Title: FOURIER SERIES
1CHAPTER 2
2Chapter 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)
3Chapter 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
4Chapter 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
5Chapter 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)
6Chapter 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
7Chapter 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.
8Chapter 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
9Chapter 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
10Chapter 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
11Chapter 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
12Chapter 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
13Chapter 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
14Chapter 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
15Chapter 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)
16Chapter 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).
17Chapter 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 )
18Chapter 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
19Chapter 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.
20Chapter 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
21Chapter 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
22Chapter 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
23Chapter 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)
24Chapter 2 FOURIER SERIES EXPONENTIAL FOURIER
SERIES
- EXERCISE
- Do Practice Problem 17.10 (Page
- 788). Find the complex FS for the
- Square Wave signal.
-
25Chapter 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)
26Chapter 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
-