Exam (elaborations)
Solution Manual for Principles of Communications 7th Edition / All Chapters 1 - 12 / Full Complete 2023
Solution Manual for Principles of Communications 7th Edition / All Chapters 1 - 12 / Full Complete 2023
[Show more]
Connected book
Book Title:
Author(s):
- Edition:
- ISBN:
- Edition:
4 reviews
By: leonsobeloff • 1 month ago
By: abello702 • 3 months ago
By: jeff38459416 • 1 year ago
typesetting is terrible
By: ma11 • 1 year ago
By: NurseGrades • 1 year ago
Thank you
Content preview
√ 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 +
The benefits of buying summaries with Stuvia:
Guaranteed quality through customer reviews
Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.
Quick and easy check-out
You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.
Focus on what matters
Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!
Frequently asked questions
What do I get when I buy this document?
You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.
Satisfaction guarantee: how does it work?
Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.
Who am I buying these notes from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller NurseGrades. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for $12.99. You're not tied to anything after your purchase.
Can Stuvia be trusted?
4.6 stars on Google & Trustpilot (+1000 reviews)
73918 documents were sold in the last 30 days
Founded in 2010, the go-to place to buy study notes for 14 years now