Continuous-Time Fourier Transform - SJTU

Continuous-Time Fourier Transform

Hongkai Xiong 熊红凯

http://min.sjtu.edu.cn

Department of Electronic Engineering Shanghai Jiao Tong University

http://ivm.sjtu.edu.cn/

𝑥𝑥𝑘𝑘[𝑛𝑛] → 𝑦𝑦𝑘𝑘[𝑛𝑛] ⟹ � 𝑎𝑎𝑘𝑘𝑥𝑥𝑘𝑘[𝑛𝑛]𝑘𝑘 → ∑ 𝑎𝑎𝑘𝑘𝑘𝑘 𝑦𝑦𝑘𝑘[𝑛𝑛]

• If we can find a set of “basic” signals, such that ▫ a rich class of signals can be represented as linear

combinations of these basic (building block) signals. ▫ the response of LTI Systems to these basic signals are both

simple and insightful

• Candidate sets of “basic” signals ▫ Unit impulse function and its delays: 𝛿𝛿(𝑡𝑡)/𝛿𝛿[𝑛𝑛] ▫ Complex exponential/sinusoid signals: 𝑒𝑒𝑠𝑠𝑡𝑡 , 𝑒𝑒𝑗𝑗𝑗𝑗𝑗/𝑧𝑧𝑛𝑛, 𝑒𝑒𝑗𝑗𝑗𝑗𝑛𝑛

LTI Systems

2

Fourier Series of CT Periodic Signals • Recall a periodic CT signal 𝑥𝑥 𝑡𝑡 ▫ with fundamental period 𝑇𝑇, and ▫ fundamental frequency 𝜔𝜔0 = 2𝜋𝜋/𝑇𝑇

• Fourier series represent 𝑥𝑥 𝑡𝑡 in terms of harmonically related complex exponentials ▫ Synthesis equation

𝑥𝑥 𝑡𝑡 = � 𝑎𝑎𝑘𝑘𝑒𝑒𝑗𝑗𝑘𝑘𝑗𝑗0𝑡𝑡∞

𝑘𝑘=−∞

= � 𝑎𝑎𝑘𝑘𝑒𝑒𝑗𝑗𝑘𝑘(2𝜋𝜋/𝑇𝑇)𝑡𝑡∞

𝑘𝑘=−∞

▫ Analysis equation

𝑎𝑎𝑘𝑘 =1𝑇𝑇�𝑥𝑥 𝑡𝑡 𝑒𝑒−𝑗𝑗𝑘𝑘𝑗𝑗0𝑡𝑡 𝑑𝑑𝑡𝑡𝑇𝑇

=1𝑇𝑇�𝑥𝑥 𝑡𝑡 𝑒𝑒−𝑗𝑗𝑘𝑘(2𝜋𝜋/𝑇𝑇)𝑡𝑡 𝑑𝑑𝑡𝑡𝑇𝑇

3

• In one period, 𝑥𝑥𝑇𝑇 𝑡𝑡 = �1, 𝑡𝑡 < 𝑇𝑇1 0, 𝑇𝑇1 < 𝑡𝑡 < 𝑇𝑇/2

• Fourier coefficients

𝑎𝑎0=1𝑇𝑇� 1𝑇𝑇1

−𝑇𝑇1𝑑𝑑𝑡𝑡 =

2𝑇𝑇1𝑇𝑇

, 𝑎𝑎𝑘𝑘 =1𝑇𝑇� 𝑒𝑒−𝑗𝑗𝑘𝑘𝑗𝑗0𝑡𝑡𝑑𝑑𝑡𝑡𝑇𝑇1

−𝑇𝑇1=

2sin (𝑘𝑘𝜔𝜔0𝑇𝑇1)𝑘𝑘𝜔𝜔0𝑇𝑇

, 𝑘𝑘 ≠ 0

• Frequency component at 𝜔𝜔𝑘𝑘 = 𝑘𝑘𝜔𝜔0 satisfies

𝑇𝑇𝑎𝑎𝑘𝑘 =2 sin 𝜔𝜔𝑇𝑇1

𝜔𝜔�𝑗𝑗=𝑘𝑘𝑗𝑗0

𝑇𝑇𝑎𝑎𝑘𝑘 thought as sampled value of envelope 𝑋𝑋 𝑗𝑗𝜔𝜔 ≜ 2 sin(𝜔𝜔𝑇𝑇1)/𝜔𝜔 at 𝜔𝜔𝑘𝑘 = 𝑘𝑘𝜔𝜔0

Motivating Example: Periodic Square Wave

4

• 𝑋𝑋 𝑗𝑗𝜔𝜔𝑘𝑘 ≜ 2 sin(𝜔𝜔𝑘𝑘𝑇𝑇1)/𝜔𝜔𝑘𝑘 for fixed 𝑇𝑇1 and increasing 𝑇𝑇

• Observation: as 𝑇𝑇 → ∞, discrete frequencies sampled more densely


5

• As 𝑻𝑻 → ∞, 𝒙𝒙𝑻𝑻 𝒕𝒕 → 𝒙𝒙 𝒕𝒕 ≜ 𝒖𝒖 𝒕𝒕 + 𝑻𝑻𝟏𝟏𝟐𝟐

− 𝒖𝒖(𝒕𝒕 − 𝑻𝑻𝟏𝟏𝟐𝟐

), a rectangular impulse

𝑥𝑥𝑇𝑇 𝑡𝑡 = � 𝑎𝑎𝑘𝑘𝑒𝑒𝑗𝑗𝑘𝑘𝑗𝑗0𝑡𝑡∞

𝑘𝑘=−∞

=1𝑇𝑇� 𝑋𝑋(𝑗𝑗𝑘𝑘𝜔𝜔0)𝑒𝑒𝑗𝑗𝑘𝑘𝑗𝑗0𝑡𝑡∞

𝑘𝑘=−∞

=12𝜋𝜋

� 𝑋𝑋 𝑗𝑗𝑘𝑘𝜔𝜔0 𝑒𝑒𝑗𝑗𝑘𝑘𝑗𝑗0𝑡𝑡𝜔𝜔0

∞

𝑘𝑘=−∞

⟹ lim𝑇𝑇→∞

𝑥𝑥𝑇𝑇 𝑡𝑡 =12𝜋𝜋

� 𝑋𝑋 𝑗𝑗𝜔𝜔 𝑒𝑒𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝜔𝜔∞

−∞

• Thus

𝑥𝑥(𝑡𝑡) =12𝜋𝜋


−∞


6

• For envelope 𝑿𝑿(𝒋𝒋𝒋𝒋)

𝑋𝑋 𝑗𝑗𝜔𝜔𝑘𝑘 = 𝑇𝑇𝑎𝑎𝑘𝑘

= � 𝑥𝑥𝑇𝑇(𝑡𝑡)𝑒𝑒−𝑗𝑗𝑘𝑘𝑗𝑗𝑡𝑡𝑑𝑑𝑡𝑡𝑇𝑇2

−𝑇𝑇2

= � 𝑥𝑥(𝑡𝑡)𝑒𝑒−𝑗𝑗𝑗𝑗𝑘𝑘𝑡𝑡𝑑𝑑𝑡𝑡𝑇𝑇2

−𝑇𝑇2

= � 𝑥𝑥(𝑡𝑡)𝑒𝑒−𝑗𝑗𝑗𝑗𝑘𝑘𝑡𝑡𝑑𝑑𝑡𝑡∞

−∞

• Thus

𝑋𝑋(𝑗𝑗𝜔𝜔) = � 𝑥𝑥(𝑡𝑡)𝑒𝑒−𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝑡𝑡∞

−∞


7


8

CT Fourier Transform of Aperiodic Signal 𝒙𝒙(𝒕𝒕)

• General strategy ▫ approximate 𝑥𝑥 𝑡𝑡 by a periodic signal 𝑥𝑥𝑇𝑇 𝑡𝑡 with infinite period 𝑇𝑇

9

𝑥𝑥 𝑡𝑡

−𝑇𝑇1 0 𝑇𝑇1

⟹ lim𝑇𝑇→∞

𝑥𝑥𝑇𝑇 𝑡𝑡 = 𝑥𝑥(𝑡𝑡) 𝑥𝑥𝑇𝑇 𝑡𝑡

−𝑇𝑇 − 𝑇𝑇2

− 𝑇𝑇1 0 𝑇𝑇1 𝑇𝑇2

𝑇𝑇

⋯ ⋯


• Step 1: for aperiodic signal 𝑥𝑥 𝑡𝑡 with sup 𝑥𝑥 𝑡𝑡 ⊂ [−𝑇𝑇1,𝑇𝑇1] construct a periodic signal 𝑥𝑥𝑇𝑇 𝑡𝑡 , for which one period is 𝑥𝑥 𝑡𝑡

𝑥𝑥𝑇𝑇 𝑡𝑡 = � 𝑥𝑥(𝑡𝑡 − 𝑘𝑘𝑇𝑇)∞

𝑘𝑘=−∞

▫ Then 𝑥𝑥 𝑡𝑡 = 𝑥𝑥𝑇𝑇 𝑡𝑡 ,∀ 𝑡𝑡 < 𝑇𝑇/2

10

𝑥𝑥𝑇𝑇 𝑡𝑡

⋯ ⋯

𝑥𝑥 𝑡𝑡



𝑇𝑇


• Step 2: determine the Fourier series representation for 𝑥𝑥𝑇𝑇 𝑡𝑡 ▫ Fourier series coefficients

𝑎𝑎𝑘𝑘 =1𝑇𝑇� 𝑥𝑥𝑇𝑇(𝑡𝑡)𝑒𝑒−𝑗𝑗𝑘𝑘𝑗𝑗0𝑡𝑡𝑑𝑑𝑡𝑡𝑇𝑇/2

−𝑇𝑇/2=

1𝑇𝑇� 𝑥𝑥(𝑡𝑡)𝑒𝑒−𝑗𝑗𝑘𝑘𝑗𝑗0𝑡𝑡𝑑𝑑𝑡𝑡∞

−∞

▫ Define the envelope of 𝑇𝑇𝑎𝑎𝑘𝑘

𝑋𝑋(𝑗𝑗𝜔𝜔) = � 𝑥𝑥(𝑡𝑡)𝑒𝑒−𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝑡𝑡∞

−∞

▫ Then

𝑥𝑥𝑇𝑇 𝑡𝑡 = � 𝑎𝑎𝑘𝑘𝑒𝑒𝑗𝑗𝑘𝑘𝑗𝑗0𝑡𝑡∞

𝑘𝑘=−∞

=1𝑇𝑇 � 𝑋𝑋 𝑗𝑗𝑘𝑘𝜔𝜔0 𝑒𝑒𝑗𝑗𝑘𝑘𝑗𝑗0𝑡𝑡 =

∞

𝑘𝑘=−∞

12𝜋𝜋 � 𝑋𝑋 𝑗𝑗𝑘𝑘𝜔𝜔0 𝑒𝑒𝑗𝑗𝑘𝑘𝑗𝑗0𝑡𝑡𝜔𝜔0

∞

𝑘𝑘=−∞

11


⋯ ⋯

𝑥𝑥 𝑡𝑡



𝑇𝑇


• Step 3: approximate 𝑥𝑥 𝑡𝑡 with 𝑥𝑥𝑇𝑇 𝑡𝑡 by increasing 𝑇𝑇 → ∞

▫ As 𝑇𝑇 → ∞ ⟹𝜔𝜔0 → 0, 𝑥𝑥𝑇𝑇 𝑡𝑡 → 𝑥𝑥 𝑡𝑡 ,𝜔𝜔0 → 𝑑𝑑𝜔𝜔,∑ → ∫

𝑥𝑥𝑇𝑇 𝑡𝑡 =12𝜋𝜋


∞

𝑘𝑘=−∞

⇓

𝑥𝑥 𝑡𝑡 = lim𝑇𝑇→∞

𝑥𝑥𝑇𝑇(𝑡𝑡) = lim𝑗𝑗0→0

12𝜋𝜋


∞

𝑘𝑘=−∞

=12𝜋𝜋


−∞

12


⋯ ⋯

𝑥𝑥 𝑡𝑡



𝑇𝑇

CT Fourier Transform Pair • Fourier transform (analysis equation)

𝑋𝑋 𝑗𝑗𝜔𝜔 = ℱ{𝑥𝑥(𝑡𝑡)} = � 𝑥𝑥(𝑡𝑡)𝑒𝑒−𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝑡𝑡∞

−∞

where 𝑋𝑋 𝑗𝑗𝜔𝜔 called spectrum of 𝑥𝑥(𝑡𝑡)

• Inverse Fourier transform (synthesis equation)

𝑥𝑥 𝑡𝑡 = ℱ−1{𝑋𝑋(𝑗𝑗𝜔𝜔)} =12𝜋𝜋� 𝑋𝑋 𝑗𝑗𝜔𝜔 𝑒𝑒𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝜔𝜔

∞

−∞

Superposition of complex exponentials at continuum of frequencies, frequency component 𝑒𝑒𝑗𝑗𝑗𝑗𝑡𝑡 has “amplitude” of 𝑋𝑋 𝑗𝑗𝜔𝜔 𝑑𝑑𝜔𝜔/2𝜋𝜋

13

CT Fourier Series of Periodic Signal 𝒙𝒙𝑻𝑻(𝒕𝒕)

• Fourier Transform of a finite duration signal 𝑥𝑥 𝑡𝑡 that is equal to 𝑥𝑥𝑇𝑇 𝑡𝑡 over one period, e.g., for 𝑡𝑡0 ≤ 𝑡𝑡 ≤ 𝑡𝑡0 + 𝑇𝑇

𝑋𝑋 𝑗𝑗𝜔𝜔 = ℱ{𝑥𝑥(𝑡𝑡)} = � 𝑥𝑥(𝑡𝑡)𝑒𝑒−𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝑡𝑡∞

−∞

• Fourier coefficients of periodic signal 𝑥𝑥𝑇𝑇 𝑡𝑡 with period 𝑇𝑇

𝑎𝑎𝑘𝑘 =1𝑇𝑇� 𝑥𝑥𝑇𝑇 𝑡𝑡 𝑒𝑒−𝑗𝑗𝑘𝑘𝑗𝑗0𝑡𝑡𝑑𝑑𝑡𝑡 =

1𝑇𝑇� 𝑥𝑥(𝑡𝑡)𝑒𝑒−𝑗𝑗𝑘𝑘𝑗𝑗0𝑡𝑡𝑑𝑑𝑡𝑡

𝑡𝑡0+𝑇𝑇

𝑡𝑡0

𝑡𝑡0+𝑇𝑇

𝑡𝑡0

=1𝑇𝑇� 𝑥𝑥(𝑡𝑡)𝑒𝑒−𝑗𝑗𝑘𝑘𝑗𝑗0𝑡𝑡𝑑𝑑𝑡𝑡∞

−∞=

1𝑇𝑇𝑋𝑋(𝑗𝑗𝜔𝜔)�

𝑗𝑗=𝑘𝑘𝑗𝑗0

⟹ proportional to samples of the Fourier transform of 𝑥𝑥 𝑡𝑡 , i.e., one period of 𝑥𝑥𝑇𝑇 𝑡𝑡

14

Convergence of Fourier Transform • Suppose from 𝑥𝑥(𝑡𝑡), we have

𝑋𝑋 𝑗𝑗𝜔𝜔 = � 𝑥𝑥(𝑡𝑡)𝑒𝑒−𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝑡𝑡∞

−∞

• Let 𝑥𝑥�(𝑡𝑡) be the Fourier transform representation of 𝑥𝑥(𝑡𝑡),

obtained using the synthesis equation

𝑥𝑥�(𝑡𝑡) =12𝜋𝜋� 𝑋𝑋 𝑗𝑗𝜔𝜔 𝑒𝑒𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝜔𝜔

∞

−∞

• Question: when can 𝑥𝑥�(𝑡𝑡) be a valid representation of the original

signal 𝑥𝑥(𝑡𝑡)?

15

Finite Energy Condition • One useful notion for engineers: ▫ no energy in approximation error

𝑒𝑒 𝑡𝑡 ≜ 𝑥𝑥 𝑡𝑡 − 𝑥𝑥� 𝑡𝑡 = 𝑥𝑥 𝑡𝑡 −12𝜋𝜋� 𝑋𝑋 𝑗𝑗𝜔𝜔 𝑒𝑒𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝜔𝜔

∞

−∞

and � 𝑒𝑒(𝑡𝑡) 2𝑑𝑑𝑡𝑡 = 0∞

−∞

• Finite energy condition (square integrable)

� 𝑥𝑥(𝑡𝑡) 2𝑑𝑑𝑡𝑡 < ∞∞

−∞

▫ Note: does not guarantee 𝑥𝑥(𝑡𝑡) equal to its Fourier representation at any time 𝑡𝑡, but guarantees no energy in approximation error

16

• Condition 1. ▫ 𝑥𝑥(𝑡𝑡) is absolutely integrable, i.e.,

� |𝑥𝑥 𝑡𝑡 |𝑑𝑑𝑡𝑡∞

−∞< ∞

• Condition 2. ▫ Within any finite interval, 𝑥𝑥(𝑡𝑡) has a finite number of maxima

and minima

• Condition 3. ▫ Within any finite interval, 𝑥𝑥(𝑡𝑡) has only a finite number of

discontinuities with finite values

• ⟹ absolutely integrable signals that are continuous or with a finite number of discontinuities have Fourier transforms

Dirichlet Conditions

17

Example: Right-sided Decaying Exponential

• 𝑥𝑥 𝑡𝑡 = 𝑒𝑒−𝑎𝑎𝑡𝑡𝑢𝑢 𝑡𝑡 ,𝑎𝑎 > 0 is a real number


−∞= � 𝑒𝑒−𝑎𝑎𝑡𝑡𝑒𝑒−𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝑡𝑡

∞

0

= −1

𝑎𝑎 + 𝑗𝑗𝜔𝜔𝑒𝑒−(𝑎𝑎+𝑗𝑗𝑗𝑗)𝑡𝑡�

0

∞

=1

𝑎𝑎 + 𝑗𝑗𝜔𝜔

𝑋𝑋 𝑗𝑗𝜔𝜔 =1

𝑎𝑎2 + 𝜔𝜔2, arg𝑋𝑋 𝑗𝑗𝜔𝜔 = − arctan

𝜔𝜔𝑎𝑎

• Observation: ▫ Magnitude spectrum: even symmetry ▫ Phase spectrum: odd symmetry

• Note: also works for complex 𝑎𝑎 with ℛℯ 𝑎𝑎 > 0

18

Example: Two-sided Decaying Exponential

• 𝑥𝑥 𝑡𝑡 = 𝑒𝑒−|𝑎𝑎|𝑡𝑡𝑢𝑢 𝑡𝑡 ,𝑎𝑎 > 0 is a real number


−∞

= � 𝑒𝑒−|𝑎𝑎|𝑡𝑡𝑒𝑒−𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝑡𝑡∞

−∞

= � 𝑒𝑒𝑎𝑎𝑡𝑡𝑒𝑒−𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝑡𝑡0

−∞+ � 𝑒𝑒−𝑎𝑎𝑡𝑡𝑒𝑒−𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝑡𝑡

∞

0

=1

𝑎𝑎 − 𝑗𝑗𝜔𝜔𝑒𝑒(𝑎𝑎−𝑗𝑗𝑗𝑗)𝑡𝑡�

−∞

0

−1

𝑎𝑎 + 𝑗𝑗𝜔𝜔𝑒𝑒−(𝑎𝑎+𝑗𝑗𝑗𝑗)𝑡𝑡�

0

∞

=1

𝑎𝑎 − 𝑗𝑗𝜔𝜔+

1𝑎𝑎 + 𝑗𝑗𝜔𝜔

=2𝑎𝑎

𝑎𝑎2 + 𝜔𝜔2

• Note: also works for complex 𝑎𝑎 with ℛℯ 𝑎𝑎 > 0

19

Example: Rectangular Pulse • 𝑥𝑥 𝑡𝑡 = 𝑢𝑢 𝑡𝑡 + 𝑇𝑇1 − 𝑢𝑢(𝑡𝑡 − 𝑇𝑇1)

• Fourier transform

𝑋𝑋 𝑗𝑗𝜔𝜔 = � 𝑒𝑒−𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝑡𝑡𝑇𝑇1

−𝑇𝑇1= 2

sin(𝜔𝜔𝑇𝑇1)𝜔𝜔

• Inverse Fourier transform

𝑥𝑥� 𝑡𝑡 =12𝜋𝜋

� 2sin(𝜔𝜔𝑇𝑇1)

𝜔𝜔𝑒𝑒𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝜔𝜔

∞

−∞

• As 𝑇𝑇1 → ∞, (define sinc 𝜃𝜃 = sin(𝜋𝜋𝜋𝜋)𝜋𝜋𝜋𝜋

)

▫ Frequency domain: 2𝜋𝜋 sin(𝑗𝑗𝑇𝑇1)𝜋𝜋𝑗𝑗

= 2𝜋𝜋 𝑇𝑇1𝜋𝜋

sinc 𝑗𝑗𝑇𝑇1𝜋𝜋

→ 2𝜋𝜋𝛿𝛿(𝜔𝜔)

▫ Time domain: 𝑥𝑥 𝑡𝑡 → 1, DC

20

1 1

1

1

Example: Ideal Lowpass Filter • Frequency response 𝐻𝐻(𝑗𝑗𝜔𝜔) = 𝑢𝑢 𝜔𝜔 + 𝜔𝜔𝑐𝑐 − 𝑢𝑢(𝜔𝜔 − 𝜔𝜔𝑐𝑐)

• Impulse response

ℎ 𝑡𝑡 =12𝜋𝜋

� 𝑒𝑒𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝜔𝜔𝑗𝑗𝑐𝑐

−𝑗𝑗𝑐𝑐

=sin(𝜔𝜔𝑐𝑐 𝑡𝑡)

𝜋𝜋𝑡𝑡=𝜔𝜔𝑐𝑐𝜋𝜋

sinc𝜔𝜔𝑐𝑐𝑡𝑡𝜋𝜋

• As 𝜔𝜔𝑐𝑐 → ∞, ▫ Time domain ℎ 𝑡𝑡 → 𝛿𝛿(𝑡𝑡), becomes identity system

▫ Frequency domain 𝐻𝐻(𝑗𝑗𝜔𝜔) → 1, passes all frequencies

21

Duality

22

Examples • Unit impulse 𝑥𝑥 𝑡𝑡 = 𝛿𝛿 𝑡𝑡

𝑋𝑋 𝑗𝑗𝜔𝜔 = � 𝛿𝛿(𝑡𝑡)𝑒𝑒−𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝑡𝑡+∞

−∞= 1

𝑥𝑥 𝑡𝑡 =12𝜋𝜋� 𝑒𝑒𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝜔𝜔

∞

−∞= lim

𝑗𝑗𝑐𝑐→∞

12𝜋𝜋� 𝑒𝑒𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝜔𝜔

𝑗𝑗𝑐𝑐

−𝑗𝑗𝑐𝑐

= lim𝑗𝑗𝑐𝑐→∞

sin(𝜔𝜔𝑐𝑐 𝑡𝑡)𝜋𝜋𝑡𝑡 = 𝛿𝛿(𝑡𝑡)

𝜹𝜹(𝒕𝒕) has “white” spectrum, equal amount of all frequencies!

• DC Signal 𝑥𝑥(𝑡𝑡) = 1

𝑋𝑋 𝑗𝑗𝜔𝜔 = lim𝑇𝑇1→∞

� 𝑒𝑒−𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝑡𝑡𝑇𝑇1

−𝑇𝑇1= lim

𝑇𝑇1→∞2𝜋𝜋

sin(𝜔𝜔𝑇𝑇1)𝜋𝜋𝜔𝜔 = 2𝜋𝜋𝛿𝛿(𝜔𝜔)

𝑥𝑥 𝑡𝑡 =12𝜋𝜋� 𝑋𝑋(𝑗𝑗𝜔𝜔)𝑒𝑒𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝜔𝜔

∞

−∞=

12𝜋𝜋� 2𝜋𝜋𝛿𝛿(𝜔𝜔)𝑒𝑒𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝜔𝜔 = 1

∞

−∞

Spectrum of DC signal is impulse at zero frequency!

23

Fourier Transform of Periodic Signals • A periodic signal 𝑥𝑥 𝑡𝑡 with fundamental frequency 𝜔𝜔0 =2𝜋𝜋/𝑇𝑇 has a Fourier series representation


𝑘𝑘=−∞

⟹ 𝑋𝑋 𝑗𝑗𝜔𝜔 = ℱ{𝑥𝑥 𝑡𝑡 } = � 𝑎𝑎𝑘𝑘ℱ{𝑒𝑒𝑗𝑗𝑘𝑘𝑗𝑗0𝑡𝑡}∞

𝑘𝑘=−∞

• Question then becomes

ℱ{𝑒𝑒𝑗𝑗𝑘𝑘𝑗𝑗0𝑡𝑡} =?

24

Fourier Transform of Periodic Signals • Using the basic Fourier transform pair

𝑒𝑒𝑗𝑗𝑗𝑗0𝑡𝑡 ℱ

2𝜋𝜋𝛿𝛿 𝜔𝜔 − 𝜔𝜔0

▫ ℱ{𝑒𝑒𝑗𝑗𝑗𝑗0𝑡𝑡} = lim𝑇𝑇1→∞

∫ 𝑒𝑒−𝑗𝑗 𝑗𝑗−𝑗𝑗0 𝑡𝑡𝑑𝑑𝑡𝑡𝑇𝑇1−𝑇𝑇1

= lim𝑇𝑇1→∞

2𝜋𝜋 sin[(𝑗𝑗−𝑗𝑗0)𝑇𝑇1]𝜋𝜋 𝑗𝑗−𝑗𝑗0

= 2𝜋𝜋𝛿𝛿(𝜔𝜔 − 𝜔𝜔0)

▫ ℱ−1{2𝜋𝜋𝛿𝛿(𝜔𝜔 − 𝜔𝜔0)} = 12𝜋𝜋 ∫ 2𝜋𝜋𝛿𝛿 𝜔𝜔 − 𝜔𝜔0 𝑒𝑒𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝜔𝜔

∞−∞ = 𝑒𝑒𝑗𝑗𝑗𝑗0𝑡𝑡

• Thus, the Fourier transform of periodic signal 𝑥𝑥(𝑡𝑡) is

𝑋𝑋 𝑗𝑗𝜔𝜔 = ℱ{𝑥𝑥 𝑡𝑡 } = � 𝑎𝑎𝑘𝑘ℱ{𝑒𝑒𝑗𝑗𝑘𝑘𝑗𝑗0𝑡𝑡}∞

𝑘𝑘=−∞

= � 2𝜋𝜋𝑎𝑎𝑘𝑘𝛿𝛿(𝜔𝜔 − 𝑘𝑘𝜔𝜔0)∞

𝑘𝑘=−∞

25

Fourier Transform of Periodic Signals • A periodic signal 𝑥𝑥 𝑡𝑡 with fundamental frequency 𝜔𝜔0 = 2𝜋𝜋/𝑇𝑇

• Fourier series


𝑘𝑘=−∞

𝑎𝑎𝑘𝑘 =1𝑇𝑇� 𝑥𝑥(𝑡𝑡)𝑒𝑒−𝑗𝑗𝑘𝑘𝑗𝑗0𝑡𝑡𝑑𝑑𝑡𝑡𝑇𝑇2

−𝑇𝑇2


𝑥𝑥(𝑡𝑡) ℱ

𝑋𝑋(𝑗𝑗𝜔𝜔） = � 2𝜋𝜋𝑎𝑎𝑘𝑘𝛿𝛿(𝜔𝜔 − 𝑘𝑘𝜔𝜔0)∞

𝑘𝑘=−∞

▫ Spectrum of periodic signal consists of a impulse train occurring at harmonically related frequencies!

▫ Areas of impulses are 𝟐𝟐𝟐𝟐 times Fourier series coefficients

26

Example: Cosine and Sine • 𝑥𝑥𝑐𝑐(𝑡𝑡) = cos(𝜔𝜔0𝑡𝑡) = 1

2𝑒𝑒𝑗𝑗𝑗𝑗0𝑡𝑡 + 1

2𝑒𝑒−𝑗𝑗𝑗𝑗0𝑡𝑡

⟹ 𝑎𝑎1= 1,𝑎𝑎−1 = 1, 𝑎𝑎𝑘𝑘 = 0，𝑘𝑘 ≠ ±1

⟹ 𝑋𝑋𝑐𝑐(𝑗𝑗𝜔𝜔) = 𝜋𝜋[𝛿𝛿(𝜔𝜔 − 𝜔𝜔0) + 𝛿𝛿(𝜔𝜔 + 𝜔𝜔0)]

• 𝑥𝑥𝑠𝑠(𝑡𝑡) = sin(𝜔𝜔0𝑡𝑡) = 12𝑗𝑗𝑒𝑒𝑗𝑗𝑗𝑗0𝑡𝑡 − 1

2𝑗𝑗𝑒𝑒−𝑗𝑗𝑗𝑗0𝑡𝑡

⟹ 𝑎𝑎1= 12𝑗𝑗

,𝑎𝑎−1 = − 12𝑗𝑗

, 𝑎𝑎𝑘𝑘 = 0，𝑘𝑘 ≠ ±1

⟹ 𝑋𝑋𝑠𝑠 𝑗𝑗𝜔𝜔 =𝜋𝜋𝑗𝑗𝛿𝛿 𝜔𝜔 − 𝜔𝜔0 −

𝜋𝜋𝑗𝑗𝛿𝛿(𝜔𝜔 + 𝜔𝜔0)

27

Example: Periodic Square Wave

𝑎𝑎0=1𝑇𝑇� 1𝑇𝑇1

−𝑇𝑇1𝑑𝑑𝑡𝑡 =

2𝑇𝑇1𝑇𝑇

, 𝑎𝑎𝑘𝑘 =1𝑇𝑇� 𝑒𝑒−𝑗𝑗𝑘𝑘𝑗𝑗0𝑡𝑡𝑑𝑑𝑡𝑡𝑇𝑇1

−𝑇𝑇1=

2sin (𝑘𝑘𝜔𝜔0𝑇𝑇1)𝑘𝑘𝜔𝜔0𝑇𝑇

, 𝑘𝑘 ≠ 0

𝑋𝑋 𝑗𝑗𝜔𝜔 = � 2𝜋𝜋2 sin 𝑘𝑘𝜔𝜔0𝑇𝑇1

𝑘𝑘𝜔𝜔0𝑇𝑇

∞

𝑘𝑘=−∞

𝛿𝛿 𝜔𝜔 − 𝑘𝑘𝜔𝜔0 = �2 sin 𝑘𝑘𝜔𝜔0𝑇𝑇1

𝑘𝑘

∞

𝑘𝑘=−∞

𝛿𝛿 𝜔𝜔 − 𝑘𝑘𝜔𝜔0

28

Example: Periodic Impulse Train • 𝑥𝑥(𝑡𝑡) = ∑ 𝛿𝛿(𝑡𝑡 − 𝑘𝑘𝑇𝑇)∞

𝑘𝑘=−∞

⟹ 𝑎𝑎𝑘𝑘=1𝑇𝑇� 𝑥𝑥 𝑡𝑡𝑇𝑇2

−𝑇𝑇2

𝑒𝑒−𝑗𝑗𝑘𝑘𝑗𝑗0𝑡𝑡𝑑𝑑𝑡𝑡 =1𝑇𝑇

⟹ 𝑋𝑋 𝑗𝑗𝜔𝜔 = �2𝜋𝜋𝑇𝑇𝛿𝛿(𝜔𝜔 − 𝑘𝑘

2𝜋𝜋𝑇𝑇

)∞

𝑘𝑘=−∞

29

Properties of CT Fourier Transform

𝑥𝑥 𝑡𝑡 ℱ

𝑋𝑋 𝑗𝑗𝜔𝜔 𝑦𝑦(𝑡𝑡) ℱ

𝑌𝑌(𝑗𝑗𝜔𝜔)

• Linearity

𝑎𝑎𝑥𝑥(𝑡𝑡) + 𝑏𝑏𝑦𝑦(𝑡𝑡) ℱ

𝑎𝑎𝑋𝑋(𝑗𝑗𝜔𝜔) + 𝑏𝑏𝑌𝑌(𝑗𝑗𝜔𝜔)

• Time shifting

𝑥𝑥(𝑡𝑡 − 𝑡𝑡0) ℱ

𝑒𝑒−𝑗𝑗𝑗𝑗𝑡𝑡0𝑋𝑋(𝑗𝑗𝜔𝜔) • Frequency shifting

𝑒𝑒𝑗𝑗𝑗𝑗0𝑡𝑡𝑥𝑥(𝑡𝑡) ℱ

𝑋𝑋(𝑗𝑗 𝜔𝜔 − 𝜔𝜔0 )

30

Example: Multipath Fading Effect Transmitter: 𝑥𝑥 𝑡𝑡 Receiver: 𝑦𝑦 𝑡𝑡 = 𝑎𝑎0𝑥𝑥 𝑡𝑡 + 𝑎𝑎1𝑥𝑥(𝑡𝑡 − 𝜏𝜏)

• 𝑌𝑌 𝑗𝑗𝜔𝜔 = 𝑎𝑎0𝑋𝑋 𝑗𝑗𝜔𝜔 + 𝑒𝑒−𝑗𝑗𝑗𝑗𝑗𝑗𝑎𝑎1𝑋𝑋 𝑗𝑗𝜔𝜔 = (𝑎𝑎0 + 𝑒𝑒−𝑗𝑗𝑗𝑗𝑗𝑗𝑎𝑎1)𝑋𝑋 𝑗𝑗𝜔𝜔 • Let 𝑥𝑥 𝑡𝑡 = 𝑢𝑢 𝑡𝑡 + 𝑇𝑇 − 𝑢𝑢 𝑡𝑡 − 𝑇𝑇

𝑋𝑋 𝑗𝑗𝜔𝜔 =2 sin(𝜔𝜔𝑇𝑇)

𝜔𝜔

𝑌𝑌 𝑗𝑗𝜔𝜔 = (𝑎𝑎0 + 𝑒𝑒−𝑗𝑗𝑗𝑗𝑗𝑗𝑎𝑎1)2 sin(𝜔𝜔𝑇𝑇)

𝜔𝜔

• When 𝑎𝑎0 = 𝑎𝑎1 = 1

𝑌𝑌 𝑗𝑗𝜔𝜔 = 4𝑒𝑒−𝑗𝑗𝑗𝑗𝑗𝑗/2 cos 𝜔𝜔𝜏𝜏/22 sin(𝜔𝜔𝑇𝑇)

𝜔𝜔

31

Example: Modulation Baseband signal: 𝑥𝑥 𝑡𝑡 Modulated signal: 𝑦𝑦 𝑡𝑡 = 𝑥𝑥 𝑡𝑡 cos(𝜔𝜔𝑐𝑐𝑡𝑡)

• 𝑦𝑦 𝑡𝑡 = 12𝑥𝑥(𝑡𝑡)(𝑒𝑒𝑗𝑗𝑗𝑗𝑐𝑐𝑡𝑡 + 𝑒𝑒−𝑗𝑗𝑗𝑗𝑐𝑐𝑡𝑡 )

𝑌𝑌 𝑗𝑗𝜔𝜔 =12𝑋𝑋 𝑗𝑗(𝜔𝜔 − 𝜔𝜔𝑐𝑐 +

12𝑋𝑋 𝑗𝑗(𝜔𝜔 + 𝜔𝜔𝑐𝑐

• Let 𝑥𝑥 𝑡𝑡 = 𝑢𝑢 𝑡𝑡 + 𝑇𝑇 − 𝑢𝑢 𝑡𝑡 − 𝑇𝑇

𝑌𝑌 𝑗𝑗𝜔𝜔 =sin((𝜔𝜔 − 𝜔𝜔𝑐𝑐)𝑇𝑇)

𝜔𝜔 − 𝜔𝜔𝑐𝑐+

sin((𝜔𝜔 + 𝜔𝜔𝑐𝑐)𝑇𝑇)𝜔𝜔 + 𝜔𝜔𝑐𝑐

32

Properties of CT Fourier Transform • Time reversal

𝑥𝑥(−𝑡𝑡) ℱ

𝑋𝑋(−𝑗𝑗𝜔𝜔)

• Conjugation

𝑥𝑥∗ 𝑡𝑡 ℱ

𝑋𝑋∗ −𝑗𝑗𝜔𝜔

Proof: ℱ 𝑥𝑥∗ 𝑡𝑡 = ∫ 𝑥𝑥∗ 𝑡𝑡 𝑒𝑒−𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝑡𝑡 =+∞−∞ ∫ 𝑥𝑥 𝑡𝑡 𝑒𝑒𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝑡𝑡+∞

−∞

∗= 𝑋𝑋∗ (−𝑗𝑗𝜔𝜔)

• Conjugate Symmetry ▫ 𝑥𝑥 𝑡𝑡 real ⟺ 𝑋𝑋(−𝑗𝑗𝜔𝜔) = 𝑋𝑋∗(𝑗𝑗𝜔𝜔) ▫ 𝑥𝑥 𝑡𝑡 even ⟺ 𝑋𝑋(𝑗𝑗𝜔𝜔) even, 𝑥𝑥 𝑡𝑡 odd ⟺ 𝑋𝑋(𝑗𝑗𝜔𝜔) odd ▫ 𝑥𝑥 𝑡𝑡 real and even ⟺ 𝑋𝑋(𝑗𝑗𝜔𝜔) real and even ▫ 𝑥𝑥 𝑡𝑡 real and odd ⟺ 𝑋𝑋(𝑗𝑗𝜔𝜔) purely imaginary and odd

33

• For a real signal: 𝑥𝑥 𝑡𝑡 ℱ

𝑋𝑋(𝑗𝑗𝜔𝜔)

𝑥𝑥𝑒𝑒 𝑡𝑡 =12 𝑥𝑥 𝑡𝑡 + 𝑥𝑥 −𝑡𝑡

ℱ ℛℯ{𝑋𝑋(𝑗𝑗𝜔𝜔)}

𝑥𝑥𝑜𝑜 𝑡𝑡 =12 𝑥𝑥 𝑡𝑡 − 𝑥𝑥 −𝑡𝑡

ℱ 𝑗𝑗 ⋅ ℐ𝓂𝓂{𝑋𝑋(𝑗𝑗𝜔𝜔)}

• Proof: ▫ Since 𝑥𝑥 𝑡𝑡 is real, then 𝑋𝑋(𝑗𝑗𝜔𝜔) is conjugate symmetric

ℱ 𝑥𝑥𝑒𝑒 𝑡𝑡 =12𝑋𝑋(𝑗𝑗𝜔𝜔) + 𝑋𝑋(−𝑗𝑗𝜔𝜔) =

12𝑋𝑋(𝑗𝑗𝜔𝜔) + 𝑋𝑋∗(𝑗𝑗𝜔𝜔) = ℛℯ 𝑋𝑋(𝑗𝑗𝜔𝜔)

ℱ 𝑥𝑥𝑜𝑜 𝑡𝑡 =12𝑋𝑋 𝑗𝑗𝜔𝜔 − 𝑋𝑋(−𝑗𝑗𝜔𝜔) =

12𝑋𝑋 𝑗𝑗𝜔𝜔 − 𝑋𝑋∗(𝑗𝑗𝜔𝜔) = 𝑗𝑗 ⋅ ℐ𝓂𝓂{𝑋𝑋(𝑗𝑗𝜔𝜔)}

Even-Odd Decomposition

34

Example For 𝑎𝑎 > 0, let

𝑥𝑥 𝑡𝑡 = �𝑒𝑒−𝑎𝑎𝑡𝑡, 𝑡𝑡 > 0−𝑒𝑒𝑎𝑎𝑡𝑡 𝑡𝑡 < 0

• Let 𝑦𝑦 𝑡𝑡 = 𝑒𝑒−𝑎𝑎𝑡𝑡𝑢𝑢(𝑡𝑡), with 𝑌𝑌 𝑗𝑗𝜔𝜔 =

1𝑎𝑎 + 𝑗𝑗𝜔𝜔

• Since 𝑥𝑥 𝑡𝑡 = 𝑦𝑦 𝑡𝑡 − 𝑦𝑦(−𝑡𝑡)

𝑋𝑋 𝑗𝑗𝜔𝜔 = 𝑌𝑌 𝑗𝑗𝜔𝜔 − 𝑌𝑌(−𝑗𝑗𝜔𝜔)

=1

𝑎𝑎 + 𝑗𝑗𝜔𝜔−

1𝑎𝑎 − 𝑗𝑗𝜔𝜔

=−2𝑗𝑗𝜔𝜔𝑎𝑎2 + 𝜔𝜔2

• 𝑥𝑥 𝑡𝑡 real and odd ⟹ 𝑋𝑋 𝑗𝑗𝜔𝜔 purely imaginary and odd

35

Time and Frequency Scaling


𝑋𝑋 𝑗𝑗𝜔𝜔

𝑥𝑥 𝑎𝑎𝑡𝑡 ℱ 1

𝑎𝑎 𝑋𝑋𝑗𝑗𝜔𝜔𝑎𝑎 ,𝑎𝑎 ≠ 0

• Proof: change variables by 𝜏𝜏 = 𝑎𝑎𝑡𝑡 ▫ For 𝑎𝑎 > 0,

� 𝑥𝑥 𝑎𝑎𝑡𝑡 𝑒𝑒−𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝑡𝑡 = � 𝑥𝑥 𝜏𝜏 𝑒𝑒−𝑗𝑗𝑗𝑗𝑎𝑎𝑗𝑗𝑑𝑑(

𝜏𝜏𝑎𝑎

) =1𝑎𝑎𝑋𝑋(𝑗𝑗𝜔𝜔𝑎𝑎

) +∞

−∞

+∞

−∞

▫ For 𝑎𝑎 < 0,

� 𝑥𝑥 𝑎𝑎𝑡𝑡 𝑒𝑒−𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝑡𝑡 = � 𝑥𝑥 𝜏𝜏 𝑒𝑒−𝑗𝑗𝑗𝑗𝑎𝑎𝑗𝑗𝑑𝑑

𝜏𝜏𝑎𝑎

= −1𝑎𝑎𝑋𝑋(𝑗𝑗𝜔𝜔𝑎𝑎

) −∞

∞

+∞

−∞

36

Time and Frequency Scaling


𝑋𝑋 𝑗𝑗𝜔𝜔 𝑥𝑥 𝑎𝑎𝑡𝑡 ℱ 1

𝑎𝑎 𝑋𝑋𝑗𝑗𝜔𝜔𝑎𝑎 ,𝑎𝑎 ≠ 0

compression in time ⟺ stretching in frequency stretching in time ⟺ compression in frequency

• Example: faster playback an audio clip ⟹ higher pitch

𝑥𝑥 𝑡𝑡 = 𝑒𝑒− 𝑎𝑎𝑡𝑡 2 ℱ

𝑋𝑋 𝑗𝑗𝜔𝜔 =𝜋𝜋𝑎𝑎𝑒𝑒−

14𝑗𝑗𝑎𝑎

2

37

Exercise • Problem: If we have the Fourier transform pair

𝑥𝑥 𝑡𝑡 ↔ 𝑋𝑋 𝑗𝑗𝜔𝜔 ,

determine the Fourier transform for the signal 𝑥𝑥(𝑎𝑎𝑡𝑡 + 𝑏𝑏).

• Answer:

ℱ 𝑥𝑥 𝑎𝑎𝑡𝑡 + 𝑏𝑏 =1

|𝑎𝑎|𝑋𝑋(𝑗𝑗𝜔𝜔𝑎𝑎 )𝑒𝑒𝑗𝑗

𝑗𝑗𝜔𝜔𝑎𝑎

38

𝑥𝑥′ 𝑡𝑡 ℱ

𝑗𝑗𝜔𝜔𝑋𝑋 𝑗𝑗𝜔𝜔 , 𝑥𝑥 𝑘𝑘 𝑡𝑡 ℱ

𝑗𝑗𝜔𝜔 𝑘𝑘𝑋𝑋 𝑗𝑗𝜔𝜔

• Proof: differentiate both sides of FT synthesis equation

𝑥𝑥 𝑡𝑡 =12𝜋𝜋

� 𝑋𝑋 𝑗𝑗𝜔𝜔 𝑒𝑒𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝜔𝜔 ⟹ ∞

−∞𝑥𝑥′ 𝑡𝑡 =

12𝜋𝜋

� 𝑗𝑗𝜔𝜔𝑋𝑋 𝑗𝑗𝜔𝜔 𝑒𝑒𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝜔𝜔∞

−∞

• Example: differentiator 𝑦𝑦 𝑡𝑡 = 𝑥𝑥′ 𝑡𝑡 , with frequency response

𝐻𝐻 𝑗𝑗𝜔𝜔 = ℱ 𝛿𝛿′ 𝑡𝑡 = � 𝛿𝛿′ 𝑡𝑡 𝑒𝑒−𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝑡𝑡+∞

−∞

= 𝛿𝛿(𝑡𝑡)𝑒𝑒−𝑗𝑗𝑗𝑗𝑡𝑡�−∞

+∞+ 𝑗𝑗𝜔𝜔� 𝛿𝛿(𝑡𝑡)𝑒𝑒−𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝑡𝑡

+∞

−∞= 𝑗𝑗𝜔𝜔

▫ 𝑌𝑌 𝑗𝑗𝜔𝜔 = 𝑋𝑋 𝑗𝑗𝜔𝜔 𝐻𝐻(𝑗𝑗𝜔𝜔) ▫ amplifies high frequencies ▫ suppresses low frequencies ▫ DC completely eliminated, cannot be completely recovered from 𝑥𝑥′ 𝑡𝑡

Differentiation

39

𝑦𝑦 𝑡𝑡 = � 𝑥𝑥 𝜏𝜏 𝑑𝑑𝜏𝜏𝑡𝑡

−∞

ℱ 𝑌𝑌 𝑗𝑗𝜔𝜔 =

1𝑗𝑗𝜔𝜔

𝑋𝑋 𝑗𝑗𝜔𝜔 + 𝜋𝜋𝑋𝑋 0 𝛿𝛿(𝜔𝜔)

• Since 𝑥𝑥 𝑡𝑡 = 𝑦𝑦′ 𝑡𝑡 , by differentiation property

𝑋𝑋 𝑗𝑗𝜔𝜔 = 𝑗𝑗𝜔𝜔𝑌𝑌 𝑗𝑗𝜔𝜔 ⟹ 𝑌𝑌 𝑗𝑗𝜔𝜔 =1𝑗𝑗𝜔𝜔

𝑋𝑋 𝑗𝑗𝜔𝜔 ,∀𝜔𝜔 ≠ 0

• Intuitively, 𝑦𝑦(𝑡𝑡) has a DC component

𝑦𝑦� = lim𝑇𝑇→∞

12𝑇𝑇

� 𝑦𝑦 𝑡𝑡 𝑑𝑑𝑡𝑡𝑇𝑇

−𝑇𝑇= lim

𝑇𝑇→∞

12𝑇𝑇

� � 𝑥𝑥 𝜏𝜏 𝑢𝑢 𝑡𝑡 − 𝜏𝜏 𝑑𝑑𝜏𝜏∞

−∞𝑑𝑑𝑡𝑡

𝑇𝑇

−𝑇𝑇

= � 𝑥𝑥 𝜏𝜏 lim𝑇𝑇→∞

12𝑇𝑇

� 𝑢𝑢(𝑡𝑡 − 𝜏𝜏)𝑑𝑑𝑡𝑡𝑇𝑇

−𝑇𝑇𝑑𝑑𝜏𝜏

∞

−∞=

12� 𝑥𝑥 𝜏𝜏 𝑑𝑑𝜏𝜏∞

−∞=

12𝑋𝑋(0)

• Therefore, at 𝜔𝜔 = 0, 𝑌𝑌 𝑗𝑗𝜔𝜔 has a component 2𝜋𝜋𝑦𝑦�𝛿𝛿 𝜔𝜔 = 𝜋𝜋𝑋𝑋 0 𝛿𝛿(𝜔𝜔)

Integration

40

• Sign function:

sgn 𝑡𝑡 = � 1, 𝑡𝑡 > 0−1, 𝑡𝑡 < 0

ℱ

2𝑗𝑗𝜔𝜔

• Constant function:

1 ℱ

2𝜋𝜋𝛿𝛿(𝜔𝜔) • Unit step function:

𝑢𝑢 𝑡𝑡 =sgn 𝑡𝑡 + 1

2 ℱ

𝑈𝑈 𝑗𝑗𝜔𝜔 =1𝑗𝑗𝜔𝜔

+ 𝜋𝜋𝛿𝛿 𝜔𝜔

• Integrator:


−∞= 𝑥𝑥 𝑡𝑡 ∗ 𝑢𝑢 𝑡𝑡

ℱ 𝑌𝑌 𝑗𝑗𝜔𝜔 = 𝑋𝑋 𝑗𝑗𝜔𝜔 𝑈𝑈 𝑗𝑗𝜔𝜔 =

1𝑗𝑗𝜔𝜔 𝑋𝑋 𝑗𝑗𝜔𝜔 + 𝜋𝜋𝑋𝑋 0 𝛿𝛿(𝜔𝜔)

Integration

41

• Sign function:

𝑥𝑥(𝑡𝑡) = �1 −2|𝑡𝑡|𝑇𝑇

, 𝑡𝑡 ≤𝑇𝑇2

0, otherwise

𝑋𝑋2 𝑗𝑗𝜔𝜔 =2𝑇𝑇

𝑒𝑒𝑗𝑗𝑗𝑗𝑇𝑇2 − 2 + 𝑒𝑒𝑗𝑗

𝑗𝑗𝑇𝑇2 = −

8𝑇𝑇

sin2𝜔𝜔𝑇𝑇4

𝑋𝑋1 𝑗𝑗𝜔𝜔 =𝑋𝑋2 𝑗𝑗𝜔𝜔𝑗𝑗𝜔𝜔

+ 𝜋𝜋𝑋𝑋2 0 𝛿𝛿(𝜔𝜔) = −8𝑗𝑗𝜔𝜔𝑇𝑇

sin2𝜔𝜔𝑇𝑇4

𝑋𝑋 𝑗𝑗𝜔𝜔 =𝑋𝑋1 𝑗𝑗𝜔𝜔𝑗𝑗𝜔𝜔

+ 𝜋𝜋𝑋𝑋1 0 𝛿𝛿(𝜔𝜔) =8

𝜔𝜔2𝑇𝑇sin2

𝜔𝜔𝑇𝑇4

Example: Triangular Pulse

42

Calculation of 𝑿𝑿(𝟎𝟎) • Method 1:

𝑋𝑋 0 = 𝑋𝑋 𝑗𝑗𝜔𝜔 �𝑗𝑗=0

• Method 2:

𝑋𝑋 0 = � 𝑥𝑥 𝑡𝑡 𝑑𝑑𝑡𝑡+∞

−∞

43

Example: 𝐬𝐬𝐬𝐬𝐬𝐬 𝒕𝒕 Function • In terms of exponential function in the limit (𝑎𝑎 > 0)

𝑥𝑥 𝑡𝑡 = �𝑒𝑒−𝑎𝑎𝑡𝑡, 𝑡𝑡 > 0−𝑒𝑒𝑎𝑎𝑡𝑡 𝑡𝑡 < 0

sgn 𝑡𝑡 = lim𝑎𝑎→0

𝑥𝑥 (𝑡𝑡)

sgn 𝑡𝑡 ℱ

lim𝒂𝒂→0

𝑋𝑋 (𝑗𝑗𝜔𝜔)＝2𝑗𝑗𝜔𝜔

• In terms of unit step functions

sgn 𝑡𝑡 = 𝑢𝑢 𝑡𝑡 − 𝑢𝑢(−𝑡𝑡)

sgn 𝑡𝑡 ℱ

[𝜋𝜋𝛿𝛿(𝜔𝜔) +1𝑗𝑗𝜔𝜔

] − [𝜋𝜋𝛿𝛿(−𝜔𝜔) +1

−𝑗𝑗𝜔𝜔] =

2𝑗𝑗𝜔𝜔

44

Example: 𝐬𝐬𝐬𝐬𝐬𝐬 𝒕𝒕 Function • Exploiting integration property

sgn 𝑡𝑡 = 2𝑢𝑢 𝑡𝑡 − 1 ≜ 𝑥𝑥 𝑡𝑡

sgn′ 𝑡𝑡 = 2𝛿𝛿(𝑡𝑡) ≜ 𝑥𝑥1 𝑡𝑡

𝑥𝑥1 𝑡𝑡 = 2𝛿𝛿 𝑡𝑡 ℱ

𝑋𝑋1 𝑗𝑗𝜔𝜔 = 2

sgn 𝑡𝑡 = 𝑥𝑥(𝑡𝑡) = 𝑥𝑥 −∞ + � 𝑥𝑥1(𝑡𝑡)𝑑𝑑𝑡𝑡𝑡𝑡

−∞

⟹ℱ sgn 𝑡𝑡 = 𝑋𝑋 𝑗𝑗𝜔𝜔 = 2𝜋𝜋𝑥𝑥 −∞ 𝛿𝛿 𝜔𝜔 +𝑋𝑋1(𝑗𝑗𝜔𝜔)𝑗𝑗𝜔𝜔

+ 𝜋𝜋𝑋𝑋1(0)𝛿𝛿(𝜔𝜔)

= 2𝜋𝜋 ⋅ −1 ⋅ 𝛿𝛿 𝜔𝜔 +2𝑗𝑗𝜔𝜔

+ 𝜋𝜋 ⋅ 2𝛿𝛿(𝜔𝜔) =2𝑗𝑗𝜔𝜔

45

Duality • Both time and frequency functions are continuous and aperiodic

in general, identical form except for ▫ different signs in exponent of complex exponential ▫ constant factor 1/2𝜋𝜋

𝑥𝑥 𝑡𝑡 =1

2𝜋𝜋� 𝑋𝑋 𝑗𝑗𝜔𝜔 𝑒𝑒𝑗𝑗𝑗𝑗𝑡𝑡∞

−∞𝑑𝑑𝜔𝜔

ℱ 𝑋𝑋 𝑗𝑗𝜔𝜔 = � 𝑥𝑥(𝑡𝑡)𝑒𝑒−𝑗𝑗𝑗𝑗𝑡𝑡

∞


• Suppose two functions are related by

𝑓𝑓 𝑟𝑟 = � 𝑔𝑔(𝜏𝜏)𝑒𝑒−𝑗𝑗𝑗𝑗𝑗𝑗∞

−∞𝑑𝑑𝜏𝜏

▫ Let 𝜏𝜏 = 𝑡𝑡, and 𝑟𝑟 = 𝜔𝜔, ⟹ 𝑔𝑔 𝑡𝑡 ℱ

𝑓𝑓 𝜔𝜔

▫ Let 𝜏𝜏 = −𝜔𝜔, and 𝑟𝑟 = 𝑡𝑡,⟹ 𝑓𝑓 𝑡𝑡 ℱ

2𝜋𝜋𝑔𝑔 −𝜔𝜔

• Duality: 𝑥𝑥 𝑡𝑡 ℱ

𝑋𝑋 𝑗𝑗𝜔𝜔 ⟺ 𝑋𝑋(𝑡𝑡) ℱ

2𝜋𝜋𝑥𝑥(−𝑗𝑗𝜔𝜔)

46

Duality

𝑢𝑢 𝑡𝑡 + 𝑎𝑎 − 𝑢𝑢 𝑡𝑡 − 𝑎𝑎 ℱ

2 sin(𝑎𝑎𝜔𝜔)

𝜔𝜔

2 sin(𝑎𝑎𝑡𝑡)

𝑡𝑡 ℱ

2𝜋𝜋[𝑢𝑢 𝜔𝜔 + 𝑎𝑎 − 𝑢𝑢 𝜔𝜔 − 𝑎𝑎 ]

47

Duality Example: 𝑥𝑥 𝑡𝑡 = 1

𝜋𝜋𝑡𝑡

• Known

sgn 𝑡𝑡 ℱ 2

𝑗𝑗𝜔𝜔

• By duality ▫ switch LHS and RHS, switch 𝜔𝜔 and 𝑡𝑡 ▫ substitute 𝜔𝜔 → −𝜔𝜔 or 𝑡𝑡 → −𝑡𝑡 (but not both) ▫ multiply RHS by 2𝜋𝜋

−2𝑗𝑗𝑡𝑡

ℱ 2𝜋𝜋 ⋅ sgn 𝜔𝜔

• By linearity 1𝜋𝜋𝑡𝑡

ℱ − 𝑗𝑗 ⋅ sgn 𝜔𝜔

48




Duality Example: 𝑥𝑥(𝑡𝑡) = 1

1+𝑡𝑡2

• Known

𝑒𝑒−|𝑡𝑡| ℱ 21 + 𝜔𝜔2

• By duality ▫ switch LHS and RHS, switch 𝜔𝜔 and 𝑡𝑡 ▫ substitute 𝜔𝜔 → −𝜔𝜔 or 𝑡𝑡 → −𝑡𝑡 (but not both) ▫ multiply RHS by 2𝜋𝜋

21 + 𝑡𝑡2

ℱ

2𝜋𝜋 ⋅ 𝑒𝑒−|𝑗𝑗|

• By linearity 1

1 + 𝑡𝑡2 ℱ

𝜋𝜋 ⋅ 𝑒𝑒−|𝑗𝑗|

49




Duality • Differentiation in frequency

−𝑗𝑗𝑡𝑡𝑥𝑥 𝑡𝑡 ℱ 𝑑𝑑𝑋𝑋 𝑗𝑗𝜔𝜔

𝑑𝑑𝜔𝜔, 𝑡𝑡𝑥𝑥 𝑡𝑡

ℱ 𝑗𝑗𝑑𝑑𝑋𝑋(𝑗𝑗𝜔𝜔)𝑑𝑑𝜔𝜔

• Proof:

𝑋𝑋 𝑗𝑗𝜔𝜔 = � 𝑥𝑥(𝑡𝑡)𝑒𝑒−𝑗𝑗𝑗𝑗𝑡𝑡∞

−∞𝑑𝑑𝑡𝑡 ⟹

𝑑𝑑𝑋𝑋 𝑗𝑗𝜔𝜔𝑑𝑑𝜔𝜔

= � (−𝑗𝑗𝑡𝑡)𝑥𝑥(𝑡𝑡)𝑒𝑒−𝑗𝑗𝑗𝑗𝑡𝑡∞


• Example: Unit ramp function 𝑢𝑢−2 𝑡𝑡 = ∫ 𝑢𝑢 𝜏𝜏 𝑑𝑑𝜏𝜏𝑡𝑡−∞ = 𝑡𝑡𝑢𝑢(𝑡𝑡)

ℱ 𝑢𝑢−2 𝑡𝑡 = 𝑗𝑗ℱ −𝑗𝑗𝑡𝑡𝑢𝑢(𝑡𝑡) = 𝑗𝑗𝑑𝑑𝑈𝑈(𝑗𝑗𝜔𝜔)𝑑𝑑𝜔𝜔

= −1𝜔𝜔2 + 𝑗𝑗𝜋𝜋𝛿𝛿′(𝜔𝜔)

• Integration in frequency

−1𝑗𝑗𝑡𝑡𝑥𝑥 𝑡𝑡 + 𝜋𝜋𝑥𝑥 0 𝛿𝛿 𝑡𝑡

ℱ � 𝑋𝑋 𝜂𝜂 𝑑𝑑𝜂𝜂

𝑗𝑗

−∞

50

Example

• Problem: 𝑥𝑥 𝑡𝑡 = 𝑡𝑡𝑒𝑒−2𝑡𝑡𝑢𝑢 𝑡𝑡 ℱ

?

• Solution:

𝑒𝑒−2𝑡𝑡𝑢𝑢 𝑡𝑡 ℱ 1

2 + 𝑗𝑗𝜔𝜔

𝑡𝑡𝑒𝑒−2𝑡𝑡𝑢𝑢 𝑡𝑡 ℱ

𝑗𝑗𝑑𝑑𝑑𝑑𝜔𝜔

12 + 𝑗𝑗𝜔𝜔 =

1(2 + 𝑗𝑗𝜔𝜔)2

• Exercise: 𝑥𝑥 𝑡𝑡 = 𝑡𝑡𝑒𝑒−2𝑡𝑡𝑢𝑢 𝑡𝑡 − 1 ↔?

51

Exercise • Determine 𝑥𝑥(𝑡𝑡) according to 𝑋𝑋 𝑗𝑗𝜔𝜔

• Hints: exploit integration property in frequency domain

𝑥𝑥1 𝑡𝑡 ℱ

𝑋𝑋′ 𝑗𝑗𝜔𝜔 ⟹ 𝑥𝑥 𝑡𝑡 = 𝑥𝑥1 𝑡𝑡−𝑗𝑗𝑡𝑡

+ 𝜋𝜋𝑥𝑥1 0 𝛿𝛿 𝑡𝑡

where 𝑥𝑥1 0 = ∫𝑋𝑋′(𝑗𝑗𝜔𝜔)𝑑𝑑𝜔𝜔 = 0

52

−𝜔𝜔1 𝜔𝜔1


1

𝑋𝑋′(𝑗𝑗𝜔𝜔)

𝜔𝜔1 −𝜔𝜔1

1/𝜔𝜔1

-1/𝜔𝜔1

• For a Fourier transform pair 𝑥𝑥 𝑡𝑡 ℱ


� 𝑥𝑥 𝑡𝑡 2∞

−∞𝑑𝑑𝑡𝑡 =

12𝜋𝜋

� 𝑋𝑋(𝑗𝑗𝜔𝜔) 2∞

−∞𝑑𝑑𝜔𝜔 = � 𝑋𝑋 𝑗𝑗𝜔𝜔 2

∞

−∞

𝑑𝑑𝜔𝜔2𝜋𝜋

• Note: 𝜔𝜔 is angular frequency and 𝑓𝑓 = 𝜔𝜔/2𝜋𝜋 is frequency

• Interpretation: Energy conservation ▫ 𝑥𝑥 𝑡𝑡 2: power, or energy per unit time (in second) ▫ 𝑋𝑋(𝑗𝑗𝜔𝜔) 2: energy per unit frequency (in Hertz), called energy-

density spectrum

• lim𝑇𝑇→∞

1𝑇𝑇 ∫ |𝑥𝑥(𝑡𝑡)|2𝑑𝑑𝑡𝑡 = 1

2𝜋𝜋𝑇𝑇 ∫ lim𝑇𝑇→∞

|𝑋𝑋(𝑗𝑗𝑗𝑗)|2

𝑇𝑇∞−∞ 𝑑𝑑𝜔𝜔

▫ lim𝑇𝑇→∞

|𝑋𝑋(𝑗𝑗𝑗𝑗)|2

𝑇𝑇 : called power-density spectrum

Parseval’s Relation

53

• For a Fourier transform pair 𝑥𝑥 𝑡𝑡 ℱ



−∞𝑑𝑑𝑡𝑡 =

12𝜋𝜋


−∞𝑑𝑑𝜔𝜔 = � 𝑋𝑋 𝑗𝑗𝜔𝜔 2

∞

−∞

𝑑𝑑𝜔𝜔2𝜋𝜋

• Proof:


−∞𝑑𝑑𝑡𝑡 = � 𝑥𝑥 𝑡𝑡 𝑥𝑥∗ 𝑡𝑡

∞

−∞𝑑𝑑𝑡𝑡 = � 𝑥𝑥 𝑡𝑡

12𝜋𝜋

� 𝑋𝑋 𝑗𝑗𝜔𝜔∞

−∞𝑒𝑒𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝜔𝜔

∗∞


= � 𝑥𝑥 𝑡𝑡12𝜋𝜋

� 𝑋𝑋∗ 𝑗𝑗𝜔𝜔∞

−∞𝑒𝑒−𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝜔𝜔

∞


=12𝜋𝜋

� 𝑋𝑋∗ 𝑗𝑗𝜔𝜔 � 𝑥𝑥 𝑡𝑡∞

−∞𝑒𝑒−𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝑡𝑡

∞


=12𝜋𝜋

� 𝑋𝑋∗ 𝑗𝑗𝜔𝜔 𝑋𝑋(𝑗𝑗𝜔𝜔)∞

−∞𝑑𝑑𝜔𝜔 =

12𝜋𝜋




54

• Example: determine the following time domain expressions:

𝐸𝐸 = � 𝑥𝑥 𝑡𝑡 |2𝑑𝑑𝑡𝑡, and 𝐷𝐷 = 𝑥𝑥′ 𝑡𝑡 𝑡𝑡=0

∞

−∞

• Solution:

𝐸𝐸 = � |𝑥𝑥 𝑡𝑡 |2𝑑𝑑𝑡𝑡∞

−∞

=12𝜋𝜋



12𝜋𝜋

⋅5𝜋𝜋4

=58

𝑔𝑔 𝑡𝑡 = 𝑥𝑥′ 𝑡𝑡 ℱ

𝐺𝐺 𝑗𝑗𝜔𝜔 = 𝑗𝑗𝜔𝜔𝑋𝑋(𝑗𝑗𝜔𝜔)

⟹ 𝐷𝐷 = 𝑔𝑔 0 =12𝜋𝜋

� 𝐺𝐺 𝑗𝑗𝜔𝜔∞


12𝜋𝜋

� 𝑗𝑗𝜔𝜔𝑋𝑋 𝑗𝑗𝜔𝜔 ∞

−∞𝑑𝑑𝜔𝜔 = 0


55

𝑋𝑋(𝑗𝑗𝜔𝜔) 𝜋𝜋

𝜋𝜋/2

-1 -0.5 1 0.5

• Example: sin 𝑊𝑊𝑡𝑡𝜋𝜋𝑡𝑡

ℱ

𝑢𝑢 𝜔𝜔 + 𝑊𝑊 − 𝑢𝑢(𝜔𝜔 −𝑊𝑊)

• By Parseval’s relation

�sin2 𝑊𝑊𝑡𝑡𝜋𝜋2𝑡𝑡2

𝑑𝑑𝑡𝑡 =12𝜋𝜋

∞

−∞� 𝑢𝑢 𝜔𝜔 + 𝑊𝑊 − 𝑢𝑢(𝜔𝜔 −𝑊𝑊) 2∞


=12𝜋𝜋

� 1𝑊𝑊

−𝑊𝑊𝑑𝑑𝜔𝜔

=𝑊𝑊𝜋𝜋


56

Example • Use Fourier transform of typical signals and Fourier

transform properties to determine the Fourier transform of the following signal

57

-1 -2 1 2

1

2

𝑥𝑥(𝑡𝑡)

Example

• Solution 1: 𝑥𝑥 𝑡𝑡 = 2𝑔𝑔2 𝑡𝑡 + 𝑢𝑢 −𝑡𝑡 − 2 + 𝑢𝑢(𝑡𝑡 − 2)

▫ 𝑔𝑔𝜏𝜏(𝑡𝑡) is the rectangular pulse with width of 𝜏𝜏 and unit magnitude

58

-1 -2 1 2

1

2

𝑥𝑥(𝑡𝑡)

1 -1

2

+

-2

1 +

2

1

Example

• Solution 2: ▫ Assume

𝑥𝑥1 𝑡𝑡 = 𝑥𝑥′ 𝑡𝑡 ℱ

𝑋𝑋1(𝑗𝑗𝜔𝜔)

▫ Then

𝑋𝑋 𝑗𝑗𝜔𝜔 = 2𝜋𝜋𝑥𝑥 −∞ 𝛿𝛿 𝜔𝜔 +𝑋𝑋1(𝑗𝑗𝜔𝜔)𝑗𝑗𝜔𝜔

+ 𝜋𝜋𝑋𝑋1 0 𝛿𝛿 𝜔𝜔

59

-1 -2 1 2

1

2

𝑥𝑥(𝑡𝑡)

-1

1

2

-2

1 2

𝑥𝑥′(𝑡𝑡)

-1 -2

Exercise • Determine the inverse Fourier transform

▫ Plot 1:

▫ Plot 2:

60

0ω− 0ω

A

|)(| ωjX )( ωjX∠

0ω− 0ω w

A

|)(| ωjX )( ωjX∠

0ω0ω−

2/π

2/π−

Note: different

phase spectrum

Exercise • Determine the Fourier transform of the following signals

▫ 𝑥𝑥(𝑡𝑡) = 𝑒𝑒−2𝑡𝑡𝑢𝑢(𝑡𝑡)

▫ 𝑥𝑥 𝑡𝑡 = 𝑒𝑒−2 𝑡𝑡−1 𝑢𝑢 𝑡𝑡

▫ 𝑥𝑥(𝑡𝑡) = 𝑒𝑒−2𝑡𝑡𝑢𝑢(𝑡𝑡 − 1)

61

Convolution Property

𝑦𝑦 𝑡𝑡 = 𝑥𝑥 𝑡𝑡 ∗ ℎ 𝑡𝑡 ℱ

𝑌𝑌 𝑗𝑗𝜔𝜔 = 𝑋𝑋 𝑗𝑗𝜔𝜔 𝐻𝐻(𝑗𝑗𝜔𝜔)

convolution in time ⟺ multiplication in frequency

• Proof:

𝑌𝑌 𝑗𝑗𝜔𝜔 = � 𝑦𝑦 𝑡𝑡 𝑒𝑒−𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝑡𝑡 = � 𝑥𝑥 𝑡𝑡 ∗ ℎ(𝑡𝑡)𝑒𝑒−𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝑡𝑡+∞

−∞

+∞

−∞

= � � 𝑥𝑥 𝜏𝜏 ℎ(𝑡𝑡 − 𝜏𝜏)+∞

−∞𝑑𝑑𝜏𝜏 𝑒𝑒−𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝑡𝑡

+∞

−∞

= � 𝑥𝑥 𝜏𝜏 � ℎ(𝑡𝑡 − 𝜏𝜏)+∞

−∞𝑒𝑒−𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝑡𝑡 𝑑𝑑𝜏𝜏

+∞

−∞

= � 𝑥𝑥 𝜏𝜏 𝑒𝑒−𝑗𝑗𝑗𝑗𝑗𝑗𝐻𝐻 𝑗𝑗𝜔𝜔 𝑑𝑑𝜏𝜏+∞

−∞= 𝐻𝐻 𝑗𝑗𝜔𝜔 � 𝑥𝑥 𝜏𝜏 𝑒𝑒−𝑗𝑗𝑗𝑗𝑗𝑗𝑑𝑑𝜏𝜏

+∞

−∞= 𝐻𝐻(𝑗𝑗𝜔𝜔)𝑋𝑋(𝑗𝑗𝜔𝜔)

62

Convolution Property • For an LTI system with ▫ impulse response ℎ(𝑡𝑡) ▫ frequency response 𝐻𝐻 𝑗𝑗𝜔𝜔 = ∫ ℎ 𝑡𝑡∞

−∞ 𝑒𝑒−𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝑡𝑡 = ℱ{ℎ 𝑡𝑡 }

▫ 𝑒𝑒𝑗𝑗𝑗𝑗𝑡𝑡 is eigenfunction associated with eigenvalue 𝐻𝐻 𝑗𝑗𝜔𝜔

𝑥𝑥 𝑡𝑡 = 𝑒𝑒𝑗𝑗𝑗𝑗𝑡𝑡 → 𝑦𝑦 𝑡𝑡 = � ℎ 𝜏𝜏 𝑒𝑒𝑗𝑗𝑗𝑗(𝑡𝑡−𝑗𝑗)𝑑𝑑𝜏𝜏∞

−∞= 𝐻𝐻 𝑗𝑗𝜔𝜔 𝑒𝑒𝑗𝑗𝑗𝑗𝑡𝑡

• Input 𝑥𝑥 𝑡𝑡 as a linear combination of complex exponentials

𝑥𝑥 𝑡𝑡 =12𝜋𝜋


−∞= lim

𝑗𝑗0→0

12𝜋𝜋


∞

𝑘𝑘=−∞

⇒ 𝑦𝑦 𝑡𝑡 = lim𝑗𝑗0→0

12𝜋𝜋

� 𝑋𝑋 𝑗𝑗𝑘𝑘𝜔𝜔0 𝐻𝐻 𝑗𝑗𝑘𝑘𝜔𝜔0 𝑒𝑒𝑗𝑗𝑘𝑘𝑗𝑗0𝑡𝑡𝜔𝜔0

∞

𝑘𝑘=−∞

=12𝜋𝜋

� 𝑋𝑋 𝑗𝑗𝜔𝜔 𝐻𝐻 𝑗𝑗𝜔𝜔 𝑒𝑒𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝜔𝜔∞

−∞=

12𝜋𝜋

� 𝑌𝑌 𝑗𝑗𝜔𝜔 𝑒𝑒𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝜔𝜔∞

−∞

63

• Rectangular pulse:

𝑥𝑥1 𝑡𝑡 = 2/𝑇𝑇 𝑢𝑢 𝑡𝑡 +𝑇𝑇4

− 𝑢𝑢(𝑡𝑡 −𝑇𝑇4

)

𝑋𝑋1 𝑗𝑗𝜔𝜔 = 2/𝑇𝑇 ⋅2 sin(𝜔𝜔𝑇𝑇/4)

𝜔𝜔

• Triangular pulse:

𝑥𝑥(𝑡𝑡) = 1 −2|𝑡𝑡|𝑇𝑇

𝑢𝑢 𝑡𝑡 +𝑇𝑇2

− 𝑢𝑢(𝑡𝑡 −𝑇𝑇2

)

𝑋𝑋1 𝑗𝑗𝜔𝜔 = 𝑋𝑋12 𝑗𝑗𝜔𝜔 =8 sin2(𝜔𝜔𝑇𝑇/4)

𝜔𝜔2𝑇𝑇

Example

64

Frequency Response of LTI Systems • Fully characterized by impulse response ℎ(𝑡𝑡)

𝑦𝑦 𝑡𝑡 = 𝑥𝑥 𝑡𝑡 ∗ ℎ 𝑡𝑡

• Also fully characterized by frequency response 𝐻𝐻 𝑗𝑗𝜔𝜔 =ℱ{ℎ 𝑡𝑡 }, if 𝐻𝐻 𝑗𝑗𝜔𝜔 is well defined ▫ BIBO stable system: ∫ ℎ 𝑡𝑡 𝑑𝑑𝑡𝑡 < ∞+∞

−∞ ▫ other systems: e.g., identity ℎ 𝑡𝑡 = 𝛿𝛿(𝑡𝑡), differentiator ℎ 𝑡𝑡 = 𝛿𝛿′(𝑡𝑡)

• Convolution property implies ▫ instead of computing 𝑥𝑥 𝑡𝑡 ∗ ℎ 𝑡𝑡 in time domain, we can analyze

a system in frequency domain 𝑦𝑦 𝑡𝑡 = 𝑥𝑥 𝑡𝑡 ∗ ℎ(𝑡𝑡)

⇓

𝑌𝑌 𝑗𝑗𝜔𝜔 = 𝑋𝑋 𝑗𝑗𝜔𝜔 ⋅ 𝐻𝐻 𝑗𝑗𝜔𝜔 , 𝐻𝐻 𝑗𝑗𝜔𝜔 =𝑌𝑌 𝑗𝑗𝜔𝜔𝑋𝑋 𝑗𝑗𝜔𝜔

= 𝐻𝐻 𝑗𝑗𝜔𝜔 𝑒𝑒𝑗𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗

65

Frequency Response of LTI Systems • Some typical systems:

66

• Example: Differentiation property

𝑦𝑦 𝑡𝑡 = 𝑥𝑥′ 𝑡𝑡 = 𝑥𝑥 𝑡𝑡 ∗ 𝛿𝛿′ 𝑡𝑡

𝑌𝑌 𝑗𝑗𝜔𝜔 = 𝑋𝑋 𝑗𝑗𝜔𝜔 ⋅ ℱ 𝛿𝛿′ 𝑡𝑡 = 𝑗𝑗𝜔𝜔 ⋅ 𝑋𝑋(𝑗𝑗𝜔𝜔)

• Example. Integration property


−∞= 𝑥𝑥 𝑡𝑡 ∗ 𝑢𝑢 𝑡𝑡

𝑌𝑌 𝑗𝑗𝜔𝜔 = 𝑋𝑋 𝑗𝑗𝜔𝜔 ⋅ 𝑈𝑈(𝑗𝑗𝜔𝜔) = 𝑋𝑋 𝑗𝑗𝜔𝜔1𝑗𝑗𝜔𝜔

+ 𝜋𝜋𝛿𝛿 𝜔𝜔

=1𝑗𝑗𝜔𝜔

𝑋𝑋 𝑗𝑗𝜔𝜔 + 𝜋𝜋𝑋𝑋 0 𝛿𝛿 𝜔𝜔

Examples

67

• Unit ramp function: 𝑢𝑢−2 𝑡𝑡 = 𝑢𝑢 𝑡𝑡 ∗ 𝑢𝑢(𝑡𝑡) = 𝑡𝑡𝑢𝑢(𝑡𝑡)

• Convolution property suggests

ℱ 𝑢𝑢−2 𝑡𝑡 = 𝑈𝑈2 𝑗𝑗𝜔𝜔 =1𝑗𝑗𝜔𝜔

+ 𝜋𝜋𝛿𝛿 𝜔𝜔2

= −1𝜔𝜔2 +

2𝜋𝜋𝑗𝑗𝜔𝜔

𝛿𝛿 𝜔𝜔 + 𝜋𝜋2𝛿𝛿 𝜔𝜔 𝛿𝛿(𝜔𝜔)

▫ But 𝛿𝛿(𝑗𝑗)𝑗𝑗𝑗𝑗

and 𝛿𝛿 𝜔𝜔 𝛿𝛿(𝜔𝜔) not well-defined!

▫ Convolution property not applicable here ▫ Rule of thumb: applicable when formula is well-defined

• Known from the differentiation in frequency property

ℱ 𝑢𝑢−2 𝑡𝑡 = 𝑗𝑗ℱ −𝑗𝑗𝑡𝑡𝑢𝑢(𝑡𝑡) = 𝑗𝑗𝑑𝑑𝑈𝑈(𝑗𝑗𝜔𝜔)𝑑𝑑𝜔𝜔

= −1𝜔𝜔2 + 𝑗𝑗𝜋𝜋𝛿𝛿′(𝜔𝜔)

Example

68

Example • Determine the Response of an LTI system with impulse response ℎ 𝑡𝑡 = 𝑒𝑒−𝑎𝑎𝑡𝑡𝑢𝑢(𝑡𝑡), to the input 𝑥𝑥(𝑡𝑡) = 𝑒𝑒−𝜔𝜔𝑡𝑡𝑢𝑢(𝑡𝑡), where 𝑎𝑎, 𝑏𝑏 > 0

• Method 1: convolution in time domain 𝑦𝑦 𝑡𝑡 = 𝑥𝑥 𝑡𝑡 ∗ ℎ 𝑡𝑡

• Method 2: solving the differential equation with initial rest condition 𝑦𝑦′ 𝑡𝑡 + 𝑎𝑎𝑦𝑦 𝑡𝑡 = 𝑒𝑒−𝜔𝜔𝑡𝑡𝑢𝑢(𝑡𝑡)

• Method 3: Fourier transform

𝐻𝐻 𝑗𝑗𝜔𝜔 =1

𝑎𝑎 + 𝑗𝑗𝜔𝜔,𝑋𝑋 𝑗𝑗𝜔𝜔 =

1𝑏𝑏 + 𝑗𝑗𝜔𝜔

⟹ 𝑌𝑌 𝑗𝑗𝜔𝜔 =1

(𝑎𝑎 + 𝑗𝑗𝜔𝜔)(𝑏𝑏 + 𝑗𝑗𝜔𝜔)

▫ If 𝑎𝑎 ≠ 𝑏𝑏,𝑌𝑌 𝑗𝑗𝜔𝜔 = 1𝜔𝜔−𝑎𝑎

1𝑎𝑎+𝑗𝑗𝑗𝑗

− 1𝜔𝜔+𝑗𝑗𝑗𝑗

⟹ 𝑦𝑦 𝑡𝑡 = 1𝜔𝜔−𝑎𝑎

𝑒𝑒−𝑎𝑎𝑡𝑡 − 𝑒𝑒−𝜔𝜔𝑡𝑡 𝑢𝑢(𝑡𝑡)

▫ If 𝑎𝑎 = 𝑏𝑏,𝑌𝑌 𝑗𝑗𝜔𝜔 = 𝑗𝑗 𝑑𝑑𝑑𝑑𝑗𝑗

1𝑎𝑎+𝑗𝑗𝑗𝑗

⟹ 𝑦𝑦 𝑡𝑡 = 𝑡𝑡 ⋅ ℱ−1{ 1𝑎𝑎+𝑗𝑗𝑗𝑗

} = 𝑡𝑡𝑒𝑒−𝑎𝑎𝑡𝑡𝑢𝑢(𝑡𝑡)

69

Example • Determine the Response of an LTI system with impulse response ℎ 𝑡𝑡 = 𝑒𝑒−𝑎𝑎𝑡𝑡𝑢𝑢(𝑡𝑡), to the input 𝑥𝑥 𝑡𝑡 = cos(𝜔𝜔0𝑡𝑡), where 𝑎𝑎 > 0

• Frequency response:


𝑎𝑎 + 𝑗𝑗𝜔𝜔=

𝑎𝑎 − 𝑗𝑗𝜔𝜔𝑎𝑎2 + 𝜔𝜔2 =

1𝑎𝑎2 + 𝜔𝜔2

𝑒𝑒−𝑗𝑗⋅arc𝑗an(𝑗𝑗/𝑎𝑎)

• Method 1: using eigenfunction property

𝑥𝑥 𝑡𝑡 =12𝑒𝑒𝑗𝑗𝑗𝑗0𝑡𝑡 +

12𝑒𝑒−𝑗𝑗𝑗𝑗0𝑡𝑡

𝑦𝑦 𝑡𝑡 =12𝐻𝐻 𝑗𝑗𝜔𝜔0 𝑒𝑒𝑗𝑗𝑗𝑗0𝑡𝑡 +

12𝐻𝐻 −𝑗𝑗𝜔𝜔0 𝑒𝑒−𝑗𝑗𝑗𝑗0𝑡𝑡 = ℛℯ 𝐻𝐻 𝑗𝑗𝜔𝜔0 𝑒𝑒𝑗𝑗𝑗𝑗0𝑡𝑡

=𝑎𝑎 cos(𝜔𝜔0𝑡𝑡) + 𝜔𝜔0sin(𝜔𝜔0𝑡𝑡)

𝑎𝑎2 + 𝜔𝜔02= 1/ 𝑎𝑎2 + 𝜔𝜔02 cos 𝜔𝜔0𝑡𝑡 − arctan

𝜔𝜔𝑎𝑎

70

Example • Determine the Response of an LTI system with impulse response ℎ 𝑡𝑡 = 𝑒𝑒−𝑎𝑎𝑡𝑡𝑢𝑢(𝑡𝑡), to the input 𝑥𝑥 𝑡𝑡 = cos(𝜔𝜔0𝑡𝑡), where 𝑎𝑎 > 0

• Frequency response:


𝑎𝑎 + 𝑗𝑗𝜔𝜔=

𝑎𝑎 − 𝑗𝑗𝜔𝜔𝑎𝑎2 + 𝜔𝜔2 =

1𝑎𝑎2 + 𝜔𝜔2

𝑒𝑒−𝑗𝑗⋅arc𝑗an(𝑗𝑗/𝑎𝑎)

• Method 2: using Fourier transform

𝑥𝑥 𝑡𝑡 =12𝑒𝑒𝑗𝑗𝑗𝑗0𝑡𝑡 +

12𝑒𝑒−𝑗𝑗𝑗𝑗0𝑡𝑡 ⟹ 𝑋𝑋 𝑗𝑗𝜔𝜔 = 𝜋𝜋𝛿𝛿 𝜔𝜔 − 𝜔𝜔0 + 𝜋𝜋𝛿𝛿(𝜔𝜔 + 𝜔𝜔0)

𝑌𝑌 𝑗𝑗𝜔𝜔 = 𝑋𝑋(𝑗𝑗𝜔𝜔)𝐻𝐻 𝑗𝑗𝜔𝜔 = 𝜋𝜋𝐻𝐻 𝑗𝑗𝜔𝜔0 𝛿𝛿 𝜔𝜔 − 𝜔𝜔0 + 𝜋𝜋𝐻𝐻 −𝑗𝑗𝜔𝜔0 𝛿𝛿(𝜔𝜔 + 𝜔𝜔0)

⟹ 𝑦𝑦 𝑡𝑡 =12𝐻𝐻 𝑗𝑗𝜔𝜔0 𝑒𝑒𝑗𝑗𝑗𝑗0𝑡𝑡 +

12𝐻𝐻 −𝑗𝑗𝜔𝜔0 𝑒𝑒−𝑗𝑗𝑗𝑗0𝑡𝑡

71

Example • Ideal lowpass filter with impulse response ℎ 𝑡𝑡 to input 𝑥𝑥 𝑡𝑡

ℎ 𝑡𝑡 =sin(𝜔𝜔𝑐𝑐𝑡𝑡)

𝜋𝜋𝑡𝑡, 𝑥𝑥 𝑡𝑡 =

sin(𝜔𝜔𝑖𝑖𝑡𝑡)𝜋𝜋𝑡𝑡

• Fourier transform:

𝑋𝑋 𝑗𝑗𝜔𝜔 = 𝑢𝑢 𝜔𝜔 + 𝜔𝜔𝑖𝑖 − 𝑢𝑢(𝜔𝜔 − 𝜔𝜔𝑖𝑖)

𝐻𝐻 𝑗𝑗𝜔𝜔 = 𝑢𝑢 𝜔𝜔 + 𝜔𝜔𝑐𝑐 − 𝑢𝑢(𝜔𝜔 − 𝜔𝜔𝑐𝑐)

• Use 𝑌𝑌 𝑗𝑗𝜔𝜔 = 𝑋𝑋(𝑗𝑗𝜔𝜔)𝐻𝐻 𝑗𝑗𝜔𝜔

𝑌𝑌 𝑗𝑗𝜔𝜔 = �𝑋𝑋 𝑗𝑗𝜔𝜔 , if 𝜔𝜔𝑖𝑖 ≤ 𝜔𝜔𝑐𝑐𝐻𝐻 𝑗𝑗𝜔𝜔 , if 𝜔𝜔𝑖𝑖 > 𝜔𝜔𝑐𝑐

⟹ 𝑦𝑦 𝑡𝑡 = �𝑥𝑥(𝑡𝑡), if 𝜔𝜔𝑖𝑖 ≤ 𝜔𝜔𝑐𝑐ℎ(𝑡𝑡), if 𝜔𝜔𝑖𝑖 > 𝜔𝜔𝑐𝑐

• Note: Convolution of two 𝑠𝑠𝑠𝑠𝑛𝑛𝑠𝑠() functions is another 𝑠𝑠𝑠𝑠𝑛𝑛𝑠𝑠() function

72

𝑦𝑦 = (𝑥𝑥 ∗ ℎ1) ∗ ℎ2 = 𝑥𝑥 ∗ (ℎ1 ∗ ℎ2) = (𝑥𝑥 ∗ ℎ2) ∗ ℎ1 ℎ(𝑡𝑡) = ℎ1(𝑡𝑡) ∗ ℎ2(𝑡𝑡) ℎ(𝑡𝑡) = ℎ2(𝑡𝑡) ∗ ℎ1(𝑡𝑡) Note: Order of processing usually not important for LTI systems

System Connections: Cascade

73

ℎ1(𝑡𝑡) 𝑥𝑥(𝑡𝑡) 𝑦𝑦(𝑡𝑡) ℎ2(𝑡𝑡) h(𝑡𝑡)

ℎ2(𝑡𝑡) 𝑥𝑥(𝑡𝑡) 𝑦𝑦(𝑡𝑡) ℎ1(𝑡𝑡) h(𝑡𝑡)

commutative

𝑌𝑌 = 𝑋𝑋 ⋅ 𝐻𝐻1 ⋅ 𝐻𝐻2= 𝑋𝑋 ⋅ 𝐻𝐻1 ⋅ 𝐻𝐻2 = 𝑋𝑋 ⋅ 𝐻𝐻2 ⋅ 𝐻𝐻1 𝐻𝐻(𝑗𝑗𝜔𝜔) = 𝐻𝐻1(𝑗𝑗𝜔𝜔)𝐻𝐻2(𝑗𝑗𝜔𝜔)

𝐻𝐻(𝑗𝑗𝜔𝜔) = 𝐻𝐻2(𝑗𝑗𝜔𝜔)𝐻𝐻1(𝑗𝑗𝜔𝜔) Note: Order of processing usually not important for LTI systems

System Connections: Cascade

74

𝐻𝐻1(𝑗𝑗𝜔𝜔) 𝑥𝑥(𝑡𝑡) 𝑦𝑦(𝑡𝑡) 𝐻𝐻2(𝑗𝑗𝜔𝜔) 𝐻𝐻(𝑗𝑗𝜔𝜔)

𝐻𝐻2(𝑗𝑗𝜔𝜔) 𝑥𝑥(𝑡𝑡) 𝑦𝑦(𝑡𝑡) 𝐻𝐻1(𝑗𝑗𝜔𝜔)

commutative

𝐻𝐻(𝑗𝑗𝜔𝜔)

System Connections: Parallel • Time domain

ℎ 𝑡𝑡 = ℎ1 𝑡𝑡 + ℎ2(𝑡𝑡)

• Frequency domain

𝐻𝐻 𝑗𝑗𝜔𝜔 = 𝐻𝐻1 𝑗𝑗𝜔𝜔 + 𝐻𝐻2(𝑗𝑗𝜔𝜔)

75

ℎ1(𝑡𝑡) 𝑦𝑦(𝑡𝑡)

ℎ2(𝑡𝑡) 𝑥𝑥(𝑡𝑡) +

ℎ 𝑡𝑡

𝐻𝐻1(𝑗𝑗𝜔𝜔) 𝑦𝑦(𝑡𝑡)

𝐻𝐻2(𝑗𝑗𝜔𝜔) 𝑥𝑥(𝑡𝑡) +

𝐻𝐻 𝑗𝑗𝜔𝜔

System Connections: Feedback • Time domain 𝑦𝑦 = ℎ1 ∗ 𝑥𝑥 − ℎ2 ∗ 𝑦𝑦 ℎ 𝑡𝑡 = ?

• Frequency domain

𝑌𝑌 = 𝐻𝐻1𝑋𝑋 − 𝐻𝐻1𝐻𝐻2𝑌𝑌

𝐻𝐻 =𝑌𝑌𝑋𝑋 =

𝐻𝐻11 + 𝐻𝐻1𝐻𝐻2

76

ℎ1(𝑡𝑡) 𝑦𝑦(𝑡𝑡)

ℎ2(𝑡𝑡)

𝑥𝑥(𝑡𝑡)

ℎ 𝑡𝑡

+ +

-

𝐻𝐻1(𝑗𝑗𝜔𝜔) 𝑦𝑦(𝑡𝑡)

𝐻𝐻2(𝑗𝑗𝜔𝜔)

𝑥𝑥(𝑡𝑡)

𝐻𝐻 𝑗𝑗𝜔𝜔

+ +

-

Filtering • changes relative amplitudes of frequency components, or eliminates

some frequency components entirely

• LPS:

• HPS:

• BPS:

77


𝜔𝜔𝑐𝑐 −𝜔𝜔𝑐𝑐

1

𝜔𝜔

passband stopband cutoff frequency



𝐻𝐻(𝑗𝑗𝜔𝜔) = �1, 𝜔𝜔 ≤ 𝜔𝜔𝑐𝑐0, 𝜔𝜔 > 𝜔𝜔𝑐𝑐

𝐻𝐻(𝑗𝑗𝜔𝜔) = �1, 𝜔𝜔 ≥ 𝜔𝜔𝑐𝑐0, 𝜔𝜔 < 𝜔𝜔𝑐𝑐

𝐻𝐻(𝑗𝑗𝜔𝜔) = �1, 𝜔𝜔𝑐𝑐1 < 𝜔𝜔 < 𝜔𝜔𝑐𝑐20, otherwise

𝜔𝜔𝑐𝑐 −𝜔𝜔𝑐𝑐 𝜔𝜔

𝜔𝜔𝑐𝑐1 −𝜔𝜔𝑐𝑐1 𝜔𝜔 −𝜔𝜔𝑐𝑐2

lower cutoff frequency

𝜔𝜔𝑐𝑐2

Ideal Zero Phase LPF • Ideal lowpass filter (unit magnitude, zero phase)

ℎ 𝑡𝑡 =sin(𝜔𝜔𝑐𝑐𝑡𝑡)

𝜋𝜋𝑡𝑡, 𝐻𝐻 𝑗𝑗𝜔𝜔 = �1, 𝜔𝜔 ≤ 𝜔𝜔𝑐𝑐

0, otherwise

• 𝒋𝒋𝒄𝒄: cutoff frequency

78

Simple RC Lowpass Filter • Differential Equation

𝑅𝑅𝑅𝑅𝑑𝑑𝑣𝑣𝑐𝑐 𝑡𝑡𝑑𝑑𝑡𝑡 + 𝑣𝑣𝑐𝑐 𝑡𝑡 = 𝑣𝑣𝑠𝑠(𝑡𝑡)

• Input 𝑣𝑣𝑠𝑠 𝑡𝑡 , output 𝑣𝑣𝑐𝑐 𝑡𝑡

• Taking Fourier transform 𝑗𝑗𝜔𝜔𝑅𝑅𝑅𝑅 ⋅ 𝑉𝑉𝑐𝑐 𝑗𝑗𝜔𝜔 + 𝑉𝑉𝑐𝑐 𝑗𝑗𝜔𝜔 = 𝑉𝑉𝑠𝑠(𝑗𝑗𝜔𝜔) • Frequency response:

𝐻𝐻 𝑗𝑗𝜔𝜔 =𝑉𝑉𝑐𝑐 𝑗𝑗𝜔𝜔𝑉𝑉𝑠𝑠 𝑗𝑗𝜔𝜔

=1

1 + 𝑗𝑗𝜔𝜔𝑅𝑅𝑅𝑅

79

Simple RC Lowpass Filter • Frequency response


1 + 𝑅𝑅𝑅𝑅𝑗𝑗𝜔𝜔

|𝐻𝐻 𝑗𝑗𝜔𝜔 | =1

1 + 𝑅𝑅𝑅𝑅𝜔𝜔 2

arg𝐻𝐻(𝑗𝑗𝜔𝜔) = − arctan 𝑅𝑅𝑅𝑅𝜔𝜔

• Impulse response

ℎ 𝑡𝑡 =1𝑅𝑅𝑅𝑅

𝑒𝑒−𝑡𝑡/𝑅𝑅𝑅𝑅𝑢𝑢(𝑡𝑡)

• Nonideal lowpass filter ▫ passes lower frequencies ▫ attenuates higher frequencies

• Larger RC ⟹ passes smaller range of lower frequencies

80

• Ideal lowpass filter (unit magnitude, linear phase)

𝐻𝐻 𝑗𝑗𝜔𝜔 = �𝑒𝑒−𝑗𝑗𝑗𝑗𝑡𝑡0 , 𝜔𝜔 ≤ 𝜔𝜔𝑐𝑐

0, otherwise

⟹ |𝐻𝐻(𝑗𝑗𝜔𝜔)| = �1, |𝜔𝜔| ≤ 𝜔𝜔𝑐𝑐0, |𝜔𝜔| > 𝜔𝜔𝑐𝑐

, ≮ 𝐻𝐻(𝑗𝑗𝜔𝜔) = �−𝜔𝜔𝑡𝑡0 𝜔𝜔 ≤ 𝜔𝜔𝑐𝑐 0, otherwise

• 𝑌𝑌 𝑗𝑗𝜔𝜔 = 𝑒𝑒−𝑗𝑗𝑗𝑗𝑡𝑡0𝑋𝑋𝐿𝐿𝐿𝐿 𝑗𝑗𝜔𝜔 ⟹ 𝑦𝑦 𝑡𝑡 = 𝑥𝑥𝐿𝐿𝐿𝐿(𝑡𝑡 − 𝑡𝑡0) ▫ Linear phase ⟺ simply a shift in time, no distortion ▫ Nonlinear phase ⟺ distortion in addition to time shift

Ideal Linear Phase LPF

81

𝜔𝜔𝑐𝑐 −𝜔𝜔𝑐𝑐

≮ 𝐻𝐻(𝑗𝑗𝜔𝜔) 𝜔𝜔𝑐𝑐𝑡𝑡0

−𝜔𝜔𝑐𝑐𝑡𝑡0

Example • Frequency response of an LTI system is

𝐻𝐻 𝑗𝑗𝜔𝜔 = �𝑗𝑗𝜔𝜔3𝜋𝜋

, 𝜔𝜔 ≤ 3𝜋𝜋

0, otherwise

• Considered as a cascade of an LPF and a differentiator

82

𝑥𝑥(𝑡𝑡) 𝑦𝑦(𝑡𝑡) 𝐻𝐻2(𝑗𝑗𝜔𝜔) = 𝑗𝑗𝜔𝜔


𝐻𝐻1(𝑗𝑗𝜔𝜔)

1/3𝜋𝜋

3𝜋𝜋 −3𝜋𝜋

• Dual of convolution property:

𝑥𝑥 𝑡𝑡 ⋅ 𝑦𝑦 𝑡𝑡 ℱ

12𝜋𝜋

[𝑋𝑋(𝑗𝑗𝜔𝜔) ∗ 𝑌𝑌(𝑗𝑗𝜔𝜔)]

multiplication in time ⟺ convolution in frequency

• Proof: let 𝑍𝑍 𝑗𝑗𝜔𝜔 = 12𝜋𝜋 ∫ 𝑋𝑋 𝑗𝑗𝜃𝜃 𝑌𝑌(𝑗𝑗(𝜔𝜔 − 𝜃𝜃))𝑑𝑑𝜃𝜃+∞

−∞

ℱ−1 𝑍𝑍 𝑗𝑗𝜔𝜔 =12𝜋𝜋

�12𝜋𝜋

� 𝑋𝑋 𝑗𝑗𝜃𝜃 𝑌𝑌(𝑗𝑗(𝜔𝜔 − 𝜃𝜃))𝑑𝑑𝜃𝜃+∞

−∞𝑒𝑒𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝜔𝜔

+∞

−∞

=12𝜋𝜋

� 𝑋𝑋 𝑗𝑗𝜃𝜃12𝜋𝜋

� 𝑌𝑌(𝑗𝑗(𝜔𝜔 − 𝜃𝜃))𝑒𝑒𝑗𝑗𝑗𝑗𝑡𝑡𝑑𝑑𝜔𝜔+∞

−∞𝑑𝑑𝜃𝜃

+∞

−∞

=12𝜋𝜋

� 𝑋𝑋 𝑗𝑗𝜃𝜃 𝑦𝑦 𝑡𝑡 𝑒𝑒𝑗𝑗𝜋𝜋𝑡𝑡𝑑𝑑𝜃𝜃 = 𝑥𝑥 𝑡𝑡 𝑦𝑦(𝑡𝑡)+∞

−∞

Multiplication Property

83

Example • Define a signal as

𝑥𝑥 𝑡𝑡 =sin 𝑡𝑡 ⋅ sin(𝑡𝑡/2)

𝜋𝜋𝑡𝑡2≜ 𝜋𝜋𝑥𝑥1 𝑡𝑡 𝑥𝑥2 𝑡𝑡

where,

𝑥𝑥1 𝑡𝑡 =sin 𝑡𝑡𝜋𝜋𝑡𝑡

𝑥𝑥2 𝑡𝑡 =sin(𝑡𝑡/2)

𝜋𝜋𝑡𝑡


𝑋𝑋 𝑗𝑗𝜔𝜔 =12𝑋𝑋1 𝑗𝑗𝜔𝜔 ∗ 𝑋𝑋2(𝑗𝑗𝜔𝜔)

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Example: Modulation Baseband signal: 𝑥𝑥 𝑡𝑡 Carrier:

𝑝𝑝 𝑡𝑡 = cos(𝜔𝜔𝑐𝑐𝑡𝑡) =12

(𝑒𝑒𝑗𝑗𝑗𝑗𝑐𝑐𝑡𝑡 + 𝑒𝑒−𝑗𝑗𝑗𝑗𝑐𝑐𝑡𝑡 )

𝑃𝑃 𝑗𝑗𝜔𝜔 = 𝜋𝜋𝛿𝛿 𝜔𝜔 − 𝜔𝜔𝑐𝑐 + 𝜋𝜋𝛿𝛿 𝜔𝜔 + 𝜔𝜔𝑐𝑐

• Modulated signal: 𝑦𝑦 𝑡𝑡 = 𝑥𝑥(𝑡𝑡)𝑝𝑝 𝑡𝑡

𝑌𝑌 𝑗𝑗𝜔𝜔 =12𝑋𝑋 𝑗𝑗(𝜔𝜔 − 𝜔𝜔𝑐𝑐) +

12𝑋𝑋 𝑗𝑗(𝜔𝜔 + 𝜔𝜔𝑐𝑐)

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Example: Demodulation • Modulated signal: 𝑦𝑦 𝑡𝑡 = 𝑥𝑥 𝑡𝑡 𝑝𝑝 𝑡𝑡

• Carrier: 𝑝𝑝 𝑡𝑡 = cos(𝜔𝜔𝑐𝑐𝑡𝑡)

• Demodulated signal: 𝑧𝑧 𝑡𝑡 = 𝑦𝑦 𝑡𝑡 𝑝𝑝 𝑡𝑡

𝑍𝑍 𝑗𝑗𝜔𝜔 =12𝑌𝑌 𝑗𝑗(𝜔𝜔 − 𝜔𝜔𝑐𝑐) +

12𝑌𝑌 𝑗𝑗(𝜔𝜔 + 𝜔𝜔𝑐𝑐)

=12𝑋𝑋 𝑗𝑗𝜔𝜔 +

14

[𝑋𝑋(𝑗𝑗 𝜔𝜔 − 2𝜔𝜔𝑐𝑐 ) + 𝑋𝑋(𝑗𝑗 𝜔𝜔 + 2𝜔𝜔𝑐𝑐 )] 𝑅𝑅 𝑗𝑗𝜔𝜔 = 𝑍𝑍 𝑗𝑗𝜔𝜔 𝐻𝐻lowpass(𝑗𝑗𝜔𝜔) 𝑥𝑥 𝑡𝑡 = 𝑟𝑟 𝑡𝑡

86

Ideal AM Communication System • Amplitude modulation

87

LCCDEs • Determine the frequency response of an LTI system described by

�𝑎𝑎𝑘𝑘𝑑𝑑𝑘𝑘𝑦𝑦(𝑡𝑡)𝑑𝑑𝑡𝑡𝑘𝑘

𝑁𝑁

𝑘𝑘=0

= �𝑏𝑏𝑘𝑘𝑑𝑑𝑘𝑘𝑥𝑥(𝑡𝑡)𝑑𝑑𝑡𝑡𝑘𝑘

𝑀𝑀

𝑘𝑘=0

• Method 1: using the eigenfunction property

let 𝑥𝑥 𝑡𝑡 = 𝑒𝑒𝑗𝑗𝑗𝑗𝑡𝑡 → 𝑦𝑦 𝑡𝑡 = 𝐻𝐻 𝑗𝑗𝜔𝜔 𝑒𝑒𝑗𝑗𝑗𝑗𝑡𝑡

substitution in to the differential equation yields

�𝑎𝑎𝑘𝑘𝐻𝐻 𝑗𝑗𝜔𝜔 𝑗𝑗𝜔𝜔 𝑘𝑘𝑒𝑒𝑗𝑗𝑗𝑗𝑡𝑡𝑁𝑁

𝑘𝑘=0

= �𝑏𝑏𝑘𝑘 𝑗𝑗𝜔𝜔 𝑘𝑘𝑒𝑒𝑗𝑗𝑗𝑗𝑡𝑡𝑀𝑀

𝑘𝑘=0

𝐻𝐻 𝑗𝑗𝜔𝜔 =∑ 𝑏𝑏𝑘𝑘 𝑗𝑗𝜔𝜔 𝑘𝑘𝑀𝑀𝑘𝑘=0

∑ 𝑎𝑎𝑘𝑘 𝑗𝑗𝜔𝜔 𝑘𝑘𝑁𝑁𝑘𝑘=0

88

LCCDEs • Method 2: taking the Fourier transform of both sides

ℱ �𝑎𝑎𝑘𝑘𝑑𝑑𝑘𝑘𝑦𝑦(𝑡𝑡)𝑑𝑑𝑡𝑡𝑘𝑘

𝑁𝑁

𝑘𝑘=0

= ℱ �𝑏𝑏𝑘𝑘𝑑𝑑𝑘𝑘𝑥𝑥(𝑡𝑡)𝑑𝑑𝑡𝑡𝑘𝑘

𝑀𝑀

𝑘𝑘=0

• By linearity and differentiation in time property

�𝑎𝑎𝑘𝑘 𝑗𝑗𝜔𝜔 𝑘𝑘𝑌𝑌(𝑗𝑗𝜔𝜔) =𝑁𝑁

𝑘𝑘=0

�𝑏𝑏𝑘𝑘 𝑗𝑗𝜔𝜔 𝑘𝑘𝑋𝑋(𝑗𝑗𝜔𝜔) 𝑀𝑀

𝑘𝑘=0

𝐻𝐻 𝑗𝑗𝜔𝜔 =𝑌𝑌 𝑗𝑗𝜔𝜔𝑋𝑋(𝑗𝑗𝜔𝜔)

=∑ 𝑏𝑏𝑘𝑘 𝑗𝑗𝜔𝜔 𝑘𝑘𝑀𝑀𝑘𝑘=0

∑ 𝑎𝑎𝑘𝑘 𝑗𝑗𝜔𝜔 𝑘𝑘𝑁𝑁𝑘𝑘=0

• 𝐻𝐻 𝑗𝑗𝜔𝜔 is a rational function of (𝑗𝑗𝜔𝜔), i.e., ratio of polynomials

89

Example 𝑦𝑦′′ 𝑡𝑡 + 4𝑦𝑦′′ 𝑡𝑡 + 3𝑦𝑦 𝑡𝑡 = 𝑥𝑥′ 𝑡𝑡 + 2𝑥𝑥(𝑡𝑡)

• Frequency response

𝐻𝐻 𝑗𝑗𝜔𝜔 =𝑗𝑗𝜔𝜔 + 2

𝑗𝑗𝜔𝜔 2 + 4 𝑗𝑗𝜔𝜔 + 3

▫ Using partial fraction expansion

𝐻𝐻 𝑗𝑗𝜔𝜔 =𝑗𝑗𝜔𝜔 + 2

𝑗𝑗𝜔𝜔 + 1 (𝑗𝑗𝜔𝜔 + 3)=

𝐴𝐴1𝑗𝑗𝜔𝜔 + 1

+𝐴𝐴2

𝑗𝑗𝜔𝜔 + 3

▫ Let 𝑣𝑣 = 𝑗𝑗𝜔𝜔

𝐴𝐴1= 𝑣𝑣 + 1 𝐻𝐻 𝑣𝑣 �𝑣𝑣=−1

=12

,𝐴𝐴2 = 𝑣𝑣 + 3 𝐻𝐻 𝑣𝑣 �𝑣𝑣=−3

=12

▫ Taking inverse Fourier transform to find impulse response

ℎ 𝑡𝑡 =12𝑒𝑒−𝑡𝑡 +

12𝑒𝑒−3𝑡𝑡 𝑢𝑢 𝑡𝑡

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Example 𝑦𝑦′′ 𝑡𝑡 + 4𝑦𝑦′′ 𝑡𝑡 + 3𝑦𝑦 𝑡𝑡 = 𝑥𝑥′ 𝑡𝑡 + 2𝑥𝑥(𝑡𝑡)

• Determine zero-state response to 𝑥𝑥 𝑡𝑡 = 𝑒𝑒−𝑡𝑡𝑢𝑢 𝑡𝑡

𝑌𝑌 𝑗𝑗𝜔𝜔 = 𝐻𝐻 𝑗𝑗𝜔𝜔 𝑋𝑋 𝑗𝑗𝜔𝜔 =𝑗𝑗𝜔𝜔 + 2

𝑗𝑗𝜔𝜔 + 1 𝑗𝑗𝜔𝜔 + 3 ⋅1

𝑗𝑗𝜔𝜔 + 1=

𝐴𝐴1,1𝑗𝑗𝜔𝜔 + 1 +

𝐴𝐴1,2𝑗𝑗𝜔𝜔 + 1 2 −

𝐴𝐴2𝑗𝑗𝜔𝜔 + 3

▫ Using partial fraction expansion, let 𝑣𝑣 = 𝑗𝑗𝜔𝜔

𝐴𝐴1,1=1

2 − 1 !𝑑𝑑𝑑𝑑𝑣𝑣

𝑣𝑣 + 1 2𝑌𝑌 𝑣𝑣 �𝑣𝑣=−1

=𝑑𝑑𝑑𝑑𝑣𝑣

𝑣𝑣 + 2𝑣𝑣 + 3

�𝑣𝑣=−1

=14

𝐴𝐴1,2= 𝑣𝑣 + 1 2𝑌𝑌 𝑣𝑣 �𝑣𝑣=−1

=12

,𝐴𝐴2 = 𝑣𝑣 + 3 𝑌𝑌 𝑣𝑣 �𝑣𝑣=−3

= −14

▫ Taking inverse Fourier transform to find output

𝑦𝑦 𝑡𝑡 =14𝑒𝑒−𝑡𝑡 +

12𝑡𝑡𝑒𝑒−𝑡𝑡 −

14𝑒𝑒−3𝑡𝑡 𝑢𝑢 𝑡𝑡

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Continuous-Time Fourier Transform - SJTU