CS 3413 Data Communications, Fall 2004 Practice Problems Solutions

1. 3t = &pi so t = &pi/3.
2. &lambda f = c so &lambda = .15m.
3. The speed is 3/4 of c the the wavelength is 3/4 if the one above, or .11m.
4. 2Hlog₂M = 2(4000)(4) = 32kbps.
5. 10 log₁₀S/N = 30, so S/N = 1000.
   H log₁₀(1 + S/N) = 4000*log₁₀1001 is about 40kbps.
6. By the Sampling Theorem, need 40,000 samples per second of 2(14) bits giving 1.12 Mbps.
7. 2
8. • .8
   • 1
   • 2
   • no, the Hamming distance would need to be at least 3.
9. • (.99)¹⁶
   • (.99)¹⁶ + 16(.99)¹⁵(.01)
10. • 100% (if propagation delay is 0)
    • 20 ms are spent sending and 200 ms are spent waiting after that giving an efficiency of 20/220 = 9%.
    • 2% of the time it takes an extra 1000 ms. Add (.02)(1000) to the denominator above to get 20/240 = 8.3%.