Exercise 2.10Z: Noise with DSB-AM and SSB-AM

Shared block diagram for DSB-AM and SSB-AM

Now the influence of noise on the sink-to-noise ratio $10 · \lg ρ_v$ for both DSB–AM and SSB–AM transmission will be compared. The illustration shows the underlying block diagram.

The differences between the two system variants are highlighted in red on the image, namely the modulator (DSB or SSB) as well as the dimensionless constant

$$ K = \left\{ \begin{array}{c} 2/\alpha_{\rm K} \\ 4/\alpha_{\rm K} \\ \end{array} \right.\quad \begin{array}{*{10}c} {\rm{bei}} \\ {\rm{bei}} \\ \end{array}\begin{array}{*{20}c} {\rm DSB} \hspace{0.05cm}, \\ {\rm SSB} \hspace{0.05cm} \\ \end{array}$$

of the receiver-side carrier signal $z_{\rm E}(t) = K · \cos(ω_{\rm T} · t)$, which is assumed to be frequency and phase synchronous with the carrier signal $z(t)$ at the transmitter.

The system parameters 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 $q(t)$ with frequency $B_{\rm NF}$ stands for a source signal with bandwidth $B_{\rm NF}$ composed of multiple frequencies.
DSB–AM with carrier is parameterized by a modulation depth of $m = A_{\rm N}/A_{\rm T}$ , while SSB-AM is determined by the sideband-to-carrier ratio $μ = A_{\rm N}/(2 · A_{\rm T})$ .
The frequency-independent channel transmission factor $α_{\rm K}$ is balanced by the constant $K$ , so that in the noise-free case $(N_0 = 0)$ , the sink signal $v(t)$ matches the source signal $q(t)$ .
The sink SNR can thus be given as follows $(T_0$ indicates the period of the source signal$)$:

$$ \rho_{v } = \frac{P_{q}}{P_{\varepsilon }}\hspace{0.5cm}{\rm mit}\hspace{0.5cm} P_{q} = \frac{1}{T_{\rm 0}}\cdot\int_{0}^{ T_{\rm 0}} {q^2(t)}\hspace{0.1cm}{\rm d}t, \hspace{0.5cm}P_{\varepsilon} = \int_{-B_{\rm NF}}^{ +B_{\rm NF}} \hspace{-0.1cm}{\it \Phi_{\varepsilon}}(f)\hspace{0.1cm}{\rm d}f\hspace{0.05cm}.$$

Hints:

This exercise belongs to the chapter Single-sideband Modulation.
Particular reference is made to the page Sideband-to-carrier ratio.
The results for DSB–AM can be found on the page Sink SNR and the performance parameter.

Questions

	Synchronous demodulation.
	Envelope demodulation.

	$ρ_v = 2 · ξ$.
	$ρ_v = ξ$.
	$ρ_v = ξ/2$.

	$ρ_v = 2 · ξ$.
	$ρ_v = ξ$.
	$ρ_v = ξ/2$.

$m = 0.5\text{:} \ \ 10 · \lg \ ρ_v \ = \ $

$\ \rm dB$

$m = 1.0\text{:} \ \ 10 · \lg \ ρ_v \ = \ $

$\ \rm dB$

$μ = 0.354\text{:} \ \ \ 10 · \lg \ ρ_v \ = \ $

$\ \rm dB$

$μ = 0.707\text{:} \ \ \ 10 · \lg \ ρ_v \ = \ $

$\ \rm dB$

Solution

(1) We are dealing with a synchronous demodulator. Answer 1 is correct.

(2) Answer 2 is correct:

For DSB–AM without a carrier, $P_{\rm S} = P_q/2$. This is simultaneously the power of the useful component of the sink signal $v(t)$.
The power density spectrum ${\it Φ}_ε(f)$ of the noise component of $v(t)$ results from the convolution :

Noise power density in DSB-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 $\big[$ ... $\big]$ describes the power density spectrum of a cosine signal with the signal amplitude $K = 2$.
The correction of channel damping is considered with $1/α_K^2$ .
Thus, taking ${\it \Phi}_n(f) = N_0/2$ into account, 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, it follows for the the signal-to-noise power ratio (SNR):

$$\rho_{v } = \frac{P_{q}}{P_{\varepsilon }} = \frac{2 \cdot P_{\rm S}}{2 \cdot N_0 \cdot B_{\rm NF}/\alpha_{\rm K}^2} = \frac{\alpha_{\rm K}^2 \cdot P_{\rm S}}{N_0 \cdot B_{\rm NF}}\hspace{0.15cm}\underline { = \xi} \hspace{0.05cm}.$$

Noise power density in USB-AM

(3) Answer 2 is correct:

In contrast to DSB, $P_S = P_q/4$ holds for SSB, 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 $B_{\rm HF} = B_{\rm NF}$ into account (see adjacent diagram for USB modulation), 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: when the carrier is not transmitted, single-sideband modulation demonstrates the same noise behaviour as DSB-AM.

(4) Assuming a cosine carrier with amplitude $A_{\rm T}$ and a similarly cosine message signal $q(t)$ , for DSB with carrier, we get:

$$ s(t) = \big (q(t) + A_{\rm T}\big ) \cdot \cos( \omega_{\rm T} \cdot t) = A_{\rm T} \cdot \cos( \omega_{\rm T} \cdot t) + \frac{A_{\rm N}}{2}\cdot \cos\big(( \omega_{\rm T}+ \omega_{\rm N}) \cdot t \big)+ \frac{A_{\rm N}}{2}\cdot \cos\big(( \omega_{\rm T}- \omega_{\rm N}) \cdot t\big)\hspace{0.05cm}.$$

The transmit 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 $P_q = A_{\rm N}^2/2$ and $m = A_{\rm N}/A_{\rm T}$ into account, 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 a noise power $P_ε$ according to subtask (2) 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) In SSB–AM there is only one sideband.

Therefore, considering the sideband-to-carrier ratio $μ = A_{\rm N}/(2A_{\rm T})$ 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, with the noise power from subtask (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 $m = \sqrt{2} · μ$.
From this, it further follows:

$$10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v } \hspace{0.15cm}({\rm SSB,} \hspace{0.1cm}\mu = {0.5}/{\sqrt{2}}) = 10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v } \hspace{0.15cm}({\rm DSB,} \hspace{0.1cm}m=0.5) \hspace{0.15cm}\underline {=30.46\,{\rm dB}},$$

$$10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v } \hspace{0.15cm}({\rm SSB,} \hspace{0.1cm}\mu = {1.0}/{\sqrt{2}}) = 10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v } \hspace{0.15cm}({\rm DSB,} \hspace{0.1cm}m=1.0) \hspace{0.15cm}\underline {=35.23\,{\rm dB}}\hspace{0.05cm}.$$