View source for Aufgaben:Exercise 2.10Z: Noise with DSB-AM and SSB-AM


{{quiz-Header|Buchseite=Modulation_Methods/Single-Sideband_Modulation
}}

[[File:EN_Mod_Z_2_9_c.png|right|frame|Shared block diagram for <br>DSB-AM and SSB-AM]]
Now the influence of noise on the sink-to-noise power ratio &nbsp;$10 · \lg ρ_v$&nbsp; for both&nbsp; $\rm DSB–AM$&nbsp; and&nbsp; $\rm SSB–AM$ transmission&nbsp; will be compared. &nbsp; The illustration shows the underlying block diagram.


The differences between the two system variants are highlighted in red on the image, namely the modulator&nbsp; (DSB or SSB)&nbsp; as well as the dimensionless constant
:$$ K = \left\{ \begin{array}{c} 2/\alpha_{\rm K} \\ 4/\alpha_{\rm K} \\ \end{array} \right.\quad \begin{array}{*{10}c} {\rm{for}} \\ {\rm{for}} \\ \end{array}\begin{array}{*{20}c} {\rm DSB} \hspace{0.05cm}, \\ {\rm SSB} \hspace{0.05cm} \\ \end{array}$$
of the receiver-side carrier signal  &nbsp;$z_{\rm E}(t) = K · \cos(ω_{\rm T} · t)$,&nbsp; which is assumed to be frequency and phase synchronous with the carrier signal &nbsp;$z(t)$&nbsp; at the transmitter.


The system characteristics
*frequency-independent channel transmission factor&nbsp; $α_{\rm K}$,
*transmission power&nbsp; $P_{\rm S}$,
*one&ndash;sided noise power density&nbsp; $N_{\rm 0}$,
*bandwidth &nbsp;$B_{\rm NF}$&nbsp; of the source signal,   


captured by the shared performance parameter are labelled in green: 
:$$\xi = \frac{\alpha_{\rm K}^2 \cdot P_{\rm S}}{N_0 \cdot B_{\rm NF}}$$


Further note:
* The cosine signal &nbsp;$q(t)$&nbsp; with frequency &nbsp;$B_{\rm NF}$&nbsp; stands for a source signal with bandwidth &nbsp;$B_{\rm NF}$ composed of multiple frequencies.
*The relationship between the transmission power&nbsp; $P_{\rm S}$&nbsp; and the power&nbsp; $P_{q}$&nbsp; of the source signal depends,&nbsp; among other things,&nbsp; on the modulation method.
* "DSB–AM with carrier"&nbsp; is parameterized by the modulation depth &nbsp;$m = A_{\rm N}/A_{\rm T}$,&nbsp; <br>while&nbsp; "SSB-AM"&nbsp; is determined by the sideband-to-carrier ratio &nbsp;$μ = A_{\rm N}/(2 · A_{\rm T})$.
* The frequency-independent channel transmission factor &nbsp;$α_{\rm K}$&nbsp; is balanced by the constant &nbsp;$K$,&nbsp; so that in the noise-free case &nbsp;$(N_0 = 0)$,&nbsp; <br>the sink signal &nbsp;$v(t)$&nbsp; matches the source signal &nbsp;$q(t)$.
* The sink SNR can thus be given as follows&nbsp; $(T_0$ indicates the period of the source signal$)$:
:$$ \rho_{v } = \frac{P_{q}}{P_{\varepsilon }}\hspace{0.5cm}{\rm with}\hspace{0.5cm} P_{q} = \frac{1}{T_{\rm 0}}\cdot\int_{0}^{ T_{\rm 0}} {q^2(t)}\hspace{0.1cm}{\rm d}t, \hspace{0.5cm}P_{\varepsilon} = \int_{-B_{\rm NF}}^{ +B_{\rm NF}} \hspace{-0.1cm}{\it \Phi_{\varepsilon}}(f)\hspace{0.1cm}{\rm d}f\hspace{0.05cm}.$$





Hints: 
*This exercise belongs to the chapter&nbsp; [[Modulation_Methods/Single-Sideband_Modulation|"Single-Sideband Modulation"]].
*Particular reference is made to the section&nbsp; [[Modulation_Methods/Single-Sideband_Modulation#Sideband-to-carrier_ratio|"Sideband-to-carrier ratio"]].
*The results for DSB–AM can be found in the section&nbsp; [[Modulation_Methods/Synchronous_Demodulation#Sink_SNR_and_the_performance_parameter|"Sink SNR and the performance parameter"]].
 


===Questions===

<quiz display=simple>
{Which kind of demodulation is considered here?
|type="()"}
+ Synchronous demodulation.
- Envelope demodulation.

{Which relationship holds between the quantities &nbsp;$ρ_v$&nbsp; and &nbsp;$ξ$&nbsp; for&nbsp;  <u>double-sideband AM without carrier</u>&nbsp;  $(m → ∞)$?
|type="[]"}
- &nbsp;$ρ_v = 2 · ξ$.
+ &nbsp;$ρ_v = ξ$.
- &nbsp;$ρ_v = ξ/2$.

{Which relationship holds between &nbsp;$ρ_v$&nbsp; and &nbsp;$ξ$&nbsp; for&nbsp;  <u>single-sideband AM without carrier</u>&nbsp; as $(μ → ∞)$?
|type="[]"}
- &nbsp;$ρ_v = 2 · ξ$.
+ &nbsp;$ρ_v = ξ$.
- &nbsp;$ρ_v = ξ/2$.

{Let &nbsp;$ξ = 10^4$.&nbsp; Calculate the sink-to-noise ratio of &nbsp; <u>double-sideband AM</u>&nbsp; for modulation depths &nbsp;$m = 0.5$&nbsp; and &nbsp;$m = 1$.
|type="{}"}
$m = 0.5\text{:} \ \   10 · \lg \ ρ_v \ = \ $  { 30.46 3% } $\ \rm dB$ 
$m = 1.0\text{:} \ \   10 · \lg \ ρ_v \ = \ $ { 35.23 3% } $\ \rm dB$ 
 
{Further let &nbsp;$ξ = 10^4$.&nbsp; Calculate the sink-to-noise ratio of &nbsp; <u>single-sideband AM</u>&nbsp;  for the parameters &nbsp;$μ = 0.354$&nbsp; and &nbsp;$μ = 0.707$.
|type="{}"}
$μ = 0.354\text{:} \ \  \ 10 · \lg \ ρ_v \ = \ $ { 30.46 3% } $\ \rm dB$  
$μ = 0.707\text{:} \ \  \ 10 · \lg \ ρ_v \ = \ $ { 35.23 3% } $\ \rm dB$ 


</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''&nbsp; We are dealing with a &nbsp; <u>synchronous demodulator</u>.&nbsp; <u>Answer 1</u>&nbsp; is correct.



'''(2)'''&nbsp; <u>Answer 2</u>&nbsp; is correct:
*For DSB–AM without a carrier, &nbsp; $P_{\rm S} = P_q/2$.&nbsp; This is simultaneously the power of the useful component of the sink signal&nbsp; $v(t)$. 
*The power-spectral density &nbsp; ${\it Φ}_ε(f)$&nbsp; of the &nbsp; $v(t)$&nbsp; noise component results from the convolution:
[[File:P_ID1048__Mod_Z_2_9_b.png|right|frame|Noise power density <br>in double-sideband AM]]
[[File:P_ID1049__Mod_Z_2_9_c.png|right|frame|Noise power density <br>in&nbsp; upper-sideband AM]]

:$${\it \Phi}_\varepsilon(f) = {\it \Phi}_{z{\rm E} }(f) \star {\it \Phi}_n (f) = \frac{1}{\alpha_{\rm K}^2} \cdot \big[\delta(f - f_{\rm T}) + \delta(f + f_{\rm T}) \big]\star {\it \Phi}_n (f) \hspace{0.05cm}.$$
*The expression&nbsp; $\big[$ ... $\big]$&nbsp; describes the power-spectral density of a cosine signal with the signal amplitude &nbsp; $K = 2$. 
*The correction of channel attenuation is considered with&nbsp; $1/α_K^2$&nbsp;. 
*Thus,&nbsp; taking  &nbsp; ${\it \Phi}_n(f) = N_0/2$&nbsp; into account,&nbsp; we get:
:$${\it \Phi}_\varepsilon(f) = \frac{N_0}{\alpha_{\rm K}^2} 
 \hspace{0.3cm}  \Rightarrow  \hspace{0.3cm} P_\varepsilon = \int_{-B_{\rm NF}}^{+B_{\rm NF}} {{\it \Phi}_\varepsilon(f) }\hspace{0.1cm}{\rm d}f = \frac{2 \cdot N_0 \cdot B_{\rm NF}}{\alpha_{\rm K}^2}\hspace{0.05cm}.$$
*From this,&nbsp; it follows for the the signal-to-noise power ratio&nbsp; $\rm (SNR)$:
:$$\rho_{v } = \frac{P_{q}}{P_{\varepsilon }} = \frac{2 \cdot P_{\rm S}}{2 \cdot N_0 \cdot B_{\rm NF}/\alpha_{\rm K}^2} = \frac{\alpha_{\rm K}^2 \cdot P_{\rm S}}{N_0 \cdot B_{\rm NF}}\hspace{0.15cm}\underline { = \xi} \hspace{0.05cm}.$$



'''(3)'''&nbsp; <u>Answer 2</u>&nbsp; is correct:
*In contrast to DSB,&nbsp;  $P_S = P_q/4$&nbsp; holds for SSB,&nbsp; as well as
:$${\it \Phi}_\varepsilon(f) = {\it \Phi}_{z{\rm E} }(f) \star {\it \Phi}_n (f) = \frac{4}{\alpha_{\rm K}^2} \cdot \big[\delta(f - f_{\rm T}) + \delta(f + f_{\rm T}) \big]\star {\it \Phi}_n (f) \hspace{0.05cm}.$$
*Taking &nbsp; $B_{\rm HF} = B_{\rm NF}$&nbsp; into account&nbsp; (see adjacent diagram for USB modulation),&nbsp; we now get:
:$${\it \Phi}_\varepsilon(f) = \frac{2 \cdot N_0}{\alpha_{\rm K}^2} \hspace{0.3cm}  \Rightarrow \hspace{0.3cm} P_\varepsilon = \frac{4 \cdot N_0 \cdot B_{\rm NF}}{\alpha_{\rm K}^2}\hspace{0.05cm}.$$

*This means:&nbsp; When without carrier,&nbsp; single-sideband modulation demonstrates the same noise behaviour as DSB-AM.



'''(4)'''&nbsp; Assuming a cosine carrier with amplitude&nbsp; $A_{\rm T}$&nbsp; and a similarly cosine source signal&nbsp; $q(t)$,&nbsp; we get for DSB with carrier:
:$$ s(t)  =  \big (q(t) + A_{\rm T}\big ) \cdot \cos( \omega_{\rm T} \cdot t)
 =  A_{\rm T} \cdot \cos( \omega_{\rm T} \cdot t) + \frac{A_{\rm N}}{2}\cdot \cos\big(( \omega_{\rm T}+ \omega_{\rm N}) \cdot t \big)+ \frac{A_{\rm N}}{2}\cdot \cos\big(( \omega_{\rm T}- \omega_{\rm N}) \cdot t\big)\hspace{0.05cm}.$$
*The transmission power is thus given by
:$$ P_{\rm S}= \frac{A_{\rm T}^2}{2} + 2 \cdot \frac{(A_{\rm N}/2)^2}{2} = \frac{A_{\rm T}^2}{2} + \frac{A_{\rm N}^2}{4} \hspace{0.05cm}.$$
*Taking&nbsp; $P_q = A_{\rm N}^2/2$&nbsp; and&nbsp; $m = A_{\rm N}/A_{\rm T}$&nbsp; into account,&nbsp;  this can also be written as:
:$$P_{\rm S}= \frac{A_{\rm N}^2}{4} \cdot \left[ 1 + \frac{2 \cdot A_{\rm T}^2}{A_{\rm N}^2}\right] = \frac{P_q}{2} \cdot \left[ 1 + {2 }/{m^2}\right]\hspace{0.05cm}.$$
*With the noise power&nbsp; $P_ε$&nbsp; according to subtask&nbsp; '''(2)'''&nbsp; we thus obtain:
:$$\rho_{v } = \frac{P_{q}}{P_{\varepsilon }} = \frac{2 \cdot P_{\rm S}\cdot (1 + 2/m^2)}{2 \cdot N_0 \cdot B_{\rm NF}/\alpha_{\rm K}^2} = \frac{\alpha_{\rm K}^2 \cdot P_{\rm S}}{N_0 \cdot B_{\rm NF}} \cdot \frac{1}{1 +{2 }/{m^2}} \hspace{0.05cm}.$$
*And in logarithmic representation:
:$$ 10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v } = 10 \cdot {\rm lg} \hspace{0.15cm}\xi - 10 \cdot {\rm lg} \hspace{0.15cm}\left[{1 +{2 }/{m^2}}\right] \hspace{0.05cm}.$$
:$$\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v } \ (m = 0.5)  =  40 \,{\rm dB} - 10 \cdot {\rm lg} (9) \hspace{0.15cm}\underline {= 30.46\, {\rm dB}}$$ 
:$$\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v } \ (m = 1.0)  =  40 \,{\rm dB} - 10 \cdot {\rm lg} (3) \hspace{0.15cm}\underline {= 35.23\, {\rm dB} \hspace{0.05cm}}.$$



'''(5)'''&nbsp; In&nbsp; "SSB–AM"&nbsp; there is only one sideband.
*Therefore,&nbsp; considering the sideband-to-carrier ratio &nbsp; $μ = A_{\rm N}/(2A_{\rm T})$&nbsp; gives:
:$$ P_{\rm S}= \frac{A_{\rm T}^2}{2} + \frac{(A_{\rm N}/2)^2}{2} = {A_{\rm N}^2}/{8} \cdot \big[ 1 + {4 \cdot A_{\rm T}^2}/{A_{\rm N}^2}\big] = {P_q}/{4} \cdot \big[ 1 + {1 }/{\mu^2}\big] \hspace{0.05cm}.$$
*Thus,&nbsp; with the noise power from subtask&nbsp; '''(3)''' we obtain:
:$$\rho_{v } = \frac{P_{q}}{P_{\varepsilon }} = \frac{4 \cdot P_{\rm S}\cdot (1 + 1/\mu^2)}{4 \cdot N_0 \cdot B_{\rm NF}/\alpha_{\rm K}^2} = \frac{\alpha_{\rm K}^2 \cdot P_{\rm S}}{N_0 \cdot B_{\rm NF}} \cdot \frac{1}{1 +{1 }/{\mu^2}}\hspace{0.3cm}
\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v } = 10 \cdot {\rm lg} \hspace{0.15cm}\xi - 10 \cdot {\rm lg} \hspace{0.15cm}\big[{1 +{1 }/{\mu^2}}\big] \hspace{0.05cm}.$$
*So we get the same result with SSB-AM as in DSB-AM with a modulation depth of&nbsp; $m = \sqrt{2} · μ$.&nbsp; From this,&nbsp; it further follows:
:$$10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v } \hspace{0.15cm}({\rm SSB,} \hspace{0.1cm}\mu = {0.5}/{\sqrt{2}}) = 10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v } \hspace{0.15cm}({\rm DSB,} \hspace{0.1cm}m=0.5) \hspace{0.15cm}\underline {=30.46\,{\rm dB}},$$ 
:$$10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v } \hspace{0.15cm}({\rm SSB,} \hspace{0.1cm}\mu = {1.0}/{\sqrt{2}})  = 10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v } \hspace{0.15cm}({\rm DSB,} \hspace{0.1cm}m=1.0) \hspace{0.15cm}\underline {=35.23\,{\rm dB}}\hspace{0.05cm}.$$
{{ML-Fuß}}



[[Category:Modulation Methods: Exercises|^2.4  Single Sideband Amplitude Modulation^]]