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1 Second-Order Systems 6.002x CIRCUITS AND ELECTRONICS [Review complex algebra Appendix C in textbook]

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1  

Second-Order Systems

6.002x CIRCUITS AND ELECTRONICS

[Review complex algebra Appendix C in textbook]

Motivating Example – Slow Case Our old friend, the inverter, driving another.  

3  

Observed Output - Slow

vA

5

0

vB

0

vC

0 t

t

t

C A B

5V

+!–!

5V

CGS

2KΩ 2KΩ

5

5

Demo  

Motivating Example – Fast Case

C A B

5V

+!–!

5V

CGS

2KΩ 2KΩ

Demo  

5  

Observed Output – Fast Case

vA

5

0

vB

0

vC

0 t

t

t

C A B

5V

+!–!

5V

CGS

2KΩ 50Ω

2KΩ S  

5

5

Fast Case – What’s Really Going On

C A B

5V

+!–!

5V

CGS

2KΩ 50Ω

2KΩ S  

Second-Order Systems

+!–!5V CGS

2KΩ B

L

Relevant circuit: C A B

5V

+!–!

5V

CGS

large loop

2KΩ 50Ω

2KΩ S  

L

First, let’s analyze the LC network

9  

Analyzing the LC network +!–! C

L +!–!v(t)

i(t) v

vI(t)

vI

10  

Solving

11  

Let’s solve

12  

1   Particular solution

13  

1   Particular solution

IPP Vv

dtvdLC =+2

2

14  

Homogeneous solution 2  

15  

02

2

=+ HH v

dtvdLCSolution to

Homogeneous solution 2  

Recall, vH : solution to homogeneous equation (drive set to zero)

Four-step method:

Assume solution of the form* 2A  ?s,A,Aev st

H ==*Differential equations are commonly solved by guessing solutions

16  

02

2

=+ HH v

dtvdLCHomogeneous solution 2  

17  

Total solution 3  

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Total solution 3  Find unknowns from initial conditions.

tjtjI

oo eAeAVtv ωω −++= 21)(v(t) = vP(t) = vH (t)

19  

Remember Euler relation Total solution 3   ( )tjtjI

Ioo eeVVtv ωω −+−=

2)(

20  

Plotting the Total Solution

0 0

v(t) i(t)

tVVtv oII ωcos)( −=

Demo  

21  

Summary of Method Write DE for circuit by applying node method.

Find particular solution vP by guessing and trial & error.

Find homogeneous solution vH

Total solution is vP + vH , then solve for remaining constants using initial conditions.

Assume solution of the form Aest .

Obtain characteristic equation.

Solve characteristic equation for roots si .

Form vH by summing Ai esit terms.

1

2 3

4

D

C

A

B

22  

What if we have:

Example

23  

We can obtain the answer directly from the homogeneous solution (VI = 0).

tj2

tj1C

oo eAeA)t(v ωω −+=

Example C L

+!

–!iC

vC

vC (0) = V

iC (0) = 0

vC (0) = V

iC (0) = 0

24  

Example  

π2

iC

π2

vc V

25  

Energy

toωπ2

EC

toωπ2

EL

26  

Next, introduce R: RLC Circuits

More in the next sequence! If you are impatient, see A&L Section 13.2

+!–! C

L

+!

–!vI (t)

i(t)

v(t)

v(t)

t