Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 13

Tham khảo tài liệu 'proakis j. (2002) communication systems engineering - solutions manual (299s) episode 13', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 2 The following figure is a plot of the amplitude characteristics of the RC filter C f . The values of the vertical axis indicate that C f can be considered constant for frequencies up to 2000 Hz. Since the same is true for the envelope delay we conclude that a lowpass signal of bandwidth Af 1 KHz will not be distorted if it passes the RC filter. Problem Let Gt f and GR f be the frequency response of the transmitting and receiving filter. Then the condition for zero ISI implies . f T . 0 lf T Gt f C Gr Xrc f T 1 cos 2nT If - T T lf T 0 lf rT Since the additive noise is white the optimum transmitting and receiving filter characteristics are given by see Example Thus Gt f l Xrc f 2 C f 2 Gr J Xrc f 2 C f 2 Gt f Gr j T 1 2 1 cos 2nfT T 1 cos 2nT If -T 2 1 cos 2nfT 0 fl 4T 4T f 1 4T otherwise 0 1 2 238 Problem A 4-PAM modulation can accommodate k 2 bits per transmitted symbol. Thus the symbol interval duration is T -k- sec 9600 4800 Since the channel s bandwidth is W 2400 2T in order to achieve the maximum rate of transmission Rmax 2T the spectrum of the signal pulse should be . f X f Tn U- V2wy Then the magnitude frequency response of the optimum transmitting and receiving filter is see Section and Example 1 4 Gt f Gr J f 2400 __ 1 2 4 n f A 2W J f 2400 J IfI 2400 otherwise 1 Problem 1 The equivalent discrete-time impulse response of the channel is 1 h t 2 hnỗ t - nT t T t t - T n -1 If by cn we denote the coefficients of the FIR equalizer then the equalized signal is 1 qm cnhm-n n -1 which in matrix notation is written as 0. C-1 0 I C0 I 1 I 0. y y C1 y 0 y The coefficients of the zero-force equalizer can be found by solving the previous matrix equation. Thus C-1 C0 I I C1 2 The values of qm for m 2 3 are given by I q2 Cnh2-n C1h1 n -1 I q-2 52 Cnh-2-n C-1h-1 _ 1 q3 52 Cnh3-n 0 n -1 1 q-3 52 Cnh-3-n 0 239 Problem 1 The output of the .