√ kT hT hT i hT × Principles of Communications 7th Edition Solution Manual Problem A.1 All parts of the problem are solved using the relation Vrms = √
4kT RB where k = 1.38 × 10—23 J/K B = 30 MHz = 3 × 107 Hz a. For R = 10, 000 ohms and T = T0 = 290 K Vrms = 4 (1.38 × 10—23) (290) (104) (3 × 107) = 6.93 × 10—5 V rms = 69.3 µV rms b. Vrms is smaller than the result in part (a) by a factor of √
10 = 3.16. Thus Vrms = 21.9 µV rms c. Vrms is smaller than the result in part (a) by a factor of √
100 = 10. Thus Vrms = 6.93 µV rms d. Each answer becomes smaller by factors of 2, √
10 = 3.16, and 10, respectively. Problem A.2 Use I = Is exp eV — 1 We want I > 20Is or exp eV — 1 > 20. a. At T = 290 K, e = 1.6x10—19 1.38x1023 x290 ~= 40, so we have exp (40V ) > 21 giving ln (21) V > 40 = 0.0761 volts 2 rms ~= 2eIB ' 2eBIs exp eV kT 2 or rms B eV = 2eIs exp kT b. If T = 90 K, then e = 2 1.6 × 10—19 1.5 × 10—5 exp (40 × 0.0761) = 1.0075 × 10—22 A2/Hz ~= 129, and for I > 20Is, we need exp(129V ) > 21 or V > ln (21) = 2.36 10—2 volts 129 i √
V = 4kT R BK K 0 Thus 2 rms B = 2 1.6 × 10—19 1.5 × 10—5 exp (129 × 0.0236) approximately as before. Problem A.3 = 1.0079 × 10—22 A2/Hz a. Use Nyquist's formula to get the equivalent circuit of R eq in parallel with R K, where Req is given by Req = R3 (R1 + R2) R1 + R2 + R3 The noise equivalent circuit consists of a rms noise voltage, V eq , in series with R eq and a rms noise voltage, V K, in series with with R K with these series combinations being in parallel. The equivalent noise voltages are Veq = √
4kT Req B The rms noise voltages across the parallel combination of Req and RK, by using superposition and voltage division, are V01 = Ve q RL Req + RK and V02 = VLRe q Req + RK Adding noise powers to get V 2 we obtain V 2 R2 V 2R2 V 2 = eq K + K eq 0 eq + RK)2 (Req + RK)2 = (4kT B) RKReq Req + RK = 4kT B RKR3 (R1 + R2) R1R3 + R2R3 + R1RK + R2RK + R3RK Note that we could have considered the parallel combination of R3 and RK as an equivalent load resistor and found the Thevenin equivalent. Let R|| = R3RK R3 + RK The Thevenin equivalent resistance of the whole circuit is then R (R1 + R2) R3 RL (R1 + R2) Req2 = || = R3 +RL R|| + R1 + R2 R3 RL R3 +RL + R1 + R2 = RKR3 (R1 + R2) R1R3 + R2R3 + R1RK + R2RK + R3RK i (R 0 0 and the mean-square output noise voltage is now V 2 = 4kT BReq2 which is the same result as obtained before. b. With R 1 = 2000 Ω, R2 = RK = 300 Ω, and R3 = 500 Ω, we have V 2 B = 4kT B = RKR3 (R1 + R2) R1R3 + R2R3 + R1RK + R2RK + R3RK 4 1.38 × 10—23 (290) (300) (500) (2000 + 300) 2000 (500) + 300 (500) + 2000 (300) + 300 (300) + 500 (300) = 2.775 × 10—18 V2/Hz Problem A.4 Find the equivalent resistance for the R1, R2, R3 combination and set RK equal to this to get R = R3 (R1 + R2) K R1 + R2 + R3 Problem A.5 Using Nyquist's formula, we find the equivalent resistance looking back into the terminals with Vrms across them. It is Req = 50 k ǁ 20 k ǁ (5 k + 10 k + 5 k) = 50 k ǁ 20 k ǁ 20 k = 50 k ǁ 10 k (50 k) (10 k) = 50 k + 10 k = 8, 333 Ω Thus which gives 2 rms = 4kT Req B = 4 1.38 × 10—23 (400) (8333) 2 × 106 = 3.68 × 10—10 V2 Vrms = 19.18 µV rms V 0 0 0 RK ǁ (R1 + RS) 2 1 + RS K S 1 2 S S 1 R2 (R1 + RS + RK) + R2 (R1 + R2 + RS) 2 RS + R 2 + (R 1 + R S ) ǁ R K 2 K Problem A.6 To find the noise figure, we first determine the noise power due to a source at the output, then due to the source and the network, and take the ratio of the latter to the former. Initally assume unmatched conditions. The results are V 2. due to RS , only = R2 ǁ RK 2 RS + R1 + R2 ǁ RK (4kT RSB) V 2. due to R1 and R2 = R2 ǁ RK 2 RS + R1 + R2 ǁ RK (4kT R1B) RK ǁ (R1 + RS) 2 V 2. due to RS , R1 and R2 = R2 ǁ RK 2 RS + R1 + R2 ǁ RK [4kT (RS + R1) B] + R2 The noise figure is (after some simplification) + (R1 + RS ) ǁ RK (4kT R2B) R1 F = 1 + RK ǁ (R1 + RS) 2 RS + R1 + R2 ǁ RK 2 R2 In the above, RS R2 + (R1 + RS) ǁ RK R2 ǁ RK RS Ra ǁ Rb RaRb = Ra + Rb Note that the noise due to R K 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 RS = Rin = R1 + R2 ǁ RK = R1 + R2RK R2 + RK and matching at the output requires that RK = R out = R2 ǁ (R1 + RS ) = R2 (R1 + RS) R1 + R2 + RS Next, these expressions are substituted back into the expression for F. After some simplification, this gives R 2R2 R (R + R + R ) / (R — R ) 2 R Note that if R1 >> R2 we then have matched conditi ons of RK =~ R2 and RS =~ R1. Then, the noise figure simplifies to F = 2 + 16 R1 R2 + (4kT R2B) F = 1 +
4kT RB where k = 1.38 × 10—23 J/K B = 30 MHz = 3 × 107 Hz a. For R = 10, 000 ohms and T = T0 = 290 K Vrms = 4 (1.38 × 10—23) (290) (104) (3 × 107) = 6.93 × 10—5 V rms = 69.3 µV rms b. Vrms is smaller than the result in part (a) by a factor of √
10 = 3.16. Thus Vrms = 21.9 µV rms c. Vrms is smaller than the result in part (a) by a factor of √
100 = 10. Thus Vrms = 6.93 µV rms d. Each answer becomes smaller by factors of 2, √
10 = 3.16, and 10, respectively. Problem A.2 Use I = Is exp eV — 1 We want I > 20Is or exp eV — 1 > 20. a. At T = 290 K, e = 1.6x10—19 1.38x1023 x290 ~= 40, so we have exp (40V ) > 21 giving ln (21) V > 40 = 0.0761 volts 2 rms ~= 2eIB ' 2eBIs exp eV kT 2 or rms B eV = 2eIs exp kT b. If T = 90 K, then e = 2 1.6 × 10—19 1.5 × 10—5 exp (40 × 0.0761) = 1.0075 × 10—22 A2/Hz ~= 129, and for I > 20Is, we need exp(129V ) > 21 or V > ln (21) = 2.36 10—2 volts 129 i √
V = 4kT R BK K 0 Thus 2 rms B = 2 1.6 × 10—19 1.5 × 10—5 exp (129 × 0.0236) approximately as before. Problem A.3 = 1.0079 × 10—22 A2/Hz a. Use Nyquist's formula to get the equivalent circuit of R eq in parallel with R K, where Req is given by Req = R3 (R1 + R2) R1 + R2 + R3 The noise equivalent circuit consists of a rms noise voltage, V eq , in series with R eq and a rms noise voltage, V K, in series with with R K with these series combinations being in parallel. The equivalent noise voltages are Veq = √
4kT Req B The rms noise voltages across the parallel combination of Req and RK, by using superposition and voltage division, are V01 = Ve q RL Req + RK and V02 = VLRe q Req + RK Adding noise powers to get V 2 we obtain V 2 R2 V 2R2 V 2 = eq K + K eq 0 eq + RK)2 (Req + RK)2 = (4kT B) RKReq Req + RK = 4kT B RKR3 (R1 + R2) R1R3 + R2R3 + R1RK + R2RK + R3RK Note that we could have considered the parallel combination of R3 and RK as an equivalent load resistor and found the Thevenin equivalent. Let R|| = R3RK R3 + RK The Thevenin equivalent resistance of the whole circuit is then R (R1 + R2) R3 RL (R1 + R2) Req2 = || = R3 +RL R|| + R1 + R2 R3 RL R3 +RL + R1 + R2 = RKR3 (R1 + R2) R1R3 + R2R3 + R1RK + R2RK + R3RK i (R 0 0 and the mean-square output noise voltage is now V 2 = 4kT BReq2 which is the same result as obtained before. b. With R 1 = 2000 Ω, R2 = RK = 300 Ω, and R3 = 500 Ω, we have V 2 B = 4kT B = RKR3 (R1 + R2) R1R3 + R2R3 + R1RK + R2RK + R3RK 4 1.38 × 10—23 (290) (300) (500) (2000 + 300) 2000 (500) + 300 (500) + 2000 (300) + 300 (300) + 500 (300) = 2.775 × 10—18 V2/Hz Problem A.4 Find the equivalent resistance for the R1, R2, R3 combination and set RK equal to this to get R = R3 (R1 + R2) K R1 + R2 + R3 Problem A.5 Using Nyquist's formula, we find the equivalent resistance looking back into the terminals with Vrms across them. It is Req = 50 k ǁ 20 k ǁ (5 k + 10 k + 5 k) = 50 k ǁ 20 k ǁ 20 k = 50 k ǁ 10 k (50 k) (10 k) = 50 k + 10 k = 8, 333 Ω Thus which gives 2 rms = 4kT Req B = 4 1.38 × 10—23 (400) (8333) 2 × 106 = 3.68 × 10—10 V2 Vrms = 19.18 µV rms V 0 0 0 RK ǁ (R1 + RS) 2 1 + RS K S 1 2 S S 1 R2 (R1 + RS + RK) + R2 (R1 + R2 + RS) 2 RS + R 2 + (R 1 + R S ) ǁ R K 2 K Problem A.6 To find the noise figure, we first determine the noise power due to a source at the output, then due to the source and the network, and take the ratio of the latter to the former. Initally assume unmatched conditions. The results are V 2. due to RS , only = R2 ǁ RK 2 RS + R1 + R2 ǁ RK (4kT RSB) V 2. due to R1 and R2 = R2 ǁ RK 2 RS + R1 + R2 ǁ RK (4kT R1B) RK ǁ (R1 + RS) 2 V 2. due to RS , R1 and R2 = R2 ǁ RK 2 RS + R1 + R2 ǁ RK [4kT (RS + R1) B] + R2 The noise figure is (after some simplification) + (R1 + RS ) ǁ RK (4kT R2B) R1 F = 1 + RK ǁ (R1 + RS) 2 RS + R1 + R2 ǁ RK 2 R2 In the above, RS R2 + (R1 + RS) ǁ RK R2 ǁ RK RS Ra ǁ Rb RaRb = Ra + Rb Note that the noise due to R K 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 RS = Rin = R1 + R2 ǁ RK = R1 + R2RK R2 + RK and matching at the output requires that RK = R out = R2 ǁ (R1 + RS ) = R2 (R1 + RS) R1 + R2 + RS Next, these expressions are substituted back into the expression for F. After some simplification, this gives R 2R2 R (R + R + R ) / (R — R ) 2 R Note that if R1 >> R2 we then have matched conditi ons of RK =~ R2 and RS =~ R1. Then, the noise figure simplifies to F = 2 + 16 R1 R2 + (4kT R2B) F = 1 +
