5,$$ | :$$ H_{\rm K}(f = f_{\rm T}) = 0.5,$$ | ||
:$$H_{\rm K}(f = f_{\rm T} \pm f_1) = 0.4,$$ | :$$H_{\rm K}(f = f_{\rm T} \pm f_1) = 0.4,$$ | ||
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− | + | Hints: | |
− | + | *This exercise belongs to the chapter [[Modulation_Methods/Envelope_Demodulation|Envelope Demodulation]]. | |
− | + | *Particular reference is made to the section [[Modulation_Methods/Envelope_Demodulation#Description_using_the_equivalent_low-pass_signal|Description using the equivalent low-pass signal]]. | |
− | |||
− | |||
− | * | ||
− | * | ||
− | === | + | ===Questions=== |
<quiz display=simple> | <quiz display=simple> | ||
− | { | + | {Give the amplitudes of the carrier and source signal. |
|type="{}"} | |type="{}"} | ||
$A_{\rm T} \ = \hspace{0.17cm} $ { 4 3% } $\ \rm V$ | $A_{\rm T} \ = \hspace{0.17cm} $ { 4 3% } $\ \rm V$ | ||
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$A_2 \ = \ $ { 4 3% } $\ \rm V$ | $A_2 \ = \ $ { 4 3% } $\ \rm V$ | ||
− | { | + | {Which kind of distortion would the application of an envelope demodulator in an ideal channel ⇒ $H_{\rm K}(f) = 1$ lead to? |
|type="()"} | |type="()"} | ||
− | - | + | - No distortion. |
− | - | + | - Linear distortions. |
− | + | + | + Nonlinear distortions. |
− | { | + | {Calculate the equivalent low-pass signal and answer the following questions. Is it true that... |
|type="[]"} | |type="[]"} | ||
− | + $r_{\rm TP}(t)$ | + | + $r_{\rm TP}(t)$ is always real, |
− | + $r_{\rm TP}(t)$ | + | + $r_{\rm TP}(t)$ is always greater than or equal to zero, |
− | - | + | - the phase function $ϕ(t)$ can take on the values $0^\circ$ and $180^\circ$ . |
− | { | + | {Which kind of distortion does the envelope demodulator in the observed transmission channel lead to? |
|type="()"} | |type="()"} | ||
− | - | + | - No distortion. |
− | + | + | + Linear distortions. |
− | - | + | - Nonlinear distortions.<br> |
</quiz> | </quiz> | ||
− | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''(1)''' | + | '''(1)''' On the basis of the graphs on the exercise page, the following statements can be made: |
:$${A_{\rm T}} \cdot 0.5 = 2 \,{\rm V}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}A_{\rm T} \hspace{0.15cm}\underline {= 4 \,{\rm V}},$$ | :$${A_{\rm T}} \cdot 0.5 = 2 \,{\rm V}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}A_{\rm T} \hspace{0.15cm}\underline {= 4 \,{\rm V}},$$ | ||
:$${A_{\rm 1}}/{2} \cdot 0.4 = 0.6\,{\rm V}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}A_{\rm 1} \hspace{0.15cm}\underline {= 3 \,{\rm V}},$$ | :$${A_{\rm 1}}/{2} \cdot 0.4 = 0.6\,{\rm V}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}A_{\rm 1} \hspace{0.15cm}\underline {= 3 \,{\rm V}},$$ | ||
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− | '''(2)''' | + | '''(2)''' <u>Answer 3</u> is correct: |
− | * | + | *The resulting modulation depth is $m = (A_1 + A_2)/A_T = 1.75 >1$. |
− | * | + | *This leads to strong nonlinear distortion when using an envelope demodulator. |
− | * | + | *A distortion factor cannot be specified because the source signal contains two frequency components. |
− | '''(3)''' | + | '''(3)''' <u>Answers 1 and 2</u> are correct: |
− | * | + | *The Fourier retransform of $R_{\rm TP}(f)$ gives us the result: |
:$$ r_{\rm TP}(t) = 2 \,{\rm V} + 1.2 \,{\rm V} \cdot \cos(2 \pi f_1 t ) + 0.8 \,{\rm V} \cdot \cos(2 \pi f_2 t )\hspace{0.05cm}.$$ | :$$ r_{\rm TP}(t) = 2 \,{\rm V} + 1.2 \,{\rm V} \cdot \cos(2 \pi f_1 t ) + 0.8 \,{\rm V} \cdot \cos(2 \pi f_2 t )\hspace{0.05cm}.$$ | ||
− | * | + | *This function is always real and non-negative. |
− | * | + | *Thus, $ϕ(t) = 0$ holds simultaneously, whereas $ϕ(t) = 180^\circ$ is not possible. |
− | '''(4)''' | + | '''(4)''' A comparison of the two signals |
:$$q(t) = 3 \,{\rm V} \cdot \cos(2 \pi f_1 t ) + 4 \,{\rm V} \cdot \cos(2 \pi f_2 t ),$$ | :$$q(t) = 3 \,{\rm V} \cdot \cos(2 \pi f_1 t ) + 4 \,{\rm V} \cdot \cos(2 \pi f_2 t ),$$ | ||
:$$ v(t) = 0.4 \cdot 3 \,{\rm V} \cdot \cos(2 \pi f_1 t ) + 0.2 \cdot 4 \,{\rm V} \cdot \cos(2 \pi f_2 t )$$ | :$$ v(t) = 0.4 \cdot 3 \,{\rm V} \cdot \cos(2 \pi f_1 t ) + 0.2 \cdot 4 \,{\rm V} \cdot \cos(2 \pi f_2 t )$$ | ||
− | : | + | :shows, that linear (attenuation) distortions now arise ⇒ <u>Answer 2</u>. |
− | * | + | *Here, the channel $H_{\rm K}(f)$ has the positive effect, that instead of irreversible nonlinear distortions, only linear distortions arise, and these can be eliminated by a downstream filter. |
− | * | + | *This is due to the fact that the higher attenuation of the source signal $q(t)$ compared to the carrier signal $z(t)$ lowers the modulation depth from $m = 1.75$ to |
+ | :$$m = (0.4 · 3 \ \rm V + 0.2 · 4 \ \rm V)/(0.5 · 4 \ \rm V) = 1.$$ | ||
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− | [[Category: | + | [[Category:Modulation Methods: Exercises|^2.3 Envelope Demodulation^]] |
Latest revision as of 15:20, 18 January 2023
The source signal made up of two components
- $$q(t) = A_1 \cdot \cos(2 \pi f_1 t ) + A_2 \cdot \cos(2 \pi f_2 t )$$
is amplitude modulated and transmitted through a linearly distorting transmission channel.
- The carrier frequency is $f_{\rm T}$ and the added DC component $A_{\rm T}$.
- Thus, a "double-sideband amplitude moduluation" $\rm (DSB–AM)$ with carrier" is present.
The upper graph shows the spectrum $S_{\rm TP}(f)$ of the equivalent low-pass signal in schematic form. This means that the lengths of the Dirac delta lines drawn do not correspond to the actual values of $A_{\rm T}$, $A_1/2$ and $A_2/2$.
The spectral function $R(f)$ of the received signal was measured. In the lower graph we can observe the equivalent low-pass spectrum $R_{\rm TP}(f)$ calculated from this.
The channel frequency response is characterized with sufficient accuracy with a few auxiliary values:
- $$ H_{\rm K}(f = f_{\rm T}) = 0.5,$$
- $$H_{\rm K}(f = f_{\rm T} \pm f_1) = 0.4,$$
- $$ H_{\rm K}(f = f_{\rm T} \pm f_2) = 0.2 \hspace{0.05cm}.$$
Hints:
- This exercise belongs to the chapter Envelope Demodulation.
- Particular reference is made to the section Description using the equivalent low-pass signal.
Questions
Solution
- $${A_{\rm T}} \cdot 0.5 = 2 \,{\rm V}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}A_{\rm T} \hspace{0.15cm}\underline {= 4 \,{\rm V}},$$
- $${A_{\rm 1}}/{2} \cdot 0.4 = 0.6\,{\rm V}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}A_{\rm 1} \hspace{0.15cm}\underline {= 3 \,{\rm V}},$$
- $${A_{\rm 2}}/{2} \cdot 0.2 = 0.4\,{\rm V}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}A_{\rm 2} \hspace{0.15cm}\underline {= 4 \,{\rm V}}\hspace{0.05cm}.$$
(2) Answer 3 is correct:
- The resulting modulation depth is $m = (A_1 + A_2)/A_T = 1.75 >1$.
- This leads to strong nonlinear distortion when using an envelope demodulator.
- A distortion factor cannot be specified because the source signal contains two frequency components.
(3) Answers 1 and 2 are correct:
- The Fourier retransform of $R_{\rm TP}(f)$ gives us the result:
- $$ r_{\rm TP}(t) = 2 \,{\rm V} + 1.2 \,{\rm V} \cdot \cos(2 \pi f_1 t ) + 0.8 \,{\rm V} \cdot \cos(2 \pi f_2 t )\hspace{0.05cm}.$$
- This function is always real and non-negative.
- Thus, $ϕ(t) = 0$ holds simultaneously, whereas $ϕ(t) = 180^\circ$ is not possible.
(4) A comparison of the two signals
- $$q(t) = 3 \,{\rm V} \cdot \cos(2 \pi f_1 t ) + 4 \,{\rm V} \cdot \cos(2 \pi f_2 t ),$$
- $$ v(t) = 0.4 \cdot 3 \,{\rm V} \cdot \cos(2 \pi f_1 t ) + 0.2 \cdot 4 \,{\rm V} \cdot \cos(2 \pi f_2 t )$$
- shows, that linear (attenuation) distortions now arise ⇒ Answer 2.
- Here, the channel $H_{\rm K}(f)$ has the positive effect, that instead of irreversible nonlinear distortions, only linear distortions arise, and these can be eliminated by a downstream filter.
- This is due to the fact that the higher attenuation of the source signal $q(t)$ compared to the carrier signal $z(t)$ lowers the modulation depth from $m = 1.75$ to
- $$m = (0.4 · 3 \ \rm V + 0.2 · 4 \ \rm V)/(0.5 · 4 \ \rm V) = 1.$$