Notes Assignments Tutorial problems 3.3 3.21 3.22 (a) -> Figs. (b) (d) (f) 3.24 3.25 Tutorial problems Basic Problems wish Answers 3.8 Basic Problems 3.34 Advanced Problems 3.40

Review for Chapter 2 Why to introduce unit impulse response? Why to introduce convolution? (DT or CT) Signal can be represented by a linear combination of unit impulse function When it goes through the system, the output is computed via convolution of input signal and unit impulse response

Chapter 3 Fourier Series Representation of Periodic Signals

Joseph Fourier

A Useful Analogy

Example

Overview on Frequency Analysis Fourier Series (Periodic/Discrete) Fourier Transform (Aperiodic/Continuous) CT Fourier Series CT Fourier Transform DT Fourier Series DT Fourier Transform Continuous-Time Domain Sampling Discrete-Time Domain Special case by using impulse function

Two Special Signals for LTI Systems System Function System Function

System Functions H(s) or H(z)

Fourier and Beyond Observation: if one signal can be written as the linear combination of 𝑒 𝑠𝑡 𝑜𝑟 𝑧 𝑛 , we need NOT to calculate the convolution for the LTI output. When 𝑠=𝑗𝜔, 𝑧= 𝑒 𝑗𝜔 ⇒ 𝑒 𝑗𝜔𝑡 , 𝑒 𝑗𝜔𝑛 : 𝐹𝑜𝑢𝑟𝑖𝑒𝑟 𝑆𝑒𝑟𝑖𝑒𝑠 When s or z is general complex number ⇒𝐿𝑎𝑝𝑙𝑎𝑐𝑒 𝑇𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚 & 𝑍 𝑇𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚

A “Special” Class of Periodic Signals CT Fourier Series: if one CT signal can be written as follows 𝑥 𝑡 = 𝑘=−∞ +∞ 𝑎 𝑘 𝑒 𝑗𝑘 𝜔 0 𝑡 where 𝜔 0 =2𝜋/𝑇. { 𝑎 𝑘 }: Fourier series coefficients, which represent the strength of the component 𝑒 𝑗𝑘 𝜔 0 𝑡 . 𝑒 𝑗𝑘 𝜔 0 𝑡 is a signal with pure frequency 𝑘 𝜔 0 ⇒ 𝑥 𝑡 is a periodic signal with period T, it consists of components with different frequencies 𝑘 𝜔 0 and different weights.

Convergence of CT Fourier Series What kind of periodic signals have Fourier series expansion? Define 𝑒 𝑡 =𝑥 𝑡 − 𝑘=−∞ +∞ 𝑎 𝑘 𝑒 𝑗𝑘 𝜔 0 𝑡 , Fourier series expansion exists <=> ∃ {𝑎 𝑘 }, e(t)=0 (D1) Relaxation: Fourier series expansion exists <=> ∃ {𝑎 𝑘 }, 𝑇 | ​𝑒 𝑡 2 =0 (D2) 𝐷1⊂𝐷2 Two kind of sufficient conditions for D2 𝑇 | ​𝑥 𝑡 2 𝑑𝑡<∞ Dirichlet condition

Dirichlet Conditions

Almost all the periodic signal in practice have Fourier series expansion!

Inner Product of Exponential Signals Define inner product as < 𝒆 𝒋𝒌 𝝎 𝟎 𝒕 ⋅ 𝒆 𝒋𝒏 𝝎 𝟎 𝒕 > = 𝟏 𝑻 𝑻 𝒆 𝒋𝒌 𝝎 𝟎 𝒕 𝒆 𝒋𝒏 𝝎 𝟎 𝒕 ∗ 𝒅𝒕 We have < 𝒆 𝒋𝒌 𝝎 𝟎 𝒕 ⋅ 𝒆 𝒋𝒏 𝝎 𝟎 𝒕 > = 𝟏 𝐤=𝐧 < 𝒆 𝒋𝒌 𝝎 𝟎 𝒕 ⋅ 𝒆 𝒋𝒏 𝝎 𝟎 𝒕 > = 0 𝐤≠𝐧 { 𝒆 𝒋𝒌 𝝎 𝟎 𝒕 |∀𝒊𝒏𝒕𝒆𝒈𝒆𝒓 𝒌} is similar to basis of vector space

How to Obtain Fourier Coefficients 𝑥 𝑡 = 𝑘=−∞ +∞ 𝑎 𝑘 𝑒 𝑗𝑘 𝜔 0 𝑡 ⇒ 𝐴 = 𝐴 𝑥 𝑥 + 𝐴 𝑦 𝑦 + 𝐴 𝑧 𝑧 Notice 𝐴 𝑥 = 𝐴 ⋅ 𝑥 Similarly, we guess 𝑎 𝑘 =<𝑥 𝑡 ⋅ 𝑒 𝑗𝑘 𝜔 0 𝑡 > Let’s double-check <𝑥 𝑡 ⋅ 𝑒 𝑗𝑘 𝜔 0 𝑡 > = 𝑛=−∞ +∞ 𝑎 𝑛 < 𝑒 𝑗𝑛 𝜔 0 𝑡 ⋅ 𝑒 𝑗𝑘 𝜔 0 𝑡 > = 𝑎 𝑘

CT Fourier Series Pair Fourier series expansion

Example 3.5: Periodic Square Wave T ak

Example Gibbs Phenomenon: The partial sum in the vicinity of the discontinuity exhibits ripples whose amplitude does not seem to decrease with increasing N

Periodic Impulse Train    

(A Few ) Properties of CT Fourier Series real

Time Reversal Time Scaling the effect of sign change for x(t) and ak are identical Example: x(t): … a-2 a-1 a0 a1 a2 … x(-t): … a2 a1 a0 a-1 a-2 … Time Scaling positive real number periodic with period T/α and fundamental frequency αw0 ak unchanged, but x(αt) and each harmonic component are different

Power is the same whether measured in the time-domain or the frequency-domain

More Frequency shifting Differentiation Integration Note: x(t) is periodic with fundamental frequency ω0 Frequency shifting Differentiation Integration 𝑒 𝑗𝑀 𝜔 0 𝑡 𝑥 𝑡    𝑏 𝑘 = 𝑎 𝑘−𝑀 𝑑𝑥 𝑑𝑡    𝑏 𝑘 =𝑗𝑘 𝜔 0 𝑎 𝑘 −∞ 𝑡 𝑥(𝑡)𝑑𝑡    𝑏 𝑘 =( 1 𝑗𝑘 𝜔 0 ) 𝑎 𝑘