Lecture 19Lecture 19 Bode Plot,,g , High Pass Filter ...

Lecture 19
Bode Plot, High Pass Filter,

Series Resonance

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

First-Order Low Pass Filter

H( f ) = 1 fB) = 1∠0o

1+ j( f 1+ ( f fB )2 ∠arctan⎛⎝⎜⎜ f ⎟⎟⎠⎞
fB

H(f ) = 1 ∠H ( f )= − arctan⎜⎜⎝⎛ f ⎠⎟⎟⎞
fB
1 + ( f )fB 2

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

Bode Plot for a First-Order Low-Pass
Filter

A Bode plot shows the magnitude of a network
function in decibels versus frequency using a
logarithmic scale for frequency.

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

Bode Plot for a First-Order Low-Pass
Filter

H(f ) = 1

1+ ( f f B )2

H(f ) = 20 log H(f ) = 20 log⎜⎜⎜⎝⎛ 1 ⎟⎞
dB
1+ ( f f B )2 ⎟⎟⎠

= 20 log(1) − 20 log⎛⎝⎜ 1+ ( f fB )2 ⎟⎞⎠

= −20 log⎝⎛⎜ 1+ ( f fB )2 ⎞⎠⎟

( )= −20 log 1+ ( f )fB 2 1/ 2

( )= −10 log 1+ ( f fB )2

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

Asymptotic Behavior of Magnitude
for Low and High Frequencies

⎡ ⎜⎜⎛⎝ f ⎟⎟⎞⎠ 2 ⎤
log⎢1 fB ⎥
H(f ) = −10 + ⎥⎦
dB ⎢⎣

For f << f B H ( f ) ≈ −10 log(1) = 0
dB

For f >> f B H(f ) ≈ −10 log⎜⎜⎝⎛ f ⎞⎠⎟⎟2 = −20 log⎛⎜⎜⎝ f ⎟⎟⎞⎠
dB fB fB

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

Magnitude Bode Plot for First-Order
Low Pass Filter

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

Asymptotic Behavior of Phase
for Low and High Frequencies

H( f ) = 1+ 1 fB) = 1+ ( f 1∠0o

j( f f B )2 ∠ arctan( f fB)

∠H ( f ) = − arctan( f fB )

= 0 f << f B

= −90 f >> f B

= −45 f = f B

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

1. A horizontal line at zero for f < fB /10.
2. A sloping line from zero phase at fB /10 to –90° at 10fB.
3. A horizontal line at –90° for f > 10fB.

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

Exercise 6.11

Sketch the Bode plots for the low-pass filter shown above:

fB = 1 = 1 = 1000Hz

2πRC 2π ⎜⎛ 1000 ⎞⎟(1x10 −6 )
⎝ ⎠
2π

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

Exercise 6.11

H(f ) = 1 ∠H ( f ) = − arctan⎛⎜ f ⎞⎟

1+ ( f 1000)2 ⎝ 1000 ⎠

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

First-Order High-Pass Filter

Vout = R 1 Vin = 1− j 1 1 Vin = j2πfRC
j j2πfRC +1 Vin
R−
2πfC 2πfRC

Vout = j( f / f B ) where fB = 1
Vin 1+ j( f / f B )
2πfRC

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

First-Order High-Pass Filter

H(f )= Vout = j (f f B ) = ( f fB )∠90o
Vin + j( 1+ ( f fB )2 ∠ arctan( f
1 f f B ) fB )

H(f ) = (f fB)
1+ ( f fB )2

∠H ( f ) = ∠90o fB ) = 90o − arctan( f fB )

∠ arctan( f

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

First-Order High-Pass Filter

H(f ) = (f fB) ∠H ( f ) = 90o − arctan( f fB )
1+ ( f fB )2

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

First-Order High-Pass Filter

H(f ) = f fB
1+ ( f fB )2

H(f ) = 20 log H(f ) = 20 log⎛⎜⎝⎜⎜ f f B ⎞⎟
dB
1+ ( f f B )2 ⎠⎟⎟

= 20 log( f fB ) − 20 log⎛⎝⎜ 1+ ( f fB )2 ⎟⎠⎞

( )= 20 log( f fB ) − 20 log 1+ ( f )fB 2 1/ 2

( )= 20 log( f fB ) −10 log 1+ ( f fB )2

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

Asymptotic Behavior of First-Order

High-Pass Filter

( )H ( f ) = 20 log( f fB ) −10 log 1+ ( f fB )2 f B )2 = 0

For f << fB H ( f ) ≈ 20 log( f fB )

For f >> f B H ( f ) ≈ 20 log( f fB ) −10 log( f

∠H ( f )= 90o − arctan⎜⎝⎛⎜ f ⎞⎠⎟⎟
fB

For f << fB ∠H ( f ) ≈ 90o

For f >> f B ∠H ( f ) ≈ 0o

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

Bode Plots for the First-Order High-
Pass Filter

For f << fB H ( f ) ≈ 20 log( f fB ) ( )H ( f ) = 20 log( f fB ) −10 log 1+ ( f fB )2 f B )2 = 0
For f >> fB H ( f ) ≈ 20 log( f fB ) −10 log( f fB )2 = 0
For f << fB H ( f ) ≈ 20 log( f fB )

For f >> fB H ( f ) ≈ 20 log( f fB ) −10 log( f

∠H ( f )= 90o − arctan⎜⎜⎛⎝ f ⎠⎟⎞⎟
fB

For f << fB ∠H ( f ) ≈ 90o

For f >> fB ∠H ( f ) ≈ 0o

( )H ( f ) = 20 log( f fB ) −10log 1+ ( f fB )2 ∠H ( f )= 90o − arctan⎜⎝⎛⎜ f ⎠⎟⎟⎞
fB
For f << f B

H ( f ) ≈ 20 log( f fB ) For f << f B ∠H ( f ) ≈ 90o

For f >> f B For f >> fB ∠H ( f ) ≈ 0o

H ( f ) ≈ 20 log( f fB ) −10 log( f fB )2 = 0

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

Exercise 6.13

Find the transfer function for this circuit:

Vout = j2πfL Vin = j2πfL / R Vin = j( f / fB) Vin where f B = R / 2πL
R + j2πfL 1+ j2πfL / R 1+ j( f / fB)

H ( f ) = Vout = j( f / f B )
Vin 1+ j( f / f B )

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

Exercise 6.14

R=1kΩ

Find the half power frequency and the capacitance for a high-
pass filter that reduces the amplitude of a frequency
component at 1 kHz by 50dB

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

Exercise 6.14

(20dB / decade)(# decades) = 50dB → # decades = 50dB = 2.5 decades

20dB / decade

What half-power frequency fB is 2.5 decades above 1 kHz?

Number of decades = log⎜⎜⎛⎝ f2 ⎠⎟⎟⎞
f1

2.5 = log⎛⎜ f2 ⎟⎞ → f2 = (1000Hz)102.5 = 316x103 Hz = 316kHz
⎝ 1000Hz ⎠

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

Exercise 6.14

fB = 1 = 316kHz = 316x103 Hz

2πRC

C = 1 = 1

2πRf B 2π (1000Ω)(316x103 Hz)

= 503x10−12 F = 503 pF

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

Computer-Generated Bode Plot

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

Computer-Generated Bode Plot

H(f )= R1 + R3 +1 R3 (R2 + jωL)]

[ jωC +1

Check the low and high frequency behavior to make
sure the computer generated Bode plot is correct

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

Low Frequency Behavior

At low frequency the inductance is a short circuit
The capacitance is an open circuit

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

Low Frequency Behavior

H (0) = R3 = 0.5

R1 + R2 + R3

HdB(0) = 20log(0.5) = −6 dB

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

High Frequency Behavior

At high frequency the inductance is an open circuit
The capacitance is a short circuit

H (∞) = R3 = 0.5263

R1 + R3

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

High Frequency Behavior

HdB(∞) = 20 log(0.5263) = −5.575 dB

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

MATLAB Code

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

MATLAB Output

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

Resonance

Resonance is a phenomenon that can be
observed in mechanical systems and
electrical circuits.

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

Series Resonance

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Series Resonance

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Series Resonance

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Series Resonance

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Series Resonance

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS,hFtotuprt:h/E/wditwionw, b.yinAltlaunitRo. Hr.acmoblmey,/©re20s0o8 PneaarnsocneEd/ucciartcioun,iItnsc..html

Series Resonance

Zs( f ) = j2πfL + R − j 1

2πfC

For resonance Fthoer rreeascotnaanncceeo: f the inductor and the capacitor cancel:

2πf 0 L = 1 → f 2 = 1
0
2πf 0C (2π )2 LC

f0 = 2π 1
LC

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

Series Resonance

Quality factor QS

Qs ≡ Reactance of inductance at resonance
Resistance

= 2πf0 L

R

Substitute L = 1 from f0 = 1
LC
(2π )2 ( f0 )2 C 2π

Qs = 1

2πf 0CR

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

Series Resonance

Zs( f ) = R + j2πfL − j 1

2πfC

= ⎡ j⎝⎛⎜⎜ 2πfL − 1 ⎟⎟⎠⎞⎥⎤⎦⎥
R ⎢1 +
R 2πfRC
⎣⎢

= ⎡ j 2πf0L ⎛⎜ f − 1 ⎠⎟⎟⎞⎥⎦⎥⎤
R ⎢1 + f0
R ⎜⎝ (2π )2 ff0LC
⎣⎢

Substitute f0 = 2π 1 and Qs = 2πf 0 L
LC
R

⎡ jQs ⎛⎝⎜⎜ f − f0 ⎞⎠⎟⎟⎤⎥⎦⎥
Zs ( f ) = R⎢1+ f0 f

⎣⎢

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

Series Resonance

Z 1+ Qs2 ⎝⎜⎛⎜ f − f0 ⎠⎟⎟⎞2 ∠Z s = ⎡ ⎜⎝⎜⎛ f − f0 ⎠⎟⎞⎟⎦⎥⎤⎥
= f0 f arctan⎢Qs f0 f

R ⎢⎣

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.