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schematic and parts layout diagrams when installing electrolytic capacitors to make sure that you have installed them correctly The plus (+) marking on the capacitor will point to the most positive portion of the circuit Let s go ahead and identify the non-polarized capacitors first Check the capacitor code chart and make sure that you can identify each of the nonpolarized capacitors In groups of two or three capacitors install them on the PC board When you have determined that you have the installed the correct values into their respective location on the circuit board, you can solder the capacitors in place on the board Remember to trim the excess capacitor leads from the circuit board Once you have the first grouping of capacitors installed you can move on to installing another group of non-polarized capacitors, until you have installed all of the non-polarized capacitors Next, find the electrolytic capacitors from the parts pile and we will install them in their proper locations Refer now to the schematic and parts layout diagrams so that you can orient the capacitors correctly when mounting them on the PC board Install two or three electrolytic capacitors and then solder them in place on the PC board Remember to cut the extra lead lengths flush to the PC board Identify and install the remaining electrolytic capacitors on the board and solder them in place Trim the component leads as necessary In order to avoid damaging the receiver, it is important to install the capacitors, diodes as well as the

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This two state model corresponds well with the data presented in Figure 83 The frequency error is modeled as a constant plus random walk (ie,

Question 7-7

162 Review Questions and Answers Answer 7-7

slowly changing constant) The phase error is modeled as the integral of the frequency error (ie, ramp) plus a random walk This is shown in block diagram form in Figure 84

A calculator can be used for this purpose, but it must be able to display a lot of digits! Otherwise, old-fashioned long division is the best way to find these: 1/2 = 05 1/3 = 0333333333333 1/4 = 025 1/5 = 02 1/6 = 0166666666666 1/7 = 0142857142857 1/8 = 0125 1/9 = 0111111111111 1/10 = 01

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semiconductors with respect to proper polarity marking on each of the components Electrolytic capacitors have either a plus or minus marking on them and must be installed correctly by observing the polarity of the capacitor with respect to the PC board Next, we will install the semiconductors Locate the semiconductor pin-out diagram shown in Figure 22-5, which will help you install the semiconductor devices properly There are a number of diodes in the Radio Jupiter project, including Zener and varactor diodes and they all have polarity In order for the circuit to work properly, these devices must be installed with respect to their polarity Diodes will generally have a white or black colored band at one edge of the diode body The band usually denotes the cathode end of the diode Transistors generally have three leads a Base, Collector and Emitter; pay particular attention to these leads when installing the transistors Integrated circuits always have some type of markings which indicate their orientation You will find either a plastic cut-out on the top of the integrated circuit package or an indented circle next to pin 1 of the IC Be sure to use these markings when installing the integrated circuits It is a good idea to incorporate integrated circuit sockets as a low cost insurance policy, in the event of a circuit failure, if it ever happens It is much easier to remove an old dead IC, and simply insert a new one rather than trying to undo many pins and risk damage to the circuit board There are a number of coils in this project, and although there is no polarity to the coils, you will need

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Figure 84: Continuous-time two state clock model For this clock model, 0 1 0 0 For the discrete-time sampling period Ts , the state space model is F= v1 (k + 1) v2 (k + 1) = 1 Ts 0 1 v1 (k) v2 (k) + d (852)

Question 7-8

Because F2 is identically zero, Qd can be calculated exactly using the rst three terms of eqn (4119): cov( d ) = Qd =

Answer 7-8

Table 16-2

First, we must identify the sequence of digits that repeats Here, it s 4, 5, 8, and 7 There are four digits in the sequence, so we create a fraction with a denominator having four digits, all 9s Then we make the repeating sequence into a four-digit numeral and use it as the numerator of the fraction That gives us 0458745874587 = 4,587/9,999

(853)

Question 7-9

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