Lecture-1

Lecture-1 - UNIVERSITY OF CALIFORNIA, SAN DIEGO Department...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
UNIVERSITY OF CALIFORNIA, SAN DIEGO Department of Electrical and Computer Engineering Bang-Sup Song 1 ECE163 Lecture #1: Feedback Amplifier Stability v i i i Network v i i i Network v o i o One-port network Two-port network Fig. 1.1: Driving-point and transfer concepts. Driving-Point and Transfer Functions Consider any networks with one port or two ports as shown in Fig. 1.1. On each port, we can define its terminal voltage and the current flowing into the terminal. In the one- port case shown on the left side, the ratio of the terminal voltage to the current can be defined. Rs () = v i s i i s , and Gs = i i s v i s . ( 1 . 1 ) The former is driving-point resistance, and the latter one is driving-point conductance. Their units are ± and 1/ ± , respectively. That is, if the input and output are referred to the same port, the term “driving-point” is used. On the other hand, in the two-port case shown on the right side, the following four ratios can be defined. A v s = v o s v i s , A i s = i o s i i s , = v o s i i s , and = i o s v i s , (1.2) where A v (s) and A i (s) are unit-less transfer functions defined as voltage and current gains, respectively. The latter two definitions are the same as in (1.1) for the one-port network, but they are called as trans-resistance and trans-conductance, respectively. The term “trans” is now used since the input and output ports are different. In steady-state small-signal analysis, impedances of reactive components such as inductor and capacitor are frequency-dependent as sL and 1/sC, where s is the complex frequency of j ² . The unit of ² is rad/sec. Note that the angular frequency ² is defined as the amount of angle rotation per second, while the ordinary frequency unit in Hertz (Hz) is defined as the number of rotations per second. Since one rotation of a vector covers an angle of 2 ³ radian, there is a 2 ³ difference between ² and f like ² = 2 ³ f. In general, all these transfer functions can be represented as a ratio of two polynomial functions N(s) and D(s) in steady state. Let’s consider a general transfer function H(s) as follows.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
UNIVERSITY OF CALIFORNIA, SAN DIEGO Department of Electrical and Computer Engineering Bang-Sup Song 2 ECE163 Hs () = Ns Ds = b 1 s m + b 2 s m ± 1 + ..... + b m a 1 s n + a 2 s n ± 1 + ..... + a n . ( 1 . 3 ) The solutions of D(s) = 0 are poles, and those of N(s) = 0 are zeros. These poles and zeros affect the magnitude and phase responses of the transfer function. For any networks to be (exponentially) stable, poles should be placed in the open left half complex s-plane. That is, the real part of the poles should be negative like Re [poles] < 0. Frequency Response If s is replaced by j ± , the steady-state frequency response of a transfer function is obtained. That is, the transfer function is defined as a complex function H(j ± ) at frequency ± . The complex function H(j ± ) can be represented by either of the following rectangular or polar form.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 14

Lecture-1 - UNIVERSITY OF CALIFORNIA, SAN DIEGO Department...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online