Electric Circuits, 9th Edition P77 ppsx

C.13 is equivalent to the magnetically coupled coils shown inside the box in Fig.. The decibel is also used as a unit of power when it expresses the ratio of a known power to a referenc

736 More on Magnetically Coupled Coils and Ideal Transformers

continuously, the use of a capacitor to simulate negative inductance is practically worthless

You can circumvent the problem of dealing with negative inductances

by introducing an ideal transformer into the equivalent circuit This doesn't completely solve the modeling problem, because ideal transformers can only be approximated However, in some situations the approximation is good enough to warrant a discussion of using an ideal transformer in the T- and ^-equivalent circuits of magnetically coupled coils

An ideal transformer can be used in two different ways in either the T-equivalent or the -equivalent circuit Figure C.12 shows the two arrange-ments for each type of equivalent circuit

Verifying any of the equivalent circuits in Fig C.12 requires showing

only that, for any circuit, the equations relating v x and v 2 to dijdt and

di 2 /dt are identical to Eqs C.l and C.2 Here, we validate the circuit shown

in Fig C.12(a); we leave it to you to verify the circuits in Figs C.12(b), (c), and (d).To aid the discussion, we redrew the circuit shown in Fig C.12(a)

as Fig C.13, adding the variables i {) and % From this circuit,

v 1 = [L l

M \ di } M d ,

and

v {) = hi M

a

di Q M d

(C.13)

a

(a)

L X L 2 - M :

Ma

UL X - M 2 ^L X L 2 -M 2 L-> - Ma 3 a 2 L\ - Ma

(c)

+

th

'\

+ •

»1

+

»1

lia

Ideal

a L] - Ma L 2 - Ma

(b)

a(L x L 2 - M 2 )

Ideal

1 :a

Ideal

M

a 2 (L x L 2 -M 2 U a 2 (L x L 2 -M 2 ) U-Ma a 2 L x - Ma Vt

(d)

Figure C.12 • The four ways of using an ideal transformer in the T- and 7r-equivalent circuit for magnetically coupled coils

M ^2 _ M

L x - -Q a 2 a

a

h) r^i^^ir^t

Ideal

(a)

Figure C.13 A The circuit of Fig C.12(a) with i 0 and v Q

defined

C.2 The Need for Ideal Transformers in the Equivalent Circuits 737

The ideal transformer imposes constraints o n v {) and / () :

«o

V2

a '

Substituting Eqs C.14 a n d C.15 into Eqs C.12 a n d C.13 gives

dh M d

Vi = LTTT + —-J-U12)

at a dt

(C.14)

(C.15)

(C.16)

and

Vi = Lid_

a a 1 dt

From Eqs C.16 a n d C.17,

, (ah) H —

(C.17)

di\ di-}

and

^2

di\ dU

M—- + L 2 -^

E q u a t i o n s C.18 and C.19 are identical to Eqs C.l and C.2; thus, insofar as

terminal behavior is concerned, the circuit shown in Fig C.13 is equivalent

to the magnetically coupled coils shown inside the box in Fig C l

In showing that the circuit in Fig C.13 is equivalent to the

magneti-cally coupled coils in Fig C l , w e placed n o restrictions on the turns

ratio a Therefore, an infinite n u m b e r of equivalent circuits are possible

F u r t h e r m o r e , we can always find a turns ratio to m a k e all the inductances

positive T h r e e values of a are of particular interest:

and

M_

hi

L2

(C.20)

(C.21)

(C22)

T h e value of a given by E q C.20 eliminates the inductances L x — M/a

and a 2 L x — aM from the T-equivalent circuits and the inductances

(L X L 2 - M 2 )/(a 2 L { - aM) and a 2 (L { L 2 - M 2 )/{a 2 U - aM) from the

17-equivalent circuits T h e value of a given by E q C.21

eliminates the inductances (L 2 /a 2 ) - (M/a) a n d L2 — aM from the

T-equivalent circuits and the inductances (L X L 2 - M 2 )J(L 2 - aM) and

a 2 (LiL 2 — M 2 )j(L 2 — aM) from the 7r-equivalent circuits

Also n o t e that w h e n a = M/L h the circuits in Figs C.l2(a) and (c)

b e c o m e identical, and w h e n a = L 2 /M, the circuits in Figs C.12(b) a n d (d)

b e c o m e identical Figures C.14 and C.15 summarize these observations

_ynrvnr>_

"1 -U-i

• I 1-O «

Ideal (a)

• 1:«

Ideal

(b) Figure C.14 • Two equivalent circuits when

/.,(1 - k 2 )

i/c 2 L,

• 1 : a •

Ideal (a)

HP-IJ *,

1 :a

Ideal

\U n

(b) Figure C.15 • Two equivalent circuits when

a = L 2 /M

• +

Figure C.16 A Experimental determination of the

ratio MfL x

738 More on Magnetically Coupled Coils and Ideal Transformers

In deriving the expressions for the inductances there, we used the

relationship M = kVL\L 2 Expressing the inductances as functions of the self-inductances L\ and L 2 and the coefficient of coupling k allows the values of a given by Eqs C.20 and C.21 not only to reduce the number of

inductances needed in the equivalent circuit, but also to guarantee that all the inductances will be positive We leave to you to investigate the

conse-quences of choosing the value of a given by Eq C.22

The values of a given by Eqs C.20-C.22 can be determined experi-mentally The ratio MjL x is obtained by driving the coil designated as

hav-ing N\ turns by a sinusoidal voltage source The source frequency is set high enough that coL\ 5$> R\, and the N 2 coil is left open Figure C.16 shows this arrangement

With the N 2 coil open,

Now, as /a>L] » R h the current I\ is

ja)L [

Substituting Eq C.24 into Eq C23 yields

(C.25)

in which the ratio M/L l is the terminal voltage ratio corresponding to coil 2 being open; that is, I2 = 0

We obtain the ratio L 2 /M by reversing the procedure; that is, coil 2 is

energized and coil 1 is left open Then

Finally, we observe that the value of a given by Eq C.22 is the

geo-metric mean of these two voltage ratios; thus

For coils wound on nonmagnetic cores, the voltage ratio is not the same as the turns ratio, as it very nearly is for coils wound on ferromagnetic cores Because the self-inductances vary as the square of the number of turns, Eq C27 reveals that the turns ratio is approximately equal to the geometric mean of the two voltage ratios, or

Appendix

~L '

D The Decibel

Telephone engineers who were concerned with the power loss across the

cascaded circuits used to transmit telephone signals introduced the

deci-bel Figure D.l defines the problem

There, p, is the power input to the system, p x is the power output of

circuit A, p 2 is the power output of circuit B, and p (> is the power output

of the system The power gain of each circuit is the ratio of the power out

to the power in Thus

# •

-C P<>

Figure D.l • Three cascaded circuits

P\ P2 , Pa

CTA = — , an = — , and err — —

Pi Pi Pi

The overall power gain of the system is simply the product of the

individ-ual gains, or

Po

Pi

P^PlPo

Pi Pi Pi = (T A <r B a c

The multiplication of power ratios is converted to addition by means of

the logarithm; that is,

log10— = logujo-A + log1()o-B + log1()crc,

Pi

This log ratio of the powers was named the bel, in honor of Alexander

Graham Bell Thus we calculate the overall power gain, in bels, simply by

summing the power gains, also in bels, of each segment of the transmission

system In practice, the bel is an inconveniently large quantity One-tenth

of a bel is a more useful measure of power gain; hence the decibel The

number of decibels equals 10 times the number of bels, so

Po

Number of decibels = 10 login —

Pi

When we use the decibel as a measure of power ratios, in some

situa-tions the resistance seen looking into the circuit equals the resistance

loading the circuit, as illustrated in Fig D.2

When the input resistance equals the load resistance, we can convert

the power ratio to either a voltage ratio or a current ratio:

Po

Pi

v ou\

Via

'in

'"in •* A

'out

+-+ {

R,

Figure D.2 • A circuit in which the input resistance

equals the load resistance

or

Po

Pi if R 'in

739

740 The Decibel

These equations show that the number of decibels becomes

Number of decibels = 20 log]0 'out

= 20 log 10 'out (D.l)

TABLE D.l Some dB-Ratio Pairs

dB

0

3

6

10

15

20

Ratio

1.00

1.41

2.00

3.16

5.62

10.00

dB

30

40

60

80

100

120

Ratio

31.62

100.00

10 3

10 4

10 5

10 6

The definition of the decibel used in Bode diagrams (see Appendix E)

is borrowed from the results expressed by Eq D.l, since these results apply to any transfer function involving a voltage ratio, a current ratio, a voltage-to-current ratio, or a current-to-voltage ratio You should keep the original definition of the decibel firmly in mind because it is of fundamen-tal importance in many engineering applications

When you are working with transfer function amplitudes expressed in decibels, having a table that translates the decibel value to the actual value

of the output/input ratio is helpful Table D.l gives some useful pairs The ratio corresponding to a negative decibel value is the reciprocal of the pos-itive ratio For example, - 3 dB corresponds to an output/input ratio of 1/1.41, or 0.707 Interestingly, —3 dB corresponds to the half-power fre-quencies of the filter circuits discussed in Chapters 14 and 15

The decibel is also used as a unit of power when it expresses the ratio

of a known power to a reference power Usually the reference power is

1 mW and the power unit is written dBm, which stands for "decibels rela-tive to one milliwatt." For example, a power of 20 mW corresponds to

±13 dBm

AC voltmeters commonly provide dBm readings that assume not only

a 1 mW reference power but also a 600 ft reference resistance (a value commonly used in telephone systems) Since a power of 1 mW in 600 ft corresponds to 0.7746 V (mis), that voltage is read as 0 dBm on the meter For analog meters, there usually is exactly a 10 dB difference between adjacent ranges Although the scales may be marked 0.1, 0.3,1, 3,10, and

so on, in fact 3.16 V on the 3 V scale lines up with 1 V on the 1 V scale Some voltmeters provide a switch to choose a reference resistance (50, 135,600, or 900 ft) or to select dBm or dBV (decibels relative to one volt)

Appendix

Q Bode Diagrams

As we have seen, the frequency response plot is a very important tool for

analyzing a circuit's behavior Up to this point, however, we have shown

qualitative sketches of the frequency response without discussing how to

create such diagrams The most efficient method for generating and

plot-ting the amplitude and phase data is to use a digital computer; we can rely

on it to give us accurate numerical plots of \H(jm)\ and d{ja>) versus co

However, in some situations, preliminary sketches using Bode diagrams

can help ensure the intelligent use of the computer

A Bode diagram, or plot, is a graphical technique that gives a feel

for the frequency response of a circuit These diagrams are named in

recognition of the pioneering work done by H W Bode.1 They are most

useful for circuits in which the poles and zeros of H(s) are reasonably

well separated

Like the qualitative frequency response plots seen thus far, a Bode

diagram consists of two separate plots: One shows how the amplitude of

H(jco) varies with frequency, and the other shows how the phase angle

of H(j(o) varies with frequency In Bode diagrams, the plots are made on

semilog graph paper for greater accuracy in representing the wide range

of frequency values In both the amplitude and phase plots, the frequency

is plotted on the horizontal log scale, and the amplitude and phase angle

are plotted on the linear vertical scale

E,l Real, First-Order Poles and Zeros

To simplify the development of Bode diagrams, we begin by considering

only cases where all the poles and zeros of H(s) are real and first order

Later we will present cases with complex and repeated poles and zeros

For our purposes, having a specific expression for H(s) is helpful Hence

we base the discussion on

K(s + zi)

from which

]0)(j(0 + Pi)

The first step in making Bode diagrams is to put the expression for

H(jco) in a standard form, which we derive simply by dividing out the

poles and zeros:

Kzi(l + / w / z i ) f s

H(m) = , w < , - (E.3)

p,(7a.)(l + /a>/A)

See II W Bode, Network Analysis and Feedback Design (New York: Van Nostrand, 1945)

742 Bode Diagrams

Next we let K a represent the constant quantity Kz.[/p\, and at the same time we express H(jto) in polar form:

M / 9 0 |1 + J^/Pil /j3i

K„|l + /w/zil

From Eq E.4,

\H(ja>)\ = / ; ', (E.5)

0(«) = ?Ai - 90° - /3, (E.6)

By definition, the phase angles ip\ and /3j are

The Bode diagrams consist of plotting Eq E.5 (amplitude) and Eq E.6 (phase) as functions of o>

E.2 Straight-Line Amplitude Plots

The amplitude plot involves the multiplication and division of factors

associated with the poles and zeros of H(s) We reduce this multiplication

and division to addition and subtraction by expressing the amplitude of

H(j(o) in terms of a logarithmic value: the decibel (dB).2 The amplitude

of H(ja)) in decibels is

/lt l B = 2()log1 ( )|//f>)| (E.9)

TABLE E.1 Actual Amplitudes and Their T o §i v e y °u a f e e l f o r t h e u n i t o f decibels, Table E.l provides a translation

Decibel Values between the actual value of several amplitudes and their values in

deci-bels Expressing Eq E.5 in terms of decibels gives

K 0 \l+jco/ Zl \

A dB = 20 logJ()

w|l + ja>/pi\

= 201og1()/C + 20lQgtJl + /»/*il

- 20 log10<u - 20 log10|l -f- ja/pxl (E.10)

See Appendix D for more information regarding the decibel

AiB

0

3

6

10

15

20

A

1.00

1.41

2.00

3.16

5.62

10.00

AiB

30

40

60

SO

100

120

A

31.62 100.00

10 3

10 4

105

106

The key to plotting Eq E.10 is to plot each term in the equation sepa-rately and then combine the separate plots graphically The individual fac-tors are easy to plot because they can be approximated in all cases by straight lines

The plot of 20 log10 K (> is a horizontal straight line because K 0 is not a

function of frequency The value of this term is positive for K a > 1, zero for K 0 - 1, and negative for K 0 < 1

Two straight lines approximate the plot of 20 log10| 1 +• j(o/z\\ For small values of <w, the magnitude 11 + jafz\ | is approximately 1, and therefore

201og1()|l + j(o/zi\^0 a s w - ^ 0 (E.ll)

For large values of w, the magnitude |1 + jo)/z\\ is approximately o)/z\,

and therefore

201og 1()|l + j(o/z.[\ ^20 \og u) ((o/Z]) a s w — > o c (E.12)

On a log frequency scale, 20 \og m (a)/z\) is a straight line with a slope of

20 dB/decade (a decade is a 10-to-l change in frequency).This straight

line intersects the 0 dB axis at w = z\ This value of o» is called the

corner frequency.Thus, on the basis of Eqs E l l and E.12, two straight

lines can approximate the amplitude plot of a first-order zero, as shown in Fig E.l

The plot of — 201ogioa> is a straight line having a slope of

- 2 0 dB/decade that intersects the 0 dB axis at a» = l.Two straight lines approximate the plot of - 2 0 log10|l 4- jco/p\\ Here the two straight lines

25

20

15

10

5

0

- 5

z

z ( ) I °8H.||T)

\y

20 dB/dec

- Uecade

ade

1C Z)

1 2 3 4 5 6 7 8 910

o) (rad/s)

20 30 40 50

Figure E.l • A straight-line approximation of the amplitude plot of a

first-order zero

intersect on the 0 dB axis at w = p\ For large values of w, the straight line

20 log10(o>/pi) has a slope of - 2 0 dB/decade Figure E.2 shows the straight-line approximation of the amplitude plot of a first-order pole

k l B

5

0

-5

- 1 0

-15

-20

p N

S j

201og10|j^j

•

-20 dB/decade ^ \

iQPi

1 2 3 4 5 6 7 8 9 1 0

o) (rad/s)

20 30 40 50

Figure E.2 • A straight-line approximation of the amplitude plot of a first-order pole

Figure E.3 shows a plot of Eq E.10 for K 0 = VlO, Z\ = 0.1 rad/s, and pi = 5 rad/s Each term in Eq E.10 is labeled on Fig E.3, so you can

verify that the individual terms sum to create the resultant plot, labeled 201og1()|//(/a>)|

Example E.l illustrates the construction of a straight-line amplitude plot for a transfer function characterized by first-order poles and zeros

A dB

50

40

30

20

10

0

-10

-20

\

\

\

\

S

N

yl

\ ,

>

mi

0 log, 0

\

-1

N \

/

•

\l

)1

S >

s

I I I

(/«01

Jgio^

•

< • - - •

>

s

>

/

\ s

20

s

/

C

\

logio

^

V

s \

s

/

***

*

•

K 0

V

.'1111

- 2 0 1 o

s *

\

01

I0\oi

\

ogio

\

;iol^C

^

i + / |

CO )

J \

1

j,

0.05 0.1 0.5 1.0 5 10 50 100 500

a> (rad/s)

Figure E.3 • A straight-line approximation of the amplitude plot for Eq E.10

E.2 Straight-Line Amplitude Plots 745

Example E.l

For the circuit in Fig E.4:

a) Compute the transfer function, H(s)

b) Construct a straight-line approximation of the

Bode amplitude plot

c) Calculate 20log10|.//(/cu)| at w = 50 rad/s and

co = 1000 rad/s

d) Plot the values computed in (c) on the

straight-line graph; and

e) Suppose that v t {t) = 5cos (500* + 15°) V, and

then use the Bode plot you constructed to

pre-dict the amplitude of v a (t) in the steady state

I

11 i l l V,

Figure E.4 • The circuit for Example E.l

c) We have

//(/50) = o.ii(;50)

(1 + /5)(1 + /0.5)

Solution

a) Transforming the circuit in Fig E.4 into the

s-domain and then using 5-domain voltage

divi-sion gives

= 0.9648/-15.25°,

20 log1 0|//(/50)| = 20 log10 0.9648

H(s) = (R/L)s

i •

S* + (R/L)s + £

Substituting the numerical values from the

cir-cuit, we get

= -0.311 dB;

//(/1000) = 0.11(/1000)

(1 + /100)(1 + /10)

s 2 + 110$ + 1000 (s + 10)(5 + 100)

b) We begin by writing H(jto) in standard form:

= 0.1094/-83.72°;

20 log1()0.1094 = -19.22 dB

[1 + /(a>/10)][l + / K 1 0 0 ) ] '

The expression for the amplitude of H(J<&) in

decibels is

Ad B = 201og1 ( )|//(/w)|

= 201og100.11 + 201og1 0|H

- 20 log 10 1 + / - 2 0 log10

Figure E.5 shows the straight-line plot

Each term contributing to the overall amplitude

is identified

40

30

20

10

0

Am -10 -20

- 3 0 -40 -50 -60

1 5 10 50 100 5001000

co (rad/s)

Figure E.5 • The straight-line amplitude plot for the transfer function of

the circuit in Fig E.4

• *

,*"'

•

^20 lo

T

y

y

' " ^ 2 0 ' l o g

,.—

SloO.ll

imr

-15

"'s&

-|

o \M

\c

Ml N

—0.311

WrA NfFx

CO i

2 0 1 o g 1 0 | i + y

-1 | i i r i m i

- 1 - 2

.1

1 1 MINI

0 l0S,r 1 1 + /

1 I I IIIII

10 log,

k,

) *^i -^

v

CO

10

1

»\H(jco

ill

-fn

; l ^ <

T S * S N

> i

1 N,

T

t

1""

5) |

- 1 9 2 2 ) "

V ^

Yk

2s> _ l