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Q1: 1.8 Estimate how many light pulses can be transmitted per second by a manually operated blinker-light system. What do you conclude about the information capacity of this manual system and how it compares with the capacity of modern fiber optic telephone links?See Answer
Q2: 1.9 A fiber telephone cable contains 144 fibers, each capable of carrying 672 voice messages.A conducting telephone cable contains 900 copper twisted pairs, and each pair can carry 24 messages. Compare the capacities of the fiber and conducting cables. How many of the conducting cables are required to equal the capacity of the fiber cable? Repeat the calculations if each fiber operates at the DS-4 signaling rate.See Answer
Q3: 1.1 Compute and plot the fraction of power transmitted as a function of the transmission-line loss in decibels. Do this on log paper for a loss range of 0-50 dB.See Answer
Q4: 1.2 One milliwatt of optical power enters a fiber. Compute and plot the output power as a function of fiber loss for losses in the range 0-50 dB.See Answer
Q5: 1.3 Two 1-km fibers are spliced together. Each fiber has a 5-dB loss, and the splice adds 1 dBof loss. If the power entering is 2 mW, then how much power is delivered to the end of this combined transmission line?See Answer
Q6: 1.4 A receiver requires 10 nW as input power. If all the system losses add up to 50 dB, then how much power is required from the source?See Answer
Q7: 1.5 How much does 1 mile of RG-19/U coaxial cable weigh in pounds?See Answer
Q8: 1.6 RG-19/U is used at 100 MHz with an input power of 10 mW. The receiver sensitivity is1 μW. Compute the maximum length of the communications link. Repeat the computation for the case in which a fiber with a loss of 5 dB/km replaces the coaxial cable. The fiber system's input power and receiver sensitivity are the same as those of the coaxial cable system.See Answer
Q9: 1.7 The telephone transmission rate at the T3 level is 44.7 Mb/s. Each telephone message occupies 64,000 b/s. How many simultaneous messages could be sent along this system? In the actual system only 672 message channels are used. The additional pulses are used for other functions, such as synchronization.See Answer
Q10:Lasers are classified based on their safety characteristics. There are several classifications from ANSI, IEC and CDRH. For example some laser pointers have a label Class III laser or in another it might say Class 3. Please write on to two pages explaining the classification meaning and practical methods of protection in each case. Please list reference sources of your information at the end.See Answer
Q11:1. (a) What is the equation for the received signal power at a monostatic radar in terms of: Pt transmit power, Gt= Gr=G antenna gain, o RCS, R range to target, f frequency, Ltot total losses? (b) What are the types of losses comprising Ltot? (c) How is receiver noise related to noise figure and other parameters? (d) If (S/N)min is the minimum detectable SNR, find the relationship between & and antenna gain when other parameters are held constant: o = g(G:other parameters)See Answer
Q12:5. A bi-static radar has identical antennas at 35 GHz which have gain of 30 dB. Radar parameters include: e P= 50 kW BN = 12 MHz (Sr/Nr) = 20 Tsys = 425 K LTotal 2.8 R1 = 1.63-R2 k=1.38x10-23 J/K Assume a target with a radar cross section of 5 m². What are the target ranges R1 and R2?See Answer
Q13:1. OSNR: In a Radio over Fiber link, RF input power to the laser is +27 dBm. Laser input impedance is 50 ohm and laser modulation gain Gm is 0.1 mW/mA. (a) There are two connectors between the laser and photodiode, each with 1 dB loss and three splices each with 0.5 dB loss. Fiber attenuation is 0.5 dB/km. Receiver output impedance is also 50 ohm. Detector responsivity 0.8 and avalanche gain M is 50. Find the optical link loss Lop as a function of fiber length L km. (b) What is the RF power emanating from the photo detector (before the avalanche gain)? (c) Find the mean square value of the ac component of the detector current id. (d) If the total optical link noise is 1 x 10-12 A² over the given bandwidth, find the OSNR for L 5, 10 and 20 km considering the avalanche gain. (e) Plot OSNR vs L.See Answer
Q14:2.8 (a) Consider a signal f(t) with unit energy defined over the time interval [0, 7] that is zero outside that time interval. For example, f(t) = √T/Tpr (1). Consider two signals of duration NT: so(t) = N-1 are orthogonal. Σsouf(t-iT) i-0 N-1 si(t) = Σsuf(t-iT). 132 Determine (so(1), $1(1)) in terms of the sequence Sq, i = 0, 1,...,N-1 and $₁,,i = 0, 1,..., N-1. 1-0 (b) Consider the two signals of duration 27 generated from f(t) and the two vectors so = (+1, +1) and s₁ = (+1,-1). We will form a matrix of vectors. In this case, N = 2: H₂ = 50 51 +1 (39) where the first row can be used to generate a first signal and the second row used to generate a second signal. Using part (a), show that the two signals so(t) = 50,0 f(t)pr (1) + 50,1f(t-T)pr(t-T) $₁(t) = $1,0 f(t)pr(t) + $1,1ƒ(t – T)pr(t – T)See Answer
Q15:2.12 A communication system uses BPSK modulation to transmit data bits b₁, 1 = 0, 1, 2,.... In the transmitter, a sequence of rectan- gular pulses is mixed to a carrier frequency by multiplying the rectangular pulses by √2P cos(27f₁t): s(t) = √2P bipr(t-IT) cos(2n fit). /-0,1.... At the receiver, the received signal is first mixed down to baseband by multiplying by √2/T cos(27f2t), where f₂ - f₁ = Af is the offset of the two 136 oscillators. After the signal is mixed down, it is filtered with a matched filter, that is h(t) = pr(t). The filter is sampled at time t = iT for i = 1,2,.... In addition, the signal is mixed down by multiplying by -√2/T sin(27f2t). Let ye(iT) denote the first output and y, (iT) denote the second output. Then, ye(it) = = £ $(7) √/²7/ CC cos(21f27)h(iT - t)dt ys(IT) = - = - * S(T) 7) √ sin(2727)h(iT - T)dt. 1-00 (a) In the absence of noise, evaluate the outputs y(iT) and y, (iT) in terms of bi-₁› E = PT. AfT, and į. Ignore double frequency terms in evaluating the output. That is, derive an expression for ye(iT) and y,(iT). (Useful trig identity sin(u) - sin(v) = 2 cos("") sin("").) (b) Assume you buy two crystal oscillators at a 10 MHz nominal frequency that have + 10 PPM accuracy. That is, factual = fnominal (1 ± 10/106). Assume that the data rate is 100 kbps (7 = 10-5), that the data bits are all positive ( b; = 1, i = 0, 1, 2,..., 500), and that E = 1. (i) Are the double frequency terms negligible? (ii) Plot the output of the filters y (iT) and y, (iT) as a function of į for 1 ≤i≤ 500. Assume all the data bits are +1 and f₂-fi = Af = 200 Hz and T = 10-³.See Answer
Q16:2.16 Suppose you want to implement an SRRC pulse shape. The pulse shape is continuous and the pulse shape theoretically lasts for- ever. You generate samples of the pulse in the time domain and truncate the pulse to some number of samples. Suppose you generated eight samples per seconds where I is the time between pulses. You can assume T = 1. You can use MATLAB". Do the following for a = 0.05,0.15,0.25. (a) Determine how many samples you need if the maximum sample you ignore is 40 dB down relative to the peak sample. That is, the amplitude of any sample you ignore must be 0.01 times as small as the peak sample. (b) Determine the frequency content of the signal (plot the frequency content in dB) versus f.See Answer
Q17:2.17 Consider a baseband signal: X(t) = x₁(1) + jxq(t), where x/(t) and xo(t) are baseband signals with frequency content limited to [-W, +W]. Let X₁(f) and Xo(f) be the frequency content of the signals. So X₁(f) = XQ(f) = 0 for f * [-W, W]. The energy of the lowpass complex signal is The passband signal is = f ₁8010³d₁. E₁ = x(t) = x1(1) √2 cos(2n fet) - xo(t) √2 sin(2n fet), where fe < W. The energy of the passband signal is 138 Ep = fix(0)³dt. Show that E = Ep. Hint: Derive expressions for the energy in the frequency domain for x(f). Use Parseval's theorem: fut² (1dt = fuvas,See Answer
Q18:2.25 Consider a time-shifted set of orthonormal rectangular pulses with amplitudes given by s(0) = +0.0460-0.0460j s(8)= +0.0460 -0.0460j s(1) = -0.1320-0.0020js(9)= +0.0020 +0.1320j s(2) = -0.0130 +0.0790j s(10) = -0.0790 +0.0130j s(3) = +0.1430 +0.0130j s(11) = -0.0130 -0.1430j s(4) = +0.0920 + 0.0000j s(12) = +0.0000 - 0.0920j s(5) = +0.1430+ 0.0130j s(13)= -0.0130 - 0.1430j s(6) = -0.0130 +0.0790j s(14)= -0.0790+ 0.0130j s(7) = -0.1320-0.0020j s(15)= +0.0020 +0.1320j. The signal is then 15 s(t) = s(n)pr (t-nT). #0 This signal is used in the preamble of the IEEE 802.11 system. The signal is filtered with a filter with impulse response h(t) s*(16T-1). = (a) Find and plot the magnitude of the output of the filter. (b) If the signal is repeated eight times, plot the real part, imaginary part, and magnitude of the output of the same filter.See Answer
Q19:5. Communication System Parameters [12 points] Consider the following communication system with constellation with 16 points in 16 dimensions as given in Question 1.5 (second set) of Chapter 1 of the textbook. The modulation is with the orthonormal waveforms given by (t)= Po(t-nT), for n=1,2,3,..., 15, and po(t) is the squarcroot raised cosine pulse with T = 0.0001 and a=0.5. Find the following parameters of the system: • Energy per bit Es • Rate R • Power P, Bandwidth W. The second set is attached below./nA second signal set with M = 16 signals in 16 dimensions that can transmit 4 bits of information has the following signals: So = A(+1, +1, +1, +1, +1, +1, +1, +1, +1, +1, +1, +1, +1, +1, +1, +1) $₁ = A(+1,-1, +1, −1, +1, −1, +1, −1, +1, −1, +1, −1, +1,-1, +1, -1) $₂ = A(+1, +1,-1, −1, +1, +1, −1, −1, +1, +1, − 1, −1, +1, +1,-1,-1) $3 = A(+1,-1,-1, +1, +1, −1, −1, +1, +1, −1, −1, +1, +1,-1,-1, +1) $4 = A(+1, +1, +1, +1, −1, −1, −1, −1, +1, +1, +1, +1, −1, −1, −1, −1) $5 = A(+1,-1, +1, −1, −1, +1, −1, +1, +1, −1, +1, −1, −1, +1, −1, +1) S6 = A(+1, +1,-1, −1, −1, −1, +1, +1, +1, +1, −1, −1, −1, −1, +1, +1) $7 = A(+1,-1,-1, +1, −1, +1, +1, −1, +1, −1, −1, +1, −1, +1, +1, −1) $g = A(+1, +1, +1, +1, +1, +1, +1, +1, −1, −1, −1, -1,-1,-1,-1,-1) $9 = A(+1,-1, +1, −1, +1, −1, +1, −1, −1, +1, −1, +1,-1, +1,-1, +1) S10 = A(+1, +1,−1, −1, +1, +1, −1, −1, −1, −1, +1, +1, −1, −1, +1, +1) S11 = A(+1,-1, −1, +1, +1, −1, −1, +1, −1, +1, +1, −1, −1, +1, +1, -1) $12 = A(+1, +1, +1, +1, −1, −1, −1, −1, −1, −1, −1, −1, +1, +1, +1, +1) S13 = A(+1,-1, +1, −1, −1, +1, −1, +1, −1, +1, −1, +1, +1, −1, +1, -1) S14 = A(+1, +1,−1, −1, −1, −1, +1, +1, −1, −1, +1, +1, +1, +1, −1, −1) S15 = A(+1,-1,-1, +1, −1, +1, +1, −1, −1, +1, +1, −1, +1, −1, −1, +1)See Answer
Q20:3.1 Let X be a random variable with density function £x(x) = { 8** x 20 x < 0. (a) Let y = + √x (positive square-root). Find the density function, fy(y), of y for all values of y. (b) Let Z = ax + b where a is a positive constant. Find the density function, fz(2), for all values of Z.See Answer

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