circuits - 7.38

b)Put the differential equation in state variable form and a numerical method to produce a detailed sketch of the output voltage Vo. Assume the system starts at rest, and the input is Vi=5V.

14.

a)Write the differential equations for the system pictured below.

b)Put the equations in state variable form.

c)Use numerical methods to find the ratio between input and output voltages for a range of frequencies. The general method is put in a voltage such as Vi=1sin(___t), and see what the magnitude of the output is. Divide the magnitude of the output sine wave by the input magnitude. Note: This should act as a high pass or low pass filter.

d)Plot a graph of gain against the frequency of the input.

R1

R2

Vi

-

Vo

+

C

C=1uF

R1=1K

R2=1K

7.10 PRACTICE PROBLEM SOLUTIONS

1.

Rparallel	=	R1R2	Rseries	=	R1 + R2
		------------------
		R1 + R2
Cseries	=	C1C2	Cparallel	=	C1 + C2
		------------------
		C1 + C2
Lparallel	=	L1L2	Lseries	=	L1 + L2
		-----------------
		L1 + L2

circuits - 7.39

2.

Vo

-----

Vi

3.

4.

5.

6.

7.

	1
=	5--

··		· R		+ V		1		= V		1
V		+ V	--			-------				-------
	o		o L		o LC				i LC

a)

··

·

1

+ V

1

= V

1

V

o

+ V

o

-------

o

-------

i

-------

LC

RC

LC

b)

·

= Xo

Vo

·

= X

–1

+ V

–1

+ V

1

X

o

o

-------

o

-------

i

-------

RC

LC

LC

··	· 1		1	1	·	1
Vo	+ Vo CR---------	1	+ CR---------	+ Vo LC-------	= Vi CR---------
			2			1

··	·	R2R3C + L				R 2
Vo	+ Vo LC------------------------------( R		2	+ R )	+ Vo LC------------------------------( R	2	+ R )
				3				3

· R2R3
= Vi L--------------------------( R	2	+ R )
			3

a)

··

· 1

+

1

1

· 1

Vo

+ Vo

CR---------1

---------

+ Vo

-------

= Vi

CR---------1

CR2

LC

b)

ω n

=

1

ζ =

L( R1 + R2)

LC-------

2 ------

CR---------------1R2

circuits - 7.40

8.

(a.

(b.

Sum currents at node Vo
∑IVo = (---------------------Vo – Vi)		+	( Vo) DC +		(----------Vo)	= 0
DL					R
		2		VoDL
Vo – Vi + VoD			LC +	-------------R	= 0
1				1			1
Vo'' + Vo'	-------		+ Vo	-------	= Vi		-------
	CR			LC			LC

··	·	1			1		1
Vo	+ Vo	CR-------		+ Vo	-------	= Vi	-------
					LC		LC

circuits - 7.41

(c.

··

·

L

+ V

= V

··

( 10

–2

10

–6

·

10–2

+ V

= 10

V

( LC) + V

--

V

) + V

----------

o

o R

o

i

o

o 103

o

··

–8

·

–5

= 10

Vo(

10

) + Vo( 10

) + Vo

··

·

3

8

9

)

homogeneous;

Vo

+ Vo

( 10

) + Vo( 10 ) = ( 10

A2eAt + AeAt( 103) + eAt( 108) = 0

A

2

3

(

8

= 0

– 103 ±

( 103) 2

– 4( 108)

9987j

+ A( 10 ) +

10 )

A = ------------------------------------------------------------- = – 500 ±

e–500t cos ( 9987t + C )

2

V

h

= C

1

2

particular;

guess

Vp = A

( 0) + ( 0) ( 103) + ( A) ( 108) = ( 109)

A = 10

Vp

= 10

for initial conditions,

V

o

= C

1

e–500t cos ( 9987t + C )

+ 10

2

for t=0, Vo=2V

2 = C1e–500( 0) cos ( 9987( 0) + C2)

+ 10

–8 = C1 cos ( C2)

(1)

·

= –500C1e

–500t

cos ( 9987t + C2)

– 9987C1e

–500t

sin ( 9987t + C2)

Vo

for t=0, d/dt Vo=3V

3 = –500C1 cos ( C2)

– 9987C1 sin ( C2)

3 = –500C1 cos ( C2)

– 9987C1 sin ( C2)

3 =

– 4000 – 9987

–8

------------------- sin ( C )

cos ( C )

2

sin ( C2)

2

–3997

= tan ( C2)

C2 =

–0.050

------------------- =

-------------------

8( 9987)

cos (

C2)

C1

=

–8

=

–8.01

-------------------

cos (

C2)

V

o

= –8.01e–500t cos ( 9987t – 0.050)

+ 10

circuits - 7.42

9.

a)

V

·

( R

C) + V

R1

= 0

+ V

1

-----

i

i

o R

2

b)

V

=

·

( –CR ) + V

–R2

V

--------

o

i

2

i R

1

10.

Vo

=

–R2

-----

---------------------------------

Vi

R1 + DR1R2C

11.

–R2

V

=

·

–L

+ V

V

-----

--------

o

i

R

1

i

R

1

12.

·

+ V

R2

=

·

–R2

V

-----

V

i

--------

o

o L

R

1

13.

(a.

Create a node between the inductor and resistor Va, and use the node voltage method

∑IVA

∑IV–

=	( VA – Vi)					( VA – V– )	= 0			V– = V +	= 0V
	---------------------0.001D +					-------------------------1000
1000000( VA						– Vi) + VAD = 0				XXXXADD UNITSXXXXX
					1000000
VA = Vi				-------------------------------
				1000000 + D
=	(-------------------------V–	– VA)				+ (------------------------V– – Vo)		+ ( V	–	– V ) ( 0.000001D)	= 0
	1000					1000				o
							+ ( –Vo)
-----------( –1) V			-------------------------------		1000000				+ ( –V ) ( 0.000001D) = 0
1000		i						1000		o
			1000000 + D

Vo( – 1 – 0.001D)		1000000
	= Vi	-------------------------------
		1000000 + D

Vo( – 1000000 – D – 1000D – 0.001D2) = 1000000Vi Vo''( –10–9) + Vo'( –1.001( 10–3) ) + Vo( –1) = Vi