Difference between revisions of "Aufgaben:Exercise 4.2: Rectangular Spectra"
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− | {{quiz-Header|Buchseite= | + | {{quiz-Header|Buchseite=Signal_Representation/Differences_and_Similarities_of_Low-Pass_and_Band-Pass_Signals |
}} | }} | ||
− | [[File:P_ID695__Sig_A_4_2_neu.png|250px|right|frame| | + | [[File:P_ID695__Sig_A_4_2_neu.png|250px|right|frame|Given low–pass and band-pass spectra]] |
− | + | We consider two signals $u(t)$ and $w(t)$ with rectangular spectra $U(f)$ and $W(f)$ respectively. | |
− | * | + | *It is obvious that |
:$$u(t) = u_0 \cdot {\rm si} ( \pi \cdot {t}/{T_{ u}})$$ | :$$u(t) = u_0 \cdot {\rm si} ( \pi \cdot {t}/{T_{ u}})$$ | ||
− | : | + | :is a low-pass signal whose two parameters $u_0$ and $T_u$ are to be determined in subtask '''(1)''' . |
− | * | + | *In contrast, the spectrum $W(f)$ shows that $w(t)$ describes a band-pass signal. |
− | + | This task also refers to the band-pass signal | |
:$$d(t) = 10 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 5 \pi f_2 \hspace{0.05cm}t) | :$$d(t) = 10 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 5 \pi f_2 \hspace{0.05cm}t) | ||
- 6 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 3 \pi f_2\hspace{0.05cm} t)$$ | - 6 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 3 \pi f_2\hspace{0.05cm} t)$$ | ||
− | + | whose spectrum was determined in [[Aufgaben:Exercise_4.1Z:_High-Pass_System|Exercise 4.1Z]] . Let $f_2 = 2 \ \rm kHz.$ | |
− | '' | + | |
− | * | + | |
− | * | + | |
− | * | + | ''Hints:'' |
+ | *This exercise belongs to the chapter [[Signal_Representation/Differences_and_Similarities_of_Low-Pass_and_Band-Pass_Signals|Differences and Similarities of Low-Pass and Band-Pass Signals]]. | ||
+ | *In this task, the function $\rm si(x) = \rm sin(x)/x = \rm sinc(x/π)$ is used. | ||
+ | |||
+ | *Consider the following trigonometric relationship in the solution: | ||
:$$\sin (\alpha) \cdot \cos (\beta) = {1}/{2} \cdot \big[ \sin | :$$\sin (\alpha) \cdot \cos (\beta) = {1}/{2} \cdot \big[ \sin | ||
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− | === | + | ===Questions=== |
<quiz display=simple> | <quiz display=simple> | ||
− | { | + | {What are the parameter values $u_0$ and $T_u$ of the low-pass signal? |
|type="{}"} | |type="{}"} | ||
$u_0\ = \ $ { 2 3% } $\text{V}$ | $u_0\ = \ $ { 2 3% } $\text{V}$ | ||
$T_u\ = \ $ { 0.5 3% } $\text{ms}$ | $T_u\ = \ $ { 0.5 3% } $\text{ms}$ | ||
− | { | + | {Calculate the band-pass signal $w(t)$. What are the signal values at $t = 0$ and $t = 62.5 \, {\rm µ}\text{s}$? |
|type="{}"} | |type="{}"} | ||
$w(t=0)\ = \ $ { 4 3% } $\text{V}$ | $w(t=0)\ = \ $ { 4 3% } $\text{V}$ | ||
− | $w(t=62.5 \,\ | + | $w(t=62.5 \,{\rm µ} \text{s})\ = \ $ { 0. } $\text{V}$ |
− | { | + | {Which statements are true regarding the band-pass signals $d(t)$ and $w(t)$ ? Justify your result in the time domain. |
− | |type=" | + | |type="()"} |
− | + | + | + The signals $d(t)$ and $w(t)$ are identical. |
− | - $d(t)$ | + | - $d(t)$ and $w(t)$ differ by a constant factor. |
− | - $d(t)$ und $w(t)$ | + | - $d(t)$ und $w(t)$ have different shapes. |
</quiz> | </quiz> | ||
− | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''1 | + | '''(1)''' The time $T_u$ ⇒ first zero of the low-pass signal $u(t)$ – is equal to the reciprocal of the width of the rectangular spectrum, i.e. $1/(2\, \text{kHz} ) \hspace{0.15 cm}\underline{= 0.5 \, \text{ms}}$. |
+ | *The pulse amplitude is equal to the rectangular area as shown in the sample solution for [[Aufgaben:Aufgabe_4.1:_Tiefpass-_und_Bandpass-Signale|Exercise 4.1]] . From this follows $u_0\hspace{0.15 cm}\underline{= 2 \, \text{V}}$. | ||
+ | |||
+ | |||
− | [[File:P_ID704__Sig_A_4_2_b_neu.png|250px|right| | + | [[File:P_ID704__Sig_A_4_2_b_neu.png|250px|right|frame|Multiplication with a cosine function]] |
− | '''2 | + | '''(2)''' The band-pass spectrum can be represented with $f_{\rm T} = 4\, \text{kHz}$ as follows: |
− | $$ W(f) = U(f- f_{\rm T}) + U(f+ f_{\rm T}) = U(f)\star \left[ | + | :$$ W(f) = U(f- f_{\rm T}) + U(f+ f_{\rm T}) = U(f)\star \left[ |
\delta(f- f_{\rm T})+ \delta(f+ f_{\rm T})\right].$$ | \delta(f- f_{\rm T})+ \delta(f+ f_{\rm T})\right].$$ | ||
− | + | According to the [[Signal_Representation/Fourier_Transform_Theorems#Shifting_Theorem|Shifting Theorem]], the following then applies to the associated time signal: | |
− | $$w(t) = 2 \cdot u(t) \cdot {\cos} ( 2 \pi f_{\rm T} t) = 2 u_0 | + | :$$w(t) = 2 \cdot u(t) \cdot {\cos} ( 2 \pi f_{\rm T} t) = 2 u_0 |
− | \cdot {\rm si} ( \pi | + | \cdot {\rm si} ( \pi {t}/{T_{\rm u}})\cdot {\cos} ( 2 \pi f_{\rm T} t). $$ |
− | + | The graph shows | |
− | * | + | *above the low–pass signal $u(t)$, |
− | * | + | *then the oscillation $c(t) = 2 · \cos(2 \pi f_{\rm T}t$ ), |
− | * | + | *below the band-pass signal $w(t) = u(t) \cdot c(t)$. |
− | + | ||
+ | In particular, at time $t = 0$ one obtains: | ||
− | $$w(t = 0) = 2 \cdot u_0 \hspace{0.15 cm}\underline{= 4 \hspace{0.05cm}{\rm V}}.$$ | + | :$$w(t = 0) = 2 \cdot u_0 \hspace{0.15 cm}\underline{= 4 \hspace{0.05cm}{\rm V}}.$$ |
− | + | The time $t=62.5 \,{\rm µ} \text{s}$ corresponds exactly to a quarter of the period of the signal $c(t)$: | |
− | $$ w(t = 62.5 \hspace{0.05cm}{\rm | + | :$$ w(t = 62.5 \hspace{0.05cm}{\rm µ s}) = 2 u_0 \cdot {\rm si} ( \pi \cdot \frac{62.5 \hspace{0.05cm}{\rm µ s}} |
− | {500 \hspace{0.05cm}{\rm | + | {500 \hspace{0.05cm}{\rm µ s}}) |
\cdot {\cos} ( 2 \pi \cdot 4\hspace{0.05cm}{\rm kHz}\cdot | \cdot {\cos} ( 2 \pi \cdot 4\hspace{0.05cm}{\rm kHz}\cdot | ||
− | 62.5 \hspace{0.05cm}{\rm | + | 62.5 \hspace{0.05cm}{\rm µ s}) $$ |
+ | :$$ \Rightarrow \hspace{0.3cm}w(t = | ||
4\hspace{0.05cm}{\rm V}\cdot{\rm si} ( {\pi}/{8}) \cdot \cos ( {\pi}/{4})\hspace{0.15 cm}\underline{ = 0}.$$ | 4\hspace{0.05cm}{\rm V}\cdot{\rm si} ( {\pi}/{8}) \cdot \cos ( {\pi}/{4})\hspace{0.15 cm}\underline{ = 0}.$$ | ||
− | '''3 | + | |
+ | |||
+ | '''(3)''' Proposed <u>solution 1 is correct</u>: | ||
+ | *If we compare the spectral function $W(f)$ of this task with the spectrum $D(f)$ in the sample solution to [[Aufgaben:Exercise_4.1:_Low-Pass_and_Band-Pass_Signals|Exercise 4.1]], we see that $w(t)$ and $d(t)$ are identical. | ||
+ | *This proof is somewhat more complex in the time domain. With $f_2 = 2 \,\text{kHz}$ can be written for the signal considered here: | ||
− | $$w(t ) = 4\hspace{0.05cm}{\rm V} | + | :$$w(t ) = 4\hspace{0.05cm}{\rm V} |
\cdot {\rm si} ( \pi f_2 t) | \cdot {\rm si} ( \pi f_2 t) | ||
\cdot {\cos} ( 4 \pi f_2 t) = | \cdot {\cos} ( 4 \pi f_2 t) = | ||
({4\hspace{0.05cm}{\rm V}})/({\pi f_2 t})\cdot \sin (\pi f_2 t) \cdot \cos ( 4 \pi f_2 t) .$$ | ({4\hspace{0.05cm}{\rm V}})/({\pi f_2 t})\cdot \sin (\pi f_2 t) \cdot \cos ( 4 \pi f_2 t) .$$ | ||
− | + | *Because of the trigonometric relationship | |
− | $$\sin (\alpha) \cdot \cos (\beta) = {1}/{2} \cdot \ | + | :$$\sin (\alpha) \cdot \cos (\beta) = {1}/{2} \cdot \big[ \sin |
− | (\alpha + \beta)+ \sin (\alpha - \beta)\ | + | (\alpha + \beta)+ \sin (\alpha - \beta)\big]$$ |
− | + | :the above equation can be transformed: | |
− | $$w(t ) = | + | :$$w(t ) = |
− | \frac{2\hspace{0.05cm}{\rm V}}{\pi f_2 t}\cdot \ | + | \frac{2\hspace{0.05cm}{\rm V}}{\pi f_2 t}\cdot \big [\sin (5\pi f_2 t) + \sin (-3\pi f_2 t)\big ] |
= 10\hspace{0.05cm}{\rm V} \cdot \frac{\sin (5\pi f_2 t)}{5\pi f_2 t}- | = 10\hspace{0.05cm}{\rm V} \cdot \frac{\sin (5\pi f_2 t)}{5\pi f_2 t}- | ||
6\hspace{0.05cm}{\rm V} \cdot \frac{\sin (3\pi f_2 t)}{3\pi f_2 t}.$$ | 6\hspace{0.05cm}{\rm V} \cdot \frac{\sin (3\pi f_2 t)}{3\pi f_2 t}.$$ | ||
− | + | *This shows that both signals are actually identical ⇒ Proposed solution 1: | |
− | $$w(t) = 10 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 5 \pi f_2 t) | + | :$$w(t) = 10 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 5 \pi f_2 t) |
- 6 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 3 \pi f_2 t) = d(t).$$ | - 6 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 3 \pi f_2 t) = d(t).$$ | ||
{{ML-Fuß}} | {{ML-Fuß}} | ||
__NOEDITSECTION__ | __NOEDITSECTION__ | ||
− | [[Category: | + | [[Category:Signal Representation: Exercises|^4.1 Differences between Low-Pass and Band-Pass^]] |
Latest revision as of 14:49, 5 May 2021
We consider two signals $u(t)$ and $w(t)$ with rectangular spectra $U(f)$ and $W(f)$ respectively.
- It is obvious that
- $$u(t) = u_0 \cdot {\rm si} ( \pi \cdot {t}/{T_{ u}})$$
- is a low-pass signal whose two parameters $u_0$ and $T_u$ are to be determined in subtask (1) .
- In contrast, the spectrum $W(f)$ shows that $w(t)$ describes a band-pass signal.
This task also refers to the band-pass signal
- $$d(t) = 10 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 5 \pi f_2 \hspace{0.05cm}t) - 6 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 3 \pi f_2\hspace{0.05cm} t)$$
whose spectrum was determined in Exercise 4.1Z . Let $f_2 = 2 \ \rm kHz.$
Hints:
- This exercise belongs to the chapter Differences and Similarities of Low-Pass and Band-Pass Signals.
- In this task, the function $\rm si(x) = \rm sin(x)/x = \rm sinc(x/π)$ is used.
- Consider the following trigonometric relationship in the solution:
- $$\sin (\alpha) \cdot \cos (\beta) = {1}/{2} \cdot \big[ \sin (\alpha + \beta)+ \sin (\alpha - \beta)\big].$$
Questions
Solution
(1) The time $T_u$ ⇒ first zero of the low-pass signal $u(t)$ – is equal to the reciprocal of the width of the rectangular spectrum, i.e. $1/(2\, \text{kHz} ) \hspace{0.15 cm}\underline{= 0.5 \, \text{ms}}$.
- The pulse amplitude is equal to the rectangular area as shown in the sample solution for Exercise 4.1 . From this follows $u_0\hspace{0.15 cm}\underline{= 2 \, \text{V}}$.
(2) The band-pass spectrum can be represented with $f_{\rm T} = 4\, \text{kHz}$ as follows:
- $$ W(f) = U(f- f_{\rm T}) + U(f+ f_{\rm T}) = U(f)\star \left[ \delta(f- f_{\rm T})+ \delta(f+ f_{\rm T})\right].$$
According to the Shifting Theorem, the following then applies to the associated time signal:
- $$w(t) = 2 \cdot u(t) \cdot {\cos} ( 2 \pi f_{\rm T} t) = 2 u_0 \cdot {\rm si} ( \pi {t}/{T_{\rm u}})\cdot {\cos} ( 2 \pi f_{\rm T} t). $$
The graph shows
- above the low–pass signal $u(t)$,
- then the oscillation $c(t) = 2 · \cos(2 \pi f_{\rm T}t$ ),
- below the band-pass signal $w(t) = u(t) \cdot c(t)$.
In particular, at time $t = 0$ one obtains:
- $$w(t = 0) = 2 \cdot u_0 \hspace{0.15 cm}\underline{= 4 \hspace{0.05cm}{\rm V}}.$$
The time $t=62.5 \,{\rm µ} \text{s}$ corresponds exactly to a quarter of the period of the signal $c(t)$:
- $$ w(t = 62.5 \hspace{0.05cm}{\rm µ s}) = 2 u_0 \cdot {\rm si} ( \pi \cdot \frac{62.5 \hspace{0.05cm}{\rm µ s}} {500 \hspace{0.05cm}{\rm µ s}}) \cdot {\cos} ( 2 \pi \cdot 4\hspace{0.05cm}{\rm kHz}\cdot 62.5 \hspace{0.05cm}{\rm µ s}) $$
- $$ \Rightarrow \hspace{0.3cm}w(t = 4\hspace{0.05cm}{\rm V}\cdot{\rm si} ( {\pi}/{8}) \cdot \cos ( {\pi}/{4})\hspace{0.15 cm}\underline{ = 0}.$$
(3) Proposed solution 1 is correct:
- If we compare the spectral function $W(f)$ of this task with the spectrum $D(f)$ in the sample solution to Exercise 4.1, we see that $w(t)$ and $d(t)$ are identical.
- This proof is somewhat more complex in the time domain. With $f_2 = 2 \,\text{kHz}$ can be written for the signal considered here:
- $$w(t ) = 4\hspace{0.05cm}{\rm V} \cdot {\rm si} ( \pi f_2 t) \cdot {\cos} ( 4 \pi f_2 t) = ({4\hspace{0.05cm}{\rm V}})/({\pi f_2 t})\cdot \sin (\pi f_2 t) \cdot \cos ( 4 \pi f_2 t) .$$
- Because of the trigonometric relationship
- $$\sin (\alpha) \cdot \cos (\beta) = {1}/{2} \cdot \big[ \sin (\alpha + \beta)+ \sin (\alpha - \beta)\big]$$
- the above equation can be transformed:
- $$w(t ) = \frac{2\hspace{0.05cm}{\rm V}}{\pi f_2 t}\cdot \big [\sin (5\pi f_2 t) + \sin (-3\pi f_2 t)\big ] = 10\hspace{0.05cm}{\rm V} \cdot \frac{\sin (5\pi f_2 t)}{5\pi f_2 t}- 6\hspace{0.05cm}{\rm V} \cdot \frac{\sin (3\pi f_2 t)}{3\pi f_2 t}.$$
- This shows that both signals are actually identical ⇒ Proposed solution 1:
- $$w(t) = 10 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 5 \pi f_2 t) - 6 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 3 \pi f_2 t) = d(t).$$