By Ian Hickman

This publication is meant for the working towards digital engineer explaining analog digital circuits as easily as attainable. Its goal is to take the reader inside of digital circuits explaining precisely what they do by using vector diagrams

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**Analog Electronics. Analog Circuitry Explained**

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**Extra resources for Analog Electronics. Analog Circuitry Explained**

**Example text**

So if the value at a particular frequency of a transfer function Μ Δ φ is expressed as A/j L φ , where A/j is the ratio Μ expressed in decibels, then the product M l Δ φ ι X M o z. Φ2 is simply expressed as (Μ„ + Μ,2)^(φ,+Φ2). , so / is small. e. 1h). At very high frequencies, the reactance of C is very low com pared with R; thus the current is virtually deter- 30 Analog Electronics mined solely by R. So each time we go to twice the frequency, Xc halves and so does the P D across it. 5 comes to - 6 (almost exactly), so the output is said to be falling by 6 d B per octave as the frequency increases.

1/7) = 1 0 ¿ - 9 0 ° when σ = - 1 / 7 , where we surmount the north face of the infinitely high F{s) mountain. Descending the western slope, the amplitude Μ falls as φ increases to —180°. Clearly the smaller the constant positive value of ω during our westward journey the higher the slope we surmount, and the more rapidly the phase twizzles round from 0° to - 1 8 0 ° in the vicinity of the p e a k . Keep this picture in mind, as we shall meet it again later. The infinitely high mountain is called a pole and occurs in the low-pass case of F(s) = {\/T)/[s + (1/7)] at s= - 1 / 7 + j O .

0 4 ) . When we first started learning algebra we would simply have said that the equation didn't factorize. However, j solves the problem: V ( - 2 . 0 4 ) = V ( - l ) V ( 2 . 428 286. So . = - ^ - ^ ^ ^ ^ ( ^ - ^ ^ - ^ ) m ) . 7 ± j 0 . 8a. Now you will recall that the magnitude of the transfer function at any frequency ω, denoted by a point on the ]ω axis, is proportional to the product of the reciprocals of the distances from the point to each of the poles. Consider the frequency ω = 0 . e. one-tenth of the resonant frequency.