EE361: SIGNALS AND SYSTEMS II CH4: CONTINUOUS TIME FOURIER TRANSFORM - - PowerPoint PPT Presentation

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EE361: SIGNALS AND SYSTEMS II

CH4: CONTINUOUS TIME FOURIER TRANSFORM

1

FOURIER TRANSFORM DERIVATION

CHAPTER 4.1 2

FOURIER SERIES REMINDER

 Previously, FS allowed representation of a periodic

signal as a linear combination of harmonically related exponentials

 𝑦 𝑢 = σ𝑙 𝑏𝑙𝑓𝑘𝑙𝜕0𝑢

𝑏𝑙 =

1 𝑈 ׬ 𝑈 𝑦 𝑢 𝑓−𝑘𝑙𝜕0𝑢 𝑒𝑢

 Would like to extend this (Transform Analysis) idea

to aperiodic (non-periodic) signals

3

CT FOURIER TRANSFORM DERIVATION I

 Intuition:  Consider a periodic signal with period 𝑈  Let 𝑈 → ∞

 Infinite period  no longer periodic signal

 Results in 𝜕0 =

2𝜌 𝑈 → 0

 Zero frequency space between “harmonics”  differential 𝑒𝜕

 Envelope (like we saw with rectangle wave/sinc) defines

the CTFT

4

CT FOURIER TRANSFORM DERIVATION II

 Will skip derivation for now  Please see details in the book

5

CT FOURIER TRANSFORM PAIR

 𝑦 𝑢 =

1 2𝜌 ׬ −∞ ∞ 𝑌 𝑘𝜕 𝑓𝑘𝜕𝑢𝑒𝜕

synthesis eq (inverse FT)

 𝑌 𝑘𝜕 = ׬

−∞ ∞ 𝑦 𝑢 𝑓−𝑘𝜕𝑢𝑒𝑢

analysis eq (FT)

 Denote

 𝑦 𝑢 ↔ 𝑌(𝑘𝜕)  𝑌 𝑘𝜕 = ℱ 𝑦 𝑢

𝑦 𝑢 = ℱ−1 𝑌 𝑘𝜕

6

CTFT CONVERGENCE

 There are conditions on signal 𝑦(𝑢) for FT to exist  Finite energy (square integrable)

 ׬

−∞ ∞ 𝑦 𝑢 2𝑒𝑢 < ∞

 Dirichlet Conditions

 We will not cover; see pg 290 for more discussion

7

CTFT FOR PERIODIC SIGNALS

CHAPTER 4.2 8

FT OF PERIODIC SIGNALS

 Derived FT by assuming a periodic padding of

aperiodic signal 𝑦(𝑢)

 What happens for FT of a periodic signal?

 Note: periodic signal will not have finite energy  Cannot evaluate FT integral directly

9

PERIODIC FT DERIVATION I

 From derivation of FT, 𝑌 𝑘𝜕 is the envelope of

𝑈𝑏𝑙

 FS coefficients are scaled samples of 𝑌 𝑘𝜕

 Assume 𝑦(𝑢) is periodic 𝑦 𝑢 = 𝑦 𝑢 + 𝑈  Then, 𝑦 𝑢 = σ𝑙=−∞

∞

𝑏𝑙𝑓𝑘𝑙𝜕0𝑢, with 𝜕0 =

2𝜌 𝑈

 Plug into FT integral and solve  Will not solve for now on slides  see book

10

PERIODIC FT DERIVATION II

 Important property

 𝑦 𝑢 = 𝑓𝑘𝑙𝜕0𝑢 ↔ 𝑌 𝑘𝜕 = 2𝜌𝜀 𝜕 − 𝑙𝜕0

 Transform pair  σ𝑙=−∞

∞

𝑏𝑙𝑓𝑘𝑙𝜕0𝑢 ↔ 2𝜌 σ𝑙=−∞

∞

𝑏𝑙𝜀 𝜕 − 𝑙𝜕0

 Each 𝑏𝑙 coefficient gets turned into a delta at the

harmonic frequency

11

FT OF SINUSOIDAL SIGNALS

 FT of periodic signals is important because of

sinusoidal signals (cannot solve using FT integral)

 Can use insight of complex exponential ↔ shifted delta

from periodic FT derivation

 Important examples

12

CTFT PROPERTIES AND PAIRS

CHAPTER 4.3-4.6 13

 Linearity

 𝑦 𝑢 ↔ 𝑌(𝑘𝜕)  𝑧 𝑢 ↔ 𝑍 𝑘𝜕  𝑏𝑦 𝑢 + 𝑐𝑧 𝑢 ↔ 𝑏𝑌 𝑘𝜕 + 𝑐𝑍 𝑘𝜕

 Time shifting

 𝑦 𝑢 − 𝑢0 ↔ 𝑓−𝑘𝜕𝑢0𝑌(𝑘𝜕)  Note, this is a phase shift of 𝑌 𝑘𝜕

 Conjugation

 𝑦∗ 𝑢 ↔ 𝑌∗ −𝑘𝜕

 Remember: conjugation switches sign of

imaginary part

 Time scaling

 𝑦 𝑏𝑢 ↔

1 𝑏 𝑌 𝑘𝜕 𝑏

 Differentiation in time



𝑒𝑦 𝑢 𝑒𝑢

↔ 𝑘𝜕𝑌(𝑘𝜕)

 Integration in time

 ׬

−∞ 𝑢

𝑦 𝜐 𝑒𝜐 ↔

1 𝑘𝜕 𝑌 𝑘𝜕 +

𝜌𝑌 0 𝜀 𝜕

14

PROPERTIES TABLE 4.1 (PG 328)

CONVOLUTION/MULTIPLICATION PROPERTIES

 Convolution

 𝑧 𝑢 = ℎ 𝑢 ∗ 𝑦 𝑢 ↔ 𝑍 𝑘𝜕 = 𝐼 𝑘𝜕 𝑌 𝑘𝜕

 Multiplication

 𝑠 𝑢 = 𝑡 𝑢 𝑞 𝑢 ↔ 𝑆 𝑘𝜕 =

1 2𝜌 ׬ −∞ ∞ 𝑇 𝑘𝜄 𝑄 𝑘 𝜕 − 𝜄

𝑒𝜄

 𝑆 𝑘𝜕 =

1 2𝜌 𝑇 𝑘𝜕 ∗ 𝑄 𝑘𝜕

 Dual properties – convolution ↔ multiplication

15

FT PAIRS TABLE 4.2 (PG 329)

 Be sure to bookmark this table (right next to Table

4.1 Properties)

 Note in particular some very useful pairs that

aren’t typical

 𝑢𝑓−𝑏𝑢𝑣 𝑢 ↔

1 𝑏+𝑘𝜕 2

repeated root

 𝑣 𝑢 ↔

1 𝑘𝜕 + 𝜌𝜀(𝜕)

16

CTFT AND LTI SYSTEMS

CHAPTER 4.7 17

FIRST-ORDER EXAMPLE

 Find impulse response ℎ(𝑢)

18

LTI SYSTEM ANALYSIS

 Note for 𝐼 𝑘𝜕 to exist, the LTI system must have

impulse response ℎ(𝑢) that satisfies stability conditions

 FT is only for the analysis of stable LTI systems

 For not stable systems, use Laplace Transform in Ch9

19

 Take FT of both sides  Solve for frequency response

 Rational form – ratio of

polynomials in 𝑘𝜕

 Best solved using partial fraction

expansion (Appendix A)

20

GENERAL DIFFERENTIAL EQUATION SYSTEM