Analog Communication - SNR Calculations

In this chapter, let us calculate Signal to Noise Ratios and Figure of Merits of various modulated waves, which are demodulated at the receiver.

Signal to Noise Ratio

Signal-to-Noise Ratio (SNR) is the ratio of the signal power to noise power. The higher the value of SNR, the greater will be the quapty of the received output.

Signal-to-Noise Ratio at different points can be calculated using the following formulas.

Input SNR = $left ( SNR ight )_I= frac{Average :: power ::of ::modulating ::signal}{Average:: power ::of ::noise ::at ::input}$

Output SNR = $left ( SNR ight )_O= frac{Average :: power ::of ::demodulated ::signal}{Average:: power ::of ::noise ::at ::output}$

Channel SNR = $left ( SNR ight )_C= frac{Average :: power ::of ::modulated ::signal}{Average:: power ::of ::noise ::in ::message ::bandwidth}$

Figure of Merit

The ratio of output SNR and input SNR can be termed as Figure of Merit. It is denoted by F. It describes the performance of a device.

$$F=frac {left ( SNR ight )_O}{left ( SNR ight )_I}$$

Figure of merit of a receiver is

$$F=frac {left ( SNR ight )_O}{left ( SNR ight )_C}$$

It is so because for a receiver, the channel is the input.

SNR Calculations in AM System

Consider the following receiver model of AM system to analyze noise.

We know that the Ampptude Modulated (AM) wave is

$$sleft ( t ight )=A_cleft [ 1+k_amleft ( t ight ) ight ] cosleft ( 2 pi f_ct ight )$$

$$Rightarrow sleft ( t ight )=A_c cos left ( 2 pi f_ct ight )+A_ck_amleft ( t ight ) cosleft ( 2 pi f_ct ight )$$

Average power of AM wave is

$$P_s=left ( frac{A_c}{sqrt{2}} ight )^2+left ( frac{A_ck_amleft ( t ight )}{sqrt{2}} ight )^2=frac{{A_{c}}^{2}}{2}+frac{{A_{c}}^{2}{k_{a}}^{2}P}{2}$$

$$Rightarrow P_s=frac{{A_{c}}^{2}left ( 1+{k_{a}}^{2}P ight )}{2}$$

Average power of noise in the message bandwidth is

$$P_{nc}=WN_0$$

Substitute, these values in channel SNR formula

$$left ( SNR ight )_{C,AM}=frac{Average :: Power :: of :: AM :: Wave}{Average :: Power :: of :: noise :: in :: message :: bandwidth}$$

$$Rightarrow left ( SNR ight )_{C,AM}=frac{{A_{c}}^{2}left ( 1+ {k_{a}}^{2} ight )P}{2WN_0}$$

Where,

P is the power of the message signal=$frac{{A_{m}}^{2}}{2}$

W is the message bandwidth

Assume the band pass noise is mixed with AM wave in the channel as shown in the above figure. This combination is appped at the input of AM demodulator. Hence, the input of AM demodulator is.

$$vleft ( t ight )=sleft ( t ight )+nleft ( t ight )$$

$Rightarrow vleft ( t ight )=A_cleft [ 1+k_amleft ( t ight ) ight ] cosleft ( 2 pi f_ct ight )+$

$left [ n_1left ( t ight ) cosleft ( 2 pi f_ct ight ) - n_Qleft ( t ight ) sin left ( 2 pi f_ct ight ) ight ]$

$Rightarrow vleft ( t ight )=left [ A_c+A_ck_amleft ( t ight )+n_1left ( t ight ) ight ] cosleft ( 2 pi f_ct ight )-n_Qleft ( t ight ) sinleft ( 2 pi f_ct ight )$

Where $n_I left ( t ight )$ and $n_Q left ( t ight )$ are in phase and quadrature phase components of noise.

The output of AM demodulator is nothing but the envelope of the above signal.

$$dleft ( t ight )=sqrt{left [ A_c+A_cK_amleft ( t ight )+n_Ileft ( t ight ) ight ]^2+left ( n_Qleft ( t ight ) ight )^2}$$

$$Rightarrow dleft ( t ight )approx A_c+A_ck_amleft ( t ight )+n_1left ( t ight )$$

Average power of the demodulated signal is

$$P_m=left ( frac{A_ck_amleft ( t ight )}{sqrt{2}} ight )^2=frac{{A_{c}}^{2}{k_{a}}^{2}P}{2}$$

Average power of noise at the output is

$$P_no=WN_0$$

Substitute, these values in output SNR formula.

$$left ( SNR ight )_{O,AM}= frac {Average :: Power :: of :: demodulated :: signal }{Average :: Power :: of :: noise :: at :: Output}$$

$$Rightarrow left ( SNR ight )_{O,AM}=frac{{A_{c}}^{2}{k_{a}}^{2}P}{2WN_0}$$

Substitute, the values in Figure of merit of AM receiver formula.

$$F=frac{left ( SNR ight )_{O,AM}}{left ( SNR ight )_{C,AM}}$$

$$Rightarrow F=left ( frac{{A_{c}^{2}}{k_{a}^{2}}P}{2WN_0} ight )/left ( frac{{A_{c}}^{2}left ( 1+ {k_{a}}^{2} ight )P}{2WN_0} ight )$$

$$Rightarrow F=frac{{K_{a}}^{2}P}{1+{K_{a}}^{2}P}$$

Therefore, the Figure of merit of AM receiver is less than one.

SNR Calculations in DSBSC System

Consider the following receiver model of DSBSC system to analyze noise.

We know that the DSBSC modulated wave is

$$sleft ( t ight )=A_cmleft ( t ight ) cosleft ( 2 pi f_ct ight )$$

Average power of DSBSC modulated wave is

$$P_s=left ( frac{A_cmleft ( t ight )}{sqrt{2}} ight )^2=frac{{A_{c}}^{2}P}{2}$$

Average power of noise in the message bandwidth is

$$P_{nc}=WN_0$$

Substitute, these values in channel SNR formula.

$$left ( SNR ight )_{C,DSBSC}=frac{Average :: Power :: of :: DSBSC :: modulated :: wave}{Average :: Power :: of :: noise :: in :: message :: bandwidth}$$

$$Rightarrow left ( SNR ight )_{C,DSBSC}=frac{{A_{c}}^{2}P}{2WN_0}$$

Assume the band pass noise is mixed with DSBSC modulated wave in the channel as shown in the above figure. This combination is appped as one of the input to the product modulator. Hence, the input of this product modulator is

$$v_1left ( t ight )=sleft ( t ight )+nleft ( t ight )$$

$$Rightarrow v_1left ( t ight )=A_cmleft ( t ight ) cos left ( 2 pi f_ct ight )+left [ n_Ileft ( t ight ) cosleft ( 2 pi f_ct ight ) - n_Qleft ( t ight ) sin left ( 2 pi f_ct ight ) ight ]$$

$$Rightarrow v_1left ( t ight )=left [ A_cm left ( t ight ) +n_Ileft ( t ight ) ight ] cosleft ( 2 pi f_ct ight )-n_Qleft ( t ight ) sinleft ( 2 pi f_ct ight )$$

Local oscillator generates the carrier signal $cleft ( t ight )= cosleft ( 2 pi f_ct ight )$. This signal is appped as another input to the product modulator. Therefore, the product modulator produces an output, which is the product of $v_1left ( t ight )$ and $cleft ( t ight )$.

$$v_2left ( t ight )= v_1left ( t ight )cleft ( t ight )$$

Substitute, $v_1left ( t ight )$ and $cleft ( t ight )$ values in the above equation.

$$Rightarrow v_2left ( t ight )=left ( left [ A_cmleft ( t ight ) + n_Ileft ( t ight ) ight ] cosleft ( 2 pi f_ct ight )- n_Qleft ( t ight ) sinleft ( 2 pi f_ct ight ) ight ) cosleft ( 2 pi f_ct ight )$$

$$Rightarrow v_2left ( t ight )=left [ A_c mleft ( t ight )+n_Ileft ( t ight ) ight ] cos^2left ( 2 pi f_ct ight )-n_Qleft ( t ight ) sinleft ( 2 pi f_ct ight ) cosleft ( 2 pi f_ct ight )$$

$$Rightarrow v_2left ( t ight )=left [ A_c mleft ( t ight )+n_Ileft ( t ight ) ight ] left ( frac{1+ cosleft ( 4 pi f_ct ight )}{2} ight ) -n_Qleft ( t ight )frac{ sinleft ( 4 pi f_ct ight )}{2}$$

When the above signal is appped as an input to low pass filter, we will get the output of low pass filter as

$$dleft ( t ight )=frac{left [ A_c mleft ( t ight )+n_Ileft ( t ight ) ight ]}{2}$$

Average power of the demodulated signal is

$$P_m=left ( frac{A_cmleft ( t ight )}{2sqrt{2}} ight )^2=frac{{A_{c}}^{2}P}{8}$$

Average power of noise at the output is

$$P_{no}=frac{WN_0}{4}$$

Substitute, these values in output SNR formula.

$$left ( SNR ight )_{O,DSBSC}= frac {Average :: Power :: of :: demodulated :: signal }{Average :: Power :: of :: noise :: at :: Output}$$

$$Rightarrow left ( SNR ight )_{O,DSBSC}=left ( frac{{A_{c}}^{2}P}{8} ight )/ left ( frac{WN_0}{4} ight )=frac{{A_{c}}^{2}P}{2WN_0}$$

Substitute, the values in Figure of merit of DSBSC receiver formula.

$$F=frac{left ( SNR ight )_{O,DSBSC}}{left ( SNR ight )_{C,DSBSC}}$$

$$Rightarrow F= left ( frac{{A_{c}}^{2}P}{2WN_0} ight )/ left ( frac{{A_{c}}^{2}P}{2WN_0} ight )$$

$$Rightarrow F= 1$$

Therefore, the Figure of merit of DSBSC receiver is 1.

SNR Calculations in SSBSC System

Consider the following receiver model of SSBSC system to analyze noise.

We know that the SSBSC modulated wave having lower sideband is

$$sleft ( t ight )=frac{A_mA_c}{2} cos left [ 2 pileft ( f_c-f_m ight )t ight ]$$

Average power of SSBSC modulated wave is

$$P_s=left ( frac{A_mA_c}{2sqrt{2}} ight )^2=frac{{A_{m}}^{2}{A_{c}}^{2}}{8}$$

Average power of noise in the message bandwidth is

$$P_{nc}=WN_0$$

Substitute, these values in channel SNR formula.

$$left ( SNR ight )_{C,SSBSC}= frac {Average :: Power :: of :: SSBSC :: modulated :: wave}{Average :: Power :: of :: noise :: in:: message :: bandwidth}$$

$$Rightarrow left ( SNR ight )_{C,SSBSC}=frac{{A_{m}}^{2}{A_{c}}^{2}}{8WN_0}$$

Assume the band pass noise is mixed with SSBSC modulated wave in the channel as shown in the above figure. This combination is appped as one of the input to the product modulator. Hence, the input of this product modulator is

$$v_1left ( t ight )=sleft ( t ight )+nleft ( t ight )$$

$$v_1left ( t ight )=frac{A_mA_c}{2} cosleft [ 2 pi left ( f_c-f_m ight )t ight ] + n_Ileft ( t ight ) cosleft ( 2 pi f_ct ight )-n_Qleft ( t ight ) sin left ( 2 pi f_ct ight )$$

The local oscillator generates the carrier signal $cleft ( t ight )= cos left ( 2 pi f_ct ight ) $. This signal is appped as another input to the product modulator. Therefore, the product modulator produces an output, which is the product of $v_1left ( t ight )$ and $cleft ( t ight )$.

$$v_2left ( t ight )=v_1left ( t ight )c left ( t ight )$$

Substitute, $v_1left ( t ight )$ and $ cleft ( t ight )$ values in the above equation.

$Rightarrow v_2(t)= (frac{A_mA_c}{2} cos[ 2 pi ( f_c-f_m )t ] + n_I ( t ) cos ( 2 pi f_ct )-$

$n_Q( t ) sin ( 2 pi f_ct ) )cos ( 2 pi f_ct )$

$Rightarrow v_2left ( t ight )=frac{A_mA_c}{2} cosleft [ 2 pi left ( f_c-f_m ight )t ight ] cosleft ( 2 pi f_ct ight )+$

$n_Ileft ( t ight ) cos^2left ( 2 pi f_ct ight )-n_Qleft ( t ight ) sinleft ( 2 pi f_ct ight ) cosleft ( 2 pi f_ct ight )$

$Rightarrow v_2left ( t ight )=frac{A_mA_c}{4} left { cosleft [ 2 pileft ( 2f_c-f_m ight )t ight ] + cos left ( 2 pi f_mt ight ) ight }+$

$n_Ileft ( t ight )left ( frac{1+ cosleft ( 4 pi f_ct ight )}{2} ight )- n_Qleft ( t ight )frac{sin left ( 4 pi f_ct ight )}{2}$

When the above signal is appped as an input to low pass filter, we will get the output of low pass filter as

$$dleft ( t ight )=frac{A_mA_c}{2} cosleft ( 2 pi f_mt ight )+frac{n_Ileft ( t ight )}{2}$$

Average power of the demodulated signal is

$$P_m=left ( frac{A_mA_c}{4sqrt{2}} ight )^2=frac{{A_{m}}^{2}{A_{c}}^{2}}{32}$$

Average power of noise at the output is

$$P_{no}=frac{WN_0}{4}$$

Substitute, these values in output SNR formula

$$left ( SNR ight )_{O,SSBSC}= frac {Average :: Power :: of :: demodulated :: signal}{Average :: Power :: of :: noise :: at :: output}$$

$$Rightarrow left ( SNR ight )_{O,SSBSC}= left ( frac{{A_{m}}^{2}{A_{c}}^{2}}{32} ight )/left ( frac{WN_0}{4} ight )=frac{{A_{m}}^{2}{A_{c}}^{2}}{8WN_0}$$

Substitute, the values in Figure of merit of SSBSC receiver formula

$$F=frac{left ( SNR ight )_{O,SSBSC}}{left ( SNR ight )_{C,SSBSC}}$$

$$F=left ( frac{{A_{m}}^{2}{A_{c}}^{2}}{8WN_0} ight )/left ( frac{{A_{m}}^{2}{A_{c}}^{2}}{8WN_0} ight )$$

$$F=1$$

Therefore, the Figure of merit of SSBSC receiver is 1.