370

P13.4-16 P13.4-17

We require

p

p

k C

Then R R and C

2001

5001

34 50

1 5 50 0 04

1

1 1

2

2 2

1 2

1 1 2 2

= =

= =

= = == = = =

C R

C R

dB R

Pick C F k k Fµ µ. , . .Ω Ω

H jj

j j j( )

( )

( )( )( )ω ω

ω ω ω = +

+ + +

10 50 1

2 1 50 1 80 1

Pick the appropriate circuit from Table 13.4-2.

Hk

where

R

pR

( )ωω ω

=

1+ jp

1+ jp

k = C R

p 1

C

= 1

C

1 2

1 2

11

2

−

=1

22

371

P13.4-18 P13.4-19

ϕ ω ω ω ω = − − −− − − −tan tan tan tan1 1 1 150 2 20 80

At

ω

ω

=

=+

=+

=

⇒ = = −

−

−

−

10 000

1

1

1

1 10 10 1010

20 20 10 40

22

4 3 62

2

2

,

log log

HR C

H dB

H ( )( )

( ),

ω ωω ω ω= = −

+= −

+

V

V

R R

j R C j

o

s

2 1

2110

110 000

HR R

j R CR R

j R CTotal1 22 1

21 22

2 12

2

2

11

1, , ( ) H H = −+

⇒ = =+

ω ω

At low frequency

H R R

Now also R C let C F

R k

T T

,

,

,

( ) ( )

for H need R R

so

R

= ∴ = == = =

∴ = = =−

2 12

1 2

1 2

1 26

1

1000 1 1

1 1000 10 1

ω µ

Ω

(a)

(b)

(a)

(b) 10=20dB

(c) 10,000 rad/sec

372

P13.4-20 P13.4-21

Assume V then V V V

ji o

Use s

~ ,− = ==

+ −0

ω

Voltage divider yields : V V V

o

s o o

sCR sC

sRC V

KCL at V V V mR V V R V V snC

=+

⇒ = +

− + − + − =

1 1 0

1 1 1 1

11

1

0

/

: ( )

⇒ = =− +

=

=+

H j VV j Q mnRC

mn

m

o

so o

( )ωω ω ω ω

ω where

Q

1

1

1

1

2 0

Plugging V into above yields

1

VmR

sCsC

ms nRC

V

mR

V

V s m RC nmR C s

os

o

s

1

1

1 1

2 2

2 2 2

+ + +

=

∴ =+ + +( )

373

Section 13-5: Resonant Circuits P13.5-1

Vj C

Rj C

V

V V

Rj C V V

j C R j C R C R V Vo a

a sa o

o o s

( ) ( )

( ) ( )( ( ) ( ))

( )( )

ω ω

ω

ω

ω ω ω ω ω

ω ω ω

V j

=+

= − + −

⇒ + + = +

1

1

0

1 1

2

2

11

1 1 2 2 1 1

T(V

V C R j C C R R j

der poles with

o

s

o

ω ωω ω ω ω ω

ω δ

)( )

.

.

Second or and

= =+ −

=− + +

= =

1

1

1

8 1

0 42 2

21 2 1 2

2

ω06

1 1

1120

130

10

60

x

k rad= =

=−LC

sec

Q CL

Q Qk

Q Qk

BW RC

= =×

=

= − +

+ =

= +

+ =

= =×

=

−

−

R

rad s

rad s

k rad s

10 000

130

10

1120

20

2 25852

2 26152

1 1

100001

3010

3

6

10 0

2

02

20 0

2

02

6

,

.

.

ω ω ω ω

ω ω ω ω

Notice that BW = = Q2

0ω ω ω− 1

374

P13.5-2 P13.5-3 P13.5-4 P13.5-5 P13.5-6

R H

rad s

Hk

Q

At rad s H

Q

= = =⋅

=

=

=

+ −

= = =

=

+ −

⇒ =

−

−

k

, so

Q

( )

( )

. ( ).

,

..

ω

ω

ωωω

ωω

ω ω

0 3

0

2

0

02

3

22

8

20 10400

1000

1

897 64

2010200

200400

1897 61000

1000897 6

8

Ω

Now

LC

CL

11000

400 8

0

Q

L = 50 mHC = 20 F

= =

= =

⇒ω

µ

ω05 41

101

10 10 rad s Q BW rad s= = = = = =LC R

L

C

R

L, ,

ω04 31

101

10 10 Q BW = = = = = =LC

rad sR

L

C

R

Lrad s, ,

R

CF

LmH

= =

= = ⇒ =

⋅= = ⇒ =

−

Z

BW C

L

ω

µ

ω

0

60

100

1100

500 20

1

20 102500 8

Ω

R Y

LL H

CF

= =

= = ⇒ =

= = ⇒ =

BW

C

1100

100500 0 2

1

0 22500 0 8

0

0

ω

ω µ

Ω

Ω.

..

375

P13.5-7 P13.5-8

Y C R j L R

R R CLR j L CR R

R R j L

R j L

R j L

R R R CLR L L CR R j R LCR R j L R R CLR

R R L

ω ωω

ω ω

ω

ω

ω

ω ω ω ω ω

ω

j

= ++

+

=+ − + +

+× −

−

=+ − − + + − + −

−

1 1

1 2

1 22

2 1 2

2 1

1

1

1 1 22

22

1 2 1 1 2 1 22

2

2 12 2

( )

( )

ω ω ω

ω

ω

is the fr which the imaginary part of is zero

rad sec

=

− + − =

= − =

0

1 1 2 1 2 02

2

02 1

22

22

0

12 9

equency at Y

R LCR R L R R CLR

LR CR R

CL RM

:

.

( )( )( )

( )( ) ( )100 100

5100 2 135 10 2 135100 100 1000 0 1000 0 1000 90

100 100 100 2 135 100 50 2 135100100 100

| | 1000

j

jVo j jj

j

V Vo

−° °∠− ∠−− °° °= ∠ = ∠ = = ∠° °− ∠− + ∠−+

−

∴ =

(a) Using voltage divider

(b) Do a source transformation to obtain

376

P13.5-9 P13.5-10

So have an RLC resonant circuit with ω0 1 400= =LC

Z j LG j C

R

R G LC j LG CR

G j C

~; R G

= + ++

=

=+ − + +

+

1

2

2 2

1 22

2 1

2

11

1

ωω

ω ω ω

ω

or LG CR

R G LC

CG

thus LG CR

R G LC

CG

C LG

LCC G L

tan tan

&

− −++ −

=

++ −

= ⇒ = − >

1 2 1

1 22

1

2

2 1

1 22

2

2 22

2 22

1

1

ω ωω

ω

ω ωω

ω ω

with R rad s

C L

LCC mF L mH

and C G L checks

1 2 0

02 4

2

22

1 100

10 10 5

R and

if

= = =

= = − = ⇒ =

>

Ω ω

ω ,

Z j

in

in

LR j C

R j C

ZR RLC j L

j RC

= ++

=− +

+

ω ωω

ω ω

ω

/1

1

2

∴ =− +

+ | |Z

R RLC L

RCin

ω ω

ω

2 2 2

21

at resonan Zce = Z 0~

∠ °

(a)

Therefore the circuit is operated at resonance.This means that the L - C parallel combination has an overall Z =

and hence I = I . When R is suddenly changed from 100 to 1 k , due to the resonance condition, V R I

suddenly increases by a factor of 10. This in turn causes very large (equal & opposite ) currents in the L& C as the

capacitor voltage is forced to abruptly change. Thus very large currents radiate electromagnetic radition and

thus see sparks. Clearly we need a variable capacitor that varies as R varies such that when R 100 ,

and thus have V = R I constant while both R and I change abruptly.

s o L o L o

L L

o o L o L o

∞=

≠≠

Ω Ω

ΩΩ Ω

377

P13.5-11

at ω = =+

11

1 2LC Z

C L R C Lin, | |

( )

Let V

R

ω ω

ωωω

ω ω

ωθ θ

ωθ θ

A A and V B

Y I

V

V V

V

A B

AR

A Bcos j Bsin

AR

Y A Bcos Bsin

AR

2

2

= ∠θ = = ∠θ

= =

−

= − ∠θ

= − −

=− +

°

| |

2 2

(b)

(c)

378

PSpice Problems SP 13-1

3 600 10 0.6

w 2 3.1823 20 sec

g

at the frequency

radπ

−= × =

= × =

This simulation shows that the gain is

379

SP 13-2

The phase shift is 45

2 3.1831 20 sec

as required.

at the frequency

radω π

°−

= × =

Here is a simulation of the circuit when R = 50k

380

SP 13-3

This simulation shows that the gain of the circuit is 0.1 at 10Hz and 0.995 at 10,000 Hz.

To satisfy the specifications on the corner frequencies, the gain must be ≤ 1.414 at 200 Hz and ≥ 7.07 at 2000 Hz.

Both conditions are met.

The circuit satisfies the specifications.

381

SP 13-4

16 420

dB dB

decadedB decade

− − =( )

This simulation shows

• The high-frequency gain is 33.928 ≅ 34dB

• The slope of the low-frequency asymptote is

382

SP 13-5

The peak of the frequency response is 72dB = 4000 at 2.25Hz = 14130rad / sec. So k = 4000 and

and are ident

ωω ω

ω

ω ω

0

1 2

0

2 1

0

2 1

14130

72 3 69

2250

2332 4 2172114

=− =

=−

=−

=−

=

ified as the frequencies where the gain is dB

Q f

f f . .

383

SP 13-6

R3 and the dependent source E1 model an ideal op amp. V(R is v V is v V is the an4 1 7 1 60: ) ( ). ( : ) . ( : )s t R t V t swer given in this solution manual. After the transient part dies out, V(R7:1) is identical to V(V6:t). The answer is

correct.

384

SP 13-7 SP 13-8

Is 1 0 ac 1 R1 2 0 600 C1 1 0 114n C2 2 0 37.9n L1 1 2 36.37m .ac dec 50 100 10k .probe .end

V1 1 0 ac 1 R2 1 2 1.35k L3 2 3 500m R4 3 0 47k E5 4 0 3 0 10 R6 4 6 1 R7 4 5 213m C8 5 6 353u C9 6 0 3.53m .ac dec 50 20 20k .probe .End

385

When 630 rad / sec, T which agrees with the tabulated values of T( ) corresponding to

= 200 and 400 rad sec.

ω ω ω

ω

< − ~ 1

10

SP 13-9 Verification Problems VP 13-1

VP 13-2

BWQ

= = = ≠ω0 10 00070

143 714 rad s,

.

When / sec, T which agrees with the tabulated values of |T( | corresponding to

= 12600, 25000, 50000 and 100000 rad sec

ω ω ωω

> −6300 rad 1 ~ )

.

Is 1 5 ac 1m C1 1 5 2.53n R1 1 2 1 L1 2 5 10u L2 3 0 10u R2 3 4 1 C2 4 0 2.53n K1 L1 L2 0.06 Rdummy 5 0 1000k

.ac dec 200 900k 1100k

.probe

.end

At ω = 630 we expect |T(ω)| = -3dB = 0.707. This agrees with the tabulated value of |T(ω)| corresponding

to ω = 6310.

At ω=630 we expect |T(ω)| = -20 + 3=-17dB = 0.14 which agrees with the tabulated values of |T(ω)|

corresponding to ω =400 and 795 rad/s.

This data does seem reasonable.

This report is not correct.

386

VP 13-3 VP 13-4 Design Problems DP 13-1

ω01

10 159

120

500 79 6

rad kHz

rad Hz

= = =

= =

= = =

LCk s

Q R

LC

BWRL

s

.

.

H

Z

ω

ω

ω = −+

+

=

=

=

k

jz

jp

where

k R

R

C R

p C R

1

1

1

1

2

1

1 1

2 2

The reported results are correct.

The network function indicates a zero at 200 rad/s and a pole at 800 rad/s. In contrast, the Bode plot indicates a

pole at 200 rad/s and a zero at 800 rad/s.

The Bode plot and network function don’t correspond to each other.

Pick an appropriate circuit from Table 13.4-2.

387

DP 13-2

2 1

1 2

1 1 2 2

1

11 2 1 2

1

The specifications indicate that

2 k , 5 k ,

1 12 1000 2 10,000

2 2000 rad . 0.05 .

1 1.592 k , 2 3.183 , 0.0

R p C

R z C

z and pC R C R

Try z s Pick C F Then

CR R R k C

pC z kk

π π

π µ

= = = =

⋅ < = ⋅ > =

= ⋅ =

= = Ω = = Ω = =

2 2

1

1: p 31.42 rad s 2 10, 000

F

Check kC R

µ

π= = < ⋅

HV

V

j CR

o

s

( )( )

( )ω ω

ω

ω

=

= 1

HV

V

j CR

j Lj C

R

R

j CR

j LR

j CR

LC

jRC LC

Pick LC

s

H LC

LCj

LC RC LC

o

s

( )( )

( )

||

||

( ) .

( )

ω ω

ω

ω

ωω

ω

ωω

ω ω

ω π ω ω

ω

= =

+

= +

++

=− + +

= = ⋅ =

=− + +

1

1

1

1

1

1 1

12 100 10

1

1 1 1 1

2

03

0

0

rad When

So

HCL

We require

dB HCL

CL

Finally

LC

CL

C nF

L mH

( )

. ( )

( )

.

.

.

ω

ω

π

0

0

3

3 0 707 1000

12 100 10

0 707 1000

113

2 26

R

R

=

− = = = =

= ⋅

=

⇒=

=

388

DP 13-3

R k

R k

R k

R M

R M

R k

C F

C F

1

2

3

4

5

6

1

2

10

866

8 06

1

2 37

499

0 47

01

========

ΩΩΩ

ΩΩ

Ω

.

.

.

.

µµ

⇒ = − +

V sin f

z

RV

R

V

R31

1( )

⇒ = − = −+

V o sf

isV

Z

ZV

R

R C R s5

4 1 5

11

2( )

⇒ = − V Vfs

oC R s1

36

( )

Solving (1) (3) for VV yields

VV

n the values for the resistors & capacitors, can draw

o

in→

=+ +

o

in

R

R R Cs

sR C

sR

R R R C C

plugging i

3

1 4 1

2

5 1

3

2 4 6 1 2

1

ckt A is inverting summer

ckt B is first order LPF

Continued ckt C is an integrator

V

V

dB)

0

in

(

389

DP 13-4

where

K R C , pC R1 2 1 1

1 1

1= =

We require

10

2001

5001

1 2 2 14

3

11 1

22 4

K K R CR

R

pR C

pC R

= − =

= =

= =

where

KR

R p

C R24

32

2 4

1= − =,

H Kj

j1 1

1

1( )ω ω

ω=−+

p

HK

jP

22

2

1( )ω ω=

+

Pick C F Then R p C

Pick C F Then R p C

Next R

R

R

R

Let R and R

1 1

1 1

2 4

2 2

2

3

6 3 2

3

2 3

11

5

011

20

10 10 20 10 500

500 1

= = =

= = =

= ⋅ ⇒ =

= =

−

µ

µ

.

. .

( )( )

k

k

k k

Ω

Ω

Ω Ω

390

DP 13-5

Pick C Fp C

k

Pick C Fp C

k

Next

R

R

R

R

Let R k k

1 1

1 1

2 4

2 2

2

3

6 3 2

3

2 3

201

500

11

10

10 20 10 10 10 50

200 4

= = =

= = =

= ⋅ ⋅ ⇒ =

= =

−

µ

µ

.

.

( )( )

Then R

Then R

and R

Ω

Ω

Ω Ω

H Kj

jp

HK

jp

where where

K R CC R

KR

R C R

1 1

1

22

2

1 2 1 11 1

24

32

2 2

1 1

1 1

( ) ( )

, ,

ω ωω ω ω= −

+=

+

= = = − = p p

We require

dB K K CR

R

pR C

pR C

20 10

011

1001

1 2 2 14

3

11 1

24 2

= = − =

= =

= =

R

.

391

DP 13-6 DP 13-7 DP 13-8

T

R

R

j R C( )ω

ω=

+

+

1

1

2

3

1

θ ω = − −tan 11R C

G

R

R

R C

R

R=+

+=

+

+

1

1

1

1

3

2

12

3

2

2( ) tanω θ

RC1 = −tan( )θ

ω

R G R

s R k k F3

22

1 2 3

1 1

45 2 1000 10 10 18 284 01

= ⋅ + − ⋅=− = = ⇒ = = = =

( (tan ) )

deg, , , , . , .

θθ ω µ G rad R R k CΩ Ω Ω

8001

0 5 102 5

32 40 100

2001 1

200 1000 05

20 100 5 0 5

0 05

16 1

2

1

2

2

z x

k

R k

pR C

C k

F

dB kp

z

F

F

F

= = ⇒ =

= = ⇒ =

= = ⇒ = =

= = = =

−RR

dB R

R

C

( . ).

( )( ).

. .

.

Ω

Ω

Ωµ

µ µ

µ

HR

j C

j CR

j CR

CR

( )

tan

( )

( )( . ).

( )

ω

ω

ωω

ω

ω

R tan

x k

lim H R

RR R k

w

= −

+= −

+

= + −

⇒ = − =

= = ⇒ = =

° −

−

→∞

2 2

1

11

1 6

2

12 1

11 1

195 180 90

270 195

1000 01 1037 3

10 10 373

Ω

Ω

The network function of this circuit is

The phase shift of this network function is

The gain of this network function is

Design of this circuit proceeds as follows. Since the frequency and capacitance are known, R1 is calculated from

Next pick R2 = 10kΩ (a convenient value) and calculated R3 using

From Table 13.4-2 and the Bode plot: