Electric Circuits, 9th Edition P78 doc

For a first-order zero or pole not at the origin, the straight-line approximations are as follows: • For frequencies less than one tenth the corner frequency, the phase angle is assumed

746 Bode Diagrams

d) See Fig E.5

e) As we can see from the Bode plot in Fig E.5, the

value of A dB at w = 500 rad/s is approximately

-12.5 dB Therefore,

\ A \ = io(-12-5/20> = 0.24 and

V mo = \A\V ltli = (0.24)(5) = 119 V

We can compute the actual value of \H(jo))\

by substituting w = 500 into the equation for

\H(]<o)\;

//(/500) = 0.11(/500)

(1 + /50)(1 + /5) 0.22/-77.54' Thus, the actual output voltage magnitude for the specified signal source at a frequency of

500 rad/s is

V,no = \A\V im = (0.22)(5) = 1.1 V

E.3 More Accurate Amplitude Plots

We can make the straight-line plots for first-order poles and zeros more accurate by correcting the amplitude values at the corner frequency, one half the corner frequency, and twice the corner frequency At the corner frequency, the actual value in decibels is

A dBc = ±20log1 0|l +/11

= ±20 log10V2

The actual value at one half the corner frequency is

A dBc!2>c/2 = ±20 log 10 1 +j:

±201og]()V574

At twice the corner frequency, the actual value in decibels is

Ad B 2 c= ±201og10|l + / 2 |

= ±201og10V5

In Eqs E.13-E.15, the plus sign applies to a first-order zero, and the minus sign applies to a first-order pole The straight-line approximation of the amplitude plot gives 0 dB at the corner and one half the corner frequencies, and ±6 dB at twice the corner frequency Hence the corrections are ± 3 dB

at the corner frequency and ±1 dB at both one half the corner frequency and twice the corner frequency Figure E.6 summarizes these corrections

A 2-to-l change in frequency is called an octave A slope of

20 dB/decade is equivalent to 6.02 dB/octave, which for graphical pur-poses is equivalent to 6 dB/octave Thus the corrections enumerated cor-respond to one octave below and one octave above the corner frequency

25

20

15

10

5

A d * 0

- 5

- 1 0

- 1 5

- 2 0

- 2 5

£ C 2c

2

Figure E.6 A Corrected amplitude plots for a first-order zero

and pole

If the poles and zeros of H(s) are well separated, inserting these

corrections into the overall amplitude plot and achieving a reasonably accurate curve is relatively easy However, if the poles and zeros are close together, the overlapping corrections are difficult to evaluate, and you're better off using the straight-line plot as a first estimate of the amplitude characteristic Then use a computer to refine the calculations

in the frequency range of interest

EA Straight-Line Phase Angle Plots

We can also make phase angle plots by using straight-line approximations

The phase angle associated with the constant K () is zero, and the phase angle associated with a first-order zero or pole at the origin is a constant

± 90° For a first-order zero or pole not at the origin, the straight-line approximations are as follows:

• For frequencies less than one tenth the corner frequency, the phase angle is assumed to be zero

• For frequencies greater than 10 times the corner frequency, the phase angle is assumed to be ±90°

• Between one tenth the corner frequency and 10 times the corner fre-quency, the phase angle plot is a straight line that goes through 0° at one-tenth the corner frequency, ±45° at the corner frequency, and

±90° at 10 times the corner frequency

In all these cases, the plus sign applies to the first-order zero and the minus sign to the first-order pole Figure E.7 depicts the straight-line approxima-tion for a first-order zero and pole The dashed curves show the exact vari-ation of the phase angle as the frequency varies Note how closely the

3dE

1 d B ^ -1 d B '

\ '*'

» /

* • /

> \

IB S

y / A/

i

^

/

V

ldB

1 1

1

- 1 dl

i

3

V

y r /

"-S

/

AS

N N

/, •y i//| - tan"' {(olz{)

-Actual 1

"" Straig it-line approximation

-jEJ] - t a n ' (w//J,

Actual 1(1 | | Ml

Straight-Line approxi

^ > s

' V

1

matio i

[

90°

60°

30°

d (w) 0

-30°

- 6 0 °

-90°

z x n0 /7,/1() 2 l p, 10«! lOp,

w (rad/s) Figure E.7 • Phase angle plots for a first-order zero and pole

straight-line plot approximates the actual variation in phase angle The maximum deviation between the straight-line plot and the actual plot is approximately 6°

Figure E.8 depicts the straight-line approximation of the phase angle of the transfer function given by Eq B.l Equation B.6 gives the equation for

the phase angle; the plot corresponds to z\ = 0,1 rad/s, and Pi = 5 rad/s

An illustration of a phase angle plot using a straight-line approxima-tion is given in Example E.2

90°

60°

30°

0(«) 0

-30°

- 6 0 °

- 9 0 °

y

"t)(to)

A

-»

t'

•

•

/

- • * l =

•

tan

1

1

J

~ \ i oil

jBj

s

l )

N s

an ' (

1

U

1 Oj/j

o

> i )

at (rad/s)

Figure E.8 A A straight-line approximation of the phase angle plot for Eq B.l

E.5 Bode Diagrams: Complex Poles and Zeros 7 4 9 Example E.2

a) Make a straight-line phase angle plot for the

transfer function in Example E.l

b) Compute the phase angle 0(<D) at <o = 50, 500,

and 1000 rad/s

c) Plot the values of (b) on the diagram of (a)

d) Using the results from Example E.l(e) and (b)

of this example, compute the steady-state

out-put voltage if the source voltage is given by

v&t) = 10cos(500f - 25°) V

Solution

a) From Example E.l,

0.11(/6.)

H(ja>)

[1 + j(<o/10)][l + /(«/100)]

o.iiH

|1 + / ( « / 1 0 ) | | l +/(o>/100)| /(fo-ft nfe)

Therefore,

0(a>) = ft - /3t - /3 2 ,

where i/'i = 90°, p l = tan_1(w/10), and (3 2 =

tan_1(w/100) Figure E.9 depicts the straight-line

approximation of 6(co)

b) We have

Thus

//(/50) = 0.96/-15.25%

//(/500) = 0.22/-77.54%

//(/1000) = 0.11/-83.72°

0(/50) = -15.25°, 0(/500) = -77.54°,

en

and

c) See Fig E.9

d) We have

0(/1000) = -83.72°

and

V m0 = \H(j500)\V m

= (0.22)(10)

= 2.2 V,

= -77.54° - 25'

= -102.54°

Thus,

v 0 (t) = 2.2 cos(500? - 102.54°) V

5 10 50100 5001000

co (rad/s)

Figure E.9 A A straight-line approximation of 6(a)) for Example E.2

E.5 Bode Diagrams: Complex Poles

and Zeros

Complex poles and zeros in the expression for H(s) require special

atten-tion when you make amplitude and phase angle plots Let's focus on the

contribution that a pair of complex poles makes to the amplitude and

phase angle plots Once you understand the rules for handling complex

poles, their application to a pair of complex zeros becomes apparent

The complex poles and zeros of H(s) always appear in conjugate

pairs The first step in making either an amplitude or a phase angle plot of

a transfer function that contains complex poles is to combine the

conju-gate pair into a single quadratic term Thus, for

His) = 7 — 7ZZ, (E.16)

we first rewrite the product (s + a - j(B)(s + a + y/3) as

(s + a) 2 + p 2 = s 2 + las + a 2 + jS2 (E.17) When making Bode diagrams, we write the quadratic term in a more

con-venient form:

s 2 4- las + a 2 + 0 1 = s 2 + 2£a> n s + a? n (E.18)

A direct comparison of the two forms shows that

and

fan = a (E.20)

The term <afl is the corner frequency of the quadratic factor, and £ is the

damping coefficient of the quadratic term The critical value of £ is 1 If

£ < 1, the roots of the quadratic factor are complex, and we use Eq E.18

to represent the complex poles If £ > 1, we factor the quadratic factor

into (s + pi)(s + p 2 ) and then plot amplitude and phase in accordance

with the discussion previously Assuming that £ < 1, we rewrite Eq E.16 as

s l + 2t,u> n s + cof,

We then write Eq E.21 in standard form by dividing through by the poles

and zeros For the quadratic term, we divide through by <*>„, so

K_ 1_

a) 2 1 + (s/o) n ) 2 + 2£{s/(o„)

H & = A^„ 2 - ( L 2 1 )

from which

where

1 - (or/tof,) + j(2^0)/(0,,)

K _ K

Before discussing the amplitude and phase angle diagrams associated

with Eq E.23, for convenience we replace the ratio (o/o) n by a new

vari-able, u Then

H{h) = TT^r- <E-24>

Now we write H(ico) in polar form:

from which

A dB = 20 log1()|//(;to)|

= 20 \og w K a - 20 log10|(l - u 2 ) + j2£u\> (E.26)

and

6((0) = - f t = - t a n "1- ^ ^ (E.27)

1 - ir

E.6 Amplitude Plots

The quadratic factor contributes to the amplitude of H(jco) by means of the

term —201og10|l - u 2 + j2gu\ Because u = to/to,,, M—»0 as to—»0, and

« —* oo as co —* oo To see how the term behaves as to ranges from 0 to oo,

we note that

- 2 0 log1()|(l - u 2 ) + j2£u\ = - 2 0 log1 0V(l - u 2 ) 1 + 4£ 2 u 2

= -101og10[w4 + 2u 2 (2( 2 - 1 ) + 1], (E.28)

as u —> 0,

- 1 0 Iog10[«4 + 2w2(2£2 - l ) + l ] - ^ 0 , (E.29) and as «—» oo,

-101og10[u4 + 2u2(2£2 - 1) + 1] - » - 401og10« (E.30)

From Eqs E.29 and E.30, we conclude that the approximate amplitude plot

consists of two straight lines For to < co„, the straight line lies along the

0 dB axis, and for to > co„, the straight line has a slope of —40 dB/decade

These two straight lines join on the 0 dB axis at u = 1 or to = &>„

Figure E.10 shows the straight-line approximation for a quadratic factor

with t < 1

20

10

0 -10

Am

-20

- 3 0

- 4 0

- 5 0

w„ 10w„

co (rad/s)

Figure E.10 • The amplitude plot for a pair of complex poles

- 4 1 ) dB/dea ide

i

E.7 Correcting Straight-Line

Amplitude Plots

Correcting the straight-line amplitude plot for a pair of complex poles

is not as easy as correcting a first-order real pole, because the

correc-tions depend on the damping coefficient £ Figure E l l shows the

effect of £ on the amplitude plot Note that as £ becomes very small, a

large peak in the amplitude occurs in the neighborhood of the corner

frequency a> n (u = 1) When £ > 1/V2, the corrected amplitude plot

lies entirely below the straight-line approximation For sketching

pur-poses, the straight-line amplitude plot can be corrected by locating

four points on the actual curve These four points correspond to

(1) one half the corner frequency, (2) the frequency at which the

ampli-tude reaches its peak value, (3) the corner frequency, and (4) the

fre-quency at which the amplitude is zero Figure E.12 shows these

four points

At one half the corner frequency (point 1), the actual amplitude is

A U B K / 2 ) = - 1 0 log1()(£2 + 0.5625) (E.31)

The amplitude peaks (point 2) at a frequency of

and it has a peak amplitude of

A dB (co p ) = - 1 0 log10[4£2(l - ft] (E.33)

At the corner frequency (point 3), the actual amplitude is

The corrected amplitude plot crosses the 0 dB axis (point 4) at

w 0 = a),(V2(l - 2 f t = V2co p (E.35)

The derivations of Eqs E.31, E.34, and E.35 follow from Eq E.28

Evaluating Eq E.28 at u = 0.5 and u = 1.0, respectively, yields Eqs E.31

and E.34 Equation E.35 corresponds to finding the value of u that makes

u 4 + 2u 2 (2£ 2 - 1) + 1 = l.The derivation of Eq E.32 requires

differen-tiating Eq E.28 with respect to u and then finding the value of u where the

derivative is zero Equation E.33 is the evaluation of Eq E.28 at the value

of u found in Eq E.32

Example E.3 illustrates the amplitude plot for a transfer function with

a pair of complex poles

E.7 Correcting Straight-Line Amplitude Plots 753

20

10

A dB

- 1 0

- 2 0

- 3 0

-40

- 5 0

c - )

/

•

707

l\i =0.1

} 1~

\

\\\\

VvW

V

= ( 3

\

\

\

a) (rad/s)

Figure E.ll • The effect of £ on the amplitude plot

3

2

1

0

- 1

- 3

- 4

- 5

- 6

- 7

1

/

X "* \

"

.1

L 4

I

I

!

a)J2

co (rad/s)

w ;, co n co ()

Figure E.12 • Four points on the corrected amplitude plot for a pair of

complex poles

754 Bode Diagrams

Example E.3

Compute the transfer function for the circuit shown

in Fig E.13

a) What is the value of the corner frequency in

radi-ans per second?

b) What is the value of K () l

c) What is the value of the damping coefficient?

d) Make a straight-line amplitude plot ranging from

10to500rad/s

e) Calculate and sketch the actual amplitude in

decibels at a>„/2, to p , co n , and co ()

f) From the straight-line amplitude plot, describe

the type of filter represented by the circuit in

Fig E.13 and estimate its cutoff frequency, m e ,

d) See Fig E.14

e) The actual amplitudes are

^ J B K / 2 ) = - 1 0 log10(0.6025) = 2.2 dB,

o) p = 50VO92 = 47.96 rad/s,

A I B K ) = -101og I0 (0.16)(0.96) = 8.14 dB,

^ W O = -20Iog1()(0.4) = 7.96 dB,

w 0 = V2(t) p = 67.82 rad/s,

^ B K ) = 0 dB

-'WV-8mF

Figure E.13 • The circuit for Example E.3

Solution

Transform the circuit in Fig E.13 to the ^-domain

and then use s-domain voltage division to get

Figure E.14 shows the corrected plot

f) It is clear from the amplitude plot in Fig E.14 that this circuit acts as a low-pass filter At the cutoff frequency, the magnitude of the transfer

function, \H(Jm c )] t is 3 dB less than the maximum magnitude From the corrected plot, the cutoff frequency appears to be about 55 rad/s, almost the same as that predicted by the straight-line Bode diagram

H(s) = LC

s2 + ({)*+ ] LC

Substituting the component values

H(s) 2500

s 2 + 20s + 2500

a) From the expression for H(s), o) 2n = 2500;

there-fore, o) u = 50 rad/s

b) By definition, K a is 2500/^,, or 1

c) The coefficient of s equals 2£a> n; therefore

20

C = ^ - = 0.20

2(i)„

15

10

5

0

- 5 -10 /IdB -15 -20 -25 -30 -35 -40 -45

2

to (rad/s)

Figure E.14 • The amplitude plot for Example E.3

(2 1«

2)

14 /

„ (7.9

\ «

\ \

6) J)

V

\

E.8 Phase Angle Plots 755

E.8 Phase Angle Plots

The phase angle plot for a pair of complex poles is a plot of Eq E.27 The

phase angle is zero at zero frequency and is -90° at the corner frequency

It approaches —180° as co(u) becomes large As in the case of the

ampli-tude plot, I is important in determining the exact shape of the phase angle

plot For small values of £, the phase angle changes rapidly in the vicinity

of the corner frequency Figure E.15 shows the effect of £ on the phase

angle plot

We can also make a straight-line approximation of the phase angle plot

for a pair of complex poles We do so by drawing a line tangent to the phase

angle curve at the corner frequency and extending this line until it

inter-sects with the 0° and -180° lines The line tangent to the phase angle curve

at -90° has a slope of -23/(, rad/decade (—132/£ degrees/decade), and it

intersects the 0° and -180° lines at u x = 4.81_f and u 2 = 4.81^,

respec-tively Figure E.16 depicts the straight-line approximation for £ = 0.3 and

shows the actual phase angle plot Comparing the straight-line

approxima-tion to the actual curve indicates that the approximaapproxima-tion is reasonable in

the vicinity of the corner frequency However, in the neighborhood of ii\

and u 2 , the error is quite large In Example E.4, we summarize our

discus-sion of Bode diagrams

0(o»)

15°

0

-15°

-30°

-45°

-60°

-75°

- 9 0 °

-105°

-120°

-135°

-150°

-165°

-180°

i

X s

).1

\

\ v

0 7 \

\ i

n - £ = 0.3

\

0

s \

i

\ \

\ \

V

v X \

—

•30°

-60°

440 (7.67

—

7dt rac

0.62 =

\

/d

N S

d e

4.81 ~'c

Act

\

il

\

\

1.6 =

ml

4.8

cu

1^

•ve

0.2 0.4 1.0 2 4

M

Figure E.15 • The effect of ( on the phase angle plot

6 (w) - 9 0 °

120°

150c

180 c

1.0 2.0

a) (rad/s)

Figure E.16 • A straight-line approximation of the phase

angle for a pair of complex poles