lecture 18.pdf

Lecture 18Lecture 18Fourier Analysis, Low Pass y ,

Filters, Decibels

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

Chapter 6F R B d PlFrequency Response, Bode Plots,

and Resonance

1. State the fundamental concepts of Fourier1. State the fundamental concepts of Fourier analysis.

2. Determine the output of a filter for a given input consisting of sinusoidal componentsinput consisting of sinusoidal components using the filter’s transfer function.


3. Use circuit analysis to determine theytransfer functions of simple circuits.

4. Draw first-order lowpass or highpass filter circuits and sketch their transfer functions.

5. Understand decibels, logarithmic frequency scales, and Bode plots.

6. Draw the Bode plots for transfer functions of first-order filters.


Fourier Analysisy


Fourier Analysisy

All real-world signals are sums of sinusoidal t h i i f icomponents having various frequencies,

amplitudes, and phases.


Fourier Analysisy

....)5sin(4)3sin(4)sin(4)( 000 +++= tAtAtAtvsq ωωω


....)5sin(5

)3sin(3

)sin()( 000 +++ ttttvsq ωπ

ωπ

ωπ

Filters

Filters process the sinusoid components of an inputsignal differently depending of the frequency ofsignal differently depending of the frequency of each component. Often, the goal of the filter is to retain the components in certain frequency rangesretain the components in certain frequency ranges and to reject components in other ranges.


Filters


Transfer Functions

The transfer function H(f ) of the two-port filter is defined to be the ratio of the phasor output voltage to the phasor input voltage as a function of frequency:

( ) outV=fH ( )

inVf


Transfer Functions

The magnitude of the transfer function showsThe magnitude of the transfer function shows how the amplitude of each frequency component is affected by the filter Similarly the phase of theis affected by the filter. Similarly, the phase of the transfer function shows how the phase of eachfrequency component is affected by the filterfrequency component is affected by the filter.


Transfer Functions

Magnitude Phase


Example 6.1

402

10002

2000)402000cos(2)(

o

o

∠

==→+= Hzfttvin πππ

V

706)402)(303())(303(303)1000(

402

ooooo ∠=∠∠=∠=→=∠=

∠=

H inoutin

out

in

VVVV

V


)702000cos(6 o+= tvout π

Exercise 6.1

02

20002

4000)4000cos(2)(

o∠

==→= Hzfttvin πππ

V

604)02)(602())(602(602)2000(

02

ooooo ∠=∠∠=∠=→=∠=

∠=

H inoutin

out

in

VVVV

V


)604000cos(4 o+= tvout π

Exercise 6.1

201

30002

6000)206000cos(1)(

∠

==→−=

o

o Hzfttvin πππ

V

1100)201)(900())(900(900)3000(

201

∠=∠∠=∠=→=∠=

∠=

oooooH inoutin

out

in

VVVV

V


0)1106000cos(0 =+= otvout π

Graphic Equalizer

Filter with an adjustable transfer function


Filter with an adjustable transfer function

Determining the output of a filterDetermining the output of a filter for an input with multiple

components:

1 D i h f d h1. Determine the frequency and phasor representation for each input component.

2. Determine the (complex) value of the transfer function for each component.p


3 Obtain the phasor for each output3. Obtain the phasor for each output component by multiplying the phasor for each input component by the correspondingeach input component by the corresponding transfer-function value.

4. Convert the phasors for the output components into time functions of various frequencies. Add these time functions to produce the output.


Example 6.2

o

o

033)(

)704000cos()2000cos(23)(

11 ∠==

−++=

inin

in

Vtv

tttv ππ

oo

o

701)704000cos()(

02)2000cos(2)(

)(

3

2

1

3

2

1

−∠=−=

∠==

inin

inin

inin

Vttv

Vttv

π

π


33 inin

Example 6.2

102)701)(602()2000(602)2000(

306)02)(303()1000(303)1000(

012)03)(04()0(4)0(

22

11

oooo

oooo

ooo

−∠=−∠∠==∠=

∠=∠∠==∠=

∠=∠∠===

VHVH

VHVH

VHVH

inout

inout

inout

)104000(2

)302000cos(6

12

0)70)(60()000(60)000(

2

1

33

o

o+=

=

tv

v

VV

out

out

inout

π


)104000cos(2)302000cos(612

)104000cos(2

321

3

oo

o

−+++=++=

−=

ttvvvv

tv

outoutoutout

out

ππ

π

Linear circuits behave as if they:

1. Separate the input signal into components p p g phaving various frequencies.

2. Alter the amplitude and phase of each component depending on its frequency.p p g q y

3. Add the altered components to produce p pthe output signal.


Determination of the Transfer Function


First-Order Low Pass Filter

R

in1

+=

VI

fCjR

in

π

11

2

⎟⎟⎞

⎜⎜⎛

⎟⎟⎞

⎜⎜⎛

+

VIV

fCjRfCjfCj

t

inout

πππ

1112122

⎟⎟⎟

⎠⎜⎜⎜

⎝+

⎟⎟⎠

⎜⎜⎝

==

V

IV

Half power


RCf

ffjfRCjfH B

Bin

out

ππ 21

)/(11

211)( =

+=

+==

VV Half power

frequency


( ) ∠fH 011 o

( ) ( ) ( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛∠+

=+

=

BB

B

ffff

ffjfH

arctan11 2

⎠⎝

( )( )21

1

fffH = ( ) ⎟⎟

⎠

⎞⎜⎜⎝

⎛−=∠

fffH arctan


( )21 Bff+ ⎠⎝ Bf


For low frequency signals the magnitude of the transfer function is unity and the phase is 0°. Low frequency signals are passed while high frequency signals are attenuated and


are passed while high frequency signals are attenuated and phase shifted.

Calculation of RC Low-Pass Output

tttvin )2000cos(5)200cos(5)20cos(5 ++= πππ

HzxRC

fB 100)1010(10002

12

16

=⎟⎞

⎜⎛

==−ππ


x )1010(2

2 ⎟⎠

⎜⎝ π

π

Calculation of RC Low-Pass Output

( )( )2

1

ffH = ( ) ⎟

⎠⎞

⎜⎝⎛−=∠100

arctan ffH

71.59950.0)100/10(1

1)10(05)20cos(511

oo −∠==∠==j

Htv inin π V

( )21001 f+ ⎠⎝100

2984099501)1000(05)2000cos(5

457071.0)100/100(1

1)100(05)200cos(5

)100/10(1)()(

22

11

oo

oo

∠∠

−∠=+

=∠==

+

Htv

jHtv

j

inin

inin

π

π

V

V

)71.520cos(975.471.5975.4)05)(71.59950.0()10(

29.840995.0)100/1000(1

)1000(05)2000cos(5

111

33

oooo −=−∠=∠−∠==

−∠=+

=∠==

tvH

jHtv

outinout

inin

π

π

VV

V

)29.842000cos(4975.029.844975.0)05)(29.840995.0()1000(

)45200cos(535.345535.3)05)(457071.0()100(

333

222

oooo

oooo

−=−∠=∠−∠==

−=−∠=∠−∠==

tvH

tvH

outinout

outinout

π

π

VV

VV


)29.842000cos(4975.0)45200cos(535.3)71.520cos(975.4)()()()(321

ooo −+−+−=++= ttttvtvtvtv outoutoutout πππ

Exercise 6.4

jfLRin

)2(+=

VI

RjfLR

R

jfLR

inout π

π

)2(

)2(

⎟⎟⎠

⎞⎜⎜⎝

⎛+

==

+

VIV

LRf

ffjLfjfLjRRfH

jfLR

BBin

out

πππ

π

2)/(11

21

12

)(

)2(

=+

=+

=+

==

⎠⎝ +

VV


Rj1+

Exercise 6.5

111

)200/(11

)/(11)(

200102.01051

101025002

12

1)102cos(5)1000cos(5)40cos(10)( 236

4

+=

+=

=====++=−−

jfjfffH

HzxxxRC

fttttv

B

Bin

πππ

πππ

1

2.68857.1)2.683714.0)(05(2.683714.05.21

1)500(05)1000cos(5)(

711.595.9)711.5995.0)(010(711.5995.01.01

1)20(010)40cos(10)(

222

111

ooooo

ooooo

−∠=−∠∠=−∠=+

=∠==

−∠=−∠∠=−∠=+

=∠==

jHttv

jHttv

out

out

π

π

VV

VV

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.)9.88102cos(10.0)2.681000cos(85.1)71.540cos(95.9

85.881.0)85.881020)(05(85.8810205011)000,10(05)102cos(5)(

4

33

33

43

ooo

ooooo

−+−+−=

−∠=−∠∠=−∠=+

=∠== −−

tttv

xxj

Httv

out

out

πππ

π VV

Decibels, the Cascade Connection, and Logarithmic Frequency Scales

( ) ( )fHfH log20dB=

dB → decibelsdB → decibels


Numbers greater than 1 are positivethan 1 are positive

Numbers smaller th 1 tithan 1 are negative


VVVVV)()()( 21

2

2

1

1

1

2

1

1

1

2 fHfHVV

VV

VV

VV

VV

fHin

out

in

out

out

out

in

out

in

out ====


Cascaded Two-Port Networks

( ) ( ) ( )fHfHfH 21 ×=( ) ( ) ( )fff 21

( ) ( ) ( )fHfHfH +=( ) ( ) ( )dB2dB1dB

fHfHfH +=


10=h

ff

One decade:

2=h

l

ff

One octave:


2=lf

One octave:

Logarithmic Frequency ScalesLogarithmic Frequency Scales

On a logarithmic scale, the variable is multiplied by a given factor for equalincrements of length along the axis.


A decade is a range of frequencies for which the ratio of the highest frequency to the the ratio of the highest frequency to the lowest is 10.

⎟⎟⎠

⎞⎜⎜⎝

⎛=

1

2log decades ofNumber ff

⎠⎝ 1f

An octave is a two-to-one change in frequencyAn octave is a two-to-one change in frequency.

( )( ) ⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

2logloglog octaves ofNumber 12

1

22

ffff


( ) ⎠⎝⎠⎝ g1f

Exercise 6.6

Convert the magnitude of the transfer function H(f) into dB’s:

fH 50)( =

( ) dBfH dB 3450log20)( ==


Exercise 6.7

623.510)()(log2015)()( 2015

==→== fHfHdBfHa dB )()(g)()(

30

fff dB

62.3110)()(log2030)()( 2030

==→== fHfHdBfHb dB


Exercise 6.8

(a) What frequency is two octaves higher than 1000 Hz?(a) What frequency is two octaves higher than 1000 Hz?

1000log 2 ⎟

⎠⎞

⎜⎝⎛

⎞⎛f

f)2log(

1000g

1000log2 2

2

⎞⎛

⎠⎝=⎟⎠⎞

⎜⎝⎛=

f

f

)2log(21000

log

)2log(22

2 =⎟⎠⎞

⎜⎝⎛

f

f

000,4101000

101000

)2log(22

)2log(22

==

=

xf

f


Exercise 6.8

(b) What frequency is three octaves lower than 1000 Hz?(b) What frequency is three octaves lower than 1000 Hz?

1000log1000 ⎟⎟

⎠

⎞⎜⎜⎝

⎛⎞⎛ f

1000

)2log(1000log3 1

12

⎞⎛

⎟⎠

⎜⎝=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

ff

1000

)2log(31000log

)2log(3

1=⎟⎟

⎠

⎞⎜⎜⎝

⎛f

1251000

101000

1

)2log(3

1

==

=

f

f


12510 )2log(31 ==f

Exercise 6.8

(c) What frequency is one decade lower than 1000 Hz?(c) What frequency is one decade lower than 1000 Hz?

⎞⎛ff

1010001000log1 =→⎟⎟⎠

⎞⎜⎜⎝

⎛=

Hzfff

100=⎠⎝


Exercise 6.9

(a) What frequency is halfway between 100 and 1000 Hz(a) What frequency is halfway between 100 and 1000 Hz on a logarithmic frequency scale?

2)100l (3)1000log(

2)100log(==

average 5.22

23

2

=+

=

Hzff 2.316105.2)log( 5.2 ==→=


Exercise 6.9

(b) What frequency is halfway between 100 and 1000 Hz(b) What frequency is halfway between 100 and 1000 Hz on a linear frequency scale?

fHzf

Hzf1000100

2

1

==

HzHzHzf 5502

1001000=

+=

2


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