View source for Aufgaben:Exercise 4.2Z: Multiplication with a Sine Signal


{{quiz-Header|Buchseite=Signal_Representation/Differences_and_Similarities_of_Low-Pass_and_Band-Pass_Signals
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[[File:P_ID697__Sig_Z_4_2_neu.png|right|frame|Spectral functions&nbsp; $Q(f)$&nbsp; and&nbsp; $Z(f)$]]
A periodic message signal&nbsp; $q(t)$is considered, whose spectral function&nbsp; $Q(f)$&nbsp; can be seen in the upper graph.

A multiplication with the dimensionless carrier&nbsp; $z(t)$, whose spectrum&nbsp; $Z(f)$&nbsp; is also shown, leads to the signal&nbsp; $s(t) = q(t) \cdot z(t).$

In this task, the spectral function&nbsp; $S(f)$&nbsp; of this signal is to be determined, whereby the solution can be either in the time or frequency domain.







''Hint:'' 
*This exercise belongs to the task&nbsp; [[Signal_Representation/Differences_and_Similarities_of_LP_and_BP_Signals|Differences and Similiarities of LP and BP Signals]].
 




===Questions===

<quiz display=simple>
{Give the source signal&nbsp; $q(t)$&nbsp; in analytical form. Which values result for&nbsp; $t = 0$&nbsp; und&nbsp; $t = 0.125\, \text{ms}$?
|type="{}"}
$q(t = 0)\ = \ $  { 4 3% } &nbsp;$\text{V}$
$q(t = 0.125 \,\text{ms})\ = \ $ { 0.828 3% } $\text{V}$


{What is the (dimensionless) carrier signal&nbsp; $z(t)$? What is its maximum value?
|type="{}"}
$z_{\rm max}\ = \ $ { 6 3% }


{Calculate the spectrum&nbsp; $S(f)$&nbsp; separately for real and imaginary parts. At which frequencies are there lines with a non-zero real part?
|type="[]"}
+ $3\ \text{kHz},$
- $4\ \text{kHz},$
- $5\ \text{kHz},$
- $6\ \text{kHz},$
+ $7\ \text{kHz}.$



{At which frequencies do purely imaginary spectral lines occur?
|type="[]"}
- $3\ \text{kHz},$
+ $4\ \text{kHz},$
- $5\ \text{kHz},$
+ $6\ \text{kHz},$
- $7\ \text{kHz}.$



</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''&nbsp;  The message signal can be represented with the abbreviations&nbsp; $f_1 = 1\ \text{kHz}$&nbsp; and&nbsp; $T_1 = 1/f_1 = 1 \ \text{ms}$&nbsp; as follows&nbsp; $($&nbsp; $f_2 = 2f_1 applies)$:
:$$q(t ) = 4\hspace{0.05cm}{\rm V}
 \cdot  {\cos} ( 2 \pi f_1 t) - 2\hspace{0.05cm}{\rm V}
 \cdot  {\sin} ( 4 \pi f_1 t)=
 4\hspace{0.05cm}{\rm V}
 \cdot  {\cos} ( 2 \pi {t}/{T_1}) - 2\hspace{0.05cm}{\rm V}
 \cdot  {\sin} ( 4 \pi {t}/{T_1}) .$$
*At time&nbsp; $t = 0$&nbsp;, the second component disappears and&nbsp; $q(t = 0)\; \underline{= 4 \ \text{V}}$. 
*On the other hand, for&nbsp; $t = 0.125 \ \text{ms} = T_1/8$ is obtained:
:$$q(t = 0.125{\rm ms})  =  4\hspace{0.05cm}{\rm V}
 \cdot  {\cos} ( {\pi}/{4}) - 2\hspace{0.05cm}{\rm V}
 \cdot  {\sin} ( {\pi}/{2}) = \frac
 {4\hspace{0.05cm}{\rm V}}{\sqrt{2}} - 2\hspace{0.05cm}{\rm V} \hspace{0.15 cm}\underline{=
 0.828 \hspace{0.05cm}{\rm V}}.$$


'''(2)'''&nbsp; According to the purely imaginary spectrum&nbsp; $Z(f)$&nbsp; and the impulse weights&nbsp; $\pm 3$&nbsp; must hold:
:$$z(t )  = 6 \cdot  {\sin} ( 2 \pi \cdot 5\hspace{0.05cm}{\rm
kHz})\hspace{0.5cm}\Rightarrow \hspace{0.5cm} z_{\rm max}\hspace{0.15 cm}\underline{ = 6} .$$


[[File:EN_Sig_Z_4_2_c_neu.png|right|frame|Discrete BP spectrum]]
'''(3)'''&nbsp; The spectral function&nbsp; $S(f)$&nbsp; results from the convolution between&nbsp; $Q(f)$&nbsp; and&nbsp; $Z(f)$. One obtains:
:$$S(f)  = - 3{\rm j} \cdot Q(f- f_{\rm T}) + 3{\rm j} \cdot Q(f+
f_{\rm T}).$$
This results in spectral lines at 
*$3\ \text{kHz}\ (–3\ {\rm V})$, 
*$4\ \text{kHz} (–{\rm j} \cdot 6\ {\rm V})$, 
*$6\ \text{kHz} (–{\rm j} \cdot 6\ {\rm V})$,
* $7\ \text{kHz}\ (–3\ {\rm V})$.

 
Plus the conjugate-complex components at negative frequencies.

Lines with real weights at&nbsp; $\underline{\pm 3 \ \text{kHz}}$&nbsp; <u>and</u>&nbsp; $\underline{\pm 7 \ \text{kHz}}$.



'''(4)'''&nbsp;  Imaginary lines appear at&nbsp; $\underline{\pm 4 \ \text{kHz}}$&nbsp; <u>and</u>&nbsp; $\underline{\pm 6 \ \text{kHz}}$ auf.

An alternative way to solve this problem is to use trigonometric equations.

In the following, for example,&nbsp; $f_5 = 5 \text{ kHz}$. Then it applies:
:$$4\hspace{0.05cm}{\rm V}
 \cdot  {\cos} ( 2 \pi f_1 \hspace{0.03cm}t) \cdot 3
 \cdot  {\sin} ( 2 \pi f_5 \hspace{0.03cm} t)=
 \frac{12\hspace{0.05cm}{\rm V}}{2}\cdot \big[{\sin} ( 2 \pi f_4 \hspace{0.03cm}
 t)+ {\sin} ( 2 \pi f_6 \hspace{0.03cm} t)\big],$$
:$$-2\hspace{0.05cm}{\rm V}
 \cdot  {\sin} ( 2 \pi f_2 \hspace{0.03cm}t) \cdot 3
 \cdot  {\sin} ( 2 \pi f_5 \hspace{0.03cm} t)=
 \frac{-6\hspace{0.05cm}{\rm V}}{2}\cdot \big[{\cos} ( 2 \pi f_3 \hspace{0.03cm}
 t)+ {\cos} ( 2 \pi f_7 \hspace{0.03cm} t)\big].$$

*From the first equation, the following spectral lines are obtained: 

:* at&nbsp; $+f_4$&nbsp; and&nbsp; $-f_4$&nbsp; with weights&nbsp; $–{\rm j} \cdot 3\ {\rm V}$&nbsp; bzw.&nbsp; $+{\rm j}\cdot 3 \ {\rm V}$ respectively,

:* at&nbsp; $+f_6$&nbsp; and&nbsp; $-f_6$&nbsp; with weights&nbsp; $–{\rm j} \cdot 3 \ {\rm V}$&nbsp; bzw.&nbsp; $+{\rm j} \cdot 3 \ {\rm V}$ respectively.

*The second equation gives a total of four diraclines&nbsp; (all&nbsp; $6 \ {\rm V}$, real and negative) at&nbsp; $\pm f_3$&nbsp; and&nbsp; $\pm f_7$. 


A comparison with the sketch above shows that both solutions lead to the same result.
{{ML-Fuß}}


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[[Category:Signal Representation: Exercises|^4.1 Differences between Low-Pass and Band-Pass^]]