U0ut cos2pif1t where w1 2pif0 kVin2pif0 For st cos2pif0t

U_0ut = cos(2pif_1t) where w_1 = 2pif_0 + kV_in2pif_0 For s(t) = cos(2pif_0t) display will be as shown. Show that the display for the simplified Spectrum Analyser shown in Fig. PI will be |H(f)| centred around f_0. Show that the 3 dB resolution bandwidth is 2B_LPF. (This will be taken as the NEB of SA.)

Solution

vout = Vp cos(2*pi*(f+ kVin)t)

bandwidth = 3 log 6

= 2Bpf