Tài liệu Circuits & Electronics P15 docx

6.002 CIRCUITS ANDELECTRONICS Second-Order Systems... Second-Order SystemsC 5V + – 5V CGS large loop 2KΩ 50Ω 2KΩ Demo Our old friend, the inverter, driving another.. Now, let’s try to s

6.002 CIRCUITS AND

ELECTRONICS

Second-Order Systems

Second-Order Systems

C

5V

+ –

5V

CGS

large loop

2KΩ

50Ω 2KΩ

Demo

Our old friend, the inverter, driving another

The parasitic inductance of the wire and

the gate-to-source capacitance of the

MOSFET are shown

[Review complex algebra appendix for next class]

S

Second-Order Systems

C

5V

+

–

5V

CGS

large loop

2KΩ

50Ω 2KΩ

Demo

+ –

2KΩ

B L

Relevant circuit:

S

Now, let’s try to speed up our inverter by

closing the switch S to lower the effective

resistance

t

vA

5

0

vB

vC

Observed Output

2 kΩ

2kΩ

vA

5

0

vB

vC

50 Ω

Huh!

v, i state variables

+

– ( t )

v

)

(t

vI

) ( t i

Node method:

dt

dv C t

i ( ) =

dt

dv C dt

v

v L

t

I − =

∫

∞

−

) (

1

2

2

) (

1

dt

v

d C v

v

I

v

v dt

v

d

dt

di L v

i dt v

v L

t

∫

∞

−

) (

1

Recall

First, let’s analyze the LC network

Recall, the method of homogeneous and particular solutions:

( ) t v ( ) t v

1

2

3

Find the particular solution.

Find the homogeneous solution.

L

4 steps

The total solution is the sum of the particular and homogeneous.

Use initial conditions to solve for the remaining constants.

Solving

And for initial conditions

v(0) = 0 i(0) = 0 [ZSR]

I

v

0

V

Let’s solve

I

v

v dt

v

d

For input

1 Particular solution

0 2

2

V

v dt

v

d

0

V

2

2

=

dt

v

d LC

Solution to

Homogeneous solution

2

Recall, v H : solution to homogeneous

equation (drive set to zero) Four-step method:

2

t j 1

H A e o A e o

v = ω + − ω

General solution,

Roots

LC

1

o = ω

Assume solution of the form*

A

? s

, A , Ae

so, LCAs2est + Aest = 0

* Differential equations are commonly

1

−

=

j LC

j

B

LC

s2 = − 1 characteristicequation

Total solution

3

Find unknowns from initial conditions.

t

j 2

t

j 1

0 A e o A e o V

) t (

) ( )

( )

v = P + H

0 )

0

v

2 1

0

0 )

0

i

t j o 2

t j o

1 j e o CA j e o CA

) t (

dt

dv C t

i ( ) =

so, 0 = CA1 j ωo − CA2 j ωo

A

V0 = 2

−

2

0 1

V

Remember Euler relation

(verify using Taylor’s expansion)

x j

x

e jx = cos + sin

x

e

cos

t sin CV

) t (

t cos V

V )

t (

LC

1

o = ω

Total solution

3

The output looks sinusoidal

(t v

0

2V

0

V

0

2

π

2

3 π

)

(t

i

o 0

0

2

π

2

3 π

Plotting the Total Solution

Summary of Method

1

2

3

Write DE for circuit by applying

node method.

and trial & error.

solve for remaining constants using

initial conditions.

Obtain characteristic equation Solve characteristic equation

for roots si .

terms.

D C A B

What if we have:

We can obtain the answer directly from

V

vC ( 0 ) =

0 )

0

C

i

C

–C

v

Example

We can obtain the answer directly from

t

j 2

t

j 1

C( t ) A e o A e o

V

vC ( 0 ) =

0 )

0

C

i

2

1 A A

V = +

o 2

o

1 j CA j CA

or

2

2 1

V A

( j t j t )

2

V

or

t cos V

t sin CV

V

vC ( 0 ) =

0 )

0

C

i

C

–C

v

Example

o

ω π

2

C

v V

C

i

t

o

ω π

2

o

−

o

Example

2 2

2

2

1 2

1 2

1

CV Li

Notice

Energy

2

2

1

C

2

2

1

L

t

o

ω π

2

C

E

2

2

1

CV

t

o

ω π

2

L

E

2

2

1

CV

Total energy in the system is a constant,

but it sloshes back and forth between the

Capacitor and the inductor

RLC Circuits

add R

no R

+

+ – ( t )

v

)

(t

vI

) ( t i

) ( t v

t

Damped sinusoids with R – remember demo!