Problem 1

Consider a communicaion system operating at below bandwidth and using following parameters.

• Let the transmission bandwidth be 1 MHz. Say a rolloff factor of r = 0.25 is used. Determine
  • Achievable rate of data traffic
  • Delay spread that no ISI occurs
• Repeat for following cases:
  • Let OFDM with N = 16 equally spaced carriers be used.
  • If 16-QAM is now utilized for the OFDM system.

Solution

PSK case: B = 1MHz, r = 0.25

• $B = R_0(1+r) \Leftrightarrow 10^6 = R(1 + 0.25)$

⇒ $R_s = 800 \ Kbps = R_b$

• Delay spread: $T_{av} < \frac{T_S}{2 \pi} < \frac{1}{2 \pi 10^6 \times 0.8} \approx 0.2 \ (\mu s)$

OFDM, N = 16 carriers

$$ R_S = N . \triangle t \Rightarrow \triangle t = \frac{10 \times 0.8}{16} = 50 \ KHz $$

$T_{av} < \frac{T_S}{2 \pi} = \frac{20 \mu s}{2 \pi} = 2.5 \mu s$

$T_S = \frac{1}{\triangle f} = \frac{1}{50 \ KHz} = 20 \mu s$

16 - QAM is utilized → 4 successive bits started at the same time → 4 bits/symbol

$R_b = 4R_S = 4 * 800 \ Kbps = 3200 \ Kbps = 3.2 \ Mbps$

Other Solution

Transmission bandwidth: $B_T = 10^6 Hz$

Rolloff factor: $r = 0.25$

Rate of Data traffic is $R_S$ (symbol transmission rate)