Electrical Circuits II (ECE233b)

The University of Western Ontario Faculty of Engineering Science

Anestis Dounavis

Fourier Analysis Techniques (Part 1)

BackgroundWe have discussed the following solution methodologies

Phasor transforms

Laplace transforms

Steady state; Sinusoidal analysis

Transient and steady state; Any type of forcing function

The periodic signal is expresses as a summation of sinusoids with harmonically related frequencies.

Fourier transforms Steady state; non-sinusoidal but periodic signals.

Fourier analysis techniques

Fourier SeriesA periodic function satisfies the following relationship

)nTf(tf(t) 0

where T0 is the period 0 is the fundamental frequency (rad/s) 0=2/T0

Examples of periodic signals

3,......2,1,n

T0 2T0 3T0 t

A

f(t)

T0 2T0 3T0 t

A

f(t)sawtooth waveform square waveform

Can be expressed as

1n

)tcos(nDaf(t) n0n0 (phasor notation)

Fourier SeriesFourier series can be expressed in the following forms

Exponential Fourier Series

Trigonometric Fourier Series

n

tjnω

0nn

tjnω0

00 eeaf(t) nn cc

1n

t))sin(nbt)cos(n(aaf(t) 0n0n0

The two forms are similar since sincos jje

where n is an integer a0, an, bn and cn are the Fourier coefficients 20, 30, …. , n0 are the harmonic frequencies 0 is the fundamental frequency

Fourier SeriesRelationship between exponential and trigonometric form

0nn

tjnω0

0eaf(t) nc

11 nn

tjnωtjnω0

00 eea nn cc

tjnω-tjnω0

00 eea *nn cc

1n

1n

)tjnω0

0eRe(2a nc

1n

)tjnωnn0

0)eRe((Da

1n

)tjnωnn0

0)ejb-Re((aa

1n

t))sin(nbt)cos(n(aa 0n0n0

* = complex conjugate

where

nnn jba nc2nD

Fourier Series

tt 2*3sin312sin

t2*3sin31

t2sin

Approximating a square wave

Fourier Series

tnnn

2*)12(sin)12(

18

1

Approximating a square wave

Fourier SeriesDetermining the coefficients for exponential Fourier seriesAny physical realizable periodic signal may be represented over the interval t1<t<t1+T0 by the exponential Fourier series

n

tjnω0ef(t) nc

Multiplying both sides by e-jk 0t and integrating over the interval t1 to t1+T0 yields

0k

Tt

t

tjkω-

n

tjnωn

Tt

t

tjkω- Tcdteecdtf(t)e01

1

00

01

1

0

since

kn forTkn for0

dte0

Tt

t

tk)ω-j(n-01

1

0

The Fourier coefficients are defined as

01

1

0

Tt

t

tjnω-

0n dtf(t)e

T1c

Example 1Determine the exponential Fourier series for the periodic voltage waveform shown

T/4

V

v(t)

T/2-T/2

-T/4

-V

Fourier SeriesStrategy to determine Fourier Coefficients

1. Identify T, 0 and defining equation of function2. Determine coefficient by appropriate integration

Fourier SeriesDetermining the coefficients for trigonometric Fourier series

1n

t))sin(nbt)cos(n(aaf(t) 0n0n0

01

1

0

Tt

t

tjnω-

0n dtf(t)e

T1c

Using the result obtained for the exponential Fourier series

andnn jba nc2

)sin()cos( tjtt00

-jn nne 0

yields the following Fourier coefficients for a0, an and bn

01

1

Tt

t0

0n t)dtf(t)cos(nω

T2a

01

1

Tt

t0

0n t)dtf(t)sin(nω

T2b

01

1

Tt

t00 f(t)dt

T1a

Fourier Series

If a signal exhibits certain symmetrical properties, we can take advantage of these properties to simplify the calculations of the Fourier coefficients

There are 3 types of symmetry: 1) Even-function symmetry2) Odd-function symmetry3) Half-wave symmetry

Trigonometric Fourier series

Fourier Series Even-function symmetry: A function is said to be even if

f(-t)f(t) Example of even functions

Even functions can only be approximated with even basis functions eg) cos(not)

1) Therefore can only have “an” coefficients since bn=02) Can integrate over half cycle

2/0T

00

0n t)dtf(t)cos(nω

T4a

2/0T

000 f(t)dt

T2a

Fourier SeriesOdd-function symmetry: A function is said to be odd if

f(-t)f(t) Example of odd functions

Odd functions can only be approximated with odd basis functions eg) sin(not)

1) Therefore can only have “bn” coefficients since an=02) Can integrate over half cycle

2/0T

00

0n t)dtf(t)sin(nω

T4b

Fourier SeriesHalf-Wave symmetry: A function is said to posses half-wave symmetry if

2T-tff(t) 0

Example of Half-Wave symmetry

This equation states that each half-cycle is an inverted version of the adjacent half-cycle; that is, if the waveform from –T0/2 to 0 is inverted, it is identical to the waveform from 0 to T0/2.

Symmetric around horizontal axis Independent of where t=0

Fourier SeriesExample of Half-Wave symmetry

1. The average of the function is zero, therefore a0=02. an=bn=0 if n is even and

2/0T

00

0n t)dtf(t)cos(nω

T4a

2/0T

00

0n t)dtf(t)sin(nω

T4b

if n is odd

Also if the function is odd then an=0and if the function is even then bn=0

Example 2Determine the trigonometric Fourier series for the periodic voltage waveform shown and compare results with example 1.

T/4

V

v(t)

T/2-T/2

-T/4

-V

Example 3Determine the trigonometric Fourier series for the periodic voltage waveform shown.

V

-T T 2T

Example 4Determine the type of symmetry exhibited by the waveform.

2

1

-1

-2

-1

Time-ShiftingTime shifting of a periodic waveform f(t) is defined as

n

tjnω0ef(t) nc

n

t-(tjnω0

00e)t-f(t )nc

time shifting f(t)

n

tjnω

n

tjnωtjnω0

0000 e)ee)t-f(t nn k(c

where 00tjnωenn ck

Therefore, time shift in the time domain corresponds to a phase shift in the frequency domain

To compute the phase shift in degrees

Phase shift (degrees) = 0td = (3600)td/T0

td=time delay

Example 5

T/4

V

v(t)

T/2-T/2

-T/4

-V

If the wave form in example 1 is shifted by a quarter period as shown below, determine the Fourier coefficients

Frequency SpectrumFrequency Spectrum: The frequency spectrum of f(t) expressed as a Fourier series consists of an amplitude spectrum and a phase spectrum

Amplitude Spectrum: Plot of the amplitude of the harmonics versus frequency.

Phase Spectrum: Plot of the phase of the harmonics versus frequency.

Since the frequency components are discrete, the spectra is called line spectra

The plots are based on the following equations

0nn

tjnω0

0eaf(t) nc

1n

)tjnωnn0

0)eRe((Da

where nnn jba nc2nD

Example 6The Fourier series for the waveform shown is given by

10

T/2-T/2

-10

odd nn

v(t)1

0220 )cos(40)sin(20 tnn

tnn

Plot the first four terms of the amplitude and phase spectra for this signal

Steady-State Network ResponseIf a periodic signal is applied to a network the steady state response of the circuit can be found as follows:

If the input forcing function for a network is a voltage, the input can be expressed as

1. Represent the periodic forcing function by a Fourier series

....(t)v(t)vv v(t) 210

+-

+-

+-

Network(t)v1

0v

(t)v2

2. Use phasor analysis in the frequency domain to determine the network response due to each source3. The network response due to each source is transformed to the time domain

4. Add time domain solutions due to each source (Principle of Superposition) to obtain Fourier series for the total steady- state network response

Example 7Determine the steady-state voltage vo(t) in for the circuit shown if the input voltage v(t) is given by the expression

odd nn

v(t)1

0220 )cos(40)sin(20 tnn

tnn

+ - 11F

+vo(t)

-

2

v(t)

1

Average PowerRecall that average power is defined as

T

0

v(t)i(t)dtT1P

If v(t) and i(t) are periodic functions, the signals can be expressed as:

1n

v0ndc )θtcos(nVVv(t)n

1n

i0ndc )θtcos(nIIi(t)n

Note to calculate the average power involves the product of two infinite series

However, only products at the same frequency survive integration over one period (due to orthogonal properties of basis functions)

Average PowerNote:

T

000 nmT/2

nm0t)dtt)cos(mcos(n

Therefore we can show that the average power can be expressed as:

1n

ivnn

dcdc )θcos(θ2IVIVP

nn

Example 8For the network shown, the input voltage is

+-v(t)

100F 20mH

16

i(t)

)00 20-12cos(754t)3016cos(377t42v(t)

Compute the current i(t) and determine the average power absorbed by the network