Homework & Tutorial

Homework: 5.2, 5.5, 5.15, 5.21(a-f, h)Tutorial Problems: 5.1, 5.3, 5.4, 5.41

Signals & Systems DT Fourier Transform P1

Chapter 5: The Discrete-Time Fourier Transform

Department of Electrical & Electronic Engineering

2020-FallLast Update on: Sunday 8th November, 2020

Signals & Systems DT Fourier Transform P2

A Big Picture

Signals & Systems DT Fourier Transform P3

Outline

Definition of discrete time Fourier transform (DTFT)I From Fourier series to Fourier transformI Convergence issue

Periodic signals’ DTFT

Properties of DTFT

Signals & Systems DT Fourier Transform P4

Discrete-Time Fourier Transform

Discrete-time Fourier series (for periodic signals)

Synthesis Equation: x [n] =∑

k=<N>

akejk(2π/N)n

Analysis Equation: ak =1

N

∑n=<N>

x [n]e−jk(2π/N)n

Aperiodic signals can be treated as periodic signals with period N →∞I x [n] must be like

∑k bke

jk(2π/N)n or∫ωb(ω)e jωndω

I bk or b(ω) can be calculated from x [n]

Signals & Systems DT Fourier Transform P5

Example: From Periodic To Aperiodic

Signals & Systems DT Fourier Transform P6

Derivation (1/3)

Original signal: x [n]

Define new periodic signal with period N: x̃ [n], such that

x̃ [n] = x [n], n = −N/2, ...,N/2− 1

Notice: when N →∞, x̃ [n] becomes x [n]

Signals & Systems DT Fourier Transform P7

Derivation (2/3)Look at the Fourier series of x̃ [n]:

ak =1

N

N/2+1∑n=−N/2

x̃ [n]e−jk(2π/N)n =1

N

∞∑n=−∞

x [n]e−jk(2π/N)n

Define X (e jω) =∑∞−∞ x [n]e−jωn, so that ak = X (e jkω0)/N (ω0 = 2π/N)

Look at the synthesis equation:

x̃ [n] =∑

k=<N>

akejk(2π/N)n =

1

N

∑k=<N>

X (e jkω0)e jkω0n

=1

2π

∑k=<N>

X (e jkω0)e jkω0nω0

When N →∞, ω0 → dω Σ→∫kω0 → ω x̃ [n]→ x [n], thus,

x [n] =1

2π

∫2π

X (e jω)e jωndω

Signals & Systems DT Fourier Transform P8

Derivation (3/3)

Therefore, we get the discrete-time Fourier transform pair

Discrete-Time Fourier Transform

Synthesis Equation: x [n] =1

2π

∫2π

X (e jω)e jωndω

Analysis Equation: X (e jω) =∞∑

n=−∞x [n]e−jωn

ObservationsI Continuous spectrum: Similar to CTFTI Periodic with period 2π: Different from CTFTI Low frequency: close to 0 and 2π; high frequency: close to π

Signals & Systems DT Fourier Transform P9

Fourier Transform Examples (1/2)

x [n] = 1 (n = −N1, ..., 0, ...,N1)

X (e jω) = sinω(N1+1/2)sin(ω/2)

Width of x [n] : Wt = 2N1 + 1; width of X (e jω) : Wf = 4π2N1+1

Wt ×Wf = 4π, which is a constant

Signals & Systems DT Fourier Transform P10

Fourier Transform Examples (2/2)

See textbook, Example 5.1

What’s the shape of magnitude when a→ 1 or a→ 0?

Signals & Systems DT Fourier Transform P11

Convergence Issue of Analysis Equation

Sufficient Condition of Convergence

The analysis equation X (e jω) =∑∞−∞ x [n]e−jωn will converge either if x [n] is

absolutely summable or if the sequence has finite energy, thus,

∞∑−∞|x [n]| <∞ or

∞∑−∞|x [n]|2 <∞

Do the following signals have Fourier transform:I anu[n] (0 < a < 1)I δ[n]I u[n]I e j

25πn, cos( 2

5πn)

I anu[n] (a > 1)

Signals & Systems DT Fourier Transform P12

Can Periodic Signals Have DTFT?

Definition of DTFT:

X (e jω) =∞∑

n=−∞x [n]e−jωn

Justification of divergence:

I Let ω = 2kπ, we have X (e jω) =∞∑

n=−∞x [n]

I Since x [n] is periodic, the summation∞∑

n=−∞x [n] will never converge unless

x [n] = 0

Conclusion: Most of periodic signals do NOT have DTFT according to thedefinition

However, it’s of significant engineering importance to extend Fouriertransform to periodic signals

Signals & Systems DT Fourier Transform P13

DTFT with Periodic Signals (1/2)

Fourier Transform of e jωn

The following transform pair is actually NOT rigorously defined:

x [n] = e jω0n ←→ X (e jω) =∞∑

l=−∞

2πδ(ω − ω0 − 2πl)

Synthesis:

1

2π

∫2π

X (e jω)e jωndω =1

2π

∫2π

∞∑l=−∞

2πδ(ω − ω0 − 2πl)e jωndω = e jω0n

Analysis:∞∑

n=−∞x [n]e−jωn =

∞∑n=−∞

e j(ω0−ω)n converge??

Signals & Systems DT Fourier Transform P14

Remark

We just believe the following equation is true:

∞∑n=−∞

e jωn =∞∑

l=−∞

2πδ(ω − 2πl)

When ω = 2kπ,∞∑

n=−∞1 = 2πδ(0)

Signals & Systems DT Fourier Transform P15

DTFT with Periodic Signals (2/2)

According to the Fourier series, for a periodic signal with period N:

x [n] = a0 + a1ej(2π/N)n + ...+ ake

jk(2π/N)n + ...+ aN−1ej(N−1)(2π/N)n

e jk(2π/N)n ←→∞∑

l=−∞2πδ(ω − k(2π/N)− 2πl)

Then, due to the linearity of Fourier transform

F{x [n]

}=

N−1∑k=0

akF{e jk(2π/N)n

}=

N−1∑k=0

ak

∞∑l=−∞

2πδ(ω − k(2π/N)− 2πl)

=∞∑

l=−∞

N−1∑k=0

ak2πδ(ω − k(2π/N)− 2πl)

Fourier transform of a periodic signal is a periodic sequence of impulsesI What’s the period? How many impulses within one period?

Signals & Systems DT Fourier Transform P16

Fourier Series v.s. Fourier Transform

Signals & Systems DT Fourier Transform P17

Example: Discrete-Time Impulse Chain

What’s the Fourier transform of x [n] =+∞∑

k=−∞δ[n − kN] ?

First of all, we calculate the Fourier series:

ak =1

N

∑n=<N>

x [n]e−jk2πN n

=1

N

∑n=<N>

+∞∑k=−∞

δ[n − kN]e−jk2πN n

=1

N

∑n=<N>

δ[n]e−jk2πN n =

1

N

Then, we have

X (e jω) =∞∑

l=−∞

N−1∑k=0

2π

Nδ(ω − k(2π/N)− 2πl) =

2π

N

∞∑k=−∞

δ(ω − 2πk

N)

Time domain period × Frequency domain period = ?

Signals & Systems DT Fourier Transform P18

Periodicity, Linearity and Shifting

PeriodicityX (e j(ω+2π)) = X (e jω)

I How about CTFT? Why?

Linearityax1[n] + bx2[n]←→ aX1(e jω) + bX2(e jω)

Time Shifting and Frequency Shifting

x [n − n0] ←→ e−jωn0X (e jω)

e jω0nx [n] ←→ X (e j(ω−ω0))

I What’s the physical meaning?

Signals & Systems DT Fourier Transform P19

Illustration on Shifting

x [n − n0]←→ e−jωn0X (e jω)

e jω0nx [n]←→ X (e j(ω−ω0))

Signals & Systems DT Fourier Transform P20

Example: Shifting

Similar to CT, the DTFT of impulse response of a DT LTI system isfrequency response.

H(e jω) =+∞∑

n=−∞h[n]e−jωn

Given an ideal LPF with the following frequency response

Suppose its impulse response is hlpf [n], please predict the behavior of anotherLTI system with impulse response (−1)nhlpf [n].

Signals & Systems DT Fourier Transform P21

Conjugation, Differencing and Accumulation

Conjugationx∗[n]←→ X ∗(e−jω)

I X (e jω) = X ∗(e−jω)⇔ x [n] is realI If x [n] is real, then <{X (e jω)} is even, ={X (e jω)} is odd

Differencing

x [n]− x [n − 1] ←→ (1− e−jω)X (e jω)

I High-pass or low-pass?

Accumulation

n∑m=−∞

x [m]←→ 1

1− e−jωX (e jω) + πX (e j0)

∞∑k=−∞

δ(ω − 2πk)

I How to derive it via differencing?

Signals & Systems DT Fourier Transform P22

Problem 5.37

Problem

Let X (e jω) be the Fourier transform of x [n]. Derive expressions in terms ofX (e jω) for the Fourier transforms of the following signals.

Re{x [n]}x∗[−n]

Ev{x [n]}

Signals & Systems DT Fourier Transform P23

Effect of Differencing

J(i , j) = |M(i , j)−M(i + 1, j + 1)|+ |M(i + 1, j)−M(i , j + 1)|

Signals & Systems DT Fourier Transform P24

Fourier Transform of u[n]

How to derive the Fourier transform of u[n]?

Option 1: From definition of Fourier transform

F{u[n]} =+∞∑

n=−∞u[n]e−jωn =

+∞∑n=0

e−jωn Converge??

Option 2: Since u[n] =∑n

m=−∞ δ[m], according to the property ofaccumulation

F{u[n]} =1

1− e−jω+ π

∞∑k=−∞

δ(ω − 2πk)

Observation: Fourier transform of u[n] does not exist according to thedefinition; however, it can be expressed in terms of δ(·)

Signals & Systems DT Fourier Transform P25

Time Reversal and ExpansionTime Reversal

x [−n]←→ X (e−jω)

Time Expansion

I Define x(k)[n] =

{x [n/k], if n is a multiple of k

0, Otherwise, then

x(k)[n]←→ X (e jkω) (1)

Signals & Systems DT Fourier Transform P26

Differentiation and Parseval

Differentiation in Frequency

nx [n]←→ jdX (e jω)

dω

Parseval’s Relation

+∞∑n=−∞

|x [n]|2 =1

2π

∫2π

|X (e jω)|2dω

I Energy density spectrum — |X (ejω)|22π

Signals & Systems DT Fourier Transform P27

Example

See textbook, Example 5.10

Spectrum within [−π, π]

Is it periodic, real, even, of finite energy?

Signals & Systems DT Fourier Transform P28