Difference between revisions of "Aufgaben:Exercise 2.5Z: Linear Distortions with DSB-AM"
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[[File:P_ID1013__Mod_Z_2_5.png|right|frame|Considered system model]] | [[File:P_ID1013__Mod_Z_2_5.png|right|frame|Considered system model]] | ||
As in [[Aufgaben:Exercise_2.5:_DSB-AM_via_a_Gaussian_Channel|Exercise 2.5]] here we will also examine: | As in [[Aufgaben:Exercise_2.5:_DSB-AM_via_a_Gaussian_Channel|Exercise 2.5]] here we will also examine: | ||
| − | *the DSB–AM/synchronous demodulator combination | + | *the DSB–AM/synchronous demodulator combination, |
*considerations involving a linear distorting channel. | *considerations involving a linear distorting channel. | ||
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Let the source signal $q(t)$ be a cosine signal with amplitude $A_{\rm N}$ and frequency $f_{\rm N}$, such that the spectrum of the modulated signal is as follows: | Let the source signal $q(t)$ be a cosine signal with amplitude $A_{\rm N}$ and frequency $f_{\rm N}$, such that the spectrum of the modulated signal is as follows: | ||
:$$S(f)= \frac{A_{\rm N}}{4} \cdot \big[\delta(f + f_{\rm O}) + \delta(f + f_{\rm U}) + \delta(f - f_{\rm U}) + \delta(f - f_{\rm O}) \big]\hspace{0.05cm}.$$ | :$$S(f)= \frac{A_{\rm N}}{4} \cdot \big[\delta(f + f_{\rm O}) + \delta(f + f_{\rm U}) + \delta(f - f_{\rm U}) + \delta(f - f_{\rm O}) \big]\hspace{0.05cm}.$$ | ||
| − | The abbreviations stand for $f_{\rm O} = f_{\rm T} + f_{\rm N}$ ( | + | The abbreviations stand for |
| + | *the upper sideband (German: "oberes Seitenband" ⇒ subscript "O") $f_{\rm O} = f_{\rm T} + f_{\rm N}$, and | ||
| + | *the lower sideband (German: "unteres Seitenband" ⇒ subscript "U") $f_{\rm U} = f_{\rm T} - f_{\rm N}$. | ||
| − | The channel frequency response is only given for these two frequencies and is: | + | |
| + | The channel frequency response (German: "Kanalfrequenzgang" ⇒ subscript "K") is only given for these two frequencies and is: | ||
:$$ H_{\rm K}(f_{\rm O}) = R_{\rm O} + {\rm j} \cdot I_{\rm O},\hspace{0.2cm}H_{\rm K}(f_{\rm U}) = R_{\rm U} + {\rm j} \cdot I_{\rm U} \hspace{0.05cm}.$$ | :$$ H_{\rm K}(f_{\rm O}) = R_{\rm O} + {\rm j} \cdot I_{\rm O},\hspace{0.2cm}H_{\rm K}(f_{\rm U}) = R_{\rm U} + {\rm j} \cdot I_{\rm U} \hspace{0.05cm}.$$ | ||
| − | For negative frequencies, $H_{\rm K}(– f) = H_{\rm K}^*(f)$ always holds. | + | For negative frequencies, $H_{\rm K}(– f) = H_{\rm K}^*(f)$ always holds. |
Use the following values for numerical calculations : | Use the following values for numerical calculations : | ||
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In subtask '''(3)''' the solution should be found from the resulting frequency response of modulator, channel and demodulator: | In subtask '''(3)''' the solution should be found from the resulting frequency response of modulator, channel and demodulator: | ||
:$$H_{\rm MKD}(f) = {1}/{2} \cdot \big[ H_{\rm K}(f + f_{\rm T}) + H_{\rm K}(f - f_{\rm T})\big]\hspace{0.05cm}.$$ | :$$H_{\rm MKD}(f) = {1}/{2} \cdot \big[ H_{\rm K}(f + f_{\rm T}) + H_{\rm K}(f - f_{\rm T})\big]\hspace{0.05cm}.$$ | ||
| − | Finally, in subtask '''(4)''' the following channel frequency response is considered ( | + | Finally, in subtask '''(4)''' the following channel frequency response is considered (this equationt is only valid for positive frequencies): |
:$$ H_{\rm K}(f) = H_{\rm(4)}(f) = \frac{1}{1 + 3{\rm j} \cdot ({f}/{f_{\rm T}} - 1)}\hspace{0.05cm}.$$ | :$$ H_{\rm K}(f) = H_{\rm(4)}(f) = \frac{1}{1 + 3{\rm j} \cdot ({f}/{f_{\rm T}} - 1)}\hspace{0.05cm}.$$ | ||
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| − | + | Hints: | |
| − | |||
| − | |||
| − | |||
*This exercise belongs to the chapter [[Modulation_Methods/Synchronous_Demodulation|Synchronous Demodulation]]. | *This exercise belongs to the chapter [[Modulation_Methods/Synchronous_Demodulation|Synchronous Demodulation]]. | ||
*Particular reference is made to the page [[Modulation_Methods/Synchronous_Demodulation#Influence_of_linear_channel_distortions|Influence of linear channel distortions]]. | *Particular reference is made to the page [[Modulation_Methods/Synchronous_Demodulation#Influence_of_linear_channel_distortions|Influence of linear channel distortions]]. | ||
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<quiz display=simple> | <quiz display=simple> | ||
| − | {Let $R_{\rm U} = 0.8, \ I_{\rm U} = -0.2, \ R_{\rm O} = 0.4, I_{\rm O} = -0.2.$ Calculate and sketch the spectrum $R(f)$ at the channel output. <br>What is the spectral line at $-f_{\rm O}$? | + | {Let $R_{\rm U} = 0.8, \ I_{\rm U} = -0.2, \ R_{\rm O} = 0.4, \ I_{\rm O} = -0.2.$ Calculate and sketch the spectrum $R(f)$ at the channel output. <br>What is the spectral line at $-f_{\rm O}$? |
|type="{}"} | |type="{}"} | ||
${\rm Re}[R(-f_{\rm O})] \ = \ $ { 0.2 3% } $\ \text{V}$ | ${\rm Re}[R(-f_{\rm O})] \ = \ $ { 0.2 3% } $\ \text{V}$ | ||
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===Solution=== | ===Solution=== | ||
{{ML-Kopf}} | {{ML-Kopf}} | ||
| − | [[File:P_ID1014__Mod_Z_2_5_a.png|right|frame| | + | [[File:P_ID1014__Mod_Z_2_5_a.png|right|frame|Spectrum $R(f)$ of the received signal]] |
| − | '''(1)''' In general, $R(f) = S(f) · H_K(f)$. This gives the line spectrum as shown in the adjacent sketch (all weights still have to be supplemented by the unit " | + | '''(1)''' In general, $R(f) = S(f) · H_K(f)$. This gives the line spectrum as shown in the adjacent sketch (all weights still have to be supplemented by the unit "Volt"). |
| − | * | + | *For the weight of the spectral line at $f = -f_{\rm O}$ it is valid: |
:$${\rm Re}[R(-f_{\rm O})]\hspace{0.15cm}\underline{=0.2 \ \rm V},$$ | :$${\rm Re}[R(-f_{\rm O})]\hspace{0.15cm}\underline{=0.2 \ \rm V},$$ | ||
:$${\rm Im}[R(-f_{\rm O})]\hspace{0.15cm}\underline{=0.1 \ \rm V}.$$ | :$${\rm Im}[R(-f_{\rm O})]\hspace{0.15cm}\underline{=0.1 \ \rm V}.$$ | ||
| − | |||
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A_{\rm N}\cdot \frac{I_{\rm U} - I_{\rm O}}{2}\cdot \frac{\rm j}{2} \cdot \left[-\delta(f - f_{\rm N}) + \delta(f + f_{\rm N}) \right]$$ | A_{\rm N}\cdot \frac{I_{\rm U} - I_{\rm O}}{2}\cdot \frac{\rm j}{2} \cdot \left[-\delta(f - f_{\rm N}) + \delta(f + f_{\rm N}) \right]$$ | ||
:$$ \Rightarrow \hspace{0.3cm}v(t) = A_{\rm N}\cdot \frac{R_{\rm U} +R_{\rm O}}{2}\cdot\cos (\omega_{\rm N}\cdot t)+ A_{\rm N}\cdot \frac{I_{\rm U} -I_{\rm O}}{2}\cdot\sin (\omega_{\rm N}\cdot t)\hspace{0.05cm}.$$ | :$$ \Rightarrow \hspace{0.3cm}v(t) = A_{\rm N}\cdot \frac{R_{\rm U} +R_{\rm O}}{2}\cdot\cos (\omega_{\rm N}\cdot t)+ A_{\rm N}\cdot \frac{I_{\rm U} -I_{\rm O}}{2}\cdot\sin (\omega_{\rm N}\cdot t)\hspace{0.05cm}.$$ | ||
| − | *When $R_{\rm U} = 0.8, I_{\rm U} = -0.2, R_{\rm O} = 0.4, I_{\rm O} = -0.2$ it follows: | + | *When $R_{\rm U} = 0.8,\ I_{\rm U} = -0.2,\ R_{\rm O} = 0.4,\ I_{\rm O} = -0.2$ it follows: |
:$$v(t) = 0.6 \cdot A_{\rm N}\cdot \cos (\omega_{\rm N}\cdot t)\hspace{0.3cm}\Rightarrow \hspace{0.3cm} v(t=0) = 0.6 \cdot A_{\rm N}\hspace{0.15cm}\underline {= 1.2\,{\rm V}}\hspace{0.05cm}.$$ | :$$v(t) = 0.6 \cdot A_{\rm N}\cdot \cos (\omega_{\rm N}\cdot t)\hspace{0.3cm}\Rightarrow \hspace{0.3cm} v(t=0) = 0.6 \cdot A_{\rm N}\hspace{0.15cm}\underline {= 1.2\,{\rm V}}\hspace{0.05cm}.$$ | ||
| − | *There is attenuation by a factor of $0.6$ compared to $q(t)$ | + | *There is attenuation by a factor of $0.6$ compared to $q(t)$. |
*The synchronous demodulator receives more information about the source signal through the lower sideband than through the upper one. | *The synchronous demodulator receives more information about the source signal through the lower sideband than through the upper one. | ||
| − | *Because of the property $I_{\rm O} = I_{\rm U}$ | + | *Because of the property $I_{\rm O} = I_{\rm U}$, $v(t)$ is also cosine-shaped. |
| − | *Accordingly, either no delay occurs or the delay is an even multiple of the period. | + | *Accordingly, it is valid either no delay occurs or the delay is an even multiple of the period. |
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'''(3)''' The following equations apply here: | '''(3)''' The following equations apply here: | ||
| − | :$$ H_{\rm K}(f_{\rm N}+ f_{\rm T}) = R_{\rm O} + {\rm j} \cdot I_{\rm O} \hspace{0.05cm}, | + | :$$ H_{\rm K}(f_{\rm N}+ f_{\rm T}) = R_{\rm O} + {\rm j} \cdot I_{\rm O} \hspace{0.05cm}, $$ |
| − | H_{\rm K}(f_{\rm N}- f_{\rm T}) = H_{\rm K}^{\star}(f_{\rm T}- f_{\rm N}) = R_{\rm U} - {\rm j} \cdot I_{\rm U} $$ | + | :$$ H_{\rm K}(f_{\rm N}- f_{\rm T}) = H_{\rm K}^{\star}(f_{\rm T}- f_{\rm N}) = R_{\rm U} - {\rm j} \cdot I_{\rm U} $$ |
:$$\Rightarrow \hspace{0.2cm} H_{\rm MKD}(f_{\rm N}) = {1}/{2} \cdot \big[(R_{\rm O} +R_{\rm U}) + {\rm j} \cdot (I_{\rm O} -I_{\rm U}) \big]\hspace{0.05cm},\hspace{0.2cm} | :$$\Rightarrow \hspace{0.2cm} H_{\rm MKD}(f_{\rm N}) = {1}/{2} \cdot \big[(R_{\rm O} +R_{\rm U}) + {\rm j} \cdot (I_{\rm O} -I_{\rm U}) \big]\hspace{0.05cm},\hspace{0.2cm} | ||
H_{\rm MKD}(-f_{\rm N}) = H_{\rm MKD}^\star(f_{\rm N}) = {1}/{2} \cdot \big[(R_{\rm O} +R_{\rm U}) - {\rm j} \cdot (I_{\rm O} -I_{\rm U}) \big]\hspace{0.05cm}.$$ | H_{\rm MKD}(-f_{\rm N}) = H_{\rm MKD}^\star(f_{\rm N}) = {1}/{2} \cdot \big[(R_{\rm O} +R_{\rm U}) - {\rm j} \cdot (I_{\rm O} -I_{\rm U}) \big]\hspace{0.05cm}.$$ | ||
| − | *Thus, one obtains the same result as in (2), but faster ⇒ <u>Answer 2</u>. | + | *Thus, one obtains the same result as in '''(2)''', but faster ⇒ <u>Answer 2</u>. |
| + | |||
| − | '''(4)''' For $f > 0$ the resulting frequency response is: | + | '''(4)''' For $f > 0$ the resulting frequency response is: |
:$$H_{\rm MKD}(f) = {1}/{2} \cdot \left[ H_{\rm K}(f_{\rm T}+ f) + H_{\rm K}^\star(f_{\rm T}-f)\right]= {1}/{2} \cdot \left[ \frac{1}{1 + 3{\rm j} \cdot (\frac{f_{\rm T}+f}{f_{\rm T}} - 1)} + \frac{1}{1 - 3{\rm j} \cdot (\frac{f_{\rm T}-f}{f_{\rm T}} - 1)}\right] $$ | :$$H_{\rm MKD}(f) = {1}/{2} \cdot \left[ H_{\rm K}(f_{\rm T}+ f) + H_{\rm K}^\star(f_{\rm T}-f)\right]= {1}/{2} \cdot \left[ \frac{1}{1 + 3{\rm j} \cdot (\frac{f_{\rm T}+f}{f_{\rm T}} - 1)} + \frac{1}{1 - 3{\rm j} \cdot (\frac{f_{\rm T}-f}{f_{\rm T}} - 1)}\right] $$ | ||
:$$ \Rightarrow \hspace{0.3cm} H_{\rm MKD}(f) = \frac{1}{1 + {\rm j} \cdot {3f}/{f_{\rm T}} } \hspace{0.05cm}.$$ | :$$ \Rightarrow \hspace{0.3cm} H_{\rm MKD}(f) = \frac{1}{1 + {\rm j} \cdot {3f}/{f_{\rm T}} } \hspace{0.05cm}.$$ | ||
| − | *Inserted at the point where $f = f_{\rm N}$ this leads to the result: | + | *Inserted at the point where $f = f_{\rm N}$ this leads to the result: |
:$$H_{\rm MKD}(f_{\rm N}) = \frac{1}{1 + {\rm j} \cdot {3f_{\rm N}}/{f_{\rm T}} } \hspace{1.0cm} | :$$H_{\rm MKD}(f_{\rm N}) = \frac{1}{1 + {\rm j} \cdot {3f_{\rm N}}/{f_{\rm T}} } \hspace{1.0cm} | ||
\Rightarrow \hspace{0.3cm}{\rm magnitude} = \frac{1}{\sqrt{1 + ({3f_{\rm N}}/{f_{\rm T}} )^2}} \hspace{0.05cm}, \hspace{0.3cm} {\rm phase} = {\rm arctan}\hspace{0.1cm}({3f_{\rm N}}/{f_{\rm T}}) \hspace{0.05cm}.$$ | \Rightarrow \hspace{0.3cm}{\rm magnitude} = \frac{1}{\sqrt{1 + ({3f_{\rm N}}/{f_{\rm T}} )^2}} \hspace{0.05cm}, \hspace{0.3cm} {\rm phase} = {\rm arctan}\hspace{0.1cm}({3f_{\rm N}}/{f_{\rm T}}) \hspace{0.05cm}.$$ | ||
Latest revision as of 16:57, 8 December 2021
Considered system model
As in Exercise 2.5 here we will also examine:
- the DSB–AM/synchronous demodulator combination,
- considerations involving a linear distorting channel.
Let the source signal $q(t)$ be a cosine signal with amplitude $A_{\rm N}$ and frequency $f_{\rm N}$, such that the spectrum of the modulated signal is as follows:
- $$S(f)= \frac{A_{\rm N}}{4} \cdot \big[\delta(f + f_{\rm O}) + \delta(f + f_{\rm U}) + \delta(f - f_{\rm U}) + \delta(f - f_{\rm O}) \big]\hspace{0.05cm}.$$
The abbreviations stand for
- the upper sideband (German: "oberes Seitenband" ⇒ subscript "O") $f_{\rm O} = f_{\rm T} + f_{\rm N}$, and
- the lower sideband (German: "unteres Seitenband" ⇒ subscript "U") $f_{\rm U} = f_{\rm T} - f_{\rm N}$.
The channel frequency response (German: "Kanalfrequenzgang" ⇒ subscript "K") is only given for these two frequencies and is:
- $$ H_{\rm K}(f_{\rm O}) = R_{\rm O} + {\rm j} \cdot I_{\rm O},\hspace{0.2cm}H_{\rm K}(f_{\rm U}) = R_{\rm U} + {\rm j} \cdot I_{\rm U} \hspace{0.05cm}.$$
For negative frequencies, $H_{\rm K}(– f) = H_{\rm K}^*(f)$ always holds.
Use the following values for numerical calculations :
- $$A_{\rm N} = 2\,{\rm V}, \hspace{0.15cm}f_{\rm N} = 3\,{\rm kHz}, \hspace{0.15cm}f_{\rm T} = 30\,{\rm kHz} \hspace{0.05cm},$$
- $$R_{\rm U} = 0.8, \hspace{0.15cm}I_{\rm U} = -0.2, \hspace{0.15cm}R_{\rm O} = 0.4, \hspace{0.15cm}I_{\rm O} = -0.2 \hspace{0.05cm}.$$
In subtask (3) the solution should be found from the resulting frequency response of modulator, channel and demodulator:
- $$H_{\rm MKD}(f) = {1}/{2} \cdot \big[ H_{\rm K}(f + f_{\rm T}) + H_{\rm K}(f - f_{\rm T})\big]\hspace{0.05cm}.$$
Finally, in subtask (4) the following channel frequency response is considered (this equationt is only valid for positive frequencies):
- $$ H_{\rm K}(f) = H_{\rm(4)}(f) = \frac{1}{1 + 3{\rm j} \cdot ({f}/{f_{\rm T}} - 1)}\hspace{0.05cm}.$$
Hints:
- This exercise belongs to the chapter Synchronous Demodulation.
- Particular reference is made to the page Influence of linear channel distortions.
Questions
Solution
Spectrum $R(f)$ of the received signal
(1) In general, $R(f) = S(f) · H_K(f)$. This gives the line spectrum as shown in the adjacent sketch (all weights still have to be supplemented by the unit "Volt").
- For the weight of the spectral line at $f = -f_{\rm O}$ it is valid:
- $${\rm Re}[R(-f_{\rm O})]\hspace{0.15cm}\underline{=0.2 \ \rm V},$$
- $${\rm Im}[R(-f_{\rm O})]\hspace{0.15cm}\underline{=0.1 \ \rm V}.$$
(2) The spectral function $V(f)$ of the sink signal $v(t)$ is:
- $$V(f) = \big[ R(f) \star \left[\delta(f - f_{\rm T}) + \delta(f + f_{\rm T}) \right]\big]\cdot H_{\rm E}(f).$$
- According to the laws of the Fourier transform, this can also be written as:
- $$V(f) = \frac{A_{\rm N}}{4} \cdot (R_{\rm O} + {\rm j} \cdot I_{\rm O}) \cdot \delta(f - f_{\rm N}) + \frac{A_{\rm N}}{4} \cdot (R_{\rm U} + {\rm j} \cdot I_{\rm U}) \cdot \delta(f + f_{\rm N})+$$
- $$\hspace{2.25cm}+ \frac{A_{\rm N}}{4} \cdot (R_{\rm O} - {\rm j} \cdot I_{\rm O}) \cdot \delta(f + f_{\rm N})+ \frac{A_{\rm N}}{4} \cdot (R_{\rm U} - {\rm j} \cdot I_{\rm U}) \cdot \delta(f - f_{\rm N}) \hspace{0.05cm}.$$
- All other terms are around twice the carrier frequency and are eliminated by the low-pass filter.
- Rearranging and combining the terms results in:
- $$V(f) = A_{\rm N}\cdot \frac{R_{\rm U} +R_{\rm O}}{2}\cdot \frac{1}{2} \cdot \left[\delta(f - f_{\rm N}) + \delta(f + f_{\rm N}) \right] + A_{\rm N}\cdot \frac{I_{\rm U} - I_{\rm O}}{2}\cdot \frac{\rm j}{2} \cdot \left[-\delta(f - f_{\rm N}) + \delta(f + f_{\rm N}) \right]$$
- $$ \Rightarrow \hspace{0.3cm}v(t) = A_{\rm N}\cdot \frac{R_{\rm U} +R_{\rm O}}{2}\cdot\cos (\omega_{\rm N}\cdot t)+ A_{\rm N}\cdot \frac{I_{\rm U} -I_{\rm O}}{2}\cdot\sin (\omega_{\rm N}\cdot t)\hspace{0.05cm}.$$
- When $R_{\rm U} = 0.8,\ I_{\rm U} = -0.2,\ R_{\rm O} = 0.4,\ I_{\rm O} = -0.2$ it follows:
- $$v(t) = 0.6 \cdot A_{\rm N}\cdot \cos (\omega_{\rm N}\cdot t)\hspace{0.3cm}\Rightarrow \hspace{0.3cm} v(t=0) = 0.6 \cdot A_{\rm N}\hspace{0.15cm}\underline {= 1.2\,{\rm V}}\hspace{0.05cm}.$$
- There is attenuation by a factor of $0.6$ compared to $q(t)$.
- The synchronous demodulator receives more information about the source signal through the lower sideband than through the upper one.
- Because of the property $I_{\rm O} = I_{\rm U}$, $v(t)$ is also cosine-shaped.
- Accordingly, it is valid either no delay occurs or the delay is an even multiple of the period.
(3) The following equations apply here:
- $$ H_{\rm K}(f_{\rm N}+ f_{\rm T}) = R_{\rm O} + {\rm j} \cdot I_{\rm O} \hspace{0.05cm}, $$
- $$ H_{\rm K}(f_{\rm N}- f_{\rm T}) = H_{\rm K}^{\star}(f_{\rm T}- f_{\rm N}) = R_{\rm U} - {\rm j} \cdot I_{\rm U} $$
- $$\Rightarrow \hspace{0.2cm} H_{\rm MKD}(f_{\rm N}) = {1}/{2} \cdot \big[(R_{\rm O} +R_{\rm U}) + {\rm j} \cdot (I_{\rm O} -I_{\rm U}) \big]\hspace{0.05cm},\hspace{0.2cm} H_{\rm MKD}(-f_{\rm N}) = H_{\rm MKD}^\star(f_{\rm N}) = {1}/{2} \cdot \big[(R_{\rm O} +R_{\rm U}) - {\rm j} \cdot (I_{\rm O} -I_{\rm U}) \big]\hspace{0.05cm}.$$
- Thus, one obtains the same result as in (2), but faster ⇒ Answer 2.
(4) For $f > 0$ the resulting frequency response is:
- $$H_{\rm MKD}(f) = {1}/{2} \cdot \left[ H_{\rm K}(f_{\rm T}+ f) + H_{\rm K}^\star(f_{\rm T}-f)\right]= {1}/{2} \cdot \left[ \frac{1}{1 + 3{\rm j} \cdot (\frac{f_{\rm T}+f}{f_{\rm T}} - 1)} + \frac{1}{1 - 3{\rm j} \cdot (\frac{f_{\rm T}-f}{f_{\rm T}} - 1)}\right] $$
- $$ \Rightarrow \hspace{0.3cm} H_{\rm MKD}(f) = \frac{1}{1 + {\rm j} \cdot {3f}/{f_{\rm T}} } \hspace{0.05cm}.$$
- Inserted at the point where $f = f_{\rm N}$ this leads to the result:
- $$H_{\rm MKD}(f_{\rm N}) = \frac{1}{1 + {\rm j} \cdot {3f_{\rm N}}/{f_{\rm T}} } \hspace{1.0cm} \Rightarrow \hspace{0.3cm}{\rm magnitude} = \frac{1}{\sqrt{1 + ({3f_{\rm N}}/{f_{\rm T}} )^2}} \hspace{0.05cm}, \hspace{0.3cm} {\rm phase} = {\rm arctan}\hspace{0.1cm}({3f_{\rm N}}/{f_{\rm T}}) \hspace{0.05cm}.$$
- When $f_{\rm N}/f_{\rm T} = 0.1$ we get the magnitude $0.958$ and the phase $16.7^\circ$. Thus, the sink signal is:
- $$v(t) = 0.958 \cdot 2\,{\rm V}\cdot \cos (\omega_{\rm N}\cdot t + 16.7^\circ) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} v(t=0)= 1.916\,{\rm V}\cdot \cos ( 16.7^\circ)\hspace{0.15cm}\underline { = 1.835\,{\rm V}}\hspace{0.05cm}.$$
