Difference between revisions of "Aufgaben:Exercise 3.5Z: Phase Modulation of a Trapezoidal Signal"
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| − | {{quiz-Header|Buchseite= | + | {{quiz-Header|Buchseite=Modulation_Methods/Frequency_Modulation_(FM) |
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| − | [[File:|right|]] | + | [[File:P_ID1100__Mod_Z_3_5.png|right|frame|Trapezoidal and rectangular signals]] |
| + | A phase modulator with input signal $q_1(t)$ a modulated signal $s(t)$ at the output are described as follows: | ||
| + | :$$s(t) = A_{\rm T} \cdot \cos \hspace{-0.05cm}\big[\psi(t) \big ]= | ||
| + | A_{\rm T} \cdot \cos \hspace{-0.05cm}\big[\omega_{\rm T} \cdot t + K_{\rm PM} \cdot q_1(t) \big ] \hspace{0.05cm}.$$ | ||
| + | *The carrier angular frequency is $ω_{\rm T} = 2π · 10^5 \cdot {1}/{\rm s}$. | ||
| + | *The instantaneous angular frequency $ω_{\rm A}(t)$ is equal to the derivative of the angle function $ψ(t)$ with respect to time. | ||
| + | *The instantaneous frequency is thus $f_{\rm A}(t) = ω_{\rm A}(t)/2π$. | ||
| − | === | + | The trapezoidal signal $q_1(t)$ is applied as a test signal, where the nominated time duration is $T = 10 \ \rm µ s$ . |
| + | |||
| + | The same modulated signal $s(t)$ would result from a frequency modulator with the angular function | ||
| + | :$$\psi(t) = \omega_{\rm T} \cdot t + K_{\rm FM} \cdot \int q_2(t)\hspace{0.15cm}{\rm d}t$$ | ||
| + | if the rectangular source signal $q_2(t)$ is applied according to the lower plot. | ||
| + | |||
| + | |||
| + | |||
| + | |||
| + | |||
| + | |||
| + | |||
| + | |||
| + | ''Hints:'' | ||
| + | *This exercise belongs to the chapter [[Modulation_Methods/Frequency_Modulation_(FM)|Frequency Modulation]]. | ||
| + | *Reference is also made to the chapter [[Modulation_Methods/Phase_Modulation_(PM)|Phase Modulation]]. | ||
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| + | |||
| + | |||
| + | ===Questions=== | ||
<quiz display=simple> | <quiz display=simple> | ||
| − | { | + | {How should the modulator constant $K_{\rm PM}$ be chosen so that $ϕ_{\rm max} = 3 \ \rm rad$ ? |
| − | |type=" | + | |type="{}"} |
| − | - | + | $K_{\rm PM} \ = \ $ { 1.5 3% } $\ \rm V^{-1}$ |
| − | + | ||
| + | {What range of values does the instantaneous frequency $f_{\rm A}(t)$ take on in the time interval $0 < t < T$ ? | ||
| + | |type="{}"} | ||
| + | $f_\text{A, min} \ = \ $ { 147.7 3% } $\ \rm kHz$ | ||
| + | $f_\text{A, max} \hspace{0.06cm} = \ $ { 147.7 3% } $\ \rm kHz$ | ||
| − | { | + | {What range of values does the instantaneous frequency $f_{\rm A}(t)$ take on in the time interval $T < t < 3T$ ? |
|type="{}"} | |type="{}"} | ||
| − | $\ | + | $f_\text{A, min} \ = \ $ { 100 3% } $\ \rm kHz$ |
| + | $f_\text{A, max} \hspace{0.06cm} = \ $ { 100 3% } $\ \rm kHz$ | ||
| + | |||
| + | {What range of values does the instantaneous frequency $f_{\rm A}(t)$ take on in the time interval $3T < t < 5T$ ? | ||
| + | |type="{}"} | ||
| + | $f_\text{A, min} \ = \ $ { 52.3 3% } $\ \rm kHz$ | ||
| + | $f_\text{A, max} \hspace{0.06cm} = \ $ { 52.3 3% } $\ \rm kHz$ | ||
| + | |||
| + | {How must the modulator constant $K_{\rm FM}$ be chosen so that the signal $q_2(t)$ results in the same RF signal $s(t)$ after frequency modulation? | ||
| + | |type="{}"} | ||
| + | $K_{\rm FM} \ = \ $ { 1.5 3% } $\ \cdot 10^5 \ \rm V^{-1}s^{-1}$ | ||
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</quiz> | </quiz> | ||
| − | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
| − | '''1 | + | '''(1)''' The phase function is calculated as $ϕ(t) = K_{\rm PM} · q_1(t)$. The phase deviation $ϕ_{\rm max}$ is equal to the phase resulting from the maximum value of the source signal: |
| − | '''2 | + | :$$ \phi_{\rm max} = K_{\rm PM} \cdot 2\,{\rm V} = 3\,{\rm rad}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} K_{\rm PM} \hspace{0.15cm}\underline {= 1.5\,{\rm V^{-1}}} \hspace{0.05cm}.$$ |
| − | '''3 | + | |
| − | '''4 | + | |
| − | ''' | + | |
| − | ''' | + | '''(2)''' In the range $0 < t < T$ , the angular function can be represented as follows: |
| − | ''' | + | :$$ \psi(t) = \omega_{\rm T} \cdot t + K_{\rm PM} \cdot 2\,{\rm V} \cdot {t}/{T}\hspace{0.05cm}.$$ |
| + | *For the instantaneous angular frequency $ω_{\rm A}(t)$ or the instantaneous frequency $f_{\rm A}(t)$ , the following holds: | ||
| + | :$$\omega_{\rm A}(t) = \frac{{\rm d}\hspace{0.03cm}\psi(t)}{{\rm d}t}= \omega_{\rm T} + K_{\rm PM} \cdot \frac{2\,{\rm V}}{10\,{\rm µ s}}\hspace{0.05cm} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} f_{\rm A}(t) = f_{\rm T} + \frac{1.5\,{\rm V}^{-1}}{2 \pi} \cdot 2 \cdot 10^5 \rm {V}/{ s} = 100\,{\rm kHz}+ 47.7\,{\rm kHz}= 147.7\,{\rm kHz}\hspace{0.05cm}.$$ | ||
| + | *The instantaneous frequency is constant, so $f_\text{A, min} = f_\text{A, max}\hspace{0.15cm}\underline{ = 147.7 \ \rm kHz}$ holds. | ||
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| + | |||
| + | |||
| + | '''(3)''' Due to the constant source signal, the derivative is zero throughout the time interval $T < t < 3T$ under consideration, so the instantaneous frequency is equal to the carrier frequency: | ||
| + | :$$f_{\rm A, \hspace{0.05cm} min} =f_{\rm A, \hspace{0.05cm} max} =f_{\rm T} \hspace{0.15cm}\underline {= 100\,{\rm kHz}}\hspace{0.05cm}.$$ | ||
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| + | |||
| + | |||
| + | '''(4)''' The linear decay of $q_1(t)$ in the time interval $3T < t < 5T$ with slope as calculated in '''(2)''' leads to the result: | ||
| + | :$$f_{\rm A, \hspace{0.05cm} min} =f_{\rm A, \hspace{0.05cm} max} =f_{\rm T} - 47.7\,{\rm kHz} \hspace{0.15cm}\underline {= 52.3\,{\rm kHz}}\hspace{0.05cm}.$$ | ||
| + | |||
| + | |||
| + | |||
| + | '''(5)''' By differentiation, we arrive at the instantaneous angular frequency: | ||
| + | :$$ \omega_{\rm A}(t) = \omega_{\rm T} + K_{\rm FM} \cdot q_2(t) \Rightarrow \hspace{0.3cm} f_{\rm A}(t) = f_{\rm T}+\frac{ K_{\rm FM}}{2 \pi} \cdot q_2(t)\hspace{0.05cm}.$$ | ||
| + | *Using the result from '''(2)''' , we get: | ||
| + | :$$\frac{ K_{\rm FM}}{2 \pi} \cdot 2\,{\rm V} = \frac{ 3 \cdot 10^5}{2 \pi} \cdot {\rm s^{-1}}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} K_{\rm FM} \hspace{0.15cm}\underline {= 1.5 \cdot 10^5 \hspace{0.15cm}{\rm V^{-1}}{\rm s^{-1}}}\hspace{0.05cm}.$$ | ||
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{{ML-Fuß}} | {{ML-Fuß}} | ||
| − | [[Category: | + | [[Category:Modulation Methods: Exercises|^3.2 Frequency Modulation^]] |
Latest revision as of 16:11, 9 April 2022
Trapezoidal and rectangular signals
A phase modulator with input signal $q_1(t)$ a modulated signal $s(t)$ at the output are described as follows:
- $$s(t) = A_{\rm T} \cdot \cos \hspace{-0.05cm}\big[\psi(t) \big ]= A_{\rm T} \cdot \cos \hspace{-0.05cm}\big[\omega_{\rm T} \cdot t + K_{\rm PM} \cdot q_1(t) \big ] \hspace{0.05cm}.$$
- The carrier angular frequency is $ω_{\rm T} = 2π · 10^5 \cdot {1}/{\rm s}$.
- The instantaneous angular frequency $ω_{\rm A}(t)$ is equal to the derivative of the angle function $ψ(t)$ with respect to time.
- The instantaneous frequency is thus $f_{\rm A}(t) = ω_{\rm A}(t)/2π$.
The trapezoidal signal $q_1(t)$ is applied as a test signal, where the nominated time duration is $T = 10 \ \rm µ s$ .
The same modulated signal $s(t)$ would result from a frequency modulator with the angular function
- $$\psi(t) = \omega_{\rm T} \cdot t + K_{\rm FM} \cdot \int q_2(t)\hspace{0.15cm}{\rm d}t$$
if the rectangular source signal $q_2(t)$ is applied according to the lower plot.
Hints:
- This exercise belongs to the chapter Frequency Modulation.
- Reference is also made to the chapter Phase Modulation.
Questions
Solution
(1) The phase function is calculated as $ϕ(t) = K_{\rm PM} · q_1(t)$. The phase deviation $ϕ_{\rm max}$ is equal to the phase resulting from the maximum value of the source signal:
- $$ \phi_{\rm max} = K_{\rm PM} \cdot 2\,{\rm V} = 3\,{\rm rad}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} K_{\rm PM} \hspace{0.15cm}\underline {= 1.5\,{\rm V^{-1}}} \hspace{0.05cm}.$$
(2) In the range $0 < t < T$ , the angular function can be represented as follows:
- $$ \psi(t) = \omega_{\rm T} \cdot t + K_{\rm PM} \cdot 2\,{\rm V} \cdot {t}/{T}\hspace{0.05cm}.$$
- For the instantaneous angular frequency $ω_{\rm A}(t)$ or the instantaneous frequency $f_{\rm A}(t)$ , the following holds:
- $$\omega_{\rm A}(t) = \frac{{\rm d}\hspace{0.03cm}\psi(t)}{{\rm d}t}= \omega_{\rm T} + K_{\rm PM} \cdot \frac{2\,{\rm V}}{10\,{\rm µ s}}\hspace{0.05cm} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} f_{\rm A}(t) = f_{\rm T} + \frac{1.5\,{\rm V}^{-1}}{2 \pi} \cdot 2 \cdot 10^5 \rm {V}/{ s} = 100\,{\rm kHz}+ 47.7\,{\rm kHz}= 147.7\,{\rm kHz}\hspace{0.05cm}.$$
- The instantaneous frequency is constant, so $f_\text{A, min} = f_\text{A, max}\hspace{0.15cm}\underline{ = 147.7 \ \rm kHz}$ holds.
(3) Due to the constant source signal, the derivative is zero throughout the time interval $T < t < 3T$ under consideration, so the instantaneous frequency is equal to the carrier frequency:
- $$f_{\rm A, \hspace{0.05cm} min} =f_{\rm A, \hspace{0.05cm} max} =f_{\rm T} \hspace{0.15cm}\underline {= 100\,{\rm kHz}}\hspace{0.05cm}.$$
(4) The linear decay of $q_1(t)$ in the time interval $3T < t < 5T$ with slope as calculated in (2) leads to the result:
- $$f_{\rm A, \hspace{0.05cm} min} =f_{\rm A, \hspace{0.05cm} max} =f_{\rm T} - 47.7\,{\rm kHz} \hspace{0.15cm}\underline {= 52.3\,{\rm kHz}}\hspace{0.05cm}.$$
(5) By differentiation, we arrive at the instantaneous angular frequency:
- $$ \omega_{\rm A}(t) = \omega_{\rm T} + K_{\rm FM} \cdot q_2(t) \Rightarrow \hspace{0.3cm} f_{\rm A}(t) = f_{\rm T}+\frac{ K_{\rm FM}}{2 \pi} \cdot q_2(t)\hspace{0.05cm}.$$
- Using the result from (2) , we get:
- $$\frac{ K_{\rm FM}}{2 \pi} \cdot 2\,{\rm V} = \frac{ 3 \cdot 10^5}{2 \pi} \cdot {\rm s^{-1}}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} K_{\rm FM} \hspace{0.15cm}\underline {= 1.5 \cdot 10^5 \hspace{0.15cm}{\rm V^{-1}}{\rm s^{-1}}}\hspace{0.05cm}.$$
