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

1 EE361: SIGNALS AND SYSTEMS II CH4: CONTINUOUS TIME FOURIER TRANSFORM http://www.ee.unlv.edu/~b1morris/ee361

2 FOURIER TRANSFORM DERIVATION CHAPTER 4.1

3 FOURIER SERIES REMINDER  Previously, FS allowed representation of a periodic signal as a linear combination of harmonically related exponentials 1 𝑈 𝑦 𝑢 𝑓 −𝑘𝑙𝜕 0 𝑢 𝑒𝑢  𝑦 𝑢 = σ 𝑙 𝑏 𝑙 𝑓 𝑘𝑙𝜕 0 𝑢 𝑏 𝑙 = 𝑈 ׬  Would like to extend this (Transform Analysis) idea to aperiodic (non-periodic) signals

4 CT FOURIER TRANSFORM DERIVATION I  Intuition:  Consider a periodic signal with period 𝑈  Let 𝑈 → ∞  Infinite period  no longer periodic signal 2𝜌  Results in 𝜕 0 = 𝑈 → 0  Zero frequency space between “harmonics”  differential 𝑒𝜕  Envelope (like we saw with rectangle wave/sinc) defines the CTFT

5 CT FOURIER TRANSFORM DERIVATION II  Will skip derivation for now  Please see details in the book

6 CT FOURIER TRANSFORM PAIR ∞ 𝑌 𝑘𝜕 𝑓 𝑘𝜕𝑢 𝑒𝜕 1 synthesis eq (inverse FT)  𝑦 𝑢 = 2𝜌 ׬ −∞ ∞ 𝑦 𝑢 𝑓 −𝑘𝜕𝑢 𝑒𝑢 analysis eq (FT)  𝑌 𝑘𝜕 = ׬ −∞  Denote  𝑦 𝑢 ↔ 𝑌(𝑘𝜕) 𝑦 𝑢 = ℱ −1 𝑌 𝑘𝜕  𝑌 𝑘𝜕 = ℱ 𝑦 𝑢

7 CTFT CONVERGENCE  There are conditions on signal 𝑦(𝑢) for FT to exist  Finite energy (square integrable) ∞ 𝑦 𝑢 2 𝑒𝑢 < ∞  ׬ −∞  Dirichlet Conditions  We will not cover; see pg 290 for more discussion

8 CTFT FOR PERIODIC SIGNALS CHAPTER 4.2

9 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

10 PERIODIC FT DERIVATION I  From derivation of FT, 𝑌 𝑘𝜕 is the envelope of 𝑈𝑏 𝑙  FS coefficients are scaled samples of 𝑌 𝑘𝜕  Assume 𝑦(𝑢) is periodic 𝑦 𝑢 = 𝑦 𝑢 + 𝑈 2𝜌 ∞  Then, 𝑦 𝑢 = σ 𝑙=−∞ 𝑏 𝑙 𝑓 𝑘𝑙𝜕 0 𝑢 , with 𝜕 0 = 𝑈  Plug into FT integral and solve  Will not solve for now on slides  see book

11 PERIODIC FT DERIVATION II  Important property  𝑦 𝑢 = 𝑓 𝑘𝑙𝜕 0 𝑢 ↔ 𝑌 𝑘𝜕 = 2𝜌𝜀 𝜕 − 𝑙𝜕 0  Transform pair 𝑏 𝑙 𝑓 𝑘𝑙𝜕 0 𝑢 ↔ 2𝜌 σ 𝑙=−∞ ∞ ∞  σ 𝑙=−∞ 𝑏 𝑙 𝜀 𝜕 − 𝑙𝜕 0  Each 𝑏 𝑙 coefficient gets turned into a delta at the harmonic frequency

12 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

13 CTFT PROPERTIES AND PAIRS CHAPTER 4.3-4.6

14 PROPERTIES TABLE 4.1 (PG 328)  Linearity  Time scaling  𝑦 𝑢 ↔ 𝑌(𝑘𝜕) 1 𝑘𝜕  𝑦 𝑏𝑢 ↔ 𝑏 𝑌  𝑧 𝑢 ↔ 𝑍 𝑘𝜕 𝑏  Differentiation in time  𝑏𝑦 𝑢 + 𝑐𝑧 𝑢 ↔ 𝑏𝑌 𝑘𝜕 + 𝑐𝑍 𝑘𝜕  Time shifting 𝑒𝑦 𝑢 ↔ 𝑘𝜕𝑌(𝑘𝜕)  𝑒𝑢  𝑦 𝑢 − 𝑢 0 ↔ 𝑓 −𝑘𝜕𝑢 0 𝑌(𝑘𝜕)  Integration in time  Note, this is a phase shift of 𝑌 𝑘𝜕  Conjugation 𝑢 1  ׬ 𝑦 𝜐 𝑒𝜐 ↔ 𝑘𝜕 𝑌 𝑘𝜕 + −∞  𝑦 ∗ 𝑢 ↔ 𝑌 ∗ −𝑘𝜕 𝜌𝑌 0 𝜀 𝜕  Remember: conjugation switches sign of imaginary part

15 CONVOLUTION/MULTIPLICATION PROPERTIES  Convolution  𝑧 𝑢 = ℎ 𝑢 ∗ 𝑦 𝑢 ↔ 𝑍 𝑘𝜕 = 𝐼 𝑘𝜕 𝑌 𝑘𝜕  Multiplication ∞ 𝑇 𝑘𝜄 𝑄 𝑘 𝜕 − 𝜄 1  𝑠 𝑢 = 𝑡 𝑢 𝑞 𝑢 ↔ 𝑆 𝑘𝜕 = 𝑒𝜄 2𝜌 ׬ −∞ 1  𝑆 𝑘𝜕 = 2𝜌 𝑇 𝑘𝜕 ∗ 𝑄 𝑘𝜕  Dual properties – convolution ↔ multiplication

16 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 1  𝑢𝑓 −𝑏𝑢 𝑣 𝑢 ↔ repeated root 𝑏+𝑘𝜕 2 1  𝑣 𝑢 ↔ 𝑘𝜕 + 𝜌𝜀(𝜕)

17 CTFT AND LTI SYSTEMS CHAPTER 4.7

18 FIRST-ORDER EXAMPLE  Find impulse response ℎ(𝑢)

19 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

20 GENERAL DIFFERENTIAL EQUATION SYSTEM  Solve for frequency response  Take FT of both sides  Rational form – ratio of polynomials in 𝑘𝜕  Best solved using partial fraction expansion (Appendix A)