Homework 2 Solution

• 1) (10 pts) Suppose users share a 1Mbps link. Also suppose each user requires 100Kbps when transmitting, but each user transmits only 10 percent of the time.
  • a) Suppose there are 40 users. Find the probability that at any given time, exactly n users are transmitting simultaneously
    p(n) = C(40,n)*0.1^n*0.9^(40-n)
  • b) Find the probability that there are 11 or more users transmitting simultaneously.
    1-sum_(n=0)^(n=10)p(n)
• 2) (36 pts) Characterizing the transmission media
  • a) (8 pts) If a binary signal is sent over a 3-kHz channel whose signal-to-noise ratio (SNR) is 20dB, what is the maximum achievable data rate?
    Solution: According to Shannon theorem which specifies the maximum data rate in a noisy channel as B*log_2^(1+S/N) where B is the bandwidth, S/N is the signal-to-noise ratio. Usually, S/N is given in "decibel", not just a ratio. decibel is calculated by dB=10log_10^(S/N)
    Therefore, we get S/N first by 20dB=10log_10^(S/N) ==> S/N=10^2=100

    Substitude S/N into Shannon's theorem, we get the maximum bps as 3k*log2^(1+S/N)=3log2^101kbps=19.97kbps.

    However, the maximum data rate, assuming the channel is noise free, is only 6kbps, according to Nyquist rate. Therefore, the final answer should be 6kbps.

  • b) (5 pts) What SNR is needed to put a T1 carrier on a 50-kHz line?
    Solution: T1 carrier has a data rate of 1.544Mbps. According to Shannon's theorem, the maximum data rate = Blog_2(1+S/N) where B is the bandwidth and is equal to 50kHz, the data rate is 1.544Mbps ==> S/N=2^(1.544M/50k)-1=1.976e9=92.9581dB

    Notice that the M or k in bps uses base 10 intead of base 2.

  • c) (7 pts) Why is the downstream data rate limit for dial up modem 56 kbps? Assuming the bandwidth of a telephone channel is 4000 Hz.
    Solution: According to Nyquist theorem, the maximum data rate is equal to 2Blog_2^V where B=4000Hz, and each sample has 8 bits in U.S. Among these 8 bits, one bit is used for control, leaving 7 bits user data. Therefore 2*4000*7=56kbps
  • d) (5 pts) How much bandwidth is there in 0.1 micron of spectrum at a wavelength of 1 micron?
    Solution: According to the equation on page102,
    delta_f = c * delta_lambda / (lambda*lambda) where delta_f is the bandwidth range, c is the light speed, delta_lamba is the wavelength range, and lambda is a certain wavelength. Therefore,
    delta_f = 3*10e8 * 0.1*10e-6 / (10e-6 * 10e-6) = 3*10e13Hz = 30THz
  • e) (5 pts) Radio antennas often work best when the diameter of the antenna is equal to the wavelength of the radio wave. Reasonable antennas range from 1 cm to 5 m in diameter. What frequency range does this cover?
  • f) (6 pts) According to a report, during the 1998 Tennessee-Florida game, the crowd noise measured at Neyland stadium peaked at 117dB, twice as high as the one measured at Tiger stadium during the 2000 LSU-Alabama game (111dB). In order to transmit a voice signal over this crowd at a rate of 1000bps at Neyland stadium, what kind of minimum bandwidth is required? Can you identify the type of transmission media used in this scenario?
    T1 carrier has a data rate of 1.544Mbps. According to Shannon's theorem, the maximum data rate = Blog_2(1+S/N) where B is the bandwidth and is equal to 50kHz, the data rate is 1.544Mbps ==> S/N=2^(1.544M/50k)-1=1.976e9=92.9581dB
• 3) (25 pts) On physical media
  • a) (8 pts.) List 6 media types in order of increasing bandwidth (list the bandwidth as well)
    Solution:
    

    Media	Bandwidth (Hz)
    Twisted pair (T3)	16M
    Twisted pair (T5)	100M
    Twisted pair (T6)	250M
    Twisted pair (T7)	600M
    Coax	1G
    Fiber optics	30T

    Note that this problem is designed to give you a sense how the media performs. You don't need to memorize those numbers.

  • b) (4 pts.) What media is least suited for a bus geography? What media is typically used for RJ45 cable (10Base-T)?
    wireless transmission media; twisted pair
  • c) (5 pts.) What is the limiting factor in achieving high data rates on fiber? (or Why does single mode fiber have the potential to support a higher data rate than multimode fiber?)
    Solution: refraction, number of users
  • d) (8 pts.) Comment on the different features of satellite and fiber.
    Solution: page 117
• 4) (29 pts) On communication networks
  • a) (5) In a typical mobile phone system with hexagonal cells, it is forbidden to reuse a frequency band in an adjacent cell. If 840 frequencies are available, how many can be used in a given cell?
    Solution: Each cell has six neighbors. If the central cell uses frequency group A, its six neighbors can use B, C, B, C, B and C respectively. That is, only 3 unique cells are needed. Therefore, each cell can have 840/3=280 frequencies.

    	B
    C	A	C
    B	C	B

  • b) (5) Calculate the maximum number of users that D-AMPS can support simultaneously within a single cell. Do the same calculation for GSM. Explain the difference.
    Solution: D-AMPS uses 832 channels with at most 6 users sharing the same channel, which allows D-AMPS to support up to 832*6=4992 users simultaneously in one cell (suppose all the 832 channels can be used for this cell).

    GSM uses 124 channels with at most 8 users sharing a single channel, which allows GSM to support up to 124*8=992 users simultaneously.

  • c) (5) D-AMPS has appreciably worse speech quality than GSM. Is this due to the requirement that D-AMPS be backward compatible with AMPS, whereas GSM had no such constraint? If not, what is the cause?
    Solution: Even though both systems use about the same bandwidth (D-AMPS: 832*30kHz=24.96MHz, GSM: 124*200kHz=24.8MHz), the reason that GSM can provide a better speech quality is the per user bandwidth (or data rate) that GSM can provide.
  • d) (5) The 66 low-orbit satellites in the Iridium project are divided into six necklaces around the earth. At the altitude they are using, the period is 90 minutes. What is the average interval for handoffs for a stationary transmitter?
    Solution: every 90 minutes there are 11 going by. so the handoff interval is 90*60(sec)/11=491sec.
  • e) (9) Suppose that A, B, and C are simultaneously transmitting 0 bits, using a CDMA system with the chip sequences of Fig. 2-45(b). What is the resulting chip sequence? Suppose a CDMA receiver gets the following chips (-1+1-3+1-1-3+1+1), which stations transmitted, and which bits did each one send?
    Solution: Using the -1, 0, 1 convention,
    A transmits 0 is: +1+1+1-1-1+1-1-1
    B transmits 0 is: +1+1-1+1-1-1-1+1
    C transmits 0 is: +1-1+1-1-1-1+1+1
    D transmits nothing is: 0+0+0+0+0+0+0+0
    A+B+C+D = +3+1+1-1-3-1-1+1
    

    S.A = (-1+1-3+1-1-3+1+1).(-1-1-1+1+1-1+1+1) = (1-1+3+1-1+3+1+1)/8 = 1
    S.B = (-1+1-3+1-1-3+1+1).(-1-1+1-1+1+1+1-1) = (1-1-3-1-1-3+1-1)/8 = -1
    S.C = (-1+1-3+1-1-3+1+1).(-1+1-1+1+1+1-1-1) = (1+1+3+1-1-3-1-1)/8 = 0
    S.D = (-1+1-3+1-1-3+1+1).(-1+1-1-1-1-1+1-1) = (1+1+3-1+1+3+1-1)/8 = 1
    Therefore, A transmits 1, B transmits 0, C doesn't transmit, D transmits 1.