BODE DIAGRAMS-2 (Frequency Response)

Magnitude Bode plot of 0.1

5

10(1 )

(1 )

j

jj

-- 20log10(1+jω/0.1)

-- -20log10(1+jω/5)

-- -20log10(ω)

-- 20log10(√10)

-- 20log10|H(jω)|

EXAMPLE

+11Ω

vi

+vo

110mH

10mF

22

( ) 110( )

1 110 1000( )

110

( 10)( 100)

R L s sH s

s ss R L sLC

s

s s

10

10 10 10 10

0.11 ( )

1 /10 1 /100

20log ( )

20log 0.11 20log 20log 1 20log 110 100

dB

jH j

j j

A H j

j j j

Calculate 20log10 |H(jω)| at ω=50 rad/s and ω=1000 rad/s

0

10 10

0

10 10

0.11( 50)( 50) 0.9648 -15.25

(1 5)(1 0.5)

20log ( 50) 20log (0.9648) 0.311 dB

0.11( 1000)( 1000) 0.1094 83.72

(1 100)(1 10)

20log ( 1000) 20log (0.1094) 19.22 dB

jH j

j j

H j

jH j

j j

H j

Using the Bode diagram, calculate the amplitude of vo if vi(t)=5cos(500t+150)V.

From the Bode diagram, the value of AdB at ω=500 rad/s is approximately -12.5 dB. Therefore,

( 12.5 / 20)10 0.24A

(0.24)(5) 1.2mo miV AV V

MORE ACCURATE AMPLITUDE PLOTS

The straight-line plots for first-order poles and zeros can be made more accurate by correcting the amplitude values at the corner frequency, one half the corner frequency, and twice the corner frequency. The actual decibel values at these frequencies

/ 2

2

10 10

10 10

10 10

20log 1 1 20log 2 3 dB

20log 1 1/ 2 20log 5/ 4 1 dB

20log 1 2 20log 5 7dB

c

c

c

dB

dB

dB

A j

A j

A j

In these equations, + sign corresponds to a first-order zero, and – sign is for a first-order pole.

STRAIGHT-LINE PHASE ANGLE PLOTS

1. The phase angle for constant Ko is zero.

2. The phase angle for a first-order zero or pole at the origin is a constant ± 900.

3. For a first-order zero or pole not at the origin,

• 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 ± 900.

• Between these frequencies the plot is a straight line that goes from 00 to ± 900 with a slope of ± 450/decade.

EXAMPLE

1 1 2

1 1 2

01

11

12

0.11( )( )

[1 ( /10)][1 ( /100)]

0.11

1 ( /10) 1 ( /100)

( )

90

tan ( /10)

tan ( /100)

jH j

j j

j

j j

Compute the phase angle θ(ω) at ω=50, 500, and 1000 rad/s.

0 0

0 0

0 0

( 50) 0.96 15.25 ( 50) 15.25

( 500) 0.22 77.54 ( 500) 77.54

( 1000) 0.11 83.72 ( 1000) 83.72

H j j

H j j

H j j

Compute the steady-state output voltage if the source voltage is given by vi(t)=10cos(500t-250) V.

0 0 0

0

( 500) (0.22)(10) 2.2

( ) 77.54 25 102.54

( ) 2.2cos(500 102.54 )

mo mi

o i

o

V H j V V

v t t V

COMPLEX POLES AND ZEROS

2 2 2

2 2 22 2

2 2

00 22 2

0 02 2

1

210 0 10

1

( )( )( ) 2

, 2

1( )

1 / 2 /

( )1 / 2 /

( )1 2 (1 ) 2

20log 20log (1 ) 2

( )

n nn n

n n n

nn n

dB

K KH s

s j s j s s

K

s s

KH s

s s

K KH j K

j

K KH j

u j u u j u

A K u j u

12

2tan

1

u

u

AMPLITUDE PLOTS

2 2 2 2 210 10

4 2 210

20log (1 ) 2 20log (1 ) 4

10log 2 (2 1) 1

u j u u u

u u

, then 0 as 0, and as n

u u u

4 2 210

4 2 210 10

as 0, -10log [ 2 (2 1) 1] 0

as , -10log [ 2 (2 1) 1] 40log

u u u

u u u u

Thus, the approximate amplitude plot consists of two straight lines. For ω<ωn, the straight line lies along the 0 dB axis, and for ω>ωn, the straight line has a slope of -40 dB/decade. Thes two straight lines intersect at u=1 or ω=ωn.

Correcting Straight-Line Amplitude Plots

The straight-line amplitude plot can be corrected by locating

four points on the actual curve. 1. One half the corner frequency: At this frequency,

the actual amplitude is

2. The frequency at which the amplitude reaches its peak value. The amplitude peaks at and it has a peak amplitude

3. At the corner frequency,

4. The corrected amplitude plot crosses the 0 dB axis at

210( / 2) 10log ( 0.5625)dB nA

21 2p n

2 21010log 4 (1 )dBA

10( ) 20log 2dB nA

20 2(1 2 ) 2n p

1.0

3.0

707.0

When ζ>1/√2, the corrected amplitude plot lies below the straight line approximation. As ζ becomes very small, a large peak in the amplitude occurs around the corner frequency.

EXAMPLE

+vi

+vo

50mH 1Ω

8mf

250020

25001

1

)(2

2

ss

LCs

LR

s

LCsH

2.02

20 ,1/2500

rad/s 502500

20

2

nn

nn

K

rad/s 82.67)96.47(2

96.2)2.02(log20)50(

14.8)]2.01()2.0(4[log10)96.47(

rad/s 96.47)2.0(2150

2.2)5625.02.0(log10)25(

0

10

2210

2

210

dBA

dBA

dBA

dB

dB

p

dB

PHASE ANGLE PLOTS

For a second-order zero or pole not at the origin,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 ± 1800.

• Between these frequencies the plot is a straight line that goes from 00 to ± 1800 with a slope of ± 900/decade.

As in the case of the amplitude plot, ζ is important in determining the exact shape of the phase angle plot. For small values of ζ , the phase angle changes rapidly in the vicinity of the corner frequency.

ζ=0.1

ζ=0.3

ζ=0.707

EXAMPLE

+vi

+

vo

50mH

40mf

1Ω

)10/(4.0)10/(1

125/)(

1004

)25(41

1

)(

2

22

ss

ssH

ss

s

LCs

LR

s

LCs

LR

sH

11

21010

12

1

)(

)10/(4.0)10/(1log2025/1log20

)10/(4.0)10/(1

25/1)(

jjA

j

jjH

dB

From the straight-line plot, this circuit acts as a low-pass filter. At the cutoff frequency, the amplitude of H(jω) is 3 dB less than the amplitude in the passband. From the plot, the cutoff frecuency is predicted approximately as 13 rad/s.

To solve the actual cutoff frequency, follow the procedure as:

max

2

2 2

2 2 2

11 ( )

24( ) 100

( )( ) 4( ) 100

(4 ) 100 1( )

2(100 ) (4 )

16 rad/s

c

cc

c c

c

H H j

jH j

j j

H j

ωc

From the phase plot, the phase angle at the cutoff frequency is estimated to be -650.

The exact phase angle at the cutoff frequency can be calculated as

2

1 1 2 0

4( 16 25)( 16)

( 16) 4( 16) 100

( 16) tan (16/ 25) tan (64/(100 16 )) 125

jH j

j j

j

Note the large error in the predicted error. In general, straight-line phase angle plots do not give satisfactory results in the frequency band where the phase angle is changing.