6.003: signal processingsigproc.mit.edu/_static/spring20/lectures/lec04b.pdf6.003: signal processing...

6.003: Signal Processing

Discrete-Time Fourier Transform

• Definition

• Examples

• Properties

• Relations between Fourier series and transforms (DT and CT)

Announcements:

• No PSet 4 – practice quiz questions have been posted instead.

• Office Hours today (7-9pm) and Sunday (4-6pm)

− practice quiz problems and general questions

• Solutions to practice quiz questions will be posted Sunday at 6pm

• Quiz 1: Tuesday, 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 27, 2020

From Periodic to Aperiodic

Last time: representing arbitrary (aperiodic) CT signals as sums of sinu-

soidal components using the continuous-time Fourier transform.

Today: generalize the Fourier Transform idea to discrete-time signals.

Fourier Representions of Aperiodic Signals

How can we represent an aperiodic signal as a sum of sinusoids?

n

x[n]

−2 0 2Strategy: make a periodic version of x[n] by summing shifted copies:

xp[n] =∞∑

m=−∞x[n−mN ]

n

xp[n]

−2 0 2−N N

Since xp[n] is periodic, it has a Fourier series (which depends on N).

Find Fourier series coefficients Xp[k] and take the limit of Xp[k] as N →∞.

As N →∞, xp[n]→ x[n], and Fourier series will approach Fourier transform.


Example.

n

x[n]

−2 0 2Strategy: make a periodic version of x[n] by summing shifted copies:

xp[n] =∞∑

m=−∞x[n−mN ]

n

xp[n]

−2 0 2−N N

Calculate the Fourier series coefficients Xp[k]:

Xp[k] = 1N

∑n=〈N〉

xp[n]e−j2πNkn = 1

N+ 2N

cos 2πkN

+ 2N

cos 4πkN



Xp[k] = 1N

∑n=〈N〉


N+ 2N

cos 2πkN

+ 2N

cos 4πkN

Plot the resulting Fourier coefficients for N=8.

k

Xp[k]

What happens if you double the period N?



Xp[k] = 1N

∑n=〈N〉


N+ 2N

cos 2πkN

+ 2N

cos 4πkN

Plot the resulting Fourier coefficients for N=8.

k

Xp[k]

What happens if you double the period N?

• The amplitude will shrink by a factor of 2.

• There will be twice as many samples per period of the cosine functions.

k

Xp[k]

k

Xp[k]

The red samples are at new intermediate frequencies.


Define a new function X(Ω) = NXp[k] where Ω = 2πk/N .

NXp[k] = 1+2 cos 2πkN

+2 cos 4πkN

= 1+2 cos(Ω)+2 cos(2Ω) = X(Ω)

N=8:

Ω = 2πkN

NXp[k] = X(Ω)

N=16:

Ω = 2πkN

N=32:

Ω = 2πkN

The discrete function NXp[k] is a sampled version of the function X(Ω).


We can reconstruct x[n] from X(Ω) using Riemann sums.

xp[n] =∑k=〈N〉

Xp[k]e j2πNkn= 1

2π∑k=〈N〉

NXp[k]ej2πNkn(2πN

)x[n] = lim

N→∞xp[n] = lim

N→∞

12π

∑k=〈N〉

NXp[k]ej2πNkn(2πN

)= 1

2π

∫2πX(Ω)ejΩndΩ

N=8:

Ω = 2πkN

NXp[k] = X(Ω)

−π π−2π 2π

2πN

N=16:

Ω = 2πkN

−π π−2π 2π

2πN

N=32:

Ω = 2πkN

−π π−2π 2π

2πN

Fourier Transform relation: x[n] ft⇐⇒ X(Ω)

Fourier Series and Fourier Transform

Fourier series and transforms are similar:

both represent signals by their frequency content.

Discrete-Time Fourier Series

X[k] = X[k+N ] = 1N

∑n=〈N〉

x[n]e−jkΩon analysis equation

x[n] = x[n+N ] =∑k=〈N〉

X[k]e jkΩon synthesis equation

where Ωo = 2πN


X(Ω)= X(Ω+2π) =∞∑

n=−∞x[n]e−jΩn analysis equation

x[n] = 12π

∫2πX(Ω)ejΩn dΩ synthesis equation


All of the information in a periodic signal is contained in one period.

The information in an aperiodic signal is spread across time.


X[k] = X[k+N ] = 1N

∑n=〈N〉


x[n] = x[n+N ] =∑k=〈N〉


where Ωo = 2πN


X(Ω)= X(Ω+2π) =∞∑


x[n] = 12π



Periodic signals can be synthesized from a discrete set of k harmonics.

Aperiodic signals generally require a continuous set of frequencies Ω.


X[k] = X[k+N ] = 1N

∑n=〈N〉


x[n] = x[n+N ] =∑k=〈N〉


where Ωo = 2πN


X(Ω)= X(Ω+2π) =∞∑


x[n] = 12π



Harmonic frequencies kΩo are samples of continuous frequency Ω.


X[k] = X[k+N ] = 1N

∑n=〈N〉


x[n] = x[n+N ] =∑k=〈N〉


where Ωo = 2πN


X(Ω)= X(Ω+2π) =∞∑


x[n] = 12π


CT and DT Fourier Transforms

DT frequencies alias because adding 2π to Ω does not change e−jΩn.

Continuous-Time Fourier Transform

X(ω)=∫ ∞−∞

x(t)e−jωtdt analysis equation

x(t) = 12π

∫ ∞−∞

X(ω)ejωtdω synthesis equation

where Ωo = 2πN


X(Ω)= X(Ω+2π) =∞∑


x[n] = 12π


CT and DT Fourier Transforms

DT frequencies alias because adding 2π to Ω does not change e−jΩn.

Because of aliasing, we need only integrate dΩ over a 2π interval.

Continuous-Time Fourier Transform

X(ω)=∫ ∞−∞

x(t)e−jωtdt analysis equation

x(t) = 12π

∫ ∞−∞

X(ω)ejωtdω synthesis equation

where Ωo = 2πN


X(Ω)= X(Ω+2π) =∞∑


x[n] = 12π


Examples of Fourier Transforms

Find the Fourier Transform (FT) of a rectangular pulse of width 2S+1:

pS [n] =

1 −S ≤ N ≤ S0 otherwise

n

p2[n]

−2 0 2

n

p5[n]

−5 0 5

PS(Ω) =∞∑

n=−∞pS [n]e−jΩn =

S∑n=−S

e−jΩn

= ejΩS + ejΩ(S−1) + · · ·+ 1 + · · ·+ e−jΩ(S−1) + e−jΩS

= 1 + 2 cos(Ω) + 2 cos(2Ω) + · · ·+ 2 cos(SΩ)

We would like to reduce this expression to a closed form to better identify

trends across S.

Working with Sums

Closed form sums of exponential sequences.

A =N−1∑n=0

αn

If the series has finite length (here N terms), it will converge for finite α.

A = 1 + α + α2 + · · · + αN−1

αA = α + α2 + · · · + αN−1 + αN

A− αA = 1 − αN

A =

1− αN

1− α if α 6= 1

N if α = 1

If the series has infinite length, it will converge if |α| < 1.

∞∑n=0

αn = limN→∞

N−1∑n=0

αn = limN→∞

1− αN

1− α = 11− α if |α| < 1


Find the Fourier Transform (FT) of a rectangular pulse of width 2S+1:

pS [n] =

1 −S ≤ N ≤ S0 otherwise

n

p2[n]

−2 0 2

PS(Ω) =S∑

n=−Se−jΩn =

(ejΩS

) 2S∑n=0

e−jΩn =(ejΩ(S+ 1

2 )

ejΩ/2

)1− e−jΩ(2S+1)

1− e−jΩ

=(ejΩ(S+ 1

2 ) − e−jΩ(S+ 12 )

ejΩ/2 − e−jΩ/2

)=

sin(Ω(S+1

2))

sin (Ω/2)

Ω

P2(Ω)

0 2π−2π

5


Compare Fourier transforms of pulses with different widths.

n

p2[n]

−2 2

1

Ω

P2(Ω)

2π−2π 2π5

5

n

p4[n]

−4 4

1

Ω

P4(Ω)

2π−2π 2π9

9

n

p6[n]

−6 6

1

Ω

P6(Ω)

2π−2π 2π13

13

As the function widens in n the Fourier transform narrows in Ω.

Moments

Similar to CT, the value of X(Ω) at Ω = 0 is the sum of x[n] over time t.

X(0) =∞∑

n=−∞x[n]e−jΩn =

∞∑n=−∞

x[n]

n

p2[n]

−2 2

1

Ω

P2(Ω)

2π−2π 2π5

5”area” = 5

Moments

The value of x[0] is 12π times the integral of X(Ω) over Ω = [−π, π].

x[0] = 12π

∫2πX(Ω) e jΩndΩ = 1

2π

∫2πX(Ω) dΩ

n

p2[n]

−2 2

1

Ω

P2(Ω)

2ππ-π−2π 2π5

5

Ω

P2(Ω)

2ππ-π−2π 2π5

5equal areas

area

2π = 1++

−−

Notice that the form of this relation is similar to the one for CT at the end

of last lecture (and repeated in the next slide).

Moments

The value of x(0) is the integral of X(ω) divided by 2π.

x(0) = 12π

∫ ∞−∞

X(ω) e jωtdω = 12π

∫ ∞−∞

X(ω) dω

−1 1

x1(t)

1

t++

−− ++ −−

2

π

X1(ω) = 2 sinωω

ω

area

2π = 1

2

πω

equal areas !

Stretching Time

Stretching time compresses frequency and increases amplitude

(preserving area).

−1 1

x1(t)

1

t

2

π

X1(ω) = 2 sinωω

ω

−2 2

1

t

4

πω

Very similar in CT and DT.

Compressing Time to the Limit

Alternatively, we could compress time while keeping area = 1.

−12

12

x(t)

1

t

1

2πω

X(ω) = sinω/2ω/2

ω

−14

14

2

t

1

2πω

In the limit, the pulse has zero width but area 1!

We represent this limit with the delta function: δ(t).

1

t

1

ω

Math With Impulses

Although physically unrealizeable, the impulse (a.k.a. Dirac delta) function

is useful as a mathematically tractable approximation to a very brief signal.

Example 1: Find the Fourier transform of a unit impulse function.

X(ω) =∫ ∞−∞

δ(t)e−jωtdt

Since δ(t) is zero except near t=0, only values of e−jωt near t=0 are impor-

tant. Because e−jωt is a smooth function of t, e−jωt can be replaced by

e−jω0:

X(ω) =∫ ∞−∞

δ(t)e−jω0dt = 1

This matches our previous result which was based explicitly on a limit.

Here the limit is implicit.

Math With Impulses

Although physically unrealizeable, the impulse function is extremely useful

as a mathematically tractable approximation to a very brief signal.

Example 2: Find the function whose Fourier transform is a unit impulse.

x(t) = 12π

∫ ∞−∞

δ(ω)ejωtdω = 12π

∫ ∞−∞

δ(ω)ej0tdω = 12π

1 ctft⇐⇒ 2πδ(ω)

Notice the similarity to the previous result:

δ(t) ctft⇐⇒ 1

These relations are duals of each other.

• A constant in time consists of a single frequency at ω = 0.

• An impulse in time contains components at all frequencies.

Math With Impulses

Delta functions are similarly useful in discrete-time transforms.

Let

X(Ω) =∞∑

m=−∞δ(Ω− 2πm)

where the sum results because DT Fourier Transforms are periodic in 2π.

Then

x[n] = 12π

∫2πX(Ω)ejΩndΩ = 1

2π

∫2πδ(Ω)ejΩndΩ = 1

2π

∫2πδ(Ω) dΩ = 1

2π

Thus if x[n] = 1 for all n, the transform is a delta function in frequency.

1 dtft⇐⇒

∞∑m=−∞

2πδ(Ω− 2πm)

We previously showed a similar result for CT:

1 ctft⇐⇒ 2πδ(ω)

Math With Impulses

Using delta functions to calculate the transforms of complex exponentials.

Let

X(ω) = δ(ω−ωo)then

x(t) = 12π

∫ ∞−∞

X(ω)ejωtdω = 12π

∫ ∞−∞

δ(ω−ωo)ejωtdω

The delta function is zero except when ω = ωo. Therefore we can substitute

ejωot for ejωt without changing the result of the integration.

x(t) = 12π

∫ ∞−∞

δ(ω−ωo)ejωotdω = 12πe

jωot

∫ ∞−∞

δ(ω−ωo)dω = 12πe

jωot

Thus the transform of a complex exponential is a delta at that frequency.

ejωot ctft⇐⇒ 2πδ(ω−ωo)

The analgous expresion for DT holds using analogous reasoning.

ejΩon dtft⇐⇒

∞∑m=−∞

2πδ(Ω−Ωo−2πm)

Relations Between Fourier Series and Fourier Transforms

Continuous Time:

ej2πkT

t ctft⇐⇒ 2πδ(ω−2πk

T

)x(t) = x(t+T ) ctfs⇐⇒ X[k]

x(t) = x(t+T ) =∞∑

k=−∞X[k]ej

2πTkt ctft⇐⇒

∞∑k=−∞

2πX[k]δ(ω−2π

Tk)

Discrete Time:

ej2πkN

n dtft⇐⇒

∞∑m=−∞

2πδ(

Ω−2πkN−2πm

)x[n] = x[n+N ] dtfs⇐⇒ X[k]

x[n] = x[n+N ] =∑k=〈N〉

X[k]ej2πNkn dtft⇐⇒

∑k=〈N〉

∞∑m=−∞

2πX[k]δ(

Ω−2πNk−2πm

)

Relation between Fourier Transform and Fourier Series

Each term in the Fourier series is replaced by an impulse in ω.

· · ·· · ·

x(t) =∞∑

k=−∞xp(t− kT )

t0 T

· · ·· · ·

X[k]

k

· · ·· · ·

X(ω) =∞∑

k=−∞2πX[k] δ(ω − k2π

T)

ω0 2π

T

Summary


• Definition

• Examples

• Properties

• Relations between Fourier series and Fourier transforms (DT and CT)

Quiz 1: Tuesday, 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, ...)

6.003: signal processingsigproc.mit.edu/_static/spring20/lectures/lec04b.pdf6.003: signal processing...

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