FOURIER SERIES - PowerPoint PPT Presentation

1 / 26
About This Presentation
Title:

FOURIER SERIES

Description:

For the Sawtooth signal. 8. FKEE Norizam. 9/16/09. Chapter 2 ... is in of term of sawtooth wave (Refer. to page 779, practice problem 17.6). 17. FKEE Norizam ... – PowerPoint PPT presentation

Number of Views:396
Avg rating:3.0/5.0
Slides: 27
Provided by: NECCompute153
Category:

less

Transcript and Presenter's Notes

Title: FOURIER SERIES


1
CHAPTER 2
  • FOURIER SERIES

2
Chapter 2FOURIER SERIES - DEFINATION
  • WHAT IS FOURIER SERIES ?
  • It is a presentation that resolves the
    periodic function, f(t) into a dc component and
    an ac component comprising an infinite series of
    harmonic sinusoids.
  • Periodic function-
  • f(t) f(t nT)

3
Chapter 2 FOURIER SERIES - COMPONENTS
  • It can be defined mathematically as below
  • ?0 is a Fundamental Frequency,
  • ?0 2?f0 2?/T
  • sin n?0t or cos n?0t is nth harmonic of f(t)
  • a0 , an , bn are the Fourier Coefficient

4
Chapter 2 FOURIER SERIES - COEFFICIENT
  • Fourier Coefficients are defined as below
  • a0 gt DC Component
  • a0 1/T
  • an gt Amplitude of Sinusoid in AC Component
  • an 2/T

5
Chapter 2 FOURIER SERIES FOURIER COEFFICIENT
  • bn gt Amplitude of Sinusoid in AC Component
  • bn 2/T
  • The Fourier series equation can be written in the
    form of amplitude-phasor shown below
  • f(t) a0 An cos (n?0t ?), where
  • An an2 bn2 and ?n - tan-1 (bn /
    an)

6
Chapter 2 FOURIER SERIES Function Values
  • Cos 2n? ? 1
  • Sin 2n? ? 0
  • Cos n? ? (-1)n
  • Sin n? ? 0
  • Cos n?/2 ? (-1)n/2, n even
  • 0, n odd
  • 6. Sin n?/2 ? (-1)(n-1)/2, n odd
  • 0, n
    even
  • 7. ej2n? ? 1,
  • 8. ejn? ? (-1)n
  • 9. ejn?/2 ? (-1)n/2, n even,
  • j(-1)(n-1)/2 , n odd

7
Chapter 2 FOURIER SERIES
  • EXAMPLE 1
  • Refer to Example 17.1 (Page 762)
  • Determine the Fourier Series for
  • Square Wave
  • Exercise 1
  • Next, determine the Fourier Series
  • For the Sawtooth signal.

8
Chapter 2 FOURIER SERIES Symmetry Consideration
  • The Fourier Series will be EVEN if it
  • meets this equation f(t) f(-t)
  • Thus, the Fourier Coefficient will be
  • 1. a0 2/T
  • 2. an 4/T
  • 3. bn 0

9
Chapter 2 FOURIER SERIES Symmetry Consideration
  • The Fourier Series will be ODD if it
  • meets the following equation f(t) -f(-t)
  • Thus, The Fourier Coefficient will be
  • 1. a0 0
  • 2. an 0
  • 3. bn 4/T

10
Chapter 2 FOURIER SERIES Symmetry Consideration
  • The properties of Even Odd function are-
  • The product of 2 even function is also even
    function
  • The product of 2 odd function is also odd
    function
  • The product of even and odd function is odd
    function
  • The sum or difference of 2 even functions is also
    even function
  • The sum or difference of 2 odd functions is also
    odd function
  • The sum or difference of even and odd functions
    is neither even or odd function

11
Chapter 2 FOURIER SERIES Half-Wave Symmetry
  • The function is half-wave symmetric if
  • f(t T/2) - f(t)
  • The Coefficient is defined as below
  • 1. a0 0
  • 2. an 4/T
    for n Odd
  • 0 for n Even
  • 3. bn 4/T
    for n Odd
  • 0 for n Even

12
Chapter 2 FOURIER SERIES Symmetry Consideration
  • EXAMPLE 2
  • Refer to Example 17.3 (Page 774)
  • Exercise 2
  • Next, determine the Fourier Series
  • for the Triangular signal

13
Chapter 2 FOURIER SERIES Application in RC, RL
RLC
  • The input voltage can be expressed in term of
    Fourier Series as shown below
  • v(t) V0 Vncos(n?0t ?n)
  • The output voltage basically expressed in term of
    phasor such Impedance, Z or Admittance, Y (1/Z)
  • For DC Component, ?n 0 or n 0,
  • thus, Vs V0

14
Chapter 2 FOURIER SERIES - Application in RC, RL
RLC
  • EXAMPLE 3
  • Determine the output voltage at inductor, vo(t)
  • for the RL Circuit where the input voltage is in
  • term of square wave.(R5? L 2H)
  • Solution
  • Determine the Fourier Series for Square Wave
  • vs(t) ½ 2/? 1/n sin n?t, n 2k 1
  • 2. Since the circuit is a series circuit, use
    voltage divider to obtain output voltage, Vo,
  • Vo j?nL / (R j?nL) Vs j2n? /(5 j2n?)
    Vs

15
Chapter 2 FOURIER SERIES - Application in RC, RL
RLC
  • EXAMPLE 3 (Continue)
  • 3. For dc component, ?n 0, thus
  • Vs ½,
  • 4. For ac component, ?n ? 0
  • Vs 2/n? ?-90o
  • 5. Now, calculate output voltage, Vo
  • Vo 2n? ?-90o / sqrt(25
  • 4n2?2)?tan-1 2n?/5 (2/n? ?-90o)
  • (4?-tan-1 2n?/5)/sqrt(25 4n2?2)
  • 6. In time domain,
  • vo (t) 4 / sqrt(25 4n2?2)cos(n?t-tan-
    12n?/5)

16
Chapter 2 FOURIER SERIES -Application in RC, RL
RLC
  • Exercise 3
  • Next, determine the output voltage,
  • vo(t) of RC circuit if the input voltage
  • is in of term of sawtooth wave (Refer
  • to page 779, practice problem 17.6).

17
Chapter 2 FOURIER SERIES AVERAGE POWER RMS
VALUE
  • The power in the electrical circuit due to the
    periodic input voltage is defined as
  • P VdcIdc 1/2 VnIn cos (?n - ?n)
  • The rms value of periodic function is defined as
    -
  • Frms sqrt (1/T )

18
Chapter 2 FOURIER SERIES AVERAGE POWER RMS
VALUE
  • EXAMPLE 4
  • Determine the average power
  • supplied to the RC circuit if the input
  • current is i(t) 2 10cos(t 10o)
  • 6cos(3t 35o) A. (Refer page 782, Example
    17.8).
  • NEXT EXERCISES
  • Do practice problem 17. 8

19
Chapter 2 FOURIER SERIES AVERAGE POWER RMS
VALUE
  • Practice Problem 17.8
  • Given,
  • v(t) 80 120cos 120?t
  • 60cos (360?t 30o)
  • i(t) 5cos(120?t - 10o)
  • 2cos(360?t 60o)
  • Find the Average Power absorbed by
  • the circuit.

20
Chapter 2 FOURIER SERIES AVERAGE POWER RMS
VALUE
  • SOLUTION
  • Since Power is defined as PVI Power due to
    periodic input voltage is defined as
  • P VdcIdc 1/2 VnIn cos (?n - ?n)
  • 2. Now, Calculate Power
  • P 80(0) ½ (120)(5)cos(10o) ½
  • 60 cos(30o)(2)
  • 295.44 51.962
  • 347.4 W

21
Chapter 2 FOURIER SERIES EXPONENTIAL FOURIER
SERIES
  • Use Euler Identity to express FS in term of
    exponential
  • 1. cos n?0t ½ ejn?0t e-jn?0t
  • 2. sin n?0t ½j ejn?0t - e-jn?0t
  • Thus, periodic function, f(t) can be expressed
    as
  • f(t) cn ejn?0t
  • cn 1/T

22
Chapter 2 FOURIER SERIES EXPONENTIAL FOURIER
SERIES
  • The exponential FS Coefficient, cn can be
    expressed in term of amplitude-phase form as
    shown below
  • cn cn ? ?n sqrt (an2 bn2)/2
  • ?-tan-1 bn / an

23
Chapter 2 FOURIER SERIES EXPONENTIAL FOURIER
SERIES
  • EXAMPLE 5
  • Find the exponential FS of
  • f(t) et , 0 lt t lt 2? with
  • f(t2?) f(t)
  • Solution
  • Refer Example 17.10 (Page 787)

24
Chapter 2 FOURIER SERIES EXPONENTIAL FOURIER
SERIES
  • EXERCISE
  • Do Practice Problem 17.10 (Page
  • 788). Find the complex FS for the
  • Square Wave signal.

25
Chapter 2 FOURIER SERIES EXPONENTIAL FOURIER
SERIES
  • SOLUTION
  • 1. For the Square Wave signal,
  • f(t) 1, 0 lt t lt 1
  • 0, 1 lt t lt 2
  • 2. Now, define the period, T ?
  • T 2, ? gt 2? / 2 ?
  • 3. Next, calculate exponential FS
    Coefficient, cn
  • 4. cn 1/2
    j/(2n?)(e-jn? -1)

26
Chapter 2 FOURIER SERIES EXPONENTIAL FOURIER
SERIES
  • SOLUTION
  • 5. Since e-jn? cos(n?) jsin(n?) gt (-1)n
  • 6. cn j/2?n(-1)n 1
  • gt 0 if n even
  • gt -j/(n?) if n odd
  • 7. for n 0, c0 0.5
  • 8. Thus, the exponential FS for f(t) is -
  • f(t) ½ - j/n? ejn?t
Write a Comment
User Comments (0)
About PowerShow.com