Chapter 6new (solid state)

8/9/2019 Chapter 6new (solid state)

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Chapter 6:Circuit AnalysisUsing Fourier


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Contents: Fourier Series

Exponential Fourier Series

Fourier Series and Fourier Transform Laplace Transform and Fourier

Transform

Properties of Fourier Transform

Circuit Analysis using Fourier

Transform


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Learning Outcome Students will be able to:

- Understand how to represent a periodic function using Fourier

series (trigonometric and exponential form)

- Understand how to work with Fourier transform

-Analyze the electric circuit response and the characteristic using

Fourier series and Fourier transform


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Some Properties of Periodic Functions A function is periodic if f(t) = f(t+T).

The fundamental period is the smallest constant T for which

this is true i.e.

For sinusoidal functions, the functions are periodic in

multiples of 2T, thus

..2,1,0,nfor ss!! nTtftf

UTUUTU sin2sin;cos2cos !! nn


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Ex

ample Find the period of f(t) = cos6t

Solution

Since n=1 gives the smallest value of T, therefore

3

T

!T

3

62

6cos66cos

6cos26cos

nTTn

tTt

tnt

TT

T

!!@!

!


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6.1 FourierSeries (FS) If f(t) is a PERIODIC signal with period T, then it can be

expressed as a trigonometric FS of the form:

where an & bn are the Fourier coefficients

The lowest freq in the ac component occurs when n=1

known as fundamental frequency

defined as

The other freqs are integer multiples of the fundamental calledthe harmonics, eg.the second harmonic

? Ag

!

!1

000 2sin2cos)(n

nn tnfbtnfaatf TT

dc ac

02 f

Tf 10 !


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General representation

of a periodic signal

Fourier coefficients

Compact Fourier

series

g

!

!1 000

sincosn nn

tnbtnaatf [[

!

!!

!

g

!

n

nn

nnn

n

nn

a

b

baa

tnAAtf

1

22

00

1

00

tan

and,,where

cos

J

J[

!

!

!

T

n

T

n

T

dttntfT

b

dttntfT

a

dttfT

a

00

00

00

sin2

cos2

1

[

[Note:

The integration limits can span any convenient interval

as long as it is exactly 1 period,

e.g. from T0/2 to +T0/2 or T0/4 to+3T0/4

Note:

An is the amplitude of the nth harmonic

n is the phase angle of the nth harmonic


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Example of FS

Tt

f(t)

A

Find the Fourier Series of the sawtooth waveform below.


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Step 1

Find the function of the signal (i.e. f(t))

TtTAttf ee! 0or

Step 2

Find a0

222

112

2

0

2

00

AT

T

At

T

A

Tdt

T

At

Ta

T

T

!

-

!

-

!!

Hint: use y=mx+c formula

Step 3

Find an

0

2

0cos2cos2

2

2cos

2

2sin2

2cos

22cos

2

0

22

020

!

-

!

-

!

!

!

Tn

n

T

A

Tn

Tnt

Tn

Tntt

T

A

dtT

ntt

T

Adt

T

nt

T

At

Ta

T

TT

n

T

T

T

T

T

T

TT

Hint:

nn

dtdt

duvuvdt

dt

dvu

1cos2.

1.

!

!


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Example of FS (cont.)Plot the amplitude and phase spectra of the sawtooth waveform with A =5 and

T = 4ms, see previous example.

Step 1

Find the amplitude and phase angle of the harmonics

Therefore

f(t)=2.5 + 1.59cos(2250t+90)+0.796cos(2T500t+90)+ 0.531cos(2T750t+90)+

"

!

r!

!

"

!!!!

0or

0or

90

unde ined

tan

0or

0or2

1

22

n

n

a

b

nn

A

nA

bbaA

n

n

n

nnnn

J

T


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Step 2

Then, plot

Amplitude spectrum An against nf0 (or n0)

Phase spectrum n against nf0 (or n0)

f(Hz)

An

250 500 7500

0

1

2

3

0.53

0.80

1.59

2.50

f(Hz)

n

250 500 7500

90

45

0

f(t)=2.5

+1.59cos(2T250t+90)+0.796cos(2T500t+90)

+0.531cos(2T750t+90)+


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Waveform Symmetries

Why?

Use to simplify the calculation of FourierCoefficient

3 main types of symmetries:- Even-Function

- Odd-Function

- Half-wave


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Even-Function Symmetry

Occurs when signal is type off(t) = f(-t)

e.g. cosine waveform, rectangular waveform

FS is made up entirely of cosine terms, i.e.bn = 0 (i.e. no sine harmonics)

Why?


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Odd-Function Symmetry

Occurs when signal is type off(t) = - f(-t)

e.g. sine waveform, square waveform

FS is made up entirely of sine terms, i.e.a0 = an = 0 (i.e. no cosine harmonics)

Why?


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Half- ave Symmetry

Occurs when signal is type off(t) = - f(t T/2)

i.e. the periodic function is type of half-wave symmetry

when the function is identical to the original function after

it is shifted one-half period and inverted.

an and bn are zero foreven values

an and bn are non-zero forodd values

Why?


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Exercise 1

Determine the symmetry type of the waveform below and find

its Fourier Series. Plot the amplitude and phase spectra of

this waveform.

Ans: ODD; a0 & an=0;

bn = 0 for even n, 4/nT for odd n

oddn

tnn

tf0

sin14

[

T


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6.2 Exponential FS

The exponential form of FS is

where

g

g!

|n

tjn

neCtf0[

dtetfT

CtjTt

t

00

0

1 [

!


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Tt

f(t)

A

Example

Find the complex Fourier series of

Step 1

Find the function of the signal (i.e. f(t))

TtT

Attf ee! 0or


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Step 2

Find Cn by applying the formula

We know that T = 1/f0 and [0 = 2Tf0, therefore

-

!

-

!!

!!

11

11

0

0

00

0

00

2

00

2

0000

202

00

TjnTjn

Ttjn

Ttjn

Ttjn

tjnTT tjn

n

ejnjn

Te

T

A

dtjn

e

jn

te

T

Adtte

T

A

dteTAt

Tdtetf

TC

[[

[[[

[[

[[

[[

/

-

11 2

2

2

2

njnj

n ejnjn

Te

T

AC

T

T

[[


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Step 2 (cont.)

Simplifying the answer by using

Then, we get and

Therefore,

UUU

sincos jej

!

12 ! nje T

n

Ae

n

Aj

Tn

Aj

jn

T

T

AC

j

n TT[[

T

22

2

00

2!!!

!

2 je j !T

Final step

SubstituteCn into the formula

g

g!

|n

tjn

neCtf0[


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Exercise 2

Find the complex Fourier series of f(t) = VsinT on thefundamental interval 0 < t < 1.

Ans: Period, T = 1

n

ntjen

tf T

T2

214

12


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6.3 Fourier Transform (FT)

FT is an integral transformation of f(t) from the

time domain to the frequency domain

(i.e. function that describes the amplitude and phase of

each sinusoid that corresponds to a specific frequency)

The FT of f(t)

The inverse FT

dtetftfF

tj

g

g

!![

[ )()]([F)(

[[

T

[[ deFFtf tj

g

g!! )(

2

1)]([F)( 1-


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Example

Find the Fourier transform of f(t) where

""

!

00,for t

0for t0

EEtetf

SolutionApply the definition of FT

22

0

0

0

1

[E

[E

[E[E

[

[E

[E

[E[

!

!

-

!

!

!!

g

g

g g

g

j

jje

dte

dteedtetfF

jt

jt

tjttj


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6.4 Laplace Transform FT

Table of Laplace Transform can be used to obtain FT.

FT only exists when the Fourier integral converges, i.e. when

all the poles of F(s) lie in the left-hand of s-plane.


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Example Find the Fourier Transform of f(t) where

u

e!

0o ttcos

0o t0

0

-

[attf

SolutionFirst, find the LT of f(t)

Then, substitute s = j[ (as it satisfies case A, refer to Table on previousslide) to get the Fourier Transform

20

2 [

!

as

assF

20

2[[

[[

!

aj

ajF


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Refer to Table on next page

6.5 Properties of FT


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Name of property Function of Time Fourier Transform

1 Definition

2 Multiplication by

constant

3 Linearity

4 Time Shift

5 Time Scaling

6 Modulation

7 Differentiation

8 Convolution

9 Time Multiplication

10 Time reversal

11 Integration

tf

tAf

tbftaf 21

0ttf

0, "aatf

tfe tj 0[

n

n

dt

tfd

dxxtfxf

g

g21

tftn

tf

XX dfg

g

[F

[AF

[[ 21 bFaF

[[ Fe tj 0

aF

a[1

0[[F

[[ Fj n

[[ 21 FF

n

nn

d

Fdj

[

[

[F [HT[

[0F

j

F


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Exercise 3

Find the Fourier Transform of f(t) if

0for

0for

!

!

tetf

tetf

t

t

E

E

Ans:

[

[[

!

jF


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6.6.1 Circuit Analysis using FS

FS is used in circuit analysis to find

- the steady state response

- rms value & average power

For steady state response at frequency, [A- can be found directly from a network function, T(s)

output amplitude = input amplitude x |T(j[A)|

output phase = input phase + T(j[A)


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Circuit Analysis using FS (cont.)

For rms value & average power

- rms value of a periodic waveform is equal to the square

root of the sum of the square of the dc value and the

square of the rms value of each of the ac components.

- average powerdelivered to a resistor is related to its rms

voltage or current as

g

!

!1

2

2

02n

n

rms

AAF

=> From the definition of

rms value i.e.

? A!0

0

2

0

1 T

rmsdttf

TF

RIR

VP

rms

rms 2

2

!!


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Example 1. Circuit analysis using FS

The input to the low pass filter circuit shown in the figure

below is

a) Find vout(t) in the steady state.

b) Find the rms value of vout(t) and the average power

absorbed by the 10k; resistor

SOLUTIONSOLUTIONSOLUTIONSOLUTION


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6.6.2 Circuit Analysis using FT

Laplace transform is used widely to find the response of

a circuit than is the Fourier Transform.

Why?

1. Laplace Transform integral converges for a widerrange of driving functions

2. Laplace Transform accommodates initial conditions.

However, FT can be used to find the response.

N

ote :in certain communication theory and signal processingsituation, FT is more useful

(as FT accommodates both -ve-time and +ve-time functions

problems described in terms of events that start at t = -g )


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Circuit Analysis using FT (cont.)

FT can be used to find the type of response:

1 - the transient response

2 - the sinusoidal steady-state response


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Example1: Using FT to find theTransient Response

Use the Fourier Transform to find i0(t) in the circuit shown

below. The current source ig(t) is the signum function 20 sgn(t)

A.

SOLUTIONSOLUTION


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Example2: Using FT to find theSinusoidal steady-state Response

The current source for the circuit shown below is a type of

sinusoidal source. The expression for the current is

ig(t) = 50 cos3tA.

Use the Fourier transform method to find i0(t).

SOLUTIONSOLUTION


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Parsevals Theorem

? A

[[T

[[[T

[[T

[

[

dV

dVV

dtdeVtv

dtVtvdttveiW

tj

2

12

1

2

1

2

1

2

1

..

g

g

g

g

g

g

g

g

g

g

g

g;

!

!

!

! F

This theorem relates the energy associated with a time-

domain function of finite energy to the FT of the function.

Power dissipated by a 1-; resistor is given by

Therefore, the energy absorbed by 1-; resistor istvp

2!

This equation is called

Parsevals Theorem


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Parsevals Theorem (cont.)

The plot of |F([)|2 versus is called the energy spectrum. Example of energy spectrum

|F([)|2

[


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We can associate a portion of total energy with a specified

band of frequencies (see Figure below).

|F([)|2

[-[2 -[1 [1 [2

continue on next page


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Therefore, the 1; energy in the frequency band from [1 to [2and from -[2 to -[1 is

[[[[[

[

[

[dFdFW

22

1

2

1

1

2 2

1

2

1

;


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Example on Parsevals Theorem(cont.)

The total energy dissipated in the 40; resistor is:

The energy associated with the frequency band 0 e [ e 23rad/s is

J

dW

4000tan116000

4

40040

0

1

040

!

!

!

g

g

;

[T

[[T

J

d

3

8000tan

116000

4

40040

32

0

1

32

0240

!!

!

T

T


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Example on Parsevals Theorem(cont.)

Hence, the percentage of the total energy associated with the

frequency band 0 e [ e 23 rad/s is

%67.66100

4000

38000!v!L


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Thank You

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Chapter 6new (solid state)