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Principles of Communication 5Ed R E Zeimer, William H Tranter Solutions Manual Chapter Signal and Linear System Theory 2.1 Problem Solutions Problem 2.1 For the single-sided spectra, write the signal in terms of cosines: x(t) = 10 cos(4t + /8) + sin(8t + 3/4) = 10 cos(4t + /8) + cos(8t + 3/4 /2) = 10 cos(4t + /8) + cos(8t + /4) For the double-sided spectra, write the signal in terms of complex exponentials using Eulers theorem: x(t) = exp[(4t + /8)] + exp[j(4t + /8)] +3 exp[j(8t + 3/4)] + exp[j(8t + 3/4)] The two sets of spectra are plotted in Figures 2.1 and 2.2 Problem 2.2 The result is x(t) = 4ej(8t+/2) + 4ej(8t+/2) + 2ej(4t/4) + 2ej(4t/4) = cos (8t + /2) + cos (4t /4) = sin (8t) + cos (4t /4) CHAPTER SIGNAL AND LINEAR SYSTEM THEORY Single-sided amplitude Single-sided phase, rad 10 /4 /8 f, Hz f, Hz Figure 2.1: Problem 2.3 (a) Not periodic (b) Periodic To find the period, note that 20 = n1 f0 and = n2 f0 2 Therefore n2 10 = n1 Hence, take n1 = 3, n2 = 10, and f0 = Hz (c) Periodic Using a similar procedure as used in (b), we find that n1 = 2, n2 = 7, and f0 = Hz (d) Periodic Using a similar procedure as used in (b), we find that n1 = 2, n2 = 3, n3 = 11, and f0 = Hz Problem 2.4 (a) The single-sided amplitude spectrum consists of a single line of height at frequency Hz, and the phase spectrum consists of a single line of height -/6 radians at frequency Hz The double-sided amplitude spectrum consists of lines of height 2.5 at frequencies of and -6 Hz, and the double-sided phase spectrum consists of a line of height -/6 radians at frequency Hz and a line of height /6 at frequency -6 radians Hz (b) Write the signal as xb (t) = cos(12t /2) + cos(16t) From this it is seen that the single-sided amplitude spectrum consists of lines of height and at frequencies and Hz, respectively, and the single-sided phase spectrum consists 2.1 PROBLEM SOLUTIONS Double-sided amplitude f, Hz -6 -4 -2 Double-sided phase, rad /4 /8 -6 -4 -2 f, Hz -/8 -/4 Figure 2.2: CHAPTER SIGNAL AND LINEAR SYSTEM THEORY of a line of height -/2 radians at frequency Hz The double-sided amplitude spectrum consists of lines of height 1.5 and at frequencies of and Hz, respectively, and lines of height 1.5 and at frequencies -6 and -8 Hz, respectively The double-sided phase spectrum consists of a line of height -/2 radians at frequency Hz and a line of height /2 radians at frequency -6 Hz Problem 2.5 (a) This function has area Area = = Z Z ã ã sin(t/) (t/) sin(u) (u) á2 á2 dt du = A sketch shows that no matter how small is, the area is still With 0, the central lobe of the function becomes narrower and higher Thus, in the limit, it approximates a delta function (b) The area for the function is Area = Z exp(t/)u (t) dt = Z exp(u)du = A sketch shows that no matter how small is, the area is still With 0, the function becomes narrower and higher Thus, R in the limit, it approximates a delta function R (c) Area = (1 |t| /) dt = (t) dt = As 0, the function becomes narrower and higher, so it approximates a delta function in the limit Problem 2.6 (a) 513; (b) 183; (c) 0; (d) 95,583.8; (e) -157.9 Problem 2.7 (a), (c), (e), and (f) are periodic Their periods are s, s, s, and 2/7 s, respectively The waveform of part (c) is a periodic train of impulses extending from - to spaced by s The waveform of part (a) is a complex sum of sinusoids that repeats (plot) The waveform of part (e) is a doubly-infinite train of square pulses, each of which is one unit high and one unit wide, centered at ã ã ã, 6, 3, 0, 3, 6, ã ã ã Waveform (f) is a raised cosine of minimum and maximum amplitudes and 2, respectively 2.1 PROBLEM SOLUTIONS Problem 2.8 (a) The result is h i  à x(t) = Re ej6t + Re ej(12t/2) = Re ej6t + 6ej(12t/2) (b) The result is 1 x(t) = ej6t + ej6t + 3ej(12t/2) + 3ej(12t/2) 2 (c) The single-sided amplitude spectrum consists of lines of height and at frequencies of and Hz, respectively The single-sided phase spectrum consists of a line of height /2 at frequency Hz The double-sided amplitude spectrum consists of lines of height 3, 1/2, 1/2, and at frequencies of 6, 3, 3, and Hz, respectively The double-sided phase spectrum consists of lines of height /2 and /2 at frequencies of and Hz, respectively Problem 2.9 (a) Power Since it is a periodic signal, we obtain P1 = T0 Z T0 sin (8t + /4) dt = T0 Z T0 [1 cos (16t + /2)] dt = W where T0 = 1/8 s is the period (b) Energy The energy is Z Z E2 = e2t u2 (t)dt = e2t dt = e2t dt = (c) Energy The energy is E3 = Z e2t u2 (t)dt = Z (d) Neither energy or power E4 = lim Z T T T (2 P4 = since limT T RT dt T (2 +t2 )1/4 dt + t2 )1/4 = 0.(e) Energy = Since it is the sum of x1 (t) and x2 (t), its energy is the sum of the energies of these two signals, or E5 = 1/ CHAPTER SIGNAL AND LINEAR SYSTEM THEORY (f) Power Since it is an aperiodic signal (the sine starts at t = 0), we use Z T 1 [1 cos (10t)] dt T 2T 0 ã T 1 sin (20t) T = W = lim T 2T 2 20 P6 = T 2T lim Z T sin2 (5t) dt = lim Problem 2.10 (a) Power Since the signal is periodic with period /, we have P = Z / A2 |sin (t + )|2 dt = Z / A2 A2 {1 cos [2 (t + )]} dt = 2 (b) Neither The energy calculation gives E = lim Z T T T (A )2 dt = lim + jt jt T Z T T (A )2 dt + t2 The power calculation gives P = lim T 2T Z T T (A )2 dt (A )2 = lim ln + t2 T 2T ! p + T / p =0 + + T / 1+ (c) Energy: E= Z A2 t4 exp (2t/ ) dt = A2 (use table of integrals) (d) Energy: E=2 Z /2 22 dt + Z /2 ! 12 dt = Problem 2.11 (a) This is a periodic train of boxcars, each units in width and centered at multiples of 6: ả Z Z t 1 1.5 dt = W dt = Pa = 3 1.5 2.1 PROBLEM SOLUTIONS (b) This is a periodic train of unit-high isoceles triangles, each units wide and centered at multiples of 5: ả ả ả Z Z 2 t t 2.5 t 22 dt = W Pb = dt = = 2.5 53 15 (c) This is a backward train of sawtooths (right triangles with the right angle on the left), each units wide and spaced by units: ả ả Z t 2 12 t dt = Pc = = W 33 (d) This is a full-wave rectified cosine wave of period 1/5 (the width of each cosine pulse): Z 1/10 Z 1/10 ã 1 |cos (5t)| dt = ì Pd = + cos (10t) dt = W 2 1/10 Problem 2.12 (a) E = , P = ; (b) E = J, P = W; (c) E = , P = 49 W; (d) E = , P = W Problem 2.13 (a) The energy is E= Z cos (6t) dt = (b) The energy is Z E= Z 6ã 1 + cos (12t) dt = J 2 Z h i2 |t|/3 e cos (12t) dt = ã 2t/3 e 1 + cos (24t) dt 2 where the last integral follows by the eveness of the integrand of the first one Use a table of definte integrals to obtain Z Z 2/3 2t/3 e dt + e2t/3 cos (24t) dt = + E= 2 (2/3) + (24)2 0 Since the result is finite, this is an energy signal (c) The energy is Z Z {2 [u (t) u (t 7)]} dt = E= 4dt = 28 J CHAPTER SIGNAL AND LINEAR SYSTEM THEORY Since the result is finite, this is an energy signal (d) Note that ẵ Z t 0, t < u () d = r (t) = t, t which is called the unit ramp The energy is E= Z [r (t) 2r (t 10) + r (t 20)] dt = Z 10 t 10 ả2 dt = 20 J where the last integral follows because the integrand is a symmetrical triangle about t = 10 Since the result is finite, this is an energy signal Problem 2.14 (a) Expand the integrand, integrate term by term, and simplify making use of the orthogonality property of the orthonormal functions (b) Add and subtract the quantity suggested right above (2.34) and simplify (c) These are unit-high rectangular pulses of width T /4 They are centered at t = T /8, 3T /8, 5T /8, and 7T /8 Since they are spaced by T /4, they are adjacent to each other and fill the interval [0, T ] (d) Using the expression for the generalized Fourier series coecients, we find that X1 = 1/8, X2 = 3/8, X3 = 5/8, and X4 = 7/8 Also, cn = T /4 Thus, the ramp signal is approximated by t = (t) + (t) + (t) + (t) , t T T 8 8 where the n (t)s are given in part (c) (e) These are unit-high rectangular pulses of width T /2 and centered at t = T /4 and 3T /4 We find that X1 = 1/4 and X2 = 3/4 (f) To compute the ISE, we use N = Z T |x (t)| dt N X n=1 cn Xn2 RT R Note that T |x (t)|2 dt = (t/T )2 dt = T /3 Hence, for (d), Ă Â 49 = 5.208 ì 103 T + 64 + 25 + ISEd = T3 T4 64 Ă 64 Â64 T T For (e), ISEe = 16 + 16 = 2.083 ì 102 T 2.1 PROBLEM SOLUTIONS Problem 2.15 (a) The Fourier coecients are (note that the period = 2 ) 1 X1 = X1 = ; X0 = All other coecients are zero (b) The Fourier coecients for this case are X1 = X1 = (1 + j) All other coecients are zero (c) The Fourier coecients for this case are (note that the period is 20 ) 1 X2 = X2 = ; X1 = X1 = ; X0 = 4 All other coecients are zero (d) The Fourier coecients for this case are X3 = X3 = ; X1 = X1 = 8 All other coecients are zero Problem 2.16 The expansion interval is T0 = and the Fourier coecients are ả Z Z 2 jn(/2)t 2 nt dt 2t e dt = 2t cos Xn = 4 which follows by the eveness of the integrand Let u = nt/2 to obtain the form ả3 Z n 16 Xn = u2 cos u du = (1)n n (n) If n is odd, the Fourier coecients are zero as is evident from the eveness of the function being represented If n = 0, the integral for the coecients is Z 2 2t dt = X0 = The Fourier series is therefore x (t) = + X n=, n6=0 (1)n 16 jn(/2)t e n 10 CHAPTER SIGNAL AND LINEAR SYSTEM THEORY Problem 2.17 Parts (a) through (c) were discussed in the text For (d), break the integral for x (t) up into a part for t < and a part for t > Then use the odd half-wave symmetry contition Problem 2.18 This is a matter of integration Only the solution for part (b) will be given here integral for the Fourier coecients is (note that the period really is T0 /2) Z A T0 /2 Xn = sin ( t) ejn0 t dt T0 T0 /2 Aejn0 t = t) + cos ( t)] [jn sin ( 0 T0 (1 n2 ) Ă Â jn A 1+e , n 6= = T0 (1 n2 ) For n = 1, the integral is Z A T0 /2 jA sin ( t) [cos (jn t) j sin (jn t)] dt = = X1 X1 = T0 This is the same result as given in Table 2.1 The Problem 2.19 (a) Use Parsevals theorem to get P|nf0 | 1/ = N X n=N ả N X A |Xn | = sinc2 (nf0 ) T0 n=N where N is an appropriately chosen limit on the sum We are given that only frequences for which |nf0 | 1/ are to be included This is the same as requiring that |n| 1/ f0 = T0 / = Also, for a pulse train, Ptotal = A2 /T0 and, in this case, Ptotal = A2 /2 Thus ả P|nf0 | 1/ X A = sinc2 (nf0 ) Ptotal A2 n=2 = = = X sinc2 (nf0 ) n=2 Ă ÂÔ 1Ê + sinc2 (1/2) + sinc2 (1) 2" ả2 # 1+2 = 0.91 10.2 COMPUTER EXERCISES 43 z = 10.^(zdB/10); % convert to linear scale ber1 = qfn(sqrt(2*z)); % PSK result ber2 = qfn(sqrt(12*2*z/23)); % CSER for (23,12) Golay code ber3 = qfn(sqrt(4*z*2/7)); % CSER for (15,11) Hamming code bergolay = cer2ber(2,23,7,3,ber2); % BER for Golay code berhamming = cer2ber(2,7,3,1,ber3); % BER for Hamming code semilogy(zdB,ber1,zdB,bergolay,zdB,berhamming) xlabel(E_b/N_o in dB) ylabel(Bit Error Probability) The funcrions Q and cer2ber are given in the previous computer exercise The results are illustrated in Figure 10.20 The order of the curves, for high signal-tonoise ratio, are in order of the error correcting capability The top curve (worst performance) is the uncoded case The middle curve is for the (7,4) Hamming code and the bottom curve is for the (23,12) Golay code Computer Exercise 10.5 The MATLAB code for generating the performance curves illustrated in Figure 10.18 in the text follows zdB = 0:30; z = 10.^(zdB/10); qa = 0.5*exp(-0.5*z); qa3 = 0.5*exp(-0.5*z/3); qa7 = 0.5*exp(-0.5*z/7); pa7 = 35*((1-qa7).^3).*(qa7.^4)+21*((1-qa7).^2).*(qa7.^5) +7*(1-qa7).*(qa7.^6)+(qa7.^7); pa3 = 3*(1-qa3).*(qa3.^2)+(qa3.^3); qr = 0.5./(1+z/2); qr3 = 0.5./(1+z/(2*3)); qr7 = 0.5./(1+z/(2*7)); pr7 = 35*((1-qr7).^3).*(qr7.^4)+21*((1-qr7).^2).*(qr7.^5) +7*(1-qr7).*(qr7.^6)+(qr7.^7); pr3 = 3*(1-qr3).*(qr3.^2)+(qr3.^3); semilogy(zdB,qr,zdB,pr3,zdB,pr7,zdB,qa,zdB,pa3,zdB,pa7) axis([0 30 0.0001 1]) xlabel(Signal-to-Noise Ratio, z - dB) ylabel(Probability ) Executing the code yields the results illustrated in Figure 10.21 The curves can be identiịed from Figure 10.18 in the text 44 CHAPTER 10 INFORMATION THEORY AND CODING 10 -2 B it E rror P robability 10 -4 10 -6 10 -8 10 -10 10 E /N in dB b 10 o Figure 10.20: Computer Exercise 10.6 For this Computer Exercise we use the uncoded Rayleigh fading result from the previous Computer Exercise and also use the result for a Rayleigh fading channel with a rate 1/7 repetition code The new piece of information needed for this Computer Exercise is the result from Problem 9.17 The MATLAB code follows: zdB z = qr1 qr7 pr7 = 0:30; 10.^(zdB/10); = 0.5./(1+z/2); = 0.5./(1+z/(2*7)); = 35*((1-qr7).^3).*(qr7.^4)+21*((1-qr7).^2).*(qr7.^5) +7*(1-qr7).*(qr7.^6)+(qr7.^7); % Calculations for diversity system pd7 = zeros(1,31); N = 7; 10.2 COMPUTER EXERCISES 45 10 -1 Probability 10 -2 10 -3 10 -4 10 10 15 20 Signal-to-Noise Ratio, z - dB Figure 10.21: for i=1:31 sum = 0; a = 0.5/(1+z(i)/(2*N)); for j=0:(N-1) term = nkchoose(N+j-1,j)*((1-a)^j); sum = sum+term; end pd7(i) = (a^N)*sum; end % semilogy(zdB,qr1,zdB,pr7,zdB,pd7) axis([0 30 0.0001 1]) xlabel(Signal-to-Noise Ratio, z - dB) ylabel(Probability) 25 30 46 CHAPTER 10 INFORMATION THEORY AND CODING 10 -1 P robability 10 -2 10 -3 10 -4 10 10 15 20 Signal-to-Noise Ratio, z - dB 25 30 Figure 10.22: The function nkchoose follows: function out=nkchoose(n,k) % Computes n!/k!/(n-k)! a = sum(log(1:n)); b = sum(log(1:k)); c = sum(log(1:(n-k))); out = round(exp(a-b-c)); % End of function file % % % % ln of n! ln of k! ln of (n-k)! result Executing the MATLAB program yields the result illustrated in Figure 10.22 The curves can be identiịed by comparison with Figure 10.20 inthe text Appendix A Physical Noise Sources and Noise Calculations Problem A.1 All parts of the problem are solved using the relation Vrms = 4kT RB where k = 1.38 ì 1023 J/K B = 30 MHz = ì 107 Hz (a) For R = 10, 000 ohms and T = T0 = 290 K p Vrms = (1.38 ì 1023 ) (290) (104 ) (3 ì 107 ) = 6.93 ì 105 V rms = 69.3 àV rms (b) Vrms is smaller than the result in part (a) by a factor of 10 = 3.16 Thus Vrms = 21.9 àV rms (c) Vrms is smaller than the result in part (a) by a factor of 100 = 10 Thus Vrms = 6.93 àV rms (d) Each answer becomes smaller by factors of 2, 10 = 3.16, and 10, respectively APPENDIX A PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS Problem A.2 Use ã ả eV I = Is exp kT Ă Â We want I > 20Is or exp eV kT > 20 e (a) At T = 290 K, kT ' 40, so we have exp (40V ) > 21 giving V > i2rms ' or i2rms B = = = (b) If T = 29 K, then e kT ln (21) = 0.0761 volts 40 ả eV 2eIB ' 2eBIs exp kT ả eV 2eIs exp kT ÂĂ Â Ă 1.6 ì 1019 1.5 ì 105 exp (40 ì 0.0761) 1.0075 ì 1022 A2 /Hz = 400, and for I > 20Is , we need exp(400V ) > 21 or V > ln (21) = 7.61 ì 103 volts 400 Thus i2rms B ÂĂ Â Ă Â Ă = 1.6 ì 1019 1.5 ì 105 exp 400 ì 7.61 ì 103 = 1.0075 ì 1022 A2 /Hz as before Problem A.3 (a) Use Nyquists formula to get the equivalent circuit of Req in parallel with R3 , where Req is given by RL (R1 + R2 ) Req = R1 + R2 + RL The noise equivalent circuit is then as shown in Figure A.1 The equivalent noise voltages are p 4kT Req B p = 4kT R3 B V1 = V2 Req R3 V0 V1 V2 Figure A.1: Adding noise powers to get V0 we obtain ả2 Ă Â R3 V1 + V22 V0 = Req + R3 ả2 R3 = (4kT B) (Req + R3 ) Req + R3 (4kT B) R23 = (Req + R3 ) (b) With R1 = 2000 , R2 = RL = 300 , and R3 = 500 , we have Req = 300 (2000 + 300) = 265.4 300 + 2000 + 300 and V02 B (4kT ) R32 (Req + R3 )  à 1.38 ì 1023 (290) (500)2 = 500 + 265.4 = 5.23 ì 1018 V2 /Hz = APPENDIX A PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS Problem A.4 Find the equivalent resistance for the R1 , R2 , R3 combination and set RL equal to this to get R3 (R1 + R2 ) RL = R1 + R2 + R3 Problem A.5 Using Nyquists formula, we ịnd the equivalent resistance looking backing into the terminals with Vrms across them It is Req = 50k k 20 k k (5 k + 10 k + k) = 50k k 20 k k 20 k = 50k k 10 k (50k) (10 k) = 50k + 10 k = 8, 333 Thus = 4kT Req B Vrms Ă Â Ă Â = 1.38 ì 1023 (400) (8333) ì 106 = 3.68 ì 1010 V2 which gives Vrms = 19.18 àV rms Problem A.6 To ịnd the noise ịgure, we ịrst determine the noise power due to a source at the output, and then due to the source and the network Initally assume unmatched conditions The results are ả2 R2 k RL (4kT RS B) V0 due to R , only = S RS + R1 + R2 k RL V02 due to R1 and R2 ả2 R2 k RL = (4kT R1 B) RS + R1 + R2 k RL ả2 RL k (R1 + RS ) + (4kT R2 B) R2 + (R1 + RS ) k RL V02 due to R S, R1 and R2 = ả2 R2 k RL [4kT (RS + R1 ) B] RS + R1 + R2 k RL ả2 RL k (R1 + RS ) + (4kT R2 B) R2 + (R1 + RS ) k RL The noise ịgure is R1 + F =1+ RS RL k (R1 + RS ) R2 + (R1 + RS ) k RL ả2 RS + R1 + R2 k RL R2 k RL ả2 R2 RS In the above, Ra Rb Ra + Rb Note that the noise due to RL has been excluded because it belongs to the next stage Since this is a matching circuit, we want the input matched to the source and the output matched to the load Matching at the input requires that Ra k Rb = RS = Rin = R1 + R2 k RL = R1 + R2 RL R2 + RL and matching at the output requires that RL = Rout = R2 k (R1 + RS ) = R2 (R1 + RS ) R1 + R2 + RS Next, these expressions are substituted back into the expression for F After some simpliịcation, this gives ả2 2R2L RS (R1 + R2 + RS ) / (RS R1 ) R2 R1 + F =1+ 2 RS RS R2 (R1 + RS + RL ) + RL (R1 + R2 + RS ) Note that if R1 >> R2 we then have matched conditions of RL = R2 and RS = R1 Then, the noise ịgure simpliịes to R1 F = + 16 R2 Note that the simple resistive pad matching circuit is very poor from the standpoint of noise The equivalent noise temperature is found by using Te = T0 (F 1) ã R1 = T0 + 16 R2 APPENDIX A PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS Problem A.7 The important relationships are Fl = + Tel T0 Tel = T0 (Fl 1) Te0 = Te1 + Te2 Te3 + Ga1 Ga1 Ga2 Completion of the table gives Ampl No F not needed dB 11 dB Tei 300 K 864.5 K 3360.9 K Gai 10 dB = 10 30 dB = 1000 30 dB = 1000 Therefore, 3360.9 864.5 + 10 (10) (1000) = 386.8 K Te0 = 300 + Hence, Te0 T0 = 2.33 = 3.68 dB F0 = + (b) With ampliịers and interchanged 300 3360.9 + 10 (10) (1000) = 865.14 K Te0 = 864.5 + This gives a noise ịgure of 865.14 290 = 3.98 = dB F0 = + (c) See part (a) for the noise temperatures (d) For B = 50 kHz, TS = 1000 K, and an overall gain of Ga = 107 , we have, for the conịguration of part (a) Pna, out = Ga k (T0 + Te0 ) B Ă Â Ă Â = 107 1.38 ì 1023 (1000 + 386.8) ì 104 = 9.57 ì 109 watt We desire Psa, out Psa, out = 104 = Pna, out 9.57 ì 109 which gives Psa, out = 9.57 ì 105 watt For part (b), we have Ă Â Ă Â Pna, out = 107 1.38 ì 1023 (1000 + 865.14) ì 104 = 1.29 ì 108 watt Once again, we desire which gives Psa, out Psa, out = 104 = Pna, out 1.29 ì 108 Psa, out = 1.29 ì 104 watt and Psa, in = Psa, out = 1.29 ì 1011 watt Ga Problem A.8 (a) The noise ịgure of the cascade is Foverall = F1 + F2 F =L+ = LF Ga1 (1/L) (b) For two identicalattenuator-ampliịer stages Foverall = L + F L1 F + + = 2LF 2LF, L >> (1/L) (1/L) L (1/L) L (1/L) (c) Generalizing, for N stages we have Foverall N F L APPENDIX A PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS Problem A.9 The data for this problem is Stage (preamp) (mixer) (ampliịer) Fi dB = 1.58 dB = 6.31 dB = 3.16 Gi G1 1.5 dB = 1.41 30 dB = 1000 The overall noise ịgure is F = F1 + F2 F3 + G1 G1 G2 which gives dB = 3.16 1.58 + 6.31 3.16 + G1 1.42G1 or 5.31 (2.16/1.41) = 4.33 = 6.37 dB 1.58 (b) First, assume that G1 = 15 dB = 31.62 Then G1 F 3.16 6.31 + 31.62 (1.41) (31.62) = 1.8 = 2.54 dB = 1.58 + Then Tes = T0 (F 1) = 290 (1.8 1) = 230.95 K and Toverall = Tes + Ta = 230.95 + 300 = 530.95 K Now use G1 as found in part (a): F Tes Toverall = 3.16 = 290 (3.16 1) = 626.4 K = 300 + 626.4 = 926.4 K (c) For G1 = 15 dB = 31.62, Ga = 31.62 (1.41) (1000) = 4.46 ì 104 Thus ÂĂ Â Ă Â Ă Pna, out = 4.46 ì 104 1.38 ì 1023 (530.9) 109 = 3.27 ì 109 watts For G1 = 6.37 dB = 4.33, Ga = 4.33 (1.41) (1000) = 6.11 ì 103 Thus Ă ÂĂ Â Ă Â Pna, out = 6.11 ì 103 1.38 ì 1023 (926.4) 107 = 7.81 ì 1010 watts Note that for the second case, we get less noise power out even wth a larger Toverall This is due to the lower gain of stage 1, which more than compensates for the larger input noise power (d) A transmission line with loss L = dB connects the antenna to the preamp We ịrst ịnd TS for the transmission line/preamp/mixer/amp chain: FS = FTL + F1 F2 F3 + + , GTL GTL G1 GTL G1 G2 where GTL = 1/L = 102/10 = 0.631 and FTL = L = 102/10 = 1.58 Assume two cases for G1 : 15 dB and 6.37 dB First, for G1 = 15 dB = 31.6, we have FS = 1.58 + 1.58 6.31 3.16 + + 0.631 (0.631) (31.6) (0.631) (31.6) (1.41) = 2.84 This gives TS = 290 (2.84 1) = 534 K and Toverall = 534 + 300 = 834 K Now, for G1 = 6.37 dB = 4.33, we have FS = 1.58 + 6.31 3.16 1.58 + + 0.631 (0.631) (4.33) (0.631) (4.33) (1.41) = This gives TS = 290 (5 1) = 1160 K 10 APPENDIX A PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS and Toverall = 1160 + 300 = 1460 K Problem A.10 (a) Using Pna, out = Ga kTS B with the given values yields Pna, out = 7.45 ì 106 watts (b) We want Psa, out = 105 Pna, out or This gives Ă ÂĂ Â Psa, out = 105 7.45 ì 106 = 0.745 watts Psa, out = 0.745 ì 108 watts Ga = 51.28 dBm Psa, in = Problem A.11 (a) For A = dB, Y = 1.259 and the eective noise temperature is Te = 600 (1.259) (300) = 858.3 K 1.259 For A = 1.5 dB, Y = 1.413 and the eective noise temperature is Te = 600 (1.413) (300) = 426.4 K 1.413 For A = dB, Y = 1.585 and the eective noise temperature is Te = 600 (1.585) (300) = 212.8 K 1.585 (b) These values can be converted to noise ịgure using F =1+ Te T0 11 With T0 = 290 K, we get the following values: (1) For A = dB, F = 5.98 dB; (2) For A = 1.5 dB, F = 3.938 dB; (3) For A = dB, F = 2.39 dB Problem A.12 (a) Using the data given, we can determine the following: ả 4d GT PT GT = 0.039 m = 202.4 dB = 39.2 dB = 74.2 dBW This gives PS = 202.4 + 74.2 + = 127.2 dBW (b) Using Pn = kTe B for the noise power, we get ã ả Te B Pn = 10 log10 kT0 T0 Te T0 ả = 10 log10 [kT0 ] + 10 log10 + 10 log10 (B) ả Ă Â 1000 + 10 log10 106 = 174 + 10 log10 290 = 108.6 dBm = 138.6 dBW (c) PS Pn ả dB = 127.2 (138.6) = 11.4 dB = 101.14 = 13.8 ratio (d) Assuming the SNR = z = Eb /N0 = 13.8, we get the results for various digital signaling techniques given in the table below: Modulation type BPSK DPSK Noncoh FSK QPSK Error probability Ă Â Q 2z = 7.4 ì 108 z = 5.06 ì 107 2e z/2 = 5.03 ì 104 2e Same as BPSK [...]... exp (jf ) The latter represents the frequency response of a filter whose impulse response is a square pulse of width and implements flat top sampling If W is the bandwidth of X (f), very little distortion will result if 1 >> W Problem 2.60 (a) The sampling frequency should be large compared with the bandwidth of the signal (b) The output spectrum of the zero-order hold circuit is Y (f) = sinc (Ts f)... (t + ) d T 2T T Problem 2.37 The result is an even triangular wave with zero average value of period T0 It makes no dierence whether the original square wave is even or odd or neither 2.1 PROBLEM SOLUTIONS 21 Problem 2.38 Fourier transform both sides of the dierential equation using the dierentiation theorem of Fourier transforms to get [j2f + a] Y (f) = [j2bf + c] X (f ) Therefore, the transfer function... is the limit of one period of area times N as N (c) The condition for stability is Z Z 1 u (t 1) dt |h3 (t)| dt = t Z dt = = ln (t)| 1 t 1 24 CHAPTER 2 SIGNAL AND LINEAR SYSTEM THEORY Problem 2.45 The energy spectral density of the output is Gy (f) = |H (f )|2 |X (f)|2 where X (f) = Hence 1 2 + j2f 100 ih i Gy (f ) = h 9 + (2f)2 4 + (2f)2 Problem 2.46 Using the Fourier coecients of a half-rectified... = H0 e à f H0 2B ả ej2f t0 Use the inverse Fourier transform of a constant, the delay theorem, and the inverse Fourier transform of a rectangular pulse function to get h (t) = H0 (t t0 ) 2BH0 sinc [2B (t t0 )] Problem 2.56 (a) The Fourier transform of this signal is Ă Â X (f ) = A 2b2 exp 2 2 2 f 2 By definition, using a table of integrals, Z 1 |x (t)| dt = 2 T = x (0) Similarly, W= Therefore,... (jn0 ) XnB where the superscript A refers to xA (t) and B refers to xB (t) 2.1 PROBLEM SOLUTIONS 13 Problem 2.24 (a) This is the right half of a triangle waveform of width and height A, or A (1 t/ ) Therefore, the Fourier transform is Z X1 (f) = A (1 t/ ) ej2f t dt 0 ã á 1 A j2f 1 1e = j2f j2f where a table of integrals has been used (b) Since x2 (t) = x1 (t) we have, by the time reversal theorem,... phase shifting by /2 Multiply x (t) and x b (t) together term by term, use trigonometric identities for the product of sines and cosines, then integrate The integrand will be a sum of terms similar to that of part (a) The limit as T will be zero term-by-term (c) Use the integral definition of x b (t), take the product, integrate over time to get Z x (t) x b (t) dt = A2 = A2 Z (t/ ) "Z Z /2 /2... plot(t, exp(-t)) title([Approximation of exp(-t) over [0, ,num2str(t_max),] with contiguous rectangular pulses of width ,num2str(tau)]) elseif type_wave == 2 plot(t, x_approx), axis([0 t_max -inf inf]), xlabel(t), ylabel(x(t); x_a_p_p_r_o_x(t)) hold plot(t, sin(2*pi*t)) title([Approximation of sin(2*pi*t) over [0, ,num2str(t_max),] with contiguous rectangular pulses of width ,num2str(tau)]) end % Decaying... +1/2) , t > + 1/2 e (b) The convolution of these two signals gives y2 (t) = (t) + tr (t) where tr(t) is a trapezoidal function given by 0, t < 3/2 or t > 3/2 1, 1/2 t 1/2 tr (t) = 3/2 + t, 3/2 t < 1/2 3/2 t, 1/2 t < 3/2 (c) The convolution results in Z y3 (t) = e || ( t) d = Z t+1/2 e|| d t1/2 Sketches of the integrand for various values of t gives the following cases: R t+1/2 ... amplitude spectrum is the same as for part (a) except that X0 = 3A 4 Note that this can be viewed as having a sinc-function envelope with zeros at multiples of 3T4 0 The phase spectrum can be obtained from that of part (a) by adding a phase shift of for negative frequencies and subtracting for postitive frequencies (or vice versa) Problem 2.23 (a) There is no line at dc; otherwise it looks like a... because 2W Problem 2.41 (a) Replace the capacitors with 1/jC which is their ac-equivalent impedance Call the junction of the input resistor, feedback resistor, and capacitors 1 Call the junction at the positive input of the operational amplifier 2 Call the junction at the negative input of the operational amplifier 3 Write down the KCL equations at these three junctions Use the constraint equation for