homework & tutorial 5a.pdf · 2020. 11. 9. · discrete-time fourier transform discrete-time...
Post on 25-Feb-2021
4 Views
Preview:
TRANSCRIPT
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
top related