Question

Int. to Comm. Systems

PAM Modulation PAM Demodulation LD Moduloting Demodulated C1 freq= 1K c= 100n L0: 2n3904" 8 Sampling gnd freq= 10K gnd Figure 2: PAM transmitter

Answer 1

Start Pageques2-C.PAGE1 SCHEMATI.. R1 10k R3 deMod 4누 Mod 1k C2 100n Q1 V1 R2 VOFF-0 Q2N1893 VAMPL 3 FREQ = 1k 6 2.7k AC= VOFF = 0 (nu VAMPL 6 FREQ 3k 辱4 석气 야} 0 5 AC

freq = 10khz

QU BU SEL>> 0.4me อ. 6ns 0.8ns 1.0ms 1.2ns 1.4ns 1.8ns 2.0ns orrmequest 1spatcher

freq =20khz

1.QU GU GU SEL>> θε 0.2ns 0.Ens 1.Gms 1.2ns 1.4ns 1.6ms 1.8ns 2.0ns ' 'Ук 5r.neACK InK hed spice

freq = 25khz

HO0n SEL> -400n 0.218 〇.4llie 0.61i9 0.8I1s 1.0lne 1.2118 1.4n8 1.Gme 1.8i9 2.0118 rr oen eckage req.re orrmequestDispatcher

freq = 30khz

1.QU 1.U O U(Hod) SEL>> 0,2ns อ. Gns 1. Gms 1.2ns 1,4ns 1.6ms 1.8ns 2,0ns Tina spices

freq = 35khz

(A) jhfjhvj.dat (active) 1.0V (Mod) 200mV OV -200mV SEL> -400mV 0s 0.5ms 1.0ms 1.5ms 2.0ms Time Date January 11, 2018 Page 1 Time: 06:36:11

i would suggest you to copy these images and open separately so you can zoom it and have better analysis of these waveforms.