Question

A communication system operates in the presence of white noise with a two-sided power spectral density Sn(w)=10^-14 W/Hz and with a path loss of 20 dB. Calculate the minimum required bandwidth and the minimum required power of the transmitter for a 10 kHz sinusoidal input and a 40 dB output signal to noise ratio if the modulation is a) DSB-SC b) SSB-SC c) FM with Af = 10 kHz

Answer 1

• 1014 h Given, Sn (W) = 10 14 h min path loss = 20dB (BW) min= ? , Parmin = ? W= B.w.of input = lokH3 (SNR). = 40 dB ca) DSBB-se a For DSBSC minimum Bandwidth =2w (SHR). = PR = 20 KHz the Pn where PRE PT, de Lexo = (PT 28 2o) dB Pn = 2 Sn (w) df = 2 xlolt x 104 = axion Pn za toth ł $c-w Noise power in the range feWalt) < fetw (SNR) oido = Pryde – Proda 40 = PT, dis - 20 - 10 los, (2x1650) PTdB = to set to -37 dB _ -36.99

lologico PT = * - 37 I » Pq = 10 3.7 = 2x10t = 0.2 m watts (milliwatts) @ ssb-sc: (BW) min = w for SSB se = lokha (),se pare = Dorose Even though SSB-sc uses halt the Bandwidth for transmission compared to DSB SC, it's modulated signal power is half compared to DSBSC and also noise power is half because Bandwidth reduces by & a factor of two, therefore we can say in noise performance they are same. 7 P = 0.2 mw = 2x104 W o FM with Af = 10kHz and Given w=lokta From carson's rule, mininum bandwidth is Bc = 2 (B+1) W = 2 Aftw) : 7 Be=2(10 +10) = 2 (20) = 40kH3 -70 Also given c ode = 40 de ), = 107 okto (unitless)

For FM, ). = 3 B? Pmin where - PR le Loss = zodB Po-PR = 2048 254 (W) Be = P11100 2 x 1514x 4x104 = PT 2 x 16 x 4 x 1610 Pt tpe = 100 - PT 8x158 where Py = A for SFMlt) = AC Cos(21ifat +8 (6) where, ef(t) = 20764 st meelde. 1, we get Assuming Pun = ()= 104 filolusyo (ho = 40 dB) ( )= 104 Ž 107 = 3x12 x1 x PT 8x169 B=AF AF , - Py = lotx 8 x108 = P x 164 watts Watts