EEE223 Signals and Systems

Lecture 22 DTFT

Dr. Shadan Khattak

DTFT

𝑋(𝑒𝑗𝜔) = 𝑥,𝑛-𝑒−𝑗𝜔𝑛∞𝑛=−∞ (DTFT) (Analysis equation)

𝑥 𝑛 =1

2𝜋 𝑋(𝑒𝑗𝜔)𝑒𝑗𝑛𝜔𝜋

𝜔=−𝜋𝑑𝜔 (IDTFT) (Synthesis equation)

• 𝑋 𝑒𝑗𝜔 is called the spectrum of 𝑥 𝑛 .

• 𝑋 𝑒𝑗𝜔 =

𝑋(𝑒𝑗𝜔) . 𝑒𝑗∠𝑋(𝑒𝑗𝜔)

𝑋(𝑒𝑗𝜔) : 𝐴𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒 𝑠𝑝𝑒𝑐𝑡𝑟𝑢𝑚

∠𝑋 𝑒𝑗𝜔 : 𝑃ℎ𝑎𝑠𝑒 𝑠𝑝𝑒𝑐𝑡𝑟𝑢𝑚

Dr. Shadan Khattak

DTFT

• Two major differences between CTFT and DTFT:

1. Periodicity of the DTFT 𝑋 𝑒𝑗𝜔

2. Finite interval of integration in the synthesis equation of DTFT.

• As 𝑋 𝑒𝑗𝜔 = 𝑋 𝑒𝑗(𝜔+2𝑛𝜋) , 𝜔 ∈ *−𝜋, 𝜋+ is sufficient to describe the

spectrum.

• Many texts refer to 𝑋 𝑒𝑗𝜔 as 𝑋 Ω or 𝑋 𝑒𝑗Ω . They all mean the

same thing.

Dr. Shadan Khattak

DTFT

• Signals at frequencies near even multiple of 𝜋 are slowly varying and, therefore, are thought of as low-frequency signals.

• Signals at frequencies near odd multiples of 𝜋 are rapidly varying and, therefore, are thought of as high-frequency signals.

Dr. Shadan Khattak

DTFT

1. DTFT of a unit impulse signal

𝑥 𝑛 = 𝛿,𝑛-

𝑋(𝑒𝑗𝜔) = 𝛿,𝑛-𝑒−𝑗𝜔𝑛∞𝑛=−∞

= 1

Similarly, 𝐷𝑇𝐹𝑇 𝛿 𝑛 − 𝑛0 = 𝑒−𝑗𝜔𝑛0

Dr. Shadan Khattak

DTFT

2. DTFT of a causal exponential signal 𝑥 𝑛 = 𝛼𝑛𝑢 𝑛 , 𝛼 < 1

𝑋(𝑒𝑗𝜔) = 𝛼𝑛𝑒−𝑗𝜔𝑛∞𝑛=0

= (𝛼𝑒−𝑗𝜔)𝑛∞𝑛=0

=1

1−𝛼𝑒−𝑗𝜔

𝑋(𝑒𝑗𝜔) =1

1 + 𝛼2 − 2𝛼𝑐𝑜𝑠𝜔

∠𝑋 𝑒𝑗𝜔 = −𝑡𝑎𝑛−1𝛼𝑠𝑖𝑛𝜔

1 − 𝛼𝑐𝑜𝑠𝜔

Dr. Shadan Khattak

𝛼 > 0

DTFT

2. DTFT of a causal exponential signal 𝑥 𝑛 = 𝛼𝑛𝑢 𝑛 , 𝛼 < 1

𝑋(𝑒𝑗𝜔) = 𝛼𝑛𝑒−𝑗𝜔𝑛∞𝑛=0

= (𝛼𝑒−𝑗𝜔)𝑛∞𝑛=0

=1

1−𝛼𝑒−𝑗𝜔

𝑋(𝑒𝑗𝜔) =1

1 + 𝛼2 − 2𝛼𝑐𝑜𝑠𝜔

∠𝑋 𝑒𝑗𝜔 = −𝑡𝑎𝑛−1𝛼𝑠𝑖𝑛𝜔

1 − 𝛼𝑐𝑜𝑠𝜔

Dr. Shadan Khattak

𝛼 < 0

DTFT

3. DTFT of a non-causal exponential signal

𝑥 𝑛 = 𝛼 𝑛 , 𝛼 < 1

𝑋(𝑒𝑗𝜔) =1 − 𝛼2

1 + 𝛼2 − 2𝛼𝑐𝑜𝑠𝜔

Dr. Shadan Khattak

0 < 𝛼 < 1

DTFT

4. DTFT of a rectangular pulse signal

𝑥 𝑛 = 1, 𝑛 ≤ 20, 𝑛 > 2

𝑋(𝑒𝑗𝜔) =𝑠𝑖𝑛

5𝜔2

𝑠𝑖𝑛 𝜔2

Generally, for a rectangular pulse signal 𝑥 𝑛 = 1, 𝑛 ≤ 𝑁1

0, 𝑛 > 𝑁1,

the DTFT is 𝑋(𝑒𝑗𝜔) =

𝑠𝑖𝑛𝜔(𝑁1+1/2)

𝑠𝑖𝑛 𝜔

2

Dr. Shadan Khattak

DTFT

DTFT Analysis of DT LTI Systems

• Frequency response is the DTFT of impulse response ℎ,𝑛-.

𝐻 Ω = ℎ,𝑛-𝑒−𝑗Ω𝑛∞

𝑛=−∞

• The impulse response h[n] can be recovered from the frequency response by taking its IDTFT.

ℎ 𝑛 =1

2𝜋 𝐻 Ω 𝑒𝑗Ω𝑛

𝜋

−𝜋

𝑑Ω

Dr. Shadan Khattak

DTFT

DTFT of Periodic Signals

Follows the same principle as that of CTFT of periodic signals except that DTFT of periodic signals is periodic in 𝜔 with period 2𝜋. If

𝑥 𝑛 = 𝑒𝑗𝜔0𝑛

Then

𝑋 𝑒𝑗𝜔 = 2𝜋𝛿(𝜔 − 𝜔0 − 2𝜋𝑙)

∞

𝑙=−∞

Dr. Shadan Khattak

DTFT

DTFT of Periodic Signals

DTFT from DTFS

𝑋 𝑒𝑗𝜔 = 2𝜋𝐷𝑟𝛿(𝜔 − 𝑟𝜔0)

∞

𝑟=−∞

Dr. Shadan Khattak

DTFT

DTFT of Periodic Signals

Example: Find the DTFT of cos (𝜔0𝑛) with 𝜔0 =2𝜋

5

𝑋 𝑒𝑗𝜔 = 𝜋𝛿 𝜔 −2𝜋

5− 2𝜋𝑙 + 𝜋𝛿(𝜔 +

2𝜋

5− 2𝜋𝑙)

∞

𝑙=−∞

∞

𝑙=−∞

= 𝜋𝛿 𝜔 −2𝜋

5+ 𝜋𝛿 𝜔 +

2𝜋

5, −𝜋 ≤ 𝜔 < 𝜋

(Example 5.5 (Oppenheim)

Dr. Shadan Khattak

Dr. Shadan Khattak

Dr. Shadan Khattak

Useful Readings

• Sections 5.1 – 5.2 (Oppenheim)

Dr. Shadan Khattak

Practice Problems

• Problem 5.1 – 5.9, 5.13 – 5.14, 5.21 – 5.24 (Oppenheim)

• All Examples of Chapter 5 (Oppenheim)

Dr. Shadan Khattak