6.003 Signal Processing

Week 4, Lecture B:DTFT and Properties of FT

6.003 Fall 2020

From Fourier Series to Fourier Transform (DT)

• Last time: use continuous-time Fourier transform to represent arbitrary (aperiodic) CT signals as sums of sinusoidal components

Today: generalize the Fourier Transform idea to discrete-timesignals and look at more FT properties.

Synthesis equation

Analysis equation𝑋 𝜔 = න−∞

∞

𝑥(𝑡) ∙ 𝑒−𝑗𝜔𝑡 𝑑𝑡

𝑥 𝑡 =1

2𝜋න−∞

∞

𝑋(𝜔) ∙ 𝑒𝑗𝜔𝑡 𝑑𝜔

Fourier Representations of Aperiodic SignalsHow can we represent an aperiodic signal as a sum of sinusoids?

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

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.

1

1

Fourier Representations of Aperiodic Signals

Calculate the Fourier series coefficients 𝑋𝑝[𝑘] :

=1

𝑁+

2

𝑁cos

2𝜋𝑘

𝑁+

2

𝑁cos

4𝜋𝑘

𝑁

What happens if you double the period N?

The red samples are at new intermediate frequencies

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

𝑋𝑝 𝑘 =1

𝑁

𝑛=<𝑁>

𝑥[𝑛] ∙ 𝑒−𝑗2𝜋𝑁 𝑘𝑛

Plot the resulting Fourier Series coefficients for N=8.

𝑋𝑝 𝑘 =1

𝑁

𝑛=<𝑁>

𝑥𝑝[𝑛] ∙ 𝑒−𝑗

2𝜋𝑁 𝑘𝑛

1

Fourier Representations of Aperiodic Signals

𝑙𝑒𝑡 Ω =2𝜋𝑘

𝑁, Define a new function X Ω = 𝑁 ∙ 𝑋𝑝 𝑘 = 1 + 2 cos Ω + 2 cos 2Ω

If we consider Ω and X Ω = 1 + 2 cos Ω + 2 cos 2Ω to be continuous, 𝑁𝑋𝑝 𝑘 represents samples of X Ω .

As N increases, the resolution in Ωincreases

𝑋𝑝[𝑘] =1

𝑁+

2

𝑁cos

2𝜋𝑘

𝑁+

2

𝑁cos

4𝜋𝑘

𝑁

Fourier Representations of Aperiodic SignalsWe can reconstruct 𝑥[𝑛] from X Ω using Riemann sums (approximating an integral by a finite sum).

As N → ∞,

• 𝑘Ω0 =2𝜋𝑘

𝑁becomes a continuum,

2𝜋𝑘

𝑁→ Ω.

• The sum takes the from of an

integral,Ω0 =2𝜋

𝑁→ 𝑑Ω

• We obtain a spectrum of coefficients: 𝑋 Ω .

Discrete-Time Fourier Transform

Since X Ω = 𝑁 ∙ 𝑋𝑝 𝑘

𝑋 Ω =

𝑛=−∞

∞

𝑥[𝑛] ∙ 𝑒−𝑗Ω𝑛

𝑋𝑝 𝑘 =1

𝑁

𝑛=<𝑁>

𝑥[𝑛] ∙ 𝑒−𝑗2𝜋𝑁 𝑘𝑛

Fourier Series and Fourier Transform

Discrete-Time Fourier Transform

Synthesis equation

Analysis equation

𝑥[𝑛] =1

2𝜋න2𝜋

𝑋(Ω) ∙ 𝑒𝑗Ω𝑛 𝑑Ω

Fourier series and transforms are similar: both represent signals by their frequency content.

𝑋 Ω = 𝑋(Ω + 2𝜋) =

𝑛=−∞

∞

𝑥[𝑛] ∙ 𝑒−𝑗Ω𝑛

Discrete-Time Fourier Series

Ω0 =2𝜋

𝑁

Synthesis equation

Analysis equation

𝑥 𝑛 = 𝑥 𝑛 + 𝑁 =

𝑘=<𝑁>

𝑋 𝑘 𝑒𝑗2𝜋𝑁 𝑘𝑛

𝑋 𝑘 = 𝑋[𝑘 + 𝑁] =1

𝑁

𝑛=<𝑁>

𝑥 𝑛 𝑒−𝑗Ω0𝑘𝑛

Fourier Series and Fourier TransformPeriodic signals can be synthesized from a discrete set of harmonics. Aperiodic signals generally require all possible frequencies.

Discrete-Time Fourier Transform

Synthesis equation

Analysis equation

𝑥[𝑛] =1

2𝜋න2𝜋

𝑋(Ω) ∙ 𝑒𝑗Ω𝑛 𝑑Ω

𝑋 Ω = 𝑋(Ω + 2𝜋) =

𝑛=−∞

∞

𝑥[𝑛] ∙ 𝑒−𝑗Ω𝑛

Discrete-Time Fourier Series

Ω0 =2𝜋

𝑁

Synthesis equation

Analysis equation

𝑥 𝑛 = 𝑥 𝑛 + 𝑁 =

𝑘=<𝑁>

𝑋 𝑘 𝑒𝑗2𝜋𝑁 𝑘𝑛

𝑋 𝑘 = 𝑋[𝑘 + 𝑁] =1

𝑁

𝑛=<𝑁>

𝑥 𝑛 𝑒−𝑗Ω0𝑘𝑛

Fourier Series and Fourier TransformAll of the information in a periodic signal is contained in one period. The information in an aperiodic signal is spread across all time.

Discrete-Time Fourier Transform

Synthesis equation

Analysis equation

𝑥[𝑛] =1

2𝜋න2𝜋

𝑋(Ω) ∙ 𝑒𝑗Ω𝑛 𝑑Ω

𝑋 Ω = 𝑋(Ω + 2𝜋) =

𝑛=−∞

∞

𝑥[𝑛] ∙ 𝑒−𝑗Ω𝑛

Discrete-Time Fourier Series

Ω0 =2𝜋

𝑁

Synthesis equation

Analysis equation

𝑥 𝑛 = 𝑥 𝑛 + 𝑁 =

𝑘=<𝑁>

𝑋 𝑘 𝑒𝑗2𝜋𝑁 𝑘𝑛

𝑋 𝑘 = 𝑋[𝑘 + 𝑁] =1

𝑁

𝑛=<𝑁>

𝑥 𝑛 𝑒−𝑗Ω0𝑘𝑛

Fourier Series and Fourier TransformHarmonic frequencies 𝑘Ω0 are samples of continuous frequency Ω

Discrete-Time Fourier Transform

Synthesis equation

Analysis equation

𝑥[𝑛] =1

2𝜋න2𝜋

𝑋(Ω) ∙ 𝑒𝑗Ω𝑛 𝑑Ω

𝑋 Ω = 𝑋(Ω + 2𝜋) =

𝑛=−∞

∞

𝑥[𝑛] ∙ 𝑒−𝑗Ω𝑛

Discrete-Time Fourier Series

Ω0 =2𝜋

𝑁

Synthesis equation

Analysis equation

𝑥 𝑛 = 𝑥 𝑛 + 𝑁 =

𝑘=<𝑁>

𝑋 𝑘 𝑒𝑗Ω0𝑘𝑛

𝑋 𝑘 = 𝑋[𝑘 + 𝑁] =1

𝑁

𝑛=<𝑁>

𝑥 𝑛 𝑒−𝑗Ω0𝑘𝑛

CT and DT Fourier TransformsDT frequencies alias because adding 2π to Ω does not change 𝑒𝑗Ω𝑛.Because of aliasing, we need only integrate dΩ over a 2π interval.

Discrete-Time Fourier Transform

Synthesis equation

Analysis equation

𝑥[𝑛] =1

2𝜋න2𝜋

𝑋(Ω) ∙ 𝑒𝑗Ω𝑛 𝑑Ω

𝑋 Ω = 𝑋(Ω + 2𝜋) =

𝑛=−∞

∞

𝑥[𝑛] ∙ 𝑒−𝑗Ω𝑛

Continuous-Time Fourier Transform

Synthesis equation

Analysis equation𝑋 𝜔 = න−∞

∞

𝑥(𝑡) ∙ 𝑒−𝑗𝜔𝑡 𝑑𝑡

𝑥 𝑡 =1

2𝜋න−∞

∞

𝑋(𝜔) ∙ 𝑒𝑗𝜔𝑡 𝑑𝜔

Fourier Transform of a Rectangular Pulse (width 2S+1)

𝑃𝑆(Ω) =

𝑛=−∞

∞

𝑝𝑆 [𝑛] ∙ 𝑒−𝑗Ω𝑛 =

𝑛=−𝑆

𝑆

𝑒−𝑗Ω𝑛 = 𝑒𝑗Ω𝑆

𝑚=0

2𝑆

𝑒−𝑗Ω𝑚

𝑃𝑆(Ω) =𝑒𝑗Ω(𝑆+

12)

𝑒𝑗Ω/2∙1 − 𝑒−𝑗Ω(2𝑆+1)

1 − 𝑒−𝑗Ω=𝑒𝑗Ω(𝑆+

12) − 𝑒−𝑗Ω(𝑆+

12)

𝑒𝑗Ω/2 − 𝑒−𝑗Ω/2 =sin(Ω 𝑆 +

12)

sin(Ω2)

When Ω ≠ 0 𝑜𝑟 2𝑘𝜋

When Ω = 0, (𝑜𝑟 2𝑘𝜋), 𝑃𝑆 Ω = 2𝑆 + 1

1 1

𝑙𝑒𝑡 𝑚 = 𝑛 + 𝑆, 𝑛 = 𝑚 − 𝑆

Fourier Transform of a rectangular pulse

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

Fourier Transform of a rectangular pulse

The value of x[0] is 1/2π times the integral of X(Ω) over Ω = [−π, π].

Fourier Transforms of Pulses with Different Widths

As the function widens in n(time) the Fourier transform narrows in Ω (freq).

How about going the other way?

𝑃𝑆(Ω) =sin(Ω(𝑆 +

12)

sin(Ω2)

In the extreme of S=0, the signal becomes a unit impulse δ[n]

DT ImpulseThe DT impulse is δ[n], its CT equivalent is δ(t)

δ[n] still has the “sifting property:”

𝛿(𝑡) 𝑋 𝜔

The DTFT of δ[n]:

𝑋 Ω =

𝑛=−∞

∞

𝛿[𝑛] ∙ 𝑒−𝑗Ω𝑛 = 1

In comparison to its CT counterpart δ(t):

δ[n]

0

Unit Impulse in Frequency Domain

𝑥 𝑛 =1

2𝜋න2𝜋

𝑋 Ω 𝑒𝑗Ω𝑛𝑑Ω

Because DT Fourier Transforms are periodic in 2π, it becomes an impulse train repeated every 2π.

=1

2𝜋න−𝜋

𝜋

𝛿 Ω 𝑒𝑗Ω𝑛𝑑Ω =1

2𝜋න0−

0+

𝛿 Ω 𝑒𝑗0𝑛𝑑Ω =1

2𝜋න0−

0+

𝛿 Ω 𝑑Ω =1

2𝜋

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

Find the DT signal whose Fourier transform is the above unit impulse.

Properties of FT

6.003 Fall 2020

FT Properties: Linearity

𝑥 𝑡 = A𝑥1 𝑡 + 𝐵 𝑥2 𝑡If:

𝑋 𝜔 = 𝐴𝑋1 𝜔 + 𝐵 𝑋2 𝜔then:

𝑥 𝑛 = A𝑥1[𝑛] + 𝐵 𝑥2[𝑛]If:

then:

𝑋 𝜔 = න−∞

∞

𝑥(𝑡) ∙ 𝑒−𝑗𝜔𝑡 𝑑𝑡

= න−∞

∞

(A𝑥1 𝑡 + 𝐵 𝑥2 𝑡 ) ∙ 𝑒−𝑗𝜔𝑡 𝑑𝑡

= 𝐴න−∞

∞

𝑥1 𝑡 ∙ 𝑒−𝑗𝜔𝑡 𝑑𝑡 + 𝐵න−∞

∞

𝑥2 𝑡 ∙ 𝑒−𝑗𝜔𝑡 𝑑𝑡

= 𝐴𝑋1 𝜔 + 𝐵 𝑋2 𝜔

𝑋 Ω = 𝐴𝑋1 Ω + 𝐵 𝑋2 Ω

𝑋 Ω =

𝑛=−∞

∞

𝑥[𝑛] ∙ 𝑒−𝑗Ω𝑛

=

𝑛=−∞

∞

(A𝑥1[𝑛] + 𝐵 𝑥2[𝑛]) ∙ 𝑒−𝑗Ω𝑛

= 𝐴

𝑛=−∞

∞

𝑥1[𝑛] ∙ 𝑒−𝑗Ω𝑛 + 𝐵

𝑛=−∞

∞

𝑥2[𝑛] ∙ 𝑒−𝑗Ω𝑛

= 𝐴𝑋1 Ω + 𝐵 𝑋2 Ω

FT Properties: Time Flip(Reversal)

𝑦 𝑡 = 𝑥 −𝑡If:

𝑌 𝜔 = 𝑋 −𝜔then:

𝑦 𝑛 = 𝑥[−𝑛]If:

then: 𝑌 Ω = 𝑋(−Ω)

𝑌 𝜔 = න−∞

∞

𝑦(𝑡) ∙ 𝑒−𝑗𝜔𝑡 𝑑𝑡

= න−∞

∞

𝑥(−𝑡) ∙ 𝑒−𝑗𝜔𝑡 𝑑𝑡

𝑙𝑒𝑡 𝑢 = −𝑡, 𝑡ℎ𝑒𝑛 𝑑𝑡 = −𝑑𝑢

𝑌 𝜔 = න∞

−∞

𝑥 𝑢 ∙ 𝑒−𝑗𝜔 −𝑢 (−𝑑𝑢)

= න−∞

∞

𝑥 𝑢 ∙ 𝑒−𝑗 −𝜔 𝑢 𝑑𝑢 = 𝑋 −𝜔

𝑌 Ω =

𝑛=−∞

∞

𝑦[𝑛] ∙ 𝑒−𝑗Ω𝑛

=

𝑛=−∞

∞

𝑥[−𝑛] ∙ 𝑒−𝑗Ω𝑛 𝑙𝑒𝑡 𝑚 = −𝑛

=

𝑚=∞

−∞

𝑥[𝑚] ∙ 𝑒−𝑗Ω(−𝑚)

=

𝑚=−∞

∞

𝑥[𝑚] ∙ 𝑒−𝑗(−Ω)𝑚 = 𝑋(−Ω)

FT Properties: Time Shift(Delay)Time delay maps to linear phase delay of the Fourier transform.

𝑥 𝑡 𝑋 𝜔𝐶𝑇𝐹𝑇

If:

𝑥 𝑡 − 𝜏 𝑒−𝑗𝜔𝜏𝑋 𝜔𝐶𝑇𝐹𝑇

then:

𝑋 𝜔 = න−∞

∞

𝑥(𝑡) ∙ 𝑒−𝑗𝜔𝑡 𝑑𝑡

𝑌 𝜔 = න−∞

∞

𝑥(𝑡 − 𝜏) ∙ 𝑒−𝑗𝜔𝑡 𝑑𝑡

𝑙𝑒𝑡 𝑢 = 𝑡 − 𝜏, 𝑡ℎ𝑒𝑛 𝑡 = 𝑢 + 𝜏, 𝑑𝑡 = 𝑑𝑢

𝑌 𝜔 = න−∞

∞

𝑥(𝑢) ∙ 𝑒−𝑗𝜔(𝑢+𝜏) 𝑑𝑢

= 𝑒−𝑗𝜔𝜏න−∞

∞

𝑥(𝑢) ∙ 𝑒−𝑗𝜔𝑢 𝑑𝑢 = 𝑒−𝑗𝜔𝜏 𝑋 𝜔

𝑥[𝑛] 𝑋 Ω𝐷𝑇𝐹𝑇

If:

𝑥[𝑛 − 𝑛0] 𝑒−𝑗Ω𝑛0𝑋 Ω𝐷𝑇𝐹𝑇

then:

𝑙𝑒𝑡 𝑚 = 𝑛 − 𝑛0, 𝑡ℎ𝑒𝑛 𝑛 = 𝑚 + 𝑛0

= 𝑒−𝑗Ω𝑛0𝑋 Ω

𝑋 Ω =

𝑛=−∞

∞

𝑥[𝑛] ∙ 𝑒−𝑗Ω𝑛

𝑌 Ω =

𝑛=−∞

∞

𝑥[𝑛 − 𝑛0] ∙ 𝑒−𝑗Ω𝑛

𝑌 Ω =

𝑚=−∞

∞

𝑥[𝑚] ∙ 𝑒−𝑗Ω(𝑚+𝑛0)

= 𝑒−𝑗Ω𝑛0 ∙

𝑚=−∞

∞

𝑥[𝑚] ∙ 𝑒−𝑗Ω𝑚

FT Properties: Conjugation

𝑦 𝑡 = 𝑥∗ 𝑡If:

𝑌 𝜔 = 𝑋∗ −𝜔then:

𝑌 𝜔 = න−∞

∞

𝑦(𝑡) ∙ 𝑒−𝑗𝜔𝑡 𝑑𝑡

= න−∞

∞

𝑥∗(𝑡) ∙ 𝑒−𝑗𝜔𝑡 𝑑𝑡

𝑦 𝑛 = 𝑥∗[𝑛]If:

then:

𝑌 Ω =

𝑛=−∞

∞

𝑦[𝑛] ∙ 𝑒−𝑗Ω𝑛

=

𝑛=−∞

∞

𝑥∗[𝑛] ∙ 𝑒−𝑗Ω𝑛

𝑌 Ω = 𝑋∗ (−Ω)

= 𝑋∗ −𝜔

=

𝑛=−∞

∞

𝑥∗[𝑛] ∙ 𝑒𝑗 −Ω 𝑛

= 𝑋∗(−Ω)

= න−∞

∞

𝑥∗(𝑡) ∙ 𝑒𝑗(−𝜔)𝑡 𝑑𝑡

FT Properties: Time Derivative/Difference

𝑦 𝑡 =𝑑

𝑑𝑡𝑥 𝑡If:

𝑌 𝜔 = 𝑗𝜔 ∙ 𝑋 𝜔then:

𝑥(𝑡) = න−∞

∞

𝑋(𝜔) ∙ 𝑒𝑗𝜔𝑡 𝑑𝜔

𝑦 𝑡 = න−∞

∞

𝑋(𝜔) ∙ (𝑗𝜔) ∙ 𝑒𝑗𝜔𝑡 𝑑𝜔

𝑦 𝑛 = 𝑥 𝑛 − 𝑥 𝑛 − 1If:

then: 𝑌 Ω = (1 − 𝑒−𝑗Ω) ∙ 𝑋(Ω)

differentiate both sides w.r.t 𝑡

𝑦 𝑡 = න−∞

∞

(𝑗𝜔 ∙ 𝑋 𝜔 ) ∙ 𝑒𝑗𝜔𝑡 𝑑𝜔

𝑦 𝑡 = න−∞

∞

𝑌(𝜔) ∙ 𝑒𝑗𝜔𝑡 𝑑𝜔

since

𝑌 𝜔 = 𝑗𝜔 ∙ 𝑋 𝜔

This can be derived based on the linearity and time delay property of DTFT.

FT Properties: Frequency Derivative

𝑦 𝑡 = 𝑡𝑥 𝑡If:

𝑌 𝜔 = 𝑗𝑑

𝑑𝜔𝑋 𝜔

then:

𝑑

𝑑𝜔𝑋 𝜔 = න

−∞

∞

𝑥(𝑡) ∙ (−𝑗𝑡) ∙ 𝑒−𝑗𝜔𝑡 𝑑𝑡

𝑦 𝑛 = 𝑛𝑥 𝑛If:

then: 𝑌 Ω = 𝑗𝑑

𝑑Ω𝑋(Ω)

differentiate both sides w.r.t 𝜔

since

𝑋 𝜔 = න−∞

∞

𝑥(𝑡) ∙ 𝑒−𝑗𝜔𝑡 𝑑𝑡

𝑗𝑑

𝑑𝜔𝑋 𝜔 = න

−∞

∞

(𝑡𝑥(𝑡)) ∙ 𝑒−𝑗𝜔𝑡 𝑑𝑡

𝑌 𝜔 = න−∞

∞

𝑦(𝑡) ∙ 𝑒−𝑗𝜔𝑡 𝑑𝑡

𝑌 𝜔 = 𝑗𝑑

𝑑𝜔𝑋 𝜔

𝑋 Ω =

𝑛=−∞

∞

𝑥[𝑛] ∙ 𝑒−𝑗Ω𝑛

differentiate both sides w.r.t Ω

𝑑

𝑑Ω𝑋 Ω =

𝑛=−∞

∞

𝑥[𝑛] ∙ (−𝑗𝑛) ∙ 𝑒−𝑗Ω𝑛

𝑗𝑑

𝑑Ω𝑋 Ω =

𝑛=−∞

∞

𝑛𝑥[𝑛] ∙ 𝑒−𝑗Ω𝑛

since

𝑌 Ω =

𝑛=−∞

∞

𝑦[𝑛] ∙ 𝑒−𝑗Ω𝑛 𝑌 Ω = 𝑗𝑑

𝑑Ω𝑋(Ω)

FT Properties: Scaling Time

𝑦 𝑡 = 𝑥 𝐴𝑡 , A>0If:

𝑌 𝜔 =1

𝐴𝑋

𝜔

𝐴then:

𝑌 𝜔 = න−∞

∞

𝑦(𝑡) ∙ 𝑒−𝑗𝜔𝑡 𝑑𝑡

= න−∞

∞

𝑥(𝐴𝑡) ∙ 𝑒−𝑗𝜔𝑡 𝑑𝑡

𝑙𝑒𝑡 𝑢 = 𝐴𝑡, 𝑡ℎ𝑒𝑛 𝑡 =𝑢

𝐴, 𝑑𝑢 = 𝐴𝑑𝑡

𝑌 𝜔 = න−∞

∞

𝑥 𝑢 ∙ 𝑒−𝑗𝜔𝑢𝐴 (1

𝐴𝑑𝑢)

=1

𝐴න−∞

∞

𝑥 𝑢 ∙ 𝑒−𝑗

𝜔𝐴 𝑢

𝑑𝑢 =1

𝐴𝑋

𝜔

𝐴

Properties of Fourier Transforms

Discrete-Time Fourier TransformContinuous-Time Fourier Transform

Check yourself!Considering finding the Fourier Transform of the following signal

If we use 𝑋 𝜔 = න−∞

∞

𝑥(𝑡) ∙ 𝑒−𝑗𝜔𝑡 𝑑𝑡 the integration can be complicated

We need 𝑋 𝜔 such that :

Using the sifting property of the delta function, we find:

𝑋 𝜔 = 2𝜋𝛿(𝜔 − 𝜔0)To generalize the result: 𝑥 𝑡 = 𝑒𝑗𝜔0𝑡𝐶𝑇𝐹𝑇

We can utilize the sifting property of δ(t):

Fourier Transform of a Complex ExponentialWe can use duality and time shift to find the Fourier transform of a complex exponential.

𝑥 𝑡 = 𝑒𝑗𝜔0𝑡 𝑋 𝜔 = 2𝜋𝛿(𝜔 − 𝜔0)𝐶𝑇𝐹𝑇

Summary

• Discrete-Time Fourier Transform: Fourier representation to all DT signals!

• Very useful signals:• Rectangular pulse and its FT(sinc)• Delta function (Unit impulse) and its FT

• Various properties of FT:• Make it easier to find the FT of new signals

Synthesis equation

Analysis equation

𝑥[𝑛] =1

2𝜋න2𝜋

𝑋(Ω) ∙ 𝑒𝑗Ω𝑛 𝑑Ω

𝑋 Ω = 𝑋(Ω + 2𝜋) =

𝑛=−∞

∞

𝑥[𝑛] ∙ 𝑒−𝑗Ω𝑛

δ[n]X(Ω)=1

𝐷𝑇𝐹𝑇

Ω

𝐷𝑇𝐹𝑇