Lecture #7:

CONTINUOUS TIME FOURIER SERIES FOR PERIODIC SIGNALS

Motivation:

Outline:

I. FOURIER SERIES OF A PERIODIC FUNCTION

1/ Periodic time function

x(t) is a periodic time function with period T.

Such a periodic function can be expanded in an infinite series of exponential time functions called the Fourier series,

x ( t ) = ∑ n = − ∞ ∞ X [ n ] e j2π nt / T x ( t ) = ∑ n = − ∞ ∞ X [ n ] e j2π nt / T size 12{x \( t \) = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {X \[ n \] e rSup { size 8{j2π ital "nt"/T} } } } {}

2/ Fourier series coefficients

The coefficients of the Fourier series can be found as follows.

1 T ∫ − T / 2 T / 2 x ( t ) e − j2π nt / T dt = 1 T ∫ − T / 2 T / 2 ( ∑ k = − ∞ ∞ X [ k ] e j2π kt / T ) e − j2π nt / T dt 1 T ∫ − T / 2 T / 2 x ( t ) e − j2π nt / T dt = 1 T ∫ − T / 2 T / 2 ( ∑ k = − ∞ ∞ X [ k ] e j2π kt / T ) e − j2π nt / T dt size 12{ { {1} over {T} } Int rSub { size 8{ - T/2} } rSup { size 8{T/2} } {x \( t \) e rSup { size 8{ - j2π ital "nt"/T} } } ital "dt"= { {1} over {T} } Int rSub { size 8{ - T/2} } rSup { size 8{T/2} } {} \( Sum cSub { size 8{k= - infinity } } cSup { size 8{ infinity } } {X \[ k \] e rSup { size 8{j2π ital "kt"/T} } } \) e rSup { size 8{ - j2π ital "nt"/T} } ital "dt"} {}

1 T ∫ − T / 2 T / 2 x ( t ) e − j2π nt / T dt = ∑ k = − ∞ ∞ X [ k ] 1 T ∫ − T / 2 T / 2 e j2π ( k − n ) t / T dt 1 T ∫ − T / 2 T / 2 x ( t ) e − j2π nt / T dt = ∑ k = − ∞ ∞ X [ k ] 1 T ∫ − T / 2 T / 2 e j2π ( k − n ) t / T dt size 12{ { {1} over {T} } Int rSub { size 8{ - T/2} } rSup { size 8{T/2} } {x \( t \) e rSup { size 8{ - j2π ital "nt"/T} } } ital "dt"= Sum cSub { size 8{k= - infinity } } cSup { size 8{ infinity } } {X \[ k \] } { {1} over {T} } Int rSub { size 8{ - T/2} } rSup { size 8{T/2} } {e rSup { size 8{j2π \( k - n \) t/T} } } ital "dt"} {}

The integral can be evaluated as follows.

1 ifk = n 0 ifk ≠ n 1 T ∫ − T / 2 T / 2 e j2π ( k − n ) t / T dt = { 1 ifk = n 0 ifk ≠ n 1 T ∫ − T / 2 T / 2 e j2π ( k − n ) t / T dt = { size 12{ { {1} over {T} } Int rSub { size 8{ - T/2} } rSup { size 8{T/2} } {e rSup { size 8{j2π \( k - n \) t/T} } } ital "dt"=alignl { stack { left lbrace 1 ital "ifk"=n {} # right none left lbrace 0 ital "ifk" <> n {} # right no } } lbrace } {}

The set of exponential time functions are said to be an orthonormal basis.

The coefficients are

X [ n ) = 1 T ∫ − T / 2 T / 2 x ( t ) e − j2π nt / T dt X [ n ) = 1 T ∫ − T / 2 T / 2 x ( t ) e − j2π nt / T dt size 12{X \[ n \) = { {1} over {T} } Int rSub { size 8{ - T/2} } rSup { size 8{T/2} } {x \( t \) } e rSup { size 8{ - j2π ital "nt"/T} } ital "dt"} {}

3/ Definition of line spectra, harmonics

The fundamental frequency fo = 1/T . The Fourier series coefficients plotted as a function of n or nfo is called a Fourier spectrum.

II. EXAMPLES OF FOURIER SERIES OF PERIODIC TIME FUNCTIONS

1/ Periodic impulse train

The periodic impulse train is an important periodic time function and we derive its Fourier series coefficients.

s ( t ) = ∑ n = − ∞ ∞ δ ( t − nT ) s ( t ) = ∑ n = − ∞ ∞ δ ( t − nT ) size 12{s \( t \) = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {δ \( t - ital "nT" \) } } {}

The Fourier series coefficients are found as follows

S [ n ] = 1 T ∫ − T / 2 T / 2 s ( t ) e − j2π nt / T dt , 1 T ∫ − T / 2 T / 2 ∑ n = − ∞ ∞ δ ( t − nT ) e − j2π nt / T dt , 1 T ∫ − T / 2 T / 2 δ ( t ) e − j2π nt / T dt = 1 T S [ n ] = 1 T ∫ − T / 2 T / 2 s ( t ) e − j2π nt / T dt , 1 T ∫ − T / 2 T / 2 ∑ n = − ∞ ∞ δ ( t − nT ) e − j2π nt / T dt , 1 T ∫ − T / 2 T / 2 δ ( t ) e − j2π nt / T dt = 1 T alignl { stack { size 12{S \[ n \] = { {1} over {T} } Int rSub { size 8{ - T/2} } rSup { size 8{T/2} } {s \( t \) e rSup { size 8{ - j2π ital "nt"/T} } } ital "dt",} {} # = { {1} over {T} } Int rSub { size 8{ - T/2} } rSup { size 8{T/2} } { Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {δ} } \( t - ital "nT" \) e rSup { size 8{ - j2π ital "nt"/T} } ital "dt", {} # = { {1} over {T} } Int rSub { size 8{ - T/2} } rSup { size 8{T/2} } {δ \( t \) e rSup { size 8{ - j2π ital "nt"/T} } } ital "dt"= { {1} over {T} } {} } } {}

The Fourier series coefficients are

S [ n ] = 1 T S [ n ] = 1 T size 12{S \[ n \] = { {1} over {T} } } {}

The time function and spectrum are shown below.

To summarize, the periodic impulse train can be represented by its Fourier series,

s ( t ) = ∑ n = − ∞ ∞ δ ( t − nT ) = 1 T ∑ n = − ∞ ∞ e j2π nt / T s ( t ) = ∑ n = − ∞ ∞ δ ( t − nT ) = 1 T ∑ n = − ∞ ∞ e j2π nt / T size 12{s \( t \) = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {δ \( t - ital "nT" \) = { {1} over {T} } } Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {e rSup { size 8{j2π ital "nt"/T} } } } {}

The Fourier series of the periodic impulse train is

s ( t ) = ∑ n = − ∞ ∞ δ ( t − nT ) = 1 T ∑ n = − ∞ ∞ e j2π nt / T s ( t ) = ∑ n = − ∞ ∞ δ ( t − nT ) = 1 T ∑ n = − ∞ ∞ e j2π nt / T size 12{s \( t \) = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {δ \( t - ital "nT" \) = { {1} over {T} } } Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {e rSup { size 8{j2π ital "nt"/T} } } } {}

It is not obvious that the two expressions are equal. To investigate this, we define the partial sum of the Fourier series, sN(t),

s N ( t ) = 1 T ∑ n = − N N e j2π nt / T s N ( t ) = 1 T ∑ n = − N N e j2π nt / T size 12{s rSub { size 8{N} } \( t \) = { {1} over {T} } Sum cSub { size 8{n= - N} } cSup { size 8{N} } {e rSup { size 8{j2π ital "nt"/T} } } } {}

and investigate its behavior as N →∞.

The partial sum of the Fourier series is

S N ( t ) = 1 T ∑ n = − N N e j2π nt / T = 1 T ∑ n = − N N ( e j2πt / T ) n S N ( t ) = 1 T ∑ n = − N N e j2π nt / T = 1 T ∑ n = − N N ( e j2πt / T ) n size 12{S rSub { size 8{N} } \( t \) = { {1} over {T} } Sum cSub { size 8{n= - N} } cSup { size 8{N} } {e rSup { size 8{j2π ital "nt"/T} } } = { {1} over {T} } Sum cSub { size 8{n= - N} } cSup { size 8{N} } { \( e rSup { size 8{j2πt/T} } } \) rSup { size 8{n} } } {}

We can use the summation formula for a finite geometric series (Lecture 10) to sum this series,

s N ( t ) = 1 T ( e j2πt / T ) − N − ( e j2πt / T ) N + 1 1 − e j2πt / T s N ( t ) = 1 T ( e j2πt / T ) − N − ( e j2πt / T ) N + 1 1 − e j2πt / T size 12{s rSub { size 8{N} } \( t \) = { {1} over {T} } { { \( e rSup { size 8{j2πt/T} } \) rSup { size 8{ - N} } - \( e rSup { size 8{j2πt/T} } \) rSup { size 8{N+1} } } over {1 - e rSup { size 8{j2πt/T} } } } } {}

s N ( t ) = 1 T e j ( 2N + 1 ) πt / T − e − j ( 2N + 1 ) πt / T e jπt / T − e − jπt / T s N ( t ) = 1 T e j ( 2N + 1 ) πt / T − e − j ( 2N + 1 ) πt / T e jπt / T − e − jπt / T size 12{s rSub { size 8{N} } \( t \) = { {1} over {T} } { {e rSup { size 8{j \( 2N+1 \) πt/T} } - e rSup { size 8{ - j \( 2N+1 \) πt/T} } } over {e rSup { size 8{jπt/T} } - e rSup { size 8{ - jπt/T} } } } } {}

s N ( t ) = 1 T sin ( 2N + 1 ) πt / T sin πt / T s N ( t ) = 1 T sin ( 2N + 1 ) πt / T sin πt / T size 12{s rSub { size 8{N} } \( t \) = { {1} over {T} } { {"sin" \( 2N+1 \) πt/T} over {"sin"πt/T} } } {}

s N ( t ) = 1 T sin ( 2N + 1 ) πt / T sin πt / T s N ( t ) = 1 T sin ( 2N + 1 ) πt / T sin πt / T size 12{s rSub { size 8{N} } \( t \) = { {1} over {T} } { {"sin" \( 2N+1 \) πt/T} over {"sin"πt/T} } } {}

Note that this function is periodic with period T, and

s N ( t ) = 2N + 1 T s N ( t ) = 2N + 1 T size 12{s rSub { size 8{N} } \( t \) = { {2N+1} over {T} } } {}

The first zero of sN(t) is at

t = T 2N + 1 t = T 2N + 1 size 12{t= { {T} over {2N+1} } } {}

Thus, as N → ∞, each lobe gets larger and narrower. To determine if each lobe acts as an impulse, we need to find its area.

The area of each period of sN(t) is simply

Area = ∫ − T / 2 T / 2 S N ( t ) dt = ∫ − T / 2 T / 2 1 T ∑ n = − N N e j2π nt / T dt = ∑ n = − N N 1 T ∫ − T / 2 T / 2 e j2π nt / T dt Area = ∫ − T / 2 T / 2 S N ( t ) dt = ∫ − T / 2 T / 2 1 T ∑ n = − N N e j2π nt / T dt = ∑ n = − N N 1 T ∫ − T / 2 T / 2 e j2π nt / T dt size 12{ ital "Area"= Int rSub { size 8{ - T/2} } rSup { size 8{T/2} } {S rSub { size 8{N} } \( t \) ital "dt"= Int rSub { size 8{ - T/2} } rSup { size 8{T/2} } { { {1} over {T} } } } Sum cSub { size 8{n= - N} } cSup { size 8{N} } {e rSup { size 8{j2π ital "nt"/T} } } ital "dt"= Sum cSub { size 8{n= - N} } cSup { size 8{N} } { { {1} over {T} } } Int rSub { size 8{ - T/2} } rSup { size 8{T/2} } {e rSup { size 8{j2π ital "nt"/T} } } ital "dt"} {}

The integral is zero except when n = 0 when it equals T so that Area = 1. Thus, each lobe of sN(t) becomes: tall, height is sN(nT) = (2N + 1)/T ; narrow, width is 2T/(2N + 1); and its area is 1. Thus, the partial sum approaches an infinite impulse train of unit area,

s ( t ) = ∑ n = − ∞ ∞ δ ( t − nT ) = 1 T ∑ n = − ∞ ∞ e j2π nt / T s ( t ) = ∑ n = − ∞ ∞ δ ( t − nT ) = 1 T ∑ n = − ∞ ∞ e j2π nt / T size 12{s \( t \) = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {δ \( t - ital "nT" \) = { {1} over {T} } } Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {e rSup { size 8{j2π ital "nt"/T} } } } {}

2/ Fourier transform of a periodic impulse train

We have two expressions for a periodic impulse train,

s ( t ) = ∑ n = − ∞ ∞ δ ( t − nT ) = 1 T ∑ n = − ∞ ∞ e j2π nt / T s ( t ) = ∑ n = − ∞ ∞ δ ( t − nT ) = 1 T ∑ n = − ∞ ∞ e j2π nt / T size 12{s \( t \) = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {δ \( t - ital "nT" \) = { {1} over {T} } } Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {e rSup { size 8{j2π ital "nt"/T} } } } {}

The Fourier transform of each expression is

S ( f ) = ∑ n = − ∞ ∞ e − j2π nTf = 1 T ∑ n = − ∞ ∞ δ ( f − n T ) S ( f ) = ∑ n = − ∞ ∞ e − j2π nTf = 1 T ∑ n = − ∞ ∞ δ ( f − n T ) size 12{S \( f \) = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {e rSup { size 8{ - j2π ital "nTf"} } } = { {1} over {T} } Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {δ \( f - { {n} over {T} } } \) } {}

Therefore, the Fourier transform of a periodic impulse train in time is a periodic impulse train in frequency.

3/ Relation of Fourier series spectrum to Fourier transform of a periodic impulse train

Therefore, the Fourier transform of the periodic impulse train has an impulse at the frequency of each Fourier series component and the area of the impulse equals the Fourier series coefficient.

III. FOURIER TRANSFORM OF AN ARBITRARY PERIODIC FUNCTION

1/ Representation of a periodic function

An arbitrary periodic function can be generated by convolving a pulse, xT (t), that represents one period of the periodic function with a periodic impulse train, s(t)=∑nδ(t−nT)s(t)=∑nδ(t−nT) size 12{s \( t \) = Sum rSub { size 8{n} } {δ \( t - ital "nT" \) } } {}

x ( t ) = x T ( t ) ∗ s ( t ) = x T ( t ) ∗ ∑ n = − ∞ ∞ δ ( t − nT ) = ∑ n = − ∞ ∞ x T ( t − nT ) x ( t ) = x T ( t ) ∗ s ( t ) = x T ( t ) ∗ ∑ n = − ∞ ∞ δ ( t − nT ) = ∑ n = − ∞ ∞ x T ( t − nT ) size 12{x \( t \) =x rSub { size 8{T} } \( t \) * s \( t \) =x rSub { size 8{T} } \( t \) * Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {δ \( t - ital "nT" \) = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {x rSub { size 8{T} } } \( t - ital "nT" \) } } {}

2/ Fourier transform of a periodic function

The Fourier transform of the periodic function is

x ( t ) = x T ( t ) ∗ s ( t ) ⇔ F X ( f ) = X T ( f ) × S ( f ) x ( t ) = x T ( t ) ∗ s ( t ) ⇔ F X ( f ) = X T ( f ) × S ( f ) size 12{x \( t \) =x rSub { size 8{T} } \( t \) * s \( t \) { dlrarrow } cSup { size 8{F} } X \( f \) =X rSub { size 8{T} } \( f \) times S \( f \) } {}

X ( f ) = X T ( f ) × 1 T ∑ n = − ∞ ∞ δ ( f − n T ) = ∑ n = − ∞ ∞ ( X T ( n T ) T ) δ ( f − n T ) X ( f ) = X T ( f ) × 1 T ∑ n = − ∞ ∞ δ ( f − n T ) = ∑ n = − ∞ ∞ ( X T ( n T ) T ) δ ( f − n T ) size 12{X \( f \) =X rSub { size 8{T} } \( f \) times { {1} over {T} } Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {δ \( f - { {n} over {T} } } \) = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } { \( { {X rSub { size 8{T} } \( { {n} over {T} } \) } over {T} } \) } δ \( f - { {n} over {T} } \) } {}

An important conclusion is that the Fourier transform of a periodic function consists of impulses in frequency at multiples of the fundamental frequency. Thus, periodic continuous time functions can be represented by a countably infinite number of complex exponentials.

3/ Fourier series coefficients

The Fourier transform of the periodic function is

X f = ∑ n = − ∞ ∞ ( X T ( n T ) T ) δ ( f − n T ) = ∑ n = − ∞ ∞ 1 2 ( sin nπ / 2 nπ / 2 ) δ ( f − n T ) X f = ∑ n = − ∞ ∞ ( X T ( n T ) T ) δ ( f − n T ) = ∑ n = − ∞ ∞ 1 2 ( sin nπ / 2 nπ / 2 ) δ ( f − n T ) size 12{X left (f right )= Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } { \( { {X"" lSub { size 8{T} } \( { {n} over {T} } \) } over {T} } \) δ \( f - { {n} over {T} } \) } = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } { { {1} over {2} } \( { {"sin" left (nπ/2 right )} over {nπ/2} } \) δ \( f - { {n} over {T} } \) } } {}

Recall that

X T ( f ) = ∫ − T / 2 T / 2 X T ( t ) e − j2π ft dt X T ( f ) = ∫ − T / 2 T / 2 X T ( t ) e − j2π ft dt size 12{X"" lSub { size 8{T} } \( f \) = Int rSub { size 8{ - T/2} } rSup { size 8{T/2} } {X rSub { size 8{T} } } \( t \) e rSup { size 8{ - j2π ital "ft"} } ital "dt"} {}

Therefore,

X T ( n T ) T = 1 T ∫ − T / 2 T / 2 X T ( t ) e − j2π ft dt = X [ n ] X T ( n T ) T = 1 T ∫ − T / 2 T / 2 X T ( t ) e − j2π ft dt = X [ n ] size 12{ { {X rSub { size 8{T} } \( { {n} over {T} } \) } over {T} } = { {1} over {T} } Int rSub { size 8{ - T/2} } rSup { size 8{T/2} } {X rSub { size 8{T} } } \( t \) e rSup { size 8{ - j2π ital "ft"} } ital "dt"=X \[ n \] } {}

Therefore, for an arbitrary periodic continuous time function, the Fourier transform consists of impulses (located at the harmonic frequencies) whose areas are the Fourier series coefficients.

4/ Fourier series of a square wave — generation of the square wave

We will find the Fourier series of a square wave by finding the Fourier transform of one period.

x ( t ) = x T ( t ) ∗ s ( t ) x ( t ) = x T ( t ) ∗ s ( t ) size 12{x \( t \) =x rSub { size 8{T} } \( t \) * s \( t \) } {}

5/ Fourier series of a square wave — Fourier transform of one period of the square wave

x T ( t ) ⇔ F X T ( f ) x T ( t ) ⇔ F X T ( f ) size 12{x rSub { size 8{T} } \( t \) { dlrarrow } cSup { size 8{F} } X rSub { size 8{T} } \( f \) } {}

6/ Fourier series of a square wave — Fourier transform of square wave

To obtain the Fourier transform of the square wave, we take the Fourier transform of one period of the square wave and multiply it by the Fourier transform of the periodic impulse train.

X ( f ) = X T ( f ) × S ( f ) = T 2 ( sin π fT / 2 π fT / 2 ) × 1 T ∑ n = − ∞ ∞ δ ( f − n T ) = ∑ n = − ∞ ∞ 1 2 ( sin nπ / 2 nπ / 2 ) δ ( f − n T ) X ( f ) = X T ( f ) × S ( f ) = T 2 ( sin π fT / 2 π fT / 2 ) × 1 T ∑ n = − ∞ ∞ δ ( f − n T ) = ∑ n = − ∞ ∞ 1 2 ( sin nπ / 2 nπ / 2 ) δ ( f − n T ) size 12{X \( f \) =X rSub { size 8{T} } \( f \) times S \( f \) = { {T} over {2} } \( { {"sin" left (π ital "fT"/2 right )} over {π ital "fT"/2} } \) times { {1} over {T} } Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {δ \( f - { {n} over {T} } \) } = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } { { {1} over {2} } \( { {"sin" left (nπ/2 right )} over {nπ/2} } \) δ \( f - { {n} over {T} } \) } } {}

Two-minute miniquiz problem

Problem 18-1 — Fourier series of the square wave

a) Determine the Fourier series coefficients of the square wave.

b) From the Fourier series coefficients determine the average value of the square wave.

Solution

a) Since the Fourier transform of the square wave is

X ( f ) = ∑ n = − ∞ ∞ 1 2 ( sin nπ / 2 nπ / 2 ) δ ( f − n T ) X ( f ) = ∑ n = − ∞ ∞ 1 2 ( sin nπ / 2 nπ / 2 ) δ ( f − n T ) size 12{X \( f \) = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } { { {1} over {2} } \( { {"sin" left (nπ/2 right )} over {nπ/2} } \) δ \( f - { {n} over {T} } \) } } {}

the Fourier series coefficients are

X [ n ] = 1 2 ( sin nπ / 2 nπ / 2 ) X [ n ] = 1 2 ( sin nπ / 2 nπ / 2 ) size 12{X \[ n \] = { {1} over {2} } \( { {"sin" left (nπ/2 right )} over {nπ/2} } \) } {}

b)

X [ 0 ] = 1 T ∫ − T / 2 T / 2 x ( t ) dt = X [ 0 ] = 1 T ∫ − T / 2 T / 2 x ( t ) dt = size 12{X \[ 0 \] = { {1} over {T} } Int rSub { size 8{ - T/2} } rSup { size 8{T/2} } {x \( t \) ital "dt"} = {1} wideslash {2} } {}

7/ Fourier series of a square wave — synthesis of a square wave

We can synthesize the square wave by adding complex exponentials weighted by their Fourier series coefficients,

x ( t ) = ∑ n = − ∞ ∞ X [ n ] e j2π nt / T = ∑ n = − ∞ ∞ 1 2 ( sin nπ / 2 nπ / 2 ) e j2π nt / T x ( t ) = ∑ n = − ∞ ∞ X [ n ] e j2π nt / T = ∑ n = − ∞ ∞ 1 2 ( sin nπ / 2 nπ / 2 ) e j2π nt / T size 12{x \( t \) = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {X \[ n \] } e rSup { size 8{j2π ital "nt"/T} } = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } { { {1} over {2} } \( { {"sin" left (nπ/2 right )} over {nπ/2} } \) } e rSup { size 8{j2π ital "nt"/T} } } {}

x ( t ) = 1 2 + ∑ n = 1 ∞ 1 2 ( sin ( nπ / 2 ) nπ / 2 ) ( e j2π nt / T + e − j2π nt / T ) x ( t ) = 1 2 + ∑ n = 1 ∞ 1 2 ( sin ( nπ / 2 ) nπ / 2 ) cos ( 2π nt / T ) x ( t ) = 1 2 + ∑ n = 1 ∞ 1 2 ( sin ( nπ / 2 ) nπ / 2 ) ( e j2π nt / T + e − j2π nt / T ) x ( t ) = 1 2 + ∑ n = 1 ∞ 1 2 ( sin ( nπ / 2 ) nπ / 2 ) cos ( 2π nt / T ) alignl { stack { size 12{x \( t \) = { {1} over {2} } + Sum cSub { size 8{n=1} } cSup { size 8{ infinity } } { { {1} over {2} } \( { {"sin" \( nπ/2 \) } over {nπ/2} } \) } \( e rSup { size 8{j2π ital "nt"/T} } +e rSup { size 8{ - j2π ital "nt"/T} } \) } {} # x \( t \) = { {1} over {2} } + Sum cSub { size 8{n=1} } cSup { size 8{ infinity } } { { {1} over {2} } \( { {"sin" \( nπ/2 \) } over {nπ/2} } \) } "cos" \( 2π ital "nt"/T \) {} } } {}

Note that the coefficients are even functions of n. Hence, we can rewrite the series as follows

x ( t ) = 1 2 + ∑ n = 1 ∞ 1 2 ( sin ( nπ / 2 ) nπ / 2 ) ( e j2π nt / T + e − j2π nt / T ) x ( t ) = 1 2 + ∑ n = 1 ∞ 1 2 ( sin ( nπ / 2 ) nπ / 2 ) cos ( 2π nt / T ) x ( t ) = 1 2 + ∑ n = 1 ∞ 1 2 ( sin ( nπ / 2 ) nπ / 2 ) ( e j2π nt / T + e − j2π nt / T ) x ( t ) = 1 2 + ∑ n = 1 ∞ 1 2 ( sin ( nπ / 2 ) nπ / 2 ) cos ( 2π nt / T ) alignl { stack { size 12{x \( t \) = { {1} over {2} } + Sum cSub { size 8{n=1} } cSup { size 8{ infinity } } { { {1} over {2} } \( { {"sin" \( nπ/2 \) } over {nπ/2} } \) } \( e rSup { size 8{j2π ital "nt"/T} } +e rSup { size 8{ - j2π ital "nt"/T} } \) } {} # x \( t \) = { {1} over {2} } + Sum cSub { size 8{n=1} } cSup { size 8{ infinity } } { { {1} over {2} } \( { {"sin" \( nπ/2 \) } over {nπ/2} } \) } "cos" \( 2π ital "nt"/T \) {} } } {}

Note that all the even harmonics of x(t) are zero except for the term for n = 0.

To investigate the synthesis of the square wave, we consider the partial sum of the Fourier series,

x N ( t ) = 1 2 + ∑ n = − 1 N 1 2 ( sin ( nπ / 2 ) nπ / 2 ) cos ( 2π nt / T ) x N ( t ) = 1 2 + ∑ n = − 1 N 1 2 ( sin ( nπ / 2 ) nπ / 2 ) cos ( 2π nt / T ) size 12{x rSub { size 8{N} } \( t \) = { {1} over {2} } + Sum cSub { size 8{n= - 1} } cSup { size 8{N} } { { {1} over {2} } \( { {"sin" \( nπ/2 \) } over {nπ/2} } \) } "cos" \( 2π ital "nt"/T \) } {}

Demo illustrating the Gibbs’phenomenon.

IV. FOURIER SERIES OF A SQUARE WAVE — GIBBS’ PHENOMENON

As harmonics are added to synthesize the square wave, the partial sum of the Fourier series converges to the square wave everywhere except near the discontinuity where the partial sum takes on the value of 1/2. There are oscillations on either side of the discontinuity whose maximum over and undershoot approach 9% of the discontinuity independent of N. We can interpret the Gibbs’ phenomenon by examining the partial sum of the Fourier series as a filtering problem.

1/ Gibbs’ phenomenon interpretation as an ideal LPF of a square wave

Truncation of the Fourier series of a square wave can be interpreted in the frequency domain as passing the square wave through an ideal lowpass filter that truncates the spectrum. That can be interpreted in the time domain as the convolution of the square wave with a sinc function. The response of the ideal lowpass filter to the discontinuity of the square wave gives rise to the oscillations.

2/ Gibbs’ phenomenon — step response of an ideal LPF

The Gibbs’ phenomenon can be investigated further by examining the step response of an ideal lowpass.

Hence,

xW(t)=u(t)∗h(t)=∫−∞th(τ)dτxW(t)=u(t)∗h(t)=∫−∞th(τ)dτ size 12{x rSub { size 8{W} } \( t \) =u \( t \) * h \( t \) = Int rSub { size 8{ - infinity } } rSup { size 8{t} } {h \( τ \) dτ} } {} where {}h(t)=2Wsin(2πWt)2πWth(t)=2Wsin(2πWt)2πWt size 12{h \( t \) =2W { {"sin" \( 2π ital "Wt" \) } over {2π ital "Wt"} } } {}

We evaluate the integral

x w ( t ) = ∫ − ∞ t 2W ( sin ( 2π Wt ) 2π Wt ) dτ x w ( t ) = ∫ − ∞ t 2W ( sin ( 2π Wt ) 2π Wt ) dτ size 12{x rSub { size 8{w} } \( t \) = Int rSub { size 8{ - infinity } } rSup { size 8{t} } {2W \( { {"sin" \( 2π ital "Wt" \) } over {2π ital "Wt"} } \) dτ} } {}

by changing variables y = 2πWt which yields

x w ( t ) = ∫ − ∞ 2π Wt ( sin y y ) dy x w ( t ) = ∫ − ∞ 2π Wt ( sin y y ) dy size 12{x rSub { size 8{w} } \( t \) = Int rSub { size 8{ - infinity } } rSup { size 8{2π ital "Wt"} } { \( { {"sin"y} over {y} } \) ital "dy"} } {}

This is closely related to a tabulated function called the sine integral function Si(t) defined as

Si ( t ) = ∫ 0 t sin y y dy Si ( t ) = ∫ 0 t sin y y dy size 12{ ital "Si" \( t \) = Int rSub { size 8{0} } rSup { size 8{t} } { { {"sin"y} over {y} } ital "dy"} } {}

The sine integral function,

Si ( t ) = ∫ 0 t sin y y dy Si ( t ) = ∫ 0 t sin y y dy size 12{ ital "Si" \( t \) = Int rSub { size 8{0} } rSup { size 8{t} } { { {"sin"y} over {y} } ital "dy"} } {}

is plotted below.

We express the step with the truncated spectrum in terms of the sine integral function as

x w ( t ) = 1 2 + 1 π Si ( 2π Wt ) x w ( t ) = 1 2 + 1 π Si ( 2π Wt ) size 12{x rSub { size 8{w} } \( t \) = { {1} over {2} } + { {1} over {π} } ital "Si" \( 2π ital "Wt" \) } {}

which is shown plotted below for several values of W.

V. CONCLUSIONS

VI. HISTORICAL PERSPECTIVE

Jean Baptiste Joseph Fourier (1768-1830)

Joseph Fourier was born in Auxerre, France on March 21, 1768 and died in Paris on May 4, 1830. This is one of two portraits that has survived.

Fourier’s work and that of those that followed him have had a profound effect on science and mathematics. Fourier transforms are common tools in many different fields of science.

The image of Fourier (left) is 582 by 582 pixels. The logarithm of the magnitude of the Fourier transform of this image is plotted on the right.

Exercises .

Solutions of Exercises.

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This work is licensed by Vu Dinh Thanh, Truc Pham-Dinh, Anh Tuan Hoang, and Tam Huynh-Ngoc under a Creative Commons Attribution License (CC-BY 3.0), and is an Open Educational Resource.

Last edited by Vietnam OpenCourseWare on Jul 3, 2009 6:32 am -0500.