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Duality

Module by: Richard Baraniuk. E-mail the author

F⁢i⁢ω=∫−∞∞f⁢t⁢e−(i⁢ω⁢t)d t F ω t f t ω t
(1)
f⁢t=12⁢π⁢∫−∞∞F⁢i⁢ω⁢ei⁢ω⁢td ω f t 1 2 ω F ω ω t
(2)
Both formulas are symmetrical except for the ±± sign and 12⁢π 1 2 . Useful! Say f⁢t ↔ F⁢i⁢ω f t ↔ F ω through CTFT, and G⁢i⁢ω=f⁢i⁢ω G ω f ω , that is: the same form as some time-domain signal f f . Then what is g⁢t g t ?
g⁢t=12⁢π⁢∫−∞∞G⁢i⁢ω⁢ei⁢ω⁢td ω =12⁢π⁢∫−∞∞f⁢i⁢ω⁢ei⁢ω⁢td ω g t 1 2 ω G ω ω t 1 2 ω f ω ω t
(3)
Let τ=ω τ ω , and i in f⁢i⁢ω f ω doesn't matter, then:
g⁢t=12⁢π⁢∫−∞∞f⁢τ⁢ei⁢τ⁢td τ =∫−∞∞f⁢τ⁢e−(i⁢τ⁢(−t))d τ =12⁢π⁢F⁢−(i⁢t) g t 1 2 τ f τ τ t τ f τ τ t 1 2 F t
(4)
In other words: g⁢t g t takes the form of FT F F (reversed and scaled, Figure 1).

Example 1

Figure 1
Figure 1 (f1.png)

Example 2

CTFT of ∀ t :g⁢t=1 t g t 1 (Figure 2).

Figure 2
Figure 2 (f2.png)

Direct approach:

G⁢i⁢ω=∫−∞∞1⁢e−(i⁢ω⁢t)d t G ω t 1 ω t
(5)

Duality approach:

We know Figure 3.

Figure 3: f⁢t=δ⁢t f t δ t and ∀ ω :F⁢i⁢ω=1 ω F ω 1 .
Figure 3 (f3.png)
So by duality we get Figure 4.
Figure 4
Figure 4 (f4.png)

Exercise 1: FT of a complex sinusoid

f⁢t=ei⁢ ω 0 ⁢t f t ω 0 t

Exercise 2: FT of cosine

Figure 5
Figure 5 (f5.png)

Exercise 3: FT of sine

Figure 6
Figure 6 (f6.png)

Exercise 4: FT of a sinc

Figure 7
Figure 7 (f7.png)

Exercise 5: FT of unit step

Figure 8
Figure 8 (f8.png)

U⁢i⁢ω=π⁢δ⁢ω+1i⁢ω U ω δ ω 1 ω .

Figure 9
Figure 9 (f9.png)

See Lathi pp. 250-1 for derivation.

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