EBK FUNDAMENTALS OF APPLIED ELECTROMAGN
7th Edition
ISBN: 8220100663659
Author: ULABY
Publisher: PEARSON
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Textbook Question
Chapter 6.5, Problem 6CQ
Contrast the operation of an ac motor with that of an ac generator.
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can you please show working and steps. The answer is 8kohms.
PSD
A certain signal f(t) has the following PSD (assume 12 load):
| Sƒ(w) = π[e¯\w\ + 8(w − 2) + +8(w + 2)]
(a) What is the mean power in the bandwidth w≤ 1 rad/sec?
(b) What is the mean power in the bandwidth 0.99 to 1.01 rad/sec?
(c) What is the mean power in the bandwidth 1.99 to 2.01 rad/sec?
(d) What is the total mean power in (t)?
Pav=
+
2T
SfLw) dw
- SALW)
An AM modulation waveform signal:-
p(t)=(8+4 cos 1000πt + 4 cos 2000πt) cos 10000nt
(a) Sketch the amplitude spectrum of p(t).
(b) Find total power, sideband power and power efficiency.
(c) Find the average power containing of each sideband.
Chapter 6 Solutions
EBK FUNDAMENTALS OF APPLIED ELECTROMAGN
Ch. 6.2 - Explain Faradays law and the function of Lenzs...Ch. 6.2 - Prob. 2CQCh. 6.2 - Prob. 3CQCh. 6.2 - For the loop shown in Fig. 6-3, what is Vemftr if...Ch. 6.2 - Suppose that the loop of Example 6-1 is replaced...Ch. 6.4 - Suppose that no friction is involved in sliding...Ch. 6.4 - Is the current flowing in the rod of Fig. 6-10 a...Ch. 6.4 - For the moving loop of Fig. 6-9, find I when the...Ch. 6.4 - Suppose that we turn the loop of Fig. 6-9 so that...Ch. 6.5 - Contrast the operation of an ac motor with that of...
Ch. 6.5 - Prob. 7CQCh. 6.5 - Prob. 8CQCh. 6.7 - A poor conductor is characterized by a...Ch. 6.8 - When conduction current flows through a material,...Ch. 6.8 - Verify that the integral form of Ampres law given...Ch. 6.10 - Explain how the charge continuity equation leads...Ch. 6.10 - How long is the relaxation time constant for...Ch. 6.10 - Determine (a) the relaxation time constant and (b)...Ch. 6.11 - Prob. 7ECh. 6 - The switch in the bottom loop of Fig. P6.1 is...Ch. 6 - The loop in Fig. P6.2 is in the xy plane and B =...Ch. 6 - A coil consists of 100 turns of wire wrapped...Ch. 6 - A stationary conducting loop with an internal...Ch. 6 - A circular-loop TV antenna with 0.02 m2 area is in...Ch. 6 - The square loop shown in Fig. P6.6 is coplanar...Ch. 6 - The rectangular conducting loop shown in Fig. P6.7...Ch. 6 - Prob. 8PCh. 6 - Prob. 9PCh. 6 - A 50 cm long metal rod rotates about the z axis at...Ch. 6 - The loop shown in P6.11 moves away from a wire...Ch. 6 - The electromagnetic generator shown in Fig. 6-12...Ch. 6 - The circular, conducting, disk shown in Fig. P6.13...Ch. 6 - The plates of a parallel-plate capacitor have...Ch. 6 - A coaxial capacitor of length l = 6 cm uses an...Ch. 6 - The parallel-plate capacitor shown in Fig. P6.16...Ch. 6 - In wet soil, characterized by = 102 (S/m), r = 1,...Ch. 6 - An electromagnetic wave propagating in seawater...Ch. 6 - At t = 0, charge density v0 was introduced into...Ch. 6 - If the current density in a conducting medium is...Ch. 6 - Prob. 21PCh. 6 - If we were to characterize how good a material is...Ch. 6 - The electric field of an electromagnetic wave...Ch. 6 - The magnetic field in a dielectric material with ...Ch. 6 - Given an electric field E=xE0sinaycos(tkz), where...Ch. 6 - The electric field radiated by a short dipole...Ch. 6 - A Hertzian dipole is a short conducting wire...Ch. 6 - In free space, the magnetic field is given by...Ch. 6 - The magnetic field in a given dielectric medium is...
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- Can you rewrite the solution because it is unclear? AM (+) = 8(1+ 0.5 cos 1000kt +0.5 ros 2000ks) = cos 10000 πt. 8 cos wat + 4 cos wit + 4 cos Wat coswet. -Jet jooort J11000 t = 4 e jqooort jgoort +4e + e +e j 12000rt. 12000 kt + e +e jooxt igoo t te (w) = 8ES(W- 100007) + 8IS (W-10000) USBarrow_forwardCan you rewrite the solution because it is unclear? AM (+) = 8(1+0.5 cos 1000kt +0.5 ros 2000 thts) = cos 10000 πt. 8 cos wat + 4 cos wit + 4 cos Wat coswet. J4000 t j11000rt $14+) = 45 jqooort +4e + e + e j 12000rt. 12000 kt + e +e +e Le jsoort -; goon t te +e Dcw> = 885(W- 100007) + 8 IS (W-10000) - USBarrow_forwardCan you rewrite the solution because it is unclear? Q2 AM ①(+) = 8 (1+0.5 cos 1000πt +0.5 ros 2000kt) $4+) = 45 = *cos 10000 πt. 8 cos wat + 4 cosat + 4 cos Wat coswet. j1000016 +4e -j10000πt j11000Rt j gooort -j 9000 πt + e +e j sooort te +e J11000 t + e te j 12000rt. -J12000 kt + с = 8th S(W- 100007) + 8 IS (W-10000) <&(w) = USB -5-5 -4-5-4 b) Pc 2² = 64 PSB = 42 + 4 2 Pt Pc+ PSB = y = Pe c) Puss = PLSB = = 32 4² = 8 w 32+ 8 = × 100% = 140 (1)³×2×2 31 = 20% x 2 = 3w 302 USB 4.5 5 5.6 6 ms Ac = 4 mi = 0.5 mz Ac = 4 ५ M2 = =0.5arrow_forward
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- An FDM is used to multiplex two groups of signals using AM-SSB, the first group contains 25 speech signals, each has maximum frequency of 4 kHz, the second group contains 15 music signals, each has maximum frequency of 10 kHz. A guard bandwidth of 500 Hz is used bety each two signals and before the first one. 1. Find the BWmultiplexing 2. Find the BWtransmission if the multiplexing signal is modulated using AM-DSB-LC.arrow_forwardAn FM signal with 75 kHz deviation, has an input signal-to-noise ratio of 18 dB, with a modulating frequency of 15 kHz. 1) Find SNRO at demodulator o/p. 2) Find SNRO at demodulator o/p if AM is used with m=0.3. 3) Compare the performance in case 1) and 2).. Hint: for single tone AM-DSB-LC, SNR₁ = (2m²) (4)arrow_forwardFind Va and Vb using Nodal analysisarrow_forward
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