Difference between revisions of "Aufgaben:Exercise 2.7: Two-Way Channel once more"
From LNTwww
| (2 intermediate revisions by the same user not shown) | |||
| Line 8: | Line 8: | ||
In contrast to the general representation in exercise 2.6, the two attenuation factors are equal here: $z_1 = z_2 = 1$. | In contrast to the general representation in exercise 2.6, the two attenuation factors are equal here: $z_1 = z_2 = 1$. | ||
| − | + | For mobile communications, this corresponds for example to an echo at a distance of $T_2 - T_1$ and of the same strength as the signal on the main path. For this, the transit time $T_1$ is assumed. | |
| − | |||
| − | Using the transit times $T_1 = 0$ and $T_2 = T = 4 \ \rm ms$ considered in the subtasks '''(1)''' ... '''(4)''', the following is obtained for the frequency response of the two-way channel, whose magnitude is depicted in the upper graph: | + | Using the transit times $T_1 = 0$ and $T_2 = T = 4 \ \rm ms$ considered in the subtasks '''(1)''' ... '''(4)''', the following is obtained for the frequency response of the two-way channel, whose magnitude is depicted in the upper graph: |
:$$H(f) = 1 + {\rm e}^{-{\rm j}\hspace{0.04cm}2 \pi f T} = 1 + | :$$H(f) = 1 + {\rm e}^{-{\rm j}\hspace{0.04cm}2 \pi f T} = 1 + | ||
\cos(2 \pi f T) - {\rm j} \cdot \sin(2 \pi f T)$$ | \cos(2 \pi f T) - {\rm j} \cdot \sin(2 \pi f T)$$ | ||
| Line 18: | Line 17: | ||
T)\right)}= 2 \cdot |\cos(\pi f T)|.$$ | T)\right)}= 2 \cdot |\cos(\pi f T)|.$$ | ||
| − | The bottom graph shows the phase function: | + | The bottom graph shows the phase function for $T_1 = 0$ and $T_2 = T = 4 \ \rm ms$: |
:$$b(f) = - {\rm arc} \hspace{0.1cm}H(f) = \arctan \frac{\sin(2 \pi f | :$$b(f) = - {\rm arc} \hspace{0.1cm}H(f) = \arctan \frac{\sin(2 \pi f | ||
T)}{1 + \cos(2 \pi f T)} = \arctan \big[\tan(\pi f T)\big].$$ | T)}{1 + \cos(2 \pi f T)} = \arctan \big[\tan(\pi f T)\big].$$ | ||
| − | The following trigonometric transformation was used here: | + | *In the frequency region $|f| < 1/(2T)$, $b(f)$ increases linearly: $b(f) = \pi \cdot f \cdot T.$ |
| + | *Also in further sections of the phase function, the phase always increases linearly from $-\pi/2$ to $+\pi/2$ . | ||
| + | *The following trigonometric transformation was used here: | ||
:$$ \frac{\sin(2 \alpha)}{1 + \cos(2 \alpha)} = \tan(\alpha).$$ | :$$ \frac{\sin(2 \alpha)}{1 + \cos(2 \alpha)} = \tan(\alpha).$$ | ||
| − | |||
| − | |||
| − | |||
| − | + | In the questions, $y_i(t)$ denotes the signal at the output of the two-way channel if the signal $x_i(t)$ is applied to the input $(i = 1, 2, 3, 4)$. | |
| − | In the questions, $y_i(t)$ denotes the signal at the output of the two-way channel if the signal $x_i(t)$ is applied to the input $( i = 1, 2, 3, 4)$. | ||
These signals are examined as input signals: | These signals are examined as input signals: | ||
| − | *a rectangular pulse $x_1(t)$ with height $1$ between $t= 0$ and $t= T$; | + | *a rectangular pulse $x_1(t)$ with height $1$ between $t= 0$ and $t= T$; $x_1(t) = 0$ holds for $t < 0$ and $t > T$ $($the value $0.5$ occurs at the two jump discontinuities$)$; |
*a rectangular pulse $x_2(t)$ with height $1$ in the range of $0 \ \text{...} \ 2T$; | *a rectangular pulse $x_2(t)$ with height $1$ in the range of $0 \ \text{...} \ 2T$; | ||
| − | *a periodic rectangular | + | *a periodic rectangular signal $x_3(t)$ with period duration $T = T_0$: |
:$$x_3(t) = \left\{ \begin{array}{c} 1 \\ | :$$x_3(t) = \left\{ \begin{array}{c} 1 \\ | ||
0 \\ \end{array} \right.\quad \quad | 0 \\ \end{array} \right.\quad \quad | ||
| Line 43: | Line 40: | ||
{ T/2 < t < T,} \\ | { T/2 < t < T,} \\ | ||
\end{array}$$ | \end{array}$$ | ||
| − | *a periodic rectangular | + | *a periodic rectangular signal $x_4(t)$ with period duration $T = 2T_0$: |
:$$x_4(t) = \left\{ \begin{array}{c} 1 \\ | :$$x_4(t) = \left\{ \begin{array}{c} 1 \\ | ||
0 \\ \end{array} \right.\quad \quad | 0 \\ \end{array} \right.\quad \quad | ||
| Line 55: | Line 52: | ||
| − | + | Please note: | |
| − | + | *The exercise belongs to the chapter [[Linear_and_Time_Invariant_Systems/Linear_Distortions|Linear Distortions]]. | |
| − | + | *For subtasks '''(1)''' to '''(4)''', $T_1 = 0$ and $T_2 = T = 4 \ \rm ms$ hold. | |
| − | + | *In subtask '''(5)''', the case $T_1 = 1 \ \rm ms$ and $T_2 = 5 \ \rm ms$ is considered. | |
| − | |||
| − | *The | ||
| − | *For subtasks '''(1)''' to '''(4)''', $T_1 = 0$ and $T_2 = T = 4 \ \rm ms$ hold. | ||
| − | *In subtask '''(5)''', the case $T_1 = 1 \ \rm ms$ and $T_2 = 5 \ \rm ms$ is considered. | ||
| Line 70: | Line 63: | ||
<quiz display=simple> | <quiz display=simple> | ||
| − | {Compute the output signal $y_1(t)$ for the input signal $x_1(t)$. Which of the statements are true? | + | {Compute the output signal $y_1(t)$ for the input signal $x_1(t)$. Which of the statements are true? |
|type="[]"} | |type="[]"} | ||
| − | + $y_1(t)$ is rectangular like $x_1(t)$ | + | + $y_1(t)$ is rectangular like $x_1(t)$. |
- $y_1(t)$ is triangular in shape. | - $y_1(t)$ is triangular in shape. | ||
+ The absolute pulse duration is $2T$. | + The absolute pulse duration is $2T$. | ||
| − | + $y_1(t)$ exhibits attenuation distortions with respect to $x_1(t)$ | + | + $y_1(t)$ exhibits attenuation distortions with respect to $x_1(t)$. |
| − | + $y_1(t)$ exhibits phase distortions with respect to $x_1(t)$ | + | + $y_1(t)$ exhibits phase distortions with respect to $x_1(t)$. |
| − | {Compute the signal $y_2(t)$. What values are obtained at times $t= 0.5 T$, $t= 1.5 T$ and $t= 2.5 T$? | + | {Compute the signal $y_2(t)$. What values are obtained at times $t= 0.5 T$, $t= 1.5 T$ and $t= 2.5 T$? |
|type="{}"} | |type="{}"} | ||
$y_2(t = 0.5T) \ = \ $ { 1 3% } | $y_2(t = 0.5T) \ = \ $ { 1 3% } | ||
| Line 86: | Line 79: | ||
| − | {Compute the signal $y_3(t)$. Check which statements are true. | + | {Compute the signal $y_3(t)$. Check which statements are true. |
|type="[]"} | |type="[]"} | ||
| − | + $y_3(t)$ is undistorted with respect to $x_3(t)$ | + | + $y_3(t)$ is undistorted with respect to $x_3(t)$. |
| − | - $y_3(t)$ exhibits attenuation distortions with respect to $x_3(t)$ | + | - $y_3(t)$ exhibits attenuation distortions with respect to $x_3(t)$. |
| − | - $y_3(t)$ exhibits phase distortions with respect to $x_3(t)$ | + | - $y_3(t)$ exhibits phase distortions with respect to $x_3(t)$. |
| − | {Which statements are true for the output signal $y_4(t)$ | + | {Which statements are true for the output signal $y_4(t)$? |
|type="[]"} | |type="[]"} | ||
| − | - $y_4(t)$ is undistorted with respect to $x_4(t)$ | + | - $y_4(t)$ is undistorted with respect to $x_4(t)$. |
| − | + $y_4(t)$ exhibits attenuation distortions with respect to $x_4(t)$ | + | + $y_4(t)$ exhibits attenuation distortions with respect to $x_4(t)$. |
| − | - $y_4(t)$ exhibits phase distortions with respect to $x_4(t)$ | + | - $y_4(t)$ exhibits phase distortions with respect to $x_4(t)$. |
| − | {Now $T_1 = 1 \ \rm ms$ and $T_2 = 5 \ \rm ms$ hold. What are the changes compared to the previous results? | + | {Now $T_1 = 1 \ \rm ms$ and $T_2 = 5 \ \rm ms$ hold. What are the changes compared to the previous results? |
|type="[]"} | |type="[]"} | ||
+ The above statements regarding distortions are still valid. | + The above statements regarding distortions are still valid. | ||
| Line 118: | Line 111: | ||
*Thus, $y_1(t)$ is a rectangular pulse of height $1$ and width $2T$. | *Thus, $y_1(t)$ is a rectangular pulse of height $1$ and width $2T$. | ||
| − | *The same result – but in a more time-consuming way – is obtained by computing in the | + | *The same result – but in a more time-consuming way – is obtained by computing in the frequency domain: |
:$$Y_1(f) = X_1(f) \cdot H(f) = T \cdot \frac {\sin(\pi f T)}{\pi f T}\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} \cdot | :$$Y_1(f) = X_1(f) \cdot H(f) = T \cdot \frac {\sin(\pi f T)}{\pi f T}\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} \cdot | ||
\big[ 1 + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \big].$$ | \big[ 1 + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \big].$$ | ||
| Line 127: | Line 120: | ||
\cdot \big[ {\rm e}^{{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} \big] = {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \cdot 2 \cos(\pi f T) .$$ | \cdot \big[ {\rm e}^{{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} \big] = {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \cdot 2 \cos(\pi f T) .$$ | ||
| − | *Hence, the following can be formulated for the output spectrum: | + | *Hence, the following can be formulated for the output spectrum: |
:$$Y_1(f) = Y_{11}(f) \cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} | :$$Y_1(f) = Y_{11}(f) \cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} | ||
, \; \; {\rm mit } \; \; | , \; \; {\rm mit } \; \; | ||
| Line 133: | Line 126: | ||
f T} = 2T \cdot \frac {\sin(2\pi f T) }{2\pi f T}.$$ | f T} = 2T \cdot \frac {\sin(2\pi f T) }{2\pi f T}.$$ | ||
| − | [[File:P_ID925__LZI_A_2_7_a.png|right|frame| | + | [[File:P_ID925__LZI_A_2_7_a.png|right|frame|Solutions '''(1)''' and '''(2)''']] |
| − | Here, the relation $\sin(\alpha) \cdot \cos(\alpha) = \sin(2\alpha)/2$ is used. | + | Here, the relation $\sin(\alpha) \cdot \cos(\alpha) = \sin(2\alpha)/2$ is used. |
| − | *The inverse Fourier transform of $Y_{11}$ results in a rectangle of width $2T$, that is symmetric about $t = 0$ . | + | *The inverse Fourier transform of $Y_{11}$ results in a rectangle of width $2T$, that is symmetric about $t = 0$ . |
*Due to the phase function this is shifted into the range $0$ ... $2T$ and the result of the time domain computation is confirmed. | *Due to the phase function this is shifted into the range $0$ ... $2T$ and the result of the time domain computation is confirmed. | ||
| − | Despite the fact that $y_1(t)$ is rectangular just as $x_1(t)$, there are distortions: | + | Despite the fact that $y_1(t)$ is rectangular just as $x_1(t)$, there are distortions: |
| − | *These are linear because of $T_y > T_x$ | + | *These are linear because of $T_y > T_x$. In the frequency range of interest $($that is all frequencies for a sinc–shaped spectrum$)$, $|H(f)|$ is not constant. So, there are attenuation distortions. |
| − | *In addition, there are also phase distortions since the phase does not increase linearly with $f$ in the whole range ⇒ The <u>proposed solutions 1, 3, 4 and 5</u> are correct. | + | *In addition, there are also phase distortions since the phase does not increase linearly with $f$ in the whole range ⇒ The <u>proposed solutions 1, 3, 4 and 5</u> are correct. |
| Line 158: | Line 151: | ||
'''(3)''' The period $T_0 = T$ of the periodic signal $x_3(t)$ is exactly as large as the delay on the second path. Therefore, $y_3(t) = 2 \cdot x_3(t) $ holds and no distortions are observable. | '''(3)''' The period $T_0 = T$ of the periodic signal $x_3(t)$ is exactly as large as the delay on the second path. Therefore, $y_3(t) = 2 \cdot x_3(t) $ holds and no distortions are observable. | ||
| − | The | + | The frequency domain computation leads to the same result. |
*$X_3(f)$ is a line spectrum with components at frequencies $f = 0$, $f = \pm f_0 = \pm 1/T$, $f = \pm 3f_0$, etc.. | *$X_3(f)$ is a line spectrum with components at frequencies $f = 0$, $f = \pm f_0 = \pm 1/T$, $f = \pm 3f_0$, etc.. | ||
| − | *However, the following holds at these discrete frequencies: | + | *However, the following holds at these discrete frequencies: |
:$$|H(f)| = 2, \hspace{0.3cm} b(f) = 0 \hspace{0.3cm} \Rightarrow | :$$|H(f)| = 2, \hspace{0.3cm} b(f) = 0 \hspace{0.3cm} \Rightarrow | ||
\hspace{0.3cm}\tau_{\rm P}(f) = 0.$$ | \hspace{0.3cm}\tau_{\rm P}(f) = 0.$$ | ||
*From this it follows again that $y_3(t) = 2 \cdot x_3(t) $. | *From this it follows again that $y_3(t) = 2 \cdot x_3(t) $. | ||
| − | *Thus, only <u>proposed solution 1</u> is correct. | + | *Thus, only <u>proposed solution 1</u> is correct. |
| − | '''(4)''' It can be seen that $y_4(t) = 1$ is distorted with respect to $x_4(t)$ from the lower sketch of the second graph. These are attenuation distortions ⇒ <u>Proposed solution 2</u> is correct as the following consideration shows. | + | '''(4)''' It can be seen that $y_4(t) = 1$ is distorted with respect to $x_4(t)$ from the lower sketch of the second graph. These are attenuation distortions ⇒ <u>Proposed solution 2</u> is correct as the following consideration shows. |
| − | *Due to $T_0 = 2T$ the signal $x_4(t)$ has the | + | *Due to $T_0 = 2T$ the signal $x_4(t)$ has the basic frequency $f_0 = 1/(2T)$. |
| − | *The frequency response thus has zeros for all odd multiples of $f_0$ | + | *The frequency response thus has zeros for all odd multiples of $f_0$. |
| − | *The only remaining spectral line of $Y_4(f)$ is at $f = 0$ where the following holds: | + | *The only remaining spectral line of $Y_4(f)$ is at $f = 0$ where the following holds: |
:$$Y_4(f) = 2 \cdot 0.5 \cdot \delta (f) = 1 \cdot \delta (f) | :$$Y_4(f) = 2 \cdot 0.5 \cdot \delta (f) = 1 \cdot \delta (f) | ||
\hspace{0.5cm}\Rightarrow \hspace{0.5cm} y_4(t) = 1.$$ | \hspace{0.5cm}\Rightarrow \hspace{0.5cm} y_4(t) = 1.$$ | ||
| Line 187: | Line 180: | ||
*The expression in parentheses describes the frequency response already considered so far. | *The expression in parentheses describes the frequency response already considered so far. | ||
| − | *The second term causes an additional runtime around $ \tau = T_1$ and the following holds for all signals $(i = 1, 2, 3, 4)$: | + | *The second term causes an additional runtime around $ \tau = T_1$ and the following holds for all signals $(i = 1, 2, 3, 4)$: |
:$$y_i^{\rm (5)}(t) = y_i(t-T_1).$$ | :$$y_i^{\rm (5)}(t) = y_i(t-T_1).$$ | ||
| − | All statements regarding the distortions are still valid. This corresponds to <u>proposed solution 1</u>. | + | All statements regarding the distortions are still valid. This corresponds to <u>proposed solution 1</u>. |
{{ML-Fuß}} | {{ML-Fuß}} | ||
Latest revision as of 16:11, 6 October 2021
Magnitude frequency response and phase function of the two-way channel
As in Exercise 2.6, a two-way channel is considered whose impulse response is:
- $$h(t) = \delta ( t - T_1) + \delta ( t - T_2).$$
In contrast to the general representation in exercise 2.6, the two attenuation factors are equal here: $z_1 = z_2 = 1$. For mobile communications, this corresponds for example to an echo at a distance of $T_2 - T_1$ and of the same strength as the signal on the main path. For this, the transit time $T_1$ is assumed.
Using the transit times $T_1 = 0$ and $T_2 = T = 4 \ \rm ms$ considered in the subtasks (1) ... (4), the following is obtained for the frequency response of the two-way channel, whose magnitude is depicted in the upper graph:
- $$H(f) = 1 + {\rm e}^{-{\rm j}\hspace{0.04cm}2 \pi f T} = 1 + \cos(2 \pi f T) - {\rm j} \cdot \sin(2 \pi f T)$$
- $$\Rightarrow \hspace{0.4cm}|H(f)| = \sqrt{2\left(1 + \cos(2 \pi f T)\right)}= 2 \cdot |\cos(\pi f T)|.$$
The bottom graph shows the phase function for $T_1 = 0$ and $T_2 = T = 4 \ \rm ms$:
- $$b(f) = - {\rm arc} \hspace{0.1cm}H(f) = \arctan \frac{\sin(2 \pi f T)}{1 + \cos(2 \pi f T)} = \arctan \big[\tan(\pi f T)\big].$$
- In the frequency region $|f| < 1/(2T)$, $b(f)$ increases linearly: $b(f) = \pi \cdot f \cdot T.$
- Also in further sections of the phase function, the phase always increases linearly from $-\pi/2$ to $+\pi/2$ .
- The following trigonometric transformation was used here:
- $$ \frac{\sin(2 \alpha)}{1 + \cos(2 \alpha)} = \tan(\alpha).$$
In the questions, $y_i(t)$ denotes the signal at the output of the two-way channel if the signal $x_i(t)$ is applied to the input $(i = 1, 2, 3, 4)$.
These signals are examined as input signals:
- a rectangular pulse $x_1(t)$ with height $1$ between $t= 0$ and $t= T$; $x_1(t) = 0$ holds for $t < 0$ and $t > T$ $($the value $0.5$ occurs at the two jump discontinuities$)$;
- a rectangular pulse $x_2(t)$ with height $1$ in the range of $0 \ \text{...} \ 2T$;
- a periodic rectangular signal $x_3(t)$ with period duration $T = T_0$:
- $$x_3(t) = \left\{ \begin{array}{c} 1 \\ 0 \\ \end{array} \right.\quad \quad \begin{array}{c} {\rm{for}} \\ {\rm{for}} \\ \end{array}\begin{array}{*{20}c} { 0 < t < T/2,} \\ { T/2 < t < T,} \\ \end{array}$$
- a periodic rectangular signal $x_4(t)$ with period duration $T = 2T_0$:
- $$x_4(t) = \left\{ \begin{array}{c} 1 \\ 0 \\ \end{array} \right.\quad \quad \begin{array}{c} {\rm{for}} \\ {\rm{for}} \\ \end{array}\begin{array}{*{20}c} { 0 < t < T,} \\ { T < t < 2T.} \\ \end{array}$$
Please note:
- The exercise belongs to the chapter Linear Distortions.
- For subtasks (1) to (4), $T_1 = 0$ and $T_2 = T = 4 \ \rm ms$ hold.
- In subtask (5), the case $T_1 = 1 \ \rm ms$ and $T_2 = 5 \ \rm ms$ is considered.
Questions
Solution
(1) The solution in the time domain leads faster to the final result:
- $$y_1(t) = x_1(t) \star h(t) = x_1(t) \star \delta (t) + x_1(t) \star \delta (t - T) = x_1(t) + x_1(t-T).$$
- Thus, $y_1(t)$ is a rectangular pulse of height $1$ and width $2T$.
- The same result – but in a more time-consuming way – is obtained by computing in the frequency domain:
- $$Y_1(f) = X_1(f) \cdot H(f) = T \cdot \frac {\sin(\pi f T)}{\pi f T}\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} \cdot \big[ 1 + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \big].$$
- The complex exponential functions can be converted using the Euler theorem as follows:
- $${\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} \big[ 1 + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \big] = {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \cdot \big[ {\rm e}^{{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} \big] = {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \cdot 2 \cos(\pi f T) .$$
- Hence, the following can be formulated for the output spectrum:
- $$Y_1(f) = Y_{11}(f) \cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} , \; \; {\rm mit } \; \; Y_{11}(f) = 2T \cdot \frac {\sin(\pi f T) \cdot \cos(\pi f T)}{\pi f T} = 2T \cdot \frac {\sin(2\pi f T) }{2\pi f T}.$$
Solutions (1) and (2)
Here, the relation $\sin(\alpha) \cdot \cos(\alpha) = \sin(2\alpha)/2$ is used.
- The inverse Fourier transform of $Y_{11}$ results in a rectangle of width $2T$, that is symmetric about $t = 0$ .
- Due to the phase function this is shifted into the range $0$ ... $2T$ and the result of the time domain computation is confirmed.
Despite the fact that $y_1(t)$ is rectangular just as $x_1(t)$, there are distortions:
- These are linear because of $T_y > T_x$. In the frequency range of interest $($that is all frequencies for a sinc–shaped spectrum$)$, $|H(f)|$ is not constant. So, there are attenuation distortions.
- In addition, there are also phase distortions since the phase does not increase linearly with $f$ in the whole range ⇒ The proposed solutions 1, 3, 4 and 5 are correct.
(2) Due to the equation
- $$y_2(t) = x_2(t) + x_2(t-T)$$
already specified in (1) a step-like curve shape corresponding to the lower diagram of the upper graph is obtained.
The numerical values that are looked for are: $y_2(t = 0.5 T) \hspace{0.15cm}\underline{= 1}, \hspace{0.3cm} y_2(t = 1.5 T) \hspace{0.15cm}\underline{= 2}, \hspace{0.3cm}y_2(t = 2.5 T) \hspace{0.15cm}\underline{ = 1}.$
Solutions (3) and (4)
(3) The period $T_0 = T$ of the periodic signal $x_3(t)$ is exactly as large as the delay on the second path. Therefore, $y_3(t) = 2 \cdot x_3(t) $ holds and no distortions are observable.
The frequency domain computation leads to the same result.
- $X_3(f)$ is a line spectrum with components at frequencies $f = 0$, $f = \pm f_0 = \pm 1/T$, $f = \pm 3f_0$, etc..
- However, the following holds at these discrete frequencies:
- $$|H(f)| = 2, \hspace{0.3cm} b(f) = 0 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\tau_{\rm P}(f) = 0.$$
- From this it follows again that $y_3(t) = 2 \cdot x_3(t) $.
- Thus, only proposed solution 1 is correct.
(4) It can be seen that $y_4(t) = 1$ is distorted with respect to $x_4(t)$ from the lower sketch of the second graph. These are attenuation distortions ⇒ Proposed solution 2 is correct as the following consideration shows.
- Due to $T_0 = 2T$ the signal $x_4(t)$ has the basic frequency $f_0 = 1/(2T)$.
- The frequency response thus has zeros for all odd multiples of $f_0$.
- The only remaining spectral line of $Y_4(f)$ is at $f = 0$ where the following holds:
- $$Y_4(f) = 2 \cdot 0.5 \cdot \delta (f) = 1 \cdot \delta (f) \hspace{0.5cm}\Rightarrow \hspace{0.5cm} y_4(t) = 1.$$
(5) Now the frequency response is with $T_1 = 1 \ \rm ms$, $T_2 = 5 \ \rm ms$ and $T = T_2 -T_1 = 4 \ \rm ms$:
- $$H(f) = {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T_1}+ {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T_2}= \big[ 1 + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \big]\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T_1}.$$
- The expression in parentheses describes the frequency response already considered so far.
- The second term causes an additional runtime around $ \tau = T_1$ and the following holds for all signals $(i = 1, 2, 3, 4)$:
- $$y_i^{\rm (5)}(t) = y_i(t-T_1).$$
All statements regarding the distortions are still valid. This corresponds to proposed solution 1.
