fourier series { complex form

6.003: Signal Processing

Fourier Series – Complex Form

• complex numbers, complex exponentials, and their relation to sinusoids

• complex exponential form of Fourier series

• properties of Fourier series

February 11, 2020

Last Week

Representing a signal as a Fourier series.

Synthesis Equation

f(t) = c0 +∞∑k=1

ck cos(kωot) +∞∑k=1

dk sin(kωot) where ωo = 2πT

Analysis Equations

c0 = 1T

∫Tf(t) dt

ck = 2T

∫Tf(t) cos(kωot) dt

dk = 2T

∫Tf(t) sin(kωot) dt

Today: Simplifying the math with complex numbers.

Simplifying Math By Using Complex Numbers

Complex numbers simplify thinking about roots of numbers / polynomials:

• all numbers have two square roots, three cube roots, etc.

• all polynomials of order n have n roots (but some may be repeated).

Our biggest simplification comes from Euler’s formula, which relates com-

plex exponentials to trigonometric functions (Leonhard Euler, 1748).

ejθ = cos θ + j sin θwhere j =

√−1.

Richard Feynman called this ”the most remarkable formula in mathemat-

ics.”

Where Does Euler’s Formula Come From?

Euler showed the relation between complex exponentials and sinusoids by

solving the following differential equation two ways.

d2f(θ)dθ2 + f(θ) = 0

let f1(θ) = A cos(αθ) +B sin(βθ)df1(θ)dθ

= −αA sin(αθ) + βB cos(βθ)

d2f1(θ)dθ2 = −α2A cos(αθ)− β2B sin(βθ)

α = β = 1f1(θ) = A cos θ +B sin θ

Let f2(θ) = Ceγθ

df2(θ)dθ

= γCeγθ

d2f2(θ)dθ2 = γ2Ceγθ

γ2 = −1f2(θ) = Cejθ

If we arbitrarily set C = 1, then f2(θ) = ejθ, f2(0) = 1 and f ′2(0) = j.

To make f1(θ) = f2(θ), we must set A = 1 and B = j.

It follows that

ejθ = cos θ + j sin θThis argument presumes the existance of a constant j whose square is −1and that can be manipulated as an ordinary algebraic constant.

Where Does Euler’s Formula Come From?

Euler’s formula also follows from Maclaurin expansion of the exponential

function, assuming the j behaves like any other algebraic constant.

Start with the expansion of the real-valued function:

eθ = 1 + θ + θ2

2! + θ3

3! + θ4

4! + θ5

5! + θ6

6! + θ7

7! + · · ·

Assume that the same expansion holds for complex-valued arguments:

ejθ = 1 + jθ + j2θ2

2! + j3θ3

3! + j4θ4

4! + j5θ5

5! + j6θ6

6! + j7θ7

7! + · · ·

= 1 + jθ − θ2

2! −jθ3

3! + θ4

4! + jθ5

5! −θ6

6! −jθ7

7! + · · ·

=(

1− θ2

2! + θ4

4! −θ6

6! + · · ·)

︸︷︷︸cos θ

+j(θ − θ3

3! + θ5

5! −θ7

7! + · · ·)

︸︷︷︸sin θ

Euler’s formula results by splitting the even and odd powers of θ.

ejθ = cos θ + j sin θ

Geometric Interpretation

In 1897, Caspar Wessel was the first to describe complex numbers as points

in the complex plane. Imaginary numbers had been in use since the 1500’s.

c = a+ jb

Re

Im

c

a

b

Algebraic Addition

Addition: the real part of a sum is the sum of the real parts, and

the imaginary part of a sum is the sum of the imaginary parts.

Let c1 and c2 represent complex numbers:

c1 = a1 + jb1

c2 = a2 + jb2

Then

c1 + c2 = (a1 + jb1) + (a2 + jb2) = (a1+a2) + j(b1+b2)

Geometric Addition

Rules for adding complex numbers are same as those for adding vectors.


c1 = a1 + jb1

c2 = a2 + jb2

Then

c1 + c2 = (a1 + jb1) + (a2 + jb2) = (a1+a2) + j(b1+b2)

Re

Im

a1a2

b1

b2

Geometric Addition

Rules for adding complex numbers are same as those for adding vectors.


c1 = a1 + jb1

c2 = a2 + jb2

Then

c1 + c2 = (a1 + jb1) + (a2 + jb2) = (a1+a2) + j(b1+b2)

Re

Im

a1a2 a1+a2

b1

b2b1+b2

Algebraic Multiplication

Multiplication is more complicated.


c1 = a1 + jb1

c2 = a2 + jb2

Thenc1×c2 = (a+jb)× (c+jd)

= a×c+ a×jd+ jb×c+ jb×jd= (ac− bd) + j(ad+ bc)

Although the rules of algebra apply, the result is complicated:

• the real part of a product is NOT the product of the real parts, and

• the imaginary part is NOT the product of the imaginary parts.

Geometric Multiplication

The two-dimensional view of complex numbers allows us to think about

multiplication by an imaginary number as a rotation.

Multiplying by j

• rotates 1 to j,

• rotates j to −1,

• rotates −1 to −j, and

• rotates −j to 1.

Re

Im

1

j




Multiplying by j

• rotates 1 to j,




Re

Im

j

−1




Multiplying by j

• rotates 1 to j,




Re

Im

−1

−j




Multiplying by j

• rotates 1 to j,




Re

Im

1

−j




Multiplying by j

• rotates 1 to j,




Re

Im

1

j

−1

Multiplying by j rotates a vector by π/2.

Multiplying by j2 = −1 rotates a vector by π.


Multiplying by j rotates an arbitrary complex number by π/2.

c = a+ jb

jc = ja− b

Re

Im

jc

c

a−b

b

a

Geometric Approach: Polar Form

The magnitude of the product of complex numbers is the product of their

magnitudes. The angle of a product is the sum of the angles.

Re

Im

a

b

a×b

θ1θ2

θ1+θ2

(r1∠θ1)× (r2∠θ2) = r1(cos θ1 + j sin θ1)× r2(cos θ2 + j sin θ2)

= r1r2(cos θ1 cos θ2 − sin θ1 sin θ2︸︷︷︸cos(θ1+θ2)

+ j cos θ1 sin θ2 + j sin θ1 cos θ2)︸︷︷︸j sin(θ1+θ2)

= r1r2∠(θ1 + θ2)

Geometric Interpretation of Euler’s Formula

Euler’s formula equates polar and rectangular descriptions of a unit vector

at angle θ.


θRe

Im

cos θ

sinθ1


This interpretation came nearly 150 years after Euler’s formulation.

Simplifying Math By Using Complex Numbers

Euler’s formula allows us to represent both sine and cosine basis functions

with a single complex exponential:

f(t) =∑(

ck cos(kωot) + dk sin(kωot))

=∑

akejkωot

2πωo

t

cos(

0t)

2πωo

t

sin(0t)

2πωo

tej0t

2πωo

t

cos(ωot)

2πωo

t

sin(ω

ot)

2πωo

t

ejωot

2πωo

t

cos(

2ωot)

...2πωo

t

sin(2ωot)

...2πωo

t

ej2ωot

...

Real-valued basis functions Complex basis functions

This halves the number of coefficients, but each is now complex-valued.

More importantly, it replaces the trig functions with an exponential.

Converting From Trig Form To Complex Exponential Form

Assume that a function f(t) can be written as a Fourier series in trig form.

f(t) = f(t+ T ) = c0 +∞∑k=1

(ck cos(kωot) + dk sin(kωot)

)We can use Euler’s formula to convert sinusoids to complex exponentials.

ejkωot = cos(kωot) + j sin(kωot)cos(kωot) = Re{ejkωot} = (ejkωot + e−jkωot)/2sin(kωot) = Im{ejkωot} = −j(ejkωot − e−jkωot)/2

f(t) = c0+12

∞∑k=1

(cke

jkωot+cke−jkωot−jdkejkωot+jdke−jkωot)

= co+12

∞∑k=1

(ck−jdk)ejkωot+12

∞∑k=1

(ck+jdk)e−jkωot

= co+12

∞∑k=1

(ck−jdk)ejkωot+12

−∞∑k=−1

(c−k+jd−k)ejkωot =∞∑

k=−∞ake

jkωot

where ak ={ (ck − jdk)/2 if k > 0

c0 if k = 0(c−k + jd−k)/2 if k < 0

Negative Frequencies

The complex form of a Fourier series has both positive and negative k’s.

Only positive values of k are used in the trig form:

f(t) = c0 +∞∑k=1


dk sin(kωot)

but both positive and negative values of k are used in the exponential form:

f(t) =∞∑

k=−∞ake

jkωot

If we only included positive k in the previous sum, the result would always

have an imaginary component (unless ak = 0 for all k).

If f(t) is real-valued (as it must be for the trig form), then the complex

coefficients ak are conjugate symmetric:

a−k = a∗k

where the ∗ denotes the complex conjugate.

k : . . . −3 −2 −1 0 1 2 3 . . .

ak : . . .c3+jd3

2c2+jd2

2c1+jd1

2 c0c1−jd1

2c2−jd2

2c3−jd3

2 . . .

Fourier Series Directly From Complex Exponential Form

Assume that f(t) is periodic in T and is composed of a weighted sum of

harmonically related complex exponentials.

f(t) = f(t+ T ) =∞∑

k=−∞ake

jωokt

We can “sift” out the component at lωo by multiplying both sides by e−jlωot

and integrating over a period.

∫Tf(t)e−jωoltdt =

∫T

∞∑k=−∞

akejωokte−jωoltdt =

∞∑k=−∞

ak

∫Te jωo(k−l)tdt

={Tal if l = k

0 otherwise

Solving for al provides an explicit formula for the coefficients:

ak = 1T

∫Tf(t)e−jωoktdt ; where ωo = 2π

T.

This formulation works even if f(t) has complex values.

Orthogonality and Projection

Fourier components are separable because they are orthogonal.

Similar to separating a vector r into x and y components.

Since x and y are orthogonal, we can separate the x and y components of

r by projection:

a= r · xb= r · y

Then

r= ax+ by

x

y

a

rb

Orthogonal Decompositions

Vector representation: let r represent a vector with components a and b

in the x and y directions, respectively.

a = r · xb = r · y (“analysis” equations)

r = ax + by (“synthesis” equation)

ak=1T

∫Tf(t)e−j

2πTktdt (“analysis” equation)

f(t)= f(t+ T ) =∞∑

k=−∞ake

j 2πTkt (“synthesis” equation)


Vector representation: let r represent a vector with components a and b

in the x and y directions, respectively.

a = r · xb = r · y (“analysis” equations)

r = ax + by (“synthesis” equation)

Fourier series: let f(t) represent a signal with harmonic components

a0, a1, . . ., ak for harmonics e j0t, e j2πTt, . . ., e j

2πTkt respectively.

ak=1T

∫Tf(t)e−j

2πTktdt (“analysis” equation)

f(t)= f(t+ T ) =∞∑

k=−∞ake

j 2πTkt (“synthesis” equation)


Integrating over a period sifts out the kth component of the series.

Sifting as a dot product:

x = r · x ≡ |r||x| cos θ

Sifting as an inner product:

ak = e j2πTkt · f(t) ≡ 1

T

∫Tf(t)e−j

2πTktdt

where

a(t) · b(t) = 1T

∫Ta∗(t)b(t)dt .

The complex conjugate (∗) makes the inner product of the kth and mth

components equal to 1 iff k = m:1T

∫T

(e j

2πTkt)∗(

e j2πTmt)dt = 1

T

∫Te−j

2πTkte j

2πTmtdt =

{1 if k = m

0 otherwise

Check Yourself

How many of the following pairs of functions are

orthogonal (⊥) in T = 3?

1. cos 2πt ⊥ sin 2πt ?

2. cos 2πt ⊥ cos 4πt ?

3. cos 2πt ⊥ sin πt ?

4. cos 2πt ⊥ e j2πt ?

Check Yourself

How many of the following are orthogonal (⊥) in T = 3?

cos 2πt ⊥ sin 2πt ?

Check Yourself


cos 2πt ⊥ sin 2πt ?

1 2 3t

cos 2πt

1 2 3t

sin 2πt

1 2 3t

cos 2πt sin 2πt = 12 sin 4πt

∫ 3

0dt = 0 therefore YES

Check Yourself


cos 2πt ⊥ cos 4πt ?

Check Yourself


cos 2πt ⊥ cos 4πt ?

1 2 3t

cos 2πt

1 2 3t

cos 4πt

1 2 3t

cos 2πt cos 4πt = 12 cos 6πt+ 1

2 cos 2πt

∫ 3

0dt = 0 therefore YES

Check Yourself


cos 2πt ⊥ sin πt ?

Check Yourself


cos 2πt ⊥ sin πt ?

1 2 3t

cos 2πt

1 2 3t

sin πt

1 2 3t

cos 2πt sin πt = 12 sin 3πt− 1

2 sin πt

∫ 3

0dt 6= 0 therefore NO

Check Yourself


cos 2πt ⊥ ej2πt ?

Check Yourself


cos 2πt ⊥ ej2πt ?

e2πt = cos 2πt+ j sin 2πt

cos 2πt ⊥ sin 2πt but not cos 2πtTherefore NO

Check Yourself

How many of the following pairs of functions

are orthogonal (⊥) in T = 3? 2

1. cos 2πt ⊥ sin 2πt ?√

2. cos 2πt ⊥ cos 4πt ?√

3. cos 2πt ⊥ sin πt ? X

4. cos 2πt ⊥ e j2πt ? X

Fourier Series

Comparison of trigonometric and complex exponential forms.

Complex Exponential Form

f(t) = f(t+ T ) =∞∑

k=−∞ake

jkωot

ak = 1T

∫Tf(t)e−jkωotdt

Trigonometric Form

f(t) = f(t+ T ) = co +∞∑k=1


dk sin(kωot)

c0 = 1T

∫Tf(t) dt

ck = 2T

∫T

cos(kωot) dt; k = 1, 2, 3, . . .

dk = 2T

∫T

sin(kωot) dt; k = 1, 2, 3, . . .

Comparison of Trigonometric and Complex Exponential Forms

It seems as though it takes more numbers to characterize the complex

exponential form:

• Each harmonic frequency in the complex exponential form depends on

two complex-valued numbers: ak and a−k.

• Each harmonic frequency in the trig form depends on two real-valued

numbers: ck and dk.

Q: What is going on?

A: The complex exponential form allows f(t) to have complex values.

The trigonometic form requires that f(t) be real-valued.

Q: Isn’t it twice the work to compute both ak and a−k?

A: Only if f(t) is complex-valued.

If f(t) is real-valued, then a−k is the complex conjugate of ak.

Is the Complex Exponential Form Actually Easier?

Last week, we determined the effect of a half-period shift on the Fourier

coefficients of the trig form. The result was a bit complicated.

Assume that f(t) is periodic in time with period T :

f(t) = f(t+ T ) .Let g(t) represent a version of f(t) shifted by half a period:

g(t) = f(t− T/2) .

How many of the following statements correctly describe the

effect of this shift on the Fourier series coefficients.

• cosine coefficients ck are negated X

• sine coefficients dk are negated X

• odd-numbered coefficients c1, d1, c3, d3, . . . are negated√

• sine and cosine coefficients are swapped: ck → dk and dk → ck X

Half-Period Shift

Shifting f(t) shifts the underlying basis functions of it Fourier expansion.

f(t− T/2) =∞∑k=0

ck cos (kωo(t− T/2)) +∞∑k=1

dk sin (kωo(t− T/2))

2πωo

t

cos(

0t)

cosine basis functions

2πωo

t

delayed half a period

2πωo

t

cos(ωot)

2πωo

t

2πωo

t

cos(

2ωot)

2πωo

t

2πωo2πωo

tt

cos(

3ωot)

...2πωo...

Half-period shift inverts ck terms if k is odd. It has no effect if k is even.

Half-Period Shift


f(t− T/2) =∞∑k=0



2πωo

t

sin(0t)

sine basis functions

2πωo

t

delayed half a period

2πωo

t

sin(ω

ot)

2πωo

t

2πωo

t

sin(2ωot)

2πωo

t

2πωo2πωo

tt

sin(3ωot)

...2πωo...

Half-period shift inverts dk terms if and only if k is odd.

Quarter-Period Shift

Shifting by T/4 is even more complicated.

f(t− T/4) =∞∑k=0



2πωo

t

cos(

0t)

cosine basis functions

2πωo

t

delayed one fourth period

2πωo

t

cos(ωot)

2πωo

t

2πωo

t

cos(

2ωot)

2πωo

t

2πωo2πωo

tt

cos(

3ωot)

...2πωo...

cos(ωot)→ sin(ωot); cos(2ωot)→ − cos(2ωot); cos(3ωot)→ − sin(3ωot)

Check Yourself: Alternative (more intuitive) Approach


f(t− T/4) =∞∑k=0



2πωo

t

sin(0t)

sine basis functions

2πωo

t

delayed 1/4 period

2πωo

t

sin(ω

ot)

2πωo

t

2πωo

t

sin(2ωot)

2πωo

t

2πωo2πωo

tt

sin(3ωot)

...2πωo...

sin(ωot)→ − cos(ωot); sin(2ωot)→ − sin(2ωot); sin(3ωot)→ cos(3ωot)

Summary of Shift Results

Let ck and dk represent the Fourier series coefficients for f(t)

f(t) = f(t+ T ) = c0 +∞∑k=1


dk sin(kωot)

and c′k and d′k represent those for a half-period delay.

g(t) = f(t− T/2) = c0 +∞∑k=1

c′k cos(kωot) +∞∑k=1

d′k sin(kωot)

Then c′k = (−1)kck and d′k = (−1)kdk.

Let c′′k and d′′k represent those for a quarter-period delay.

g(t) = f(t− T/2) = c0 +∞∑k=1


d′k sin(kωot)

Then

c′′k =

ck if k = 0, 4, 8, 12, . . .dk if k = 1, 5, 9, 13, . . .−ck if k = 2, 6, 10, 14, . . .−dk if k = 3, 7, 11, 15, . . .

d′′k =

dk if k = 0, 4, 8, 12, . . .−ck if k = 1, 5, 9, 13, . . .−dk if k = 2, 6, 10, 14, . . .ck if k = 3, 7, 11, 15, . . .

Other Shifts Yield Even More Complicated Results

Let ck and dk represent the Fourier series coefficients for f(t)

f(t) = f(t+ T ) = c0 +∞∑k=1


dk sin(kωot)

and c′′′k and d′′′k represent those for n eighth-period delay.

g(t) = f(t− T/8) = c0 +∞∑k=1


d′k sin(kωot)

c′′′k =

ck if k = 0, 8, 16, 24, . . .√

22 (ck + dk) if k = 1, 9, 17, 25, . . .

dk if k = 2, 10, 18, 26, . . .√

22 (−ck + dk) if k = 3, 11, 19, 27, . . .

−ck if k = 4, 12, 20, 28, . . .√

22 (−ck − dk) if k = 5, 13, 21, 29, . . .

−dk if k = 6, 14, 22, 30, . . .√

22 (ck − dk) if k = 7, 15, 23, 31, . . .

d′′′k = . . .

Effects of Time Shifts on Complex Exponential Series

Delaying time by τ multiplies the complex exponential coefficients of a

Fourier series by a constant e−jkωoτ .

Let ak represent the complex exponential series coefficients of f(t) and a′krepresent the complex exponential series coefficents of g(t) = f(t− τ).

a′k = 1T

∫Tg(t)e−jkωotdt

= 1T

∫Tf(t− τ)e−jkωotdt

= 1T

∫Tf(s)e−jkωo(s+τ)ds

= e−jωoτ1T

∫Tf(s)e−jkωosds

= e−jkωoτak

Each coefficient a′k in the series for g(t) is a constant e−jkωot times the

corresponding coefficient ak in the series for f(t).

Summary

We introduced the complex exponential form of Fourier series.

• complex numbers, complex exponentials, and their relation to sinusoids

• analysis and synthesis with complex exponentials

• delay property: much simpler with complex exponentials

Trig Table

sin(a+b) = sin(a) cos(b) + cos(a) sin(b)sin(a-b) = sin(a) cos(b) - cos(a) sin(b)cos(a+b) = cos(a) cos(b) - sin(a) sin(b)cos(a-b) = cos(a) cos(b) + sin(a) sin(b)tan(a+b) = (tan(a)+tan(b))/(1-tan(a) tan(b))tan(a-b) = (tan(a)-tan(b))/(1+tan(a) tan(b))

sin(A) + sin(B) = 2 sin((A+B)/2) cos((A-B)/2)sin(A) - sin(B) = 2 cos((A+B)/2) sin((A-B)/2)cos(A) + cos(B) = 2 cos((A+B)/2) cos((A-B)/2)cos(A) - cos(B) = -2 sin((A+B)/2) sin((A-B)/2)

sin(a+b) + sin(a-b) = 2 sin(a) cos(b)sin(a+b) - sin(a-b) = 2 cos(a) sin(b)cos(a+b) + cos(a-b) = 2 cos(a) cos(b)cos(a+b) - cos(a-b) = -2 sin(a) sin(b)

2 cos(A) cos(B) = cos(A-B) + cos(A+B)2 sin(A) sin(B) = cos(A-B) - cos(A+B)2 sin(A) cos(B) = sin(A+B) + sin(A-B)2 cos(A) sin(B) = sin(A+B) - sin(A-B)

fourier series { complex form

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