Question:

Published on: 10 August, 2022

**Calculate the SNR at the output of a synchronous SSB-SC demodulator.**

Answer:

The receiver of an SSB-SC system is similar to that of or DSB-SC system for synchronous detection except for the fact that the bandwidth of the BPF of SSB-Sc receiver should be one- half of the required for DSB-SC.

Fig.27 Block diagram of SSB-SC demodulation with noise

Let the transmitted signal

\(S\left(t\right)=\frac{A_c}{2}\left[x\left(t\right)cos2\pi f_ct\mp p\left(t\right)sin2\pi f_ct\right]\)

Where –ve sign for the Upper side band (USB) and +ve sign for Lower side band (LSB), x(t) is the message signal, p(t) is the Hilbert Transform of x(t) and \(A_c\) is the amplitude of the carrier.

The signal power available at the input of the demodulator is

\(S_i=\frac{1}{2}\left(\frac{A_c}{2}\right)^2x^{2^\prime}\left(t\right)+12\left(\frac{A_c}{2}\right)^2p^{2^\prime}\left(t\right)=\frac{{A_c}^2}{4}\left[\frac{1}{2}x^{2^\prime}\left(t\right)+\frac{1}{2}p^{2^\prime}\left(t\right)\right]=\left(\frac{A_c}{2}\right)^2x^{2^\prime}\left(t\right)\)

It should be noted that, the average power of the signal x(t) is same as that of its Hilbert Transform. This is because p(t) is basically the signal x(t) in which all the frequency components undergo a phase shift of \(\frac{\pi}{2}\). Thus x(t) and p(t) occupy the same spectral range, that is \(\omega\) Hz in this case. The power spectral density of x(t) and p(t) is same. Thus, the average power of the message signal and its Hilbert Transform are the same.

From Fig. 27 We have,

\(S_{iL}\left(t\right)=s\left(t\right)cos2\pi f_ct\)

\(S_{iL}\left(t\right)=\frac{A_c}{2}\left[x\left(t\right)cos2\pi f_ct\mp p\left(t\right)sin2\pi f_c\left(t\right)\right]cos2\pi f_ct\)

\(S_{iL}\left(t\right)=\frac{A_c}{4}\left[x\left(t\right)\left(1+cos2\pi f_ct\right)\pm\frac{A_c}{4}p\left(t\right)sin2\pi f_ct\right]\)

The signal is finally passed through the Low Pass Filter(LPF) which rejects all the term except the term which occupies a bandwidth \(f_m\).

The final demodulated output is

\({s\left(t\right)}_0=\frac{A_c}{4}x\left(t\right)\) and input of LPF noise signal is \(n\left(t\right)=n_i\left(t\right){cos}^22\pi f_ct+{\frac{1}{2}n}_q\left(t\right)sin4\pi f_ct\)

Output noise signal of demodulator \(\frac{1}{2}n_i\left(t\right)\) because all high frequency components are rejected output signal power \(S_0=\left(\frac{A_c}{4}\right)^2x^{2^\prime}t\)

Noise power \(N_0=\frac{1}{4}n_i^{2^\prime}\left(t\right)\)

Output signal to noise ratio

\(\left(\frac{S}{N}\right)_0=\frac{\frac{1}{16}{A_c}^2x^{2^\prime}\left(t\right)}{\frac{1}{4}n_i^{2^\prime}\left(t\right)}\)

And input signal to noise ratio

\(\frac{\frac{1}{4}{A_c}^2x^{2^\prime}\left(t\right)}{n_i^{2^\prime}\left(t\right)}\)

Figure of merit ofDSB-SCsignal=\(\frac{\frac{S_0}{N_0}}{\frac{S_i}{N_i}}=\frac{\left(\frac{S_0}{S_i}\right)}{\left(\frac{N_0}{N_i}\right)}=\frac{\frac{1}{16}{A_c}^2x^{2^\prime}\left(t\right)}{\frac{1}{4}n_i^{2^\prime}\left(t\right)}X\frac{n_i^{2^\prime}\left(t\right)}{\frac{1}{4}{A_c}^2x^{2^\prime}\left(t\right)}=1\)

It is therefore, seen that the SNR at the output of SSB-SC coherent demodulator is same as that the input. Thus, there is no improvement of the SNR.

Subjects

Trending

Random questions

**Write short note on PCM ( Pulse Code Modulation).**

10 August, 2022

**State and prove maximum power transfer theorem**

21 June, 2022