Analysis of Alternating Current

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Digital Systems II

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In applying the superstition theorem, (c) - the sources are considered one at a time with all others replaced by a short. A Thevenin ac equivalent circuit always consist of an equivalent ac voltage source. ( c) and an equivalent impedance. One circuit is equivalent to another when ( d)-The circuits produce the same output voltage. The Thevenin equivalent voltage is (b) the short circuit voltage. The Thevenin equivalent impedance is the impedance looking from ( a) the source with the output shorted. A Norton ac equivalent circuit always consist of. (a) an equivalent ac current source in series with an equivalent impedance. The Norton equivalent current is ( a) the total current from the source. The complex conjugate of 50Ω+j100Ω is ( a) 50Ω- 50Ω. ⅉ In order to get maximum power transfer from a capacity source, the load must (b) have an impedance equal in magnitude to the source impedance. 1)Using the superposition method, calculate the current through R3 in figure 20-44 Solution. IR3 = IV1 + IV2 = (0.31 + j0.366 + 0.767 + j0.226)mA IR3 = (1.077 + j0.592)mA IR3 = 1.23 < 28.77mA 7)For each circuit in figure 20-49, determine the Thevenin equivalent circuit for the portion of the circuit viewed by RL VC = ( Xc ← 90 R 1 − jXc 2 ) VS ¿ ( 75 ← 90 Ω 100 Ω − j 75 Ω ) ( 25 < 0 V ) 0.6 ← 53.13 ¿ ( 25 < 0 V ) Vc = 15 ← 53.13 V VTH = 15 ← 53.13 V B) VAB = ( jXL 1 jXL 1 + JxL 2 ) VS ( J 400 Ω J 400 Ω + J 580 Ω ) ( 3 < 0 V ) = (0.4082)(3<0V) VAB = 1.22<0V VTH = 1.22<0V C) VR2 = ( R 2 R 1 + R 2 ) ( V 1 + V 2 ) = ( 100 kΩ 100 kΩ + 100 kΩ ) ( 15 < 0 + 10 < 30 V ) =(0.5)(24.2<11.93V) VR2 = 12.09<11.93V VTH= 12.09<11.93V 12)Using Norton’s theorem, Find the current through the load resistor RL in figure 20- 50 IRL = (ZN / ZN + RL)IN ( 39.63 <− 28.26 kΩ 39.63 ← 28.26 kΩ + 100 kΩ ) ( 0.402 ← 45.2 mA ) IRL = (0.3<-20.34)(0.402<-45.2mA) IRL = 0.117<-65.5mA 14)For each circuit in figure 20-53, maximum power to be transferred to the load R L .Determine the appropriate value for the load impedance in each. a) ZL = 6.8+j11.29kΩ b) ZL = 8.2+j5kΩ c) ZL = 50+j0.3Ω

EET | 310 | Digitial Electronics | homework | In applying the superstition t

Completion Complete each statement. 1. A satellite in geosynchronous orbit takes ______24______________ hours to complete one orbit. 2. The __ uplink__________________ is the signal path from the earth station to the satellite. 3. The __ _downlink_________________ is the signal path from the satellite to the earth station. 4. A satellite in a ___geostationary_________________ orbit appears to stay directly above one spot on the equator. 5. All satellite orbits are ________elliptical____________ in shape. 6. The ____ perigee________________ is the distance of a satellite's closest approach to the earth. 7. The ____ apogee________________ is a satellite's farthest distance from the earth. 8. Satellites using the _____Ku_______________ band operate on 12 GHz. 9. In a frequency-hopping CDMA system, when no two transmitters use the same frequency at the same time the PN sequences are said to be ____ orthogonal________________. 10. In FDM, each signal uses part of the bandwidth _______all_____________ of the time. 11. In TDM, each signal uses all of the bandwidth _____part_______________ of the time. 12. Using CDMA on a radio channel, all signals can transmit ______all______________ of the time. Short Answer 13. What does Hartley's Law tell us about the relationship between time and bandwidth for digital transmission? Hartley’s Law describes that with the more available bandwidth, the less time it takes to send a certain amount of information. The more bandwidth available the higher the possible bit rate. 14. How many signals could fit into 1 MHz of bandwidth if each signal required 100 kHz of bandwidth and the separation between adjacent channels was 10 kHz? Nine signals could fit into 1 MHz of bandwidth if each signal required 100 KHz of bandwidth. 15. Why is it difficult to jam a spread-spectrum signal? Jamming requires an interference signal of sufficient power in the same part of the spectrum the information signal takes up. Because a spread-spectrum signal is known as spread out over a very wide bandwidth, jamming can interfere with only a small fraction of the total signal. 16. Why is it difficult to eavesdrop on a spread-spectrum signal? In a spread-spectrum transmission, the signal power at any given frequency in its band is so low that it is hard to differ it from noise. An eavesdropper would not know a signal was being sent. And without knowing the exact sequence being used, it is virtually impossible to de-spread the signal. 17. What is meant by "orthogonal sequences" in CDMA? During the transmission the PN sequences determine which parts of the bandwidth the spread-spectrum signal will take up. A good example of this would be : If you have two sequences, PN1 and PN2. Assume that the transmission causes PN1 to occupy frequencies f11, f12, f13… Now look at PN2. Suppose the transmission causes the sequence PN2 to occupy frequencies f21,f22, f23… If PN1 and PN2 have no frequencies in common these two PN frequencies are going to be “orthogonal sequences”.

EET | 280 | Wireless | homework | Completion Complete each state

Unit 3 Review Assignment 2. The Boolean expression A + B + C is… a sum term 4. The domain of the expression A BCD + A B + CD + B is… A, B, C, and D 6. According to the associative law of multiplication… A(BC) = (AB)C 8. Which one of the following is not a valid rule of Boolean algebra? A = A 10. According to De Morgan’s theorems, the following equality(s) is (are) correct: all of these 12. An example of a sum-of-products expression is AB + AC + ABC 14. An example of a standard SOP expression is AB + AB + AB 16. In a 4-variable Karnaugh map, a 2-variable product term is produced by a 4-cell group of 1s 18. A 5-variable Karnaugh map has thirty-two cells 20. VHDL, Verilog, AHDL, and ABEL are types of hardware description languages

EET | 130 | Digital Systems II | assessment | Unit 3 Review Assignment 2. Th

Unit 1 Review Assignment Modulators are devices used to convert an information signal to a signal compatible with the medium over which it is being transmitted. A transceiver is a form of communication equipment package in which both transmitter and receiver are in a single unit.

EET | 280 | Wireless | homework | Unit 1 Review Assignment Modul

Self Test – Chapter 3: Logic Gates 2. An inverter performs an operation known as? D inversion of complementation (Floyd, pg.114) 4. The output of an OR gate with inputs A, B, and C is a 1 (HIGH) when? E The output of an OR gate is 1 (HIGH) when any OR all of the inputs is 1 (HIGH) 6. A pulse is applied to each input of a 2-input NOR gate. One pulse goes HIGH at t = 0 and goes bac

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