6.003: Signal ProcessingToday: Discrete Time Fourier Series Goal: Develop Fourier representation for discrete-time signals. While the high-level goal is the same in DT as in CT, there

6.003: Signal Processing

Discrete-Time Fourier Series

• Fourier series representations for discrete-time signals

• Comparison of Fourier series for CT and DT signals

• Properties of DT Fourier series

• Applications of Fourier analysis

Announcements:

• PSet 3 is posted and is due in one week (on Thursday, February 27).

• PSet 3 check-in due Tuesday, February 25.

• No PSet 4 – a practice quiz will be posted instead.

• Quiz 1: March 3, 2-4pm in 32-141.

− Coverage up to and including all of week 4.

− Closed book except for one page of notes (8.5”x11” both sides).

− No electronic devices (no headphones, cellphones, calculators, ...).

February 20, 2020

Continuous-Time Fourier Series

Last week, we explored the expansion of periodic CT functions as Fourier

series using either trigonometric functions or complex exponentials

f(t) = f(t+ T ) = c0 +∞∑k=1

ck cos kωot+∞∑k=1

dk sin kωot =∞∑

k=−∞ake

jkωot

where ωo = 2πT represents the fundamental frequency.

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


We found the Fourier series coefficients using two key insights.

1. Multiplying complex harmonics of ωo yields a complex harmonic of ωo:

e jkωot × e jlωot = e j(k+l)ωot

2. Integrating a complex harmonic over a period T yields zero unless the

harmonic is at DC:∫ t0+T

t0e jkωotdt ≡

∫Te jkωotdt =

T if k = 00 if k 6= 0

= Tδ[k]where δ[k] is the Kronecker delta function

δ[k] =

1 if k = 00 otherwise

→ Fourier components are orthogonal.


Use orthogonality to find the Fourier series coefficients.

f(t) = f(t+ T ) =∞∑

k=−∞ake

jkωot

Multiply f(t) by the complex conjugate of the basis function of interest,

and then integrate over T .∫Tf(t)e−jlωotdt =

∫T

( ∞∑k=−∞

akejkωot

)e−jlωotdt

=∞∑

k=−∞ak

∫Te j(k−l)ωotdt

=∞∑

k=−∞akTδ[k−l] = alT

Solving for al and then substituting k for l yields

ak = 1T

∫Tf(t)e−jkωotdt


Representing a periodic signal as a sum of harmonic sinusoids.

Synthesis Equation

f(t) = f(t+ T ) =∞∑

k=−∞ake

jkωot

Analysis Equation

ak = 1T

∫Tf(t)e−jkωot dt

where ωo = 2πT

Today: Discrete Time Fourier Series

Goal: Develop Fourier representation for discrete-time signals.

While the high-level goal is the same in DT as in CT, there are important

differences between CT and DT signals.

As we saw last time the same set of samples can represent many different

frequencies – a phenomenon called aliasing.

t

Samples (blue) of the original high-frequency signal (green) could just as

easily have come from a much lower frequency signal (red).

Aliasing

We saw that many CT frequencies ω alias to the same DT frequency Ω.

Sample a CT signal

f(t) = cos(ωt)at integer multiples of the sampling period T to generate a DT signal:

f [n] = f(t)|t=nT = f(nT ) = cos(ωnT ) = cos(Ωn) where Ω = ωT

Each point on the blue lines below indicates a pair of frequencies (ω, Ω)

for which cos(ωTn) = cos(Ωn).ωoT

ωT

Ω

π 2π 3π 4π

π

2π

Aliasing


Sample a CT signal



ωoT

ωT

Ω

π 2π 3π 4π

π

2π

Ωo

Many continuous frequencies ω produce the same samples:

f [n] = cos(Ωon) = cos(ωTn) for ω = Ωo

T,2π − Ωo

T,2π + Ωo

T,4π − Ωo

T, . . .

Aliasing


Sample a CT signal



ωoT

ωT

Ω

π 2π 3π 4π

π

2πωoT

Many discrete frequencies Ω could represent the samples produced by ωo:

f [n] = cos(ωoTn) = cos(Ωn) for Ω = ωoT, 2π−ΩoT, 2π+ΩoT, 4π−ΩoT . . .

Aliasing

Aliasing is an intrinsic property of DT sinusoids.

t

And there are other unique properties of DT sinsuoids that are important

for representing DT signals as Fourier series.

Check Yourself

What is the fundamental (shortest) period of each of the follow-

ing signals?

1. f1[n] = cos πn122. f2[n] = cos πn12 + 3 cos πn15

3. f3[n] = cos(n) + cos(2n) + cos(3n)

Check Yourself

If

f1[n] = cos πn12 = f1[n+N ]then

πN

12 = 2πm

where both N and m are integers. Solving, we find

N = 24πmπ

which is 24 if m = 1. Therefore N = 24.

Check Yourself

Similarly if

f2[n] = cos πn12 + 3 cos πn15 = f2[n+N ]then

πN

12 = 2πm1 → N = 24πm1π

andπN

15 = 2πm2 → N = 30πm2π

for integers m1 and m2.

We seek the smallest possible value of N so

N = 24m1 = 2× 2× 2× 3×m1 = 30m2 = 2× 3× 5×m2

and the smallest possible N is 2× 2× 2× 3× 5 = 120.

Check Yourself

If

f3[n] = cos(n) + cos(2n) + cos(3n) = f3[n+N ]then

N = 2πmwhere both N and m are integers.

Since the is not possible, f3[n] is not periodic.

Check Yourself


ing signals?

1. x1[n] = cos πn12 24

2. x2[n] = cos πn12 + 3 cos πn15 120

3. x3[n] = cosn+ cos 2n+ cos 3n ∞

Check Yourself


ing signals?

1. x1[n] = cos πn12 24

2. x2[n] = cos πn12 + 3 cos πn15 120

3. x3[n] = cosn+ cos 2n+ cos 3n ∞

The period of a periodic DT signal must be an integer. Therefore the

fundamental frequency Ωo = 2πN must be an integer submultiple of 2π.

No such constraints on fundamental frequencies in CT. In CT, the funda-

mental frequency ω = 2πT can be any real number.

→ This is an intrinsic difference between CT and DT signals.

Discrete-Time Sinusoids

If Ωo is a submultiple of 2π, and the harmonic frequencies alias, then there

are (only) N distinct complex exponentials with period N .

(There were an infinite number in CT!)

If f [n] = e jΩn is periodic in N then

f [n] = e jΩn = f [n+N ] = e jΩ(n+N) = e jΩne jΩN

and e jΩN must be 1. Therefore ejΩ must be one of the N th roots of 1.

Example: N = 8

Re

Im

There are only 8 unique harmonics of Ωo: 0,1π8 ,

2π8 ,

3π8 ,

4π8 ,

5π8 ,

6π8 ,

7π8 .

Discrete-Time Sinusoids

There are N distinct complex exponentials with period N .

n

Example: periodic in N=3

3 samples repeated in time 3 complex exponentials

n

Example: periodic in N=4

4 samples repeated in time 4 complex exponentials

If a DT signal is periodic with period N ,

then its Fourier series will contain just N terms.

Discrete Time Fourier Series

A DT Fourier Series has just N harmonic frequencies kΩo.

f [n] = c0 +∑k=〈N〉

ck cos(kΩon) +∑k=〈N〉

dk sin(kΩon)

where Ωo represents the fundamental frequency (radians/sample).

Otherwise, DT Fourier series are similar to CT Fourier series.

n2πΩo

cos(

0n)

n2πΩo

sin(0n

)n

2πΩo

n2πΩo

ej0n

n2πΩoco

s(Ωon

)

n2πΩosin

(Ωon

)

n2πΩo

n2πΩo

ejΩon

n2πΩoco

s(2Ω

on

)

...n

2πΩosin

(2Ωon

)

...n

2πΩo

n2πΩo

ej2Ωon

...

Real-valued basis functions Complex basis functions


The same two key insights apply to both CT and DT analysis.

1. Multiplying complex DT harmonics of Ωo yields a new harmonic of Ωo:

e jkΩon × e jlΩon = e j(k+l)Ωon

2. Summing a complex harmonic over a period N is zero unless the har-

monic is at DC:n0+N∑n=n0

e jkΩon ≡∑n=〈N〉

e jkΩon =N if k = 00 if k 6= 0

= Nδ[k]

→ DT Fourier components are orthogonal.


Using orthogonality to find the DT Fourier series coefficients.

f [n] = f [n+N ] =∑k=〈N〉

akejkΩon

Multiply f [n] by the complex conjugate of the basis function of interest,

and then sum over N .

∑n=〈N〉

f [n]e−jlΩon =∑n=〈N〉

∑k=〈N〉

akejkΩon

e−jlΩon

=∑k=〈N〉

ak∑n=〈N〉

e j(k−l)Ωon

=∑k=〈N〉

akNδ[k − l] = alN

Solving for al and then substituting k for l yields

ak = 1N

∑n=〈N〉

f [n]e−jkΩon


Both f [n] and ak are periodic in N .

The DT signal is periodic by construction:

f [n] = f [n+N ]

Its Fourier series coefficients are also periodic.

ak = 1N

∑n=〈N〉

f [n]e−jk2πNn

ak+N = 1N

∑n=〈N〉

f [n]e−j(k+N) 2πNn

ak+N =

1N

∑n=〈N〉

f [n]e−jk2πNn

︸︷︷︸

ak

(e−jN

2πNn)

︸︷︷︸1

ak+N = ak


Representing a periodic DT signal as a sum of harmonic sinusoids.

Synthesis Equation

f [n] = f [n+N ] =∑k=〈N〉

akejkΩon

Analysis Equation

ak = ak+N = 1N

∑n=〈N〉

f [n]e−jkΩon

where Ωo = 2πN

Fourier Series Summary

CT and DT Fourier Series are similar, but DT Fourier Series have just N

coefficients while CT Fourier Series have an infinite number.


ak=1T

∫Tf(t)e−jkωotdt analysis equation

f(t)= f(t+ T ) =∞∑

k=−∞ake

jkωot synthesis equation

where ωo = 2πT


ak= ak+N = 1N

∑n=〈N〉

f [n]e−jkΩon analysis equation

f [n]= f [n+N ] =∑k=〈N〉

akejkΩon synthesis equation

where Ωo = 2πN

Properties of Discrete-Time Fourier Series

Operations on the time representation of a signal can often be interpreted

as equivalent (but easier) operations on the series coefficients.

Here we will discuss four (of many) properites of Fourier series.

• linearity

• time shift

• time reversal

• conjugate symmetry

Linearity

The Fourier series coefficients of a linear combination of two signals is the

linear combination of their Fourier series coefficients.

Let

f [n] = af1[n] + bf2[n] where f1[n] = f1[n+N ] and f2[n] = f2[n+N ]

then the Fourier series coefficents for f [n] are given by

F [k] = 1N

∑n=〈N〉

f [n]e−jk2πNn = 1

N

∑n=〈N〉

(af1[n] + bf2[n])e−jk2πNn

= a1N

∑n=〈N〉

f1[n]e−jk2πNn + b

1N

∑n=〈N〉

f2[n]e−jk2πNn

= aF1[k] + bF2[k]

where F1[k] and F2[k] are Fourier series coefficients for f1[n] and f2[n].

Time Shift

Shifting time changes the phases of a signal’s Fourier coefficients.

Let

g[n] = f [n− n0] where f [n] = f [n+N ]If

F [k] = 1N

∑n=〈N〉

f [n]e−jk2πNn

then

G[k] = 1N

∑n=〈N〉

g[n]e−jk2πNn = 1

N

∑n=〈N〉

f [n− n0]e−jk2πNn

= 1N

∑m=〈N〉

f [m]e−jk2πN

(m+n0) where m = n−n0

= e−jk2πNn0 1N

∑m=〈N〉

f [m]e−jk2πNm = e−jk

2πNn0 F [k]

Time Reversal

Reversing time reverses frequency.

Let

g[n] = f [−n] where f [n] = f [n+N ]If

F [k] = 1N

∑n=〈N〉

f [n]e−jk2πNn

then

G[k] = 1N

∑n=〈N〉

g[n]e−jk2πNn = 1

N

∑n=〈N〉

f [−n]e−jk2πNn

= 1N

∑m=〈N〉

f [m]ejk2πNm where m = −n

= F [−k]

Conjugate Symmetry

If f [n] is real-valued, then its Fourier coefficients have conjugate symmetry.

If f [n] is real-valued, then f [n] = f∗[n].

F [k] = 1N

∑n=〈N〉

f [n]e−jk2πNn

F [−k] = 1N

∑n=〈N〉

f [n]ejk2πNn

= 1N

∑n=〈N〉

(f [n]e−jk

2πNn)∗

= F ∗[k]

Applications of Fourier Series

Signal processing is widely used in science and engineering to . . .

• model some aspect of the world,

• analyze the model, and

• interpret results to gain a new or better understanding.

model result

world new understanding

make model

analyze

(math, computation)

interpret results

We previously touched on applications in physics, including the wave equa-

tion and how it leads directly to Fourier analysis.

Applications of Fourier analysis in hearing.

Applications of Fourier Analysis in Hearing

What determines the pitch of a sound? This seemingly simple question has

evoked debate (sometimes fierce) for more than 150 years.

Compare two periodic signals with the same period, each played with 4000

samples per second

nf1[n]

nf2[n]

Different sounds, same pitch. We would like to understand why.

Pitch Experiments

Early experiments were based on stringed instruments and tubes, which

were known to produce not just a fundamental but also harmonic overtones.

dominant modedominant frequency

fundamental

2nd harmonic

3rd harmonic

Although different sources produced different mixtures of harmonics, it was

difficult to separate effects of one harmonic from those of others.

A breakthrough occured with the work of Seebeck who used sirens to

generate more complicated sounds.

Very clever experiment, but very controversial interpretations.

Sirens

Seebeck used a siren to generate more complicated sounds (circa 1841) by

passing a jet of compressed air through holes in a spinning disk.

nf1[n]

nf2[n]

nf3[n]

0 10 20

The pattern of holes determined the pattern of pulses in each period. The

speed of spinning controlled the number of periods per second.

Sirens

Strangely, adding a second hole per period didn’t seem to affect the pitch.

nf0[n]

nf3[n]

0 10 20

Pitch should be different if it is determined by the intervals between pulses.

Sirens

There was one very interesting exception.

nf1[n]

nf2[n]

nf3[n]

nf4[n]

nf5[n]

nf6[n]

nf7[n]

nf8[n]

nf9[n]

0 10 20

But hearing this exception required precise alignment of the siren’s holes.

Sirens and Controversy

Seebeck interpreted his results in terms of the intervals between the holes.

He held that pitch results from timing with some intervals being more

important than others. As the lengths of the two intervals in his experiment

converged, the pitch favored what had been the second harmonic and that

frequency increasingly dominated.

Georg Ohm (already known for his work on electrical conduction) in-

terpreted Seebeck’s results using Fourier’s recently described series. He

held that the pulses generated by a siren contained a fundamental and

harmonics that were physically present just as much as they are in a

stringed instrument.

A bitter controversy ensued.

Fourier Interpretation

To understand Ohm’s argument, compute the Fourier series for the siren’s

sound.

nf1[n]

nf2[n]

nf3[n]

nf4[n]

nf5[n]

nf6[n]

nf7[n]

nf8[n]

nf9[n]

0 10 20

Fourier Interpretation

Find the kth coefficient of the ith signal.

Fi[k] = 1N

∑n=〈N〉

fi[n] e−j2πkN

n = 110

9∑n=0

fi[n] e−j2πk10 n = 1

10

(1 + e−j

2πk10 i

)

DC: k = 0 term

Fi[0] = 110

(1 + e−j

2π010 i)

= 210

Fundamental: k = 1 term

Fi[1] = 110

(1 + e−j

2π10 i)

1 2 3 4 5 6 7 8 9i

|Fi[1]|θ= 2π

10

Re

Im

Fourier Series

Notice that f5[n] has no fundamental component!

nf1[n]

nf2[n]

nf3[n]

nf4[n]

nf5[n]

nf6[n]

nf7[n]

nf8[n]

nf9[n]

0 10 20

k

∣∣∣F1[k]∣∣∣

k

∣∣∣F2[k]∣∣∣

k

∣∣∣F3[k]∣∣∣

k

∣∣∣F4[k]∣∣∣

k

∣∣∣F5[k]∣∣∣

k

∣∣∣F6[k]∣∣∣

k

∣∣∣F7[k]∣∣∣

k

∣∣∣F8[k]∣∣∣

k

∣∣∣F9[k]∣∣∣

Fourier Series With and Without the Fundamental

Resynthesize each waveform without its fundamental component.

kF1[k]

kG1[k]

kF2[k]

kG2[k]

kF3[k]

kG3[k]

kF4[k]

kG4[k]

kF5[k]

kG5[k]

kF6[k]

kG6[k]

kF7[k]

kG7[k]

kF8[k]

kG8[k]

kF9[k]

kG9[k]

Although perception of the fundamental is weakened, it is not gone!

Summary

Seebeck designed an extremely clever experiment to test pitch perception.

Ohm analyzed an important theory (from Fourier) and argued that har-

monics are present even in the pulsatile sounds generated by a siren.

Neither Seebeck nor Ohm could convincingly account for experimental re-

sults that demonstrated the dominance of the fundamental, even when it

was weak or missing.

Progress in understanding the “missing fundamental” awaited Helmholtz,

who demonstrated the importance of “combination tones” in the ear.

model result

world new understanding

make model

analyze

(math, computation)

interpret results

Summary

Today we focused on discrete-time Fourier analysis.

• We developed Fourier series for discrete-time signals.

• We compared Fourier series for CT and DT signals.

• We looked at four (of many) properties of DT Fourier series.

• We looked briefly at applications of Fourier analysis in hearing.

Next week: Fourier analysis of aperiodic signals.

Documents

6.003: Signal ProcessingToday: Discrete Time Fourier Series Goal: Develop Fourier representation for discrete-time signals. While the high-level goal is the same in DT as in CT, there