{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

EE 230 Lecture 6 Spring 2010

# EE 230 Lecture 6 Spring 2010 - EE 230 Lecture 6 Linear...

This preview shows pages 1–12. Sign up to view the full content.

EE 230 Lecture 6 Linear Systems Poles/Zeros/Stability Stability

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

View Full Document
Quiz 5 A system has the transfer function T(s) Determine the poles of the system. () 1 4 s Ts 10 7s s + = ++
And the number is ? 1 3 8 4 6 7 5 2 9

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

View Full Document
And the number is ? 1 3 8 4 6 7 5 2 9 ?
Quiz 5 A system has the transfer function T(s) Determine the poles of the system. () 1 4 s Ts 10 7s s + = ++ Solution: 71 0 2 1 1 4s 4 4 s 10 ss s ⎛⎞ + + ⎜⎟ ⎝⎠ == + + 10 5 2 11 44 s7 s s 2 s + + Poles at s = -2 and s = -5

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

View Full Document
Step Response of First-Order Networks IN OUT Many times interested in the step response of a linear system when the system is first-order X OUT (t)=? 0 C t-T t OUT X = F + (I-F)e I is the intital value, F is the final value and t C is the time constant For any first-order linear system, the unit step response is given by Review from Last Time
Step Response of First-Order Networks X OUT t I F T 0 T 0 + t C 0 C t-T t OUT X = F + (I-F)e Review from Last Time

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

View Full Document
Step Response of First-Order Networks 0 C t-T t OUT V = F + (I-F)e -1 C t= - p=RC Example: () K Ts = s-p Obtain the step response of the circuit shown if the step is applied at time T=1msec and prior to V OUT (t)=0 for t<1msec. Assume R=1K, C=0.1uF Solution: 1 1+RCs 1 RC 1 s+ RC 1 p = - RC F=1V I=1V t-.001 RC OUT V = 1 + (-1)e t-.001 RC OUT V= 1 - e This is first order and of the form: Thus, the output can be expressed as: Review from Last Time
Impedance and Conductance Notation Circuit Analysis with Impedance Notation (Z) and Conductance Notation (G) Ohms Law V=I Z I=V G KCL ( ) X12 3 K 1 12 23 3 k k V G +G +G +. ..+G = V G +V G +V G +. ..+V G 1 kk Xi i i i=1 VG = V G i = ⎛⎞ ⎜⎟ ⎝⎠ ∑∑ KCL is often the fastest way to analyze electronic circuits G 2 3 k Node with conductance notation Conductance notation is often much less cumbersome than impedance notation when analyzing electronic circuits Why? Why? Formally: Review from Last Time

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

View Full Document
Poles and Zeros of Linear Networks () m i i i=0 n i i i=0 as Ts bs = For any linear system, T(s) can be expressed as where a i and b i are all real, , , and n m Numerator often termed N(s) Denominator often termed D(s) Definition: The roots of D(s) are the poles of T(s) and the roots of N(s) are the zeros of T(s) The poles of T(s) are often termed the poles of the system m i i i=0 n i i i=0 Ns Ds == Can always make b n =1 n b 0 m a 0 Linear System X IN X OUT Review from Last Time
Step Response of First-Order Networks ( ) I 0 pt -T OUT X=

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 33

EE 230 Lecture 6 Spring 2010 - EE 230 Lecture 6 Linear...

This preview shows document pages 1 - 12. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online