College of Computer & Information Science Fall 2007 Northeastern University Handout 3 CSG250: Wireless Networks 24 September 2007

Sample Solution to Problem Set 1

1. (10 points) Applying low-pass and bandpass filters to a digital signal

A square periodic signal is represented as the following sum of sinusoids:

s(t) = \frac{2}{π} ∑_{k=0}^{∞} \frac{(−1)^{k}}{2k + 1}cos((2k + 1)πt).

(Note that this is just a rewriting of the formula we discussed in class.)

(a) Suppose that the signal is applied to an ideal low-pass filter with bandwidth 15 Hz. Plot the output from the low-pass filter and compare to the original signal. Repeat for 5 Hz; for 3 Hz. What happens as the bandwidth increases.

(b) Suppose that the signal is applied to a bandpass filter that passes the frequencies from 5 to 9 Hz. Plot the output from the filter and compare to the original signal.

For your plots, use an appropriate plotting tool. One such tool is gnuplot, available in Unix.

Answer:

(a) The non-zero frequency components of the signal s(t) correspond to the frequencies 1/2 Hz, 3/2 Hz, 5/2 Hz, …. If the signal is passed through an ideal low-pass filter with bandwidth 15Hz, we obtain the output signal as:

\frac{2}{π} ∑_{k=0}^{14} \frac{(−1)^{k}}{2k + 1}cos((2k + 1)πt).

Signals can be obtained for the other cases similarly. On plotting, we obtain Figures 1, 2, and 3, respectively.

(b) See Figure 4.

2. (4 points) Bandwidth, signal element, and SNR

A digital signaling system is required to operate at 38.4 Kbps. If a signal element encodes a 8bit word, what is the minimum required bandwidth of the channel. What signal-to-noise ratio is required to achieve the desired capacity on the bandwidth that you have computed?

Answer: We need capacity C ≥ 38400 bps. We have M = 2^{8}. Nyquist’s sampling theorem says that C = 2B log_{2} M. Therefore, B ≥ 38400/(16) = 2400Hz.

Figure 1: Passing through a low-pass filter of 15Hz bandwidth.

Using Shannon’s capacity theorem, we have C = B log_{2}(1 + SNR). Therefore SNR ≥ 2^{C/B} − 1 = 65535. In decibels, we obtain SNR_{dB} = 10 · log_{10}(65535) = 48.16 dB.

3. (6 points) Effect of transmission frequency and distance on attenuation

Under the free-space path-loss model, find the transmit power required to obtain a received power of 1 dBm for a wireless system with isotropic antennas (gain = 1) and a carrier frequency f = 5 GHz, assuming a distance d = 10 m. Repeat for d = 100 m, keeping f = 5 GHz. Repeat for f = 50 GHz, keeping d = 10 m.

Answer: In free space, we have the path-loss model as:

\frac{P_{t}}{P_{r}} = \frac{(4πfd)^{2}}{(G_{r}G_{t}c^{2})}.

Expressing the same in decibels and noting that the gains are 1, we obtain:

10 log P_{t} = 10 log[(4πfd)2/c^{2}] + 10 log P_{r}

For d = 10m, f = 5 GHz:

10 log P_{t} = 10 log[(4π · 5 · 10^{9} · 10)^{2}/c^{2}] + 1 dBm
= 66.42 + 1 dBm
= 67.42 dBm

For d = 100m f = 5 GHz:

10 log P_{t} = 10log[(4π · 5 · 10^{9} · 100)^{2}/c^{2}] + 1 dBm
= 86.42 + 1 dBm
= 87.42 dBm.

For d = 10 m f = 50 GHz:

10 log P_{t} = 10 log[(4π · 50 · 10^{9} · 10)^{2}/c^{2}] + 1 dBm

= 86.42 + 1 dBm
= 87.42 dBm

4. (5 points) Received power using half-wave dipoles Problem 5.9. Answer: The relationship between transmitted power P_{t} and received power P_{r} is:

\frac{P_{t}}{P_{r}} = \frac{16π^{2}d^{2}}{G_{r}G_{t}λ^{2}} ,

where d is the distance and G_{r} and G_{t} are the gains of the receiving and transmitting antennas, respectively. The two gains are given to be 3 dB, which is a factor of 10^{3/10} = 2. The wavelength λ is given to be 3 × 10^{8}/10^{8} = 3m. The distance is 10^{4}m and P_{t} = 1W. Substituting these numbers we get:

P_{r} = \frac{2 · 2 · 3^{2} · 1}{16π^{2} · 10^{8}} = 2.28 × 10^{−9}W.

5. (7 points) Modulation schemes and E_{b}/N_{0} Problem 6.2. Answer: Let T_{s} be the signal element period, T_{b} be the bit period, and A be the amplitude of the signal.

(a) We have A = 0.005V, T_{s} = T_{b} = 10^{−5}s. We obtain the power P as follows.

P = \frac{1}{T_{s}} ∫_{0}^{T_{s}} s^{2}(t) = \frac{A^{2}}{2}

E_{b} = P × T_{b} = P × T_{s} = \frac{A^{2}}{2} T_{s}; N_{0} = 2.5 × 10^{−8} × T_{s}

\frac{E_{b}}{N_{0}} = \frac{(A^{2}/2) × T_{s}}{2.5 × 10^{−8} × T_{s}} = 500.

In dB, we obtain 27dB.

(b) For this, all the values are identical except that T_{b} = T_{s}/2. So we obtain E_{b}/N_{0} = 250. In dB, this is 24 dB.

6. (5 points) Telephone channel bandwidth Problem 6.6. Answer: The required bandwidth for 2400 bps QPSK is 2R/2 = 2400 Hz. The required bandwidth for 4800 bps MPSK with 8-level multisignaling is 2R/3 = 3200Hz. Thus, bandwidth of the telephone channel is adequate for the former but not for the latter.

7. (5 points) Sampling Problem 6.10. Answer: The bandwidth of the signal is 2700 Hz. And the sampling rate is 7000 samples per second.

Figure 4: Passing through a 5-9Hz bandpass filter.

(a) SNR (in dB) and the number n of bits in the number of quantizing levels are related as

SNR = 6.02n + 1.76dB.

So we obtain n ≥ 4.69. Thus, the number of quantizing levels needed is 32.

(b) With 7000 samples per second and 5 bits per sample, we obtain a data rate of 35000 bps.

8. (8 points) Delta modulation Problem 6.13. Answer: See Figure 5.

Figure 5: Delta modulation 6