EEE223 Signals and Systems Lecture 17
Transcript of EEE223 Signals and Systems Lecture 17
DTFT
π(πππ) = π₯,π-πβπππβπ=ββ (DTFT) (Analysis equation)
π₯ π =1
2π π(πππ)πππππ
π=βπππ (IDTFT) (Synthesis equation)
β’ π πππ is called the spectrum of π₯ π .
β’ π πππ =
π(πππ) . ππβ π(πππ)
π(πππ) : π΄πππππ‘π’ππ π ππππ‘ππ’π
β π πππ : πβππ π π ππππ‘ππ’π
Dr. Shadan Khattak
DTFT
β’ Two major differences between CTFT and DTFT:
1. Periodicity of the DTFT π πππ
2. Finite interval of integration in the synthesis equation of DTFT.
β’ As π πππ = π ππ(π+2ππ) , π β *βπ, π+ is sufficient to describe the
spectrum.
β’ Many texts refer to π πππ as π Ξ© or π ππΞ© . They all mean the
same thing.
Dr. Shadan Khattak
DTFT
β’ Signals at frequencies near even multiple of π are slowly varying and, therefore, are thought of as low-frequency signals.
β’ 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
π₯ π = πΏ,π-
π(πππ) = πΏ,π-πβπππβπ=ββ
= 1
Similarly, π·ππΉπ πΏ π β π0 = πβπππ0
Dr. Shadan Khattak
DTFT
2. DTFT of a causal exponential signal π₯ π = πΌππ’ π , πΌ < 1
π(πππ) = πΌππβπππβπ=0
= (πΌπβππ)πβπ=0
=1
1βπΌπβππ
π(πππ) =1
1 + πΌ2 β 2πΌπππ π
β π πππ = βπ‘ππβ1πΌπ πππ
1 β πΌπππ π
Dr. Shadan Khattak
πΌ > 0
DTFT
2. DTFT of a causal exponential signal π₯ π = πΌππ’ π , πΌ < 1
π(πππ) = πΌππβπππβπ=0
= (πΌπβππ)πβπ=0
=1
1βπΌπβππ
π(πππ) =1
1 + πΌ2 β 2πΌπππ π
β π πππ = βπ‘ππβ1πΌπ πππ
1 β πΌπππ π
Dr. Shadan Khattak
πΌ < 0
DTFT
3. DTFT of a non-causal exponential signal
π₯ π = πΌ π , πΌ < 1
π(πππ) =1 β πΌ2
1 + πΌ2 β 2πΌπππ π
Dr. Shadan Khattak
0 < πΌ < 1
DTFT
4. DTFT of a rectangular pulse signal
π₯ π = 1, π β€ 20, π > 2
π(πππ) =π ππ
5π2
π ππ π2
Generally, for a rectangular pulse signal π₯ π = 1, π β€ π1
0, π > π1,
the DTFT is π(πππ) =
π πππ(π1+1/2)
π ππ π
2
Dr. Shadan Khattak
DTFT
DTFT Analysis of DT LTI Systems
β’ Frequency response is the DTFT of impulse response β,π-.
π» Ξ© = β,π-πβπΞ©πβ
π=ββ
β’ The impulse response h[n] can be recovered from the frequency response by taking its IDTFT.
β π =1
2π π» Ξ© ππΞ©π
π
βπ
πΞ©
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 π with period 2π. If
π₯ π = πππ0π
Then
π πππ = 2ππΏ(π β π0 β 2ππ)
β
π=ββ
Dr. Shadan Khattak
DTFT
DTFT of Periodic Signals
DTFT from DTFS
π πππ = 2ππ·ππΏ(π β ππ0)
β
π=ββ
Dr. Shadan Khattak
DTFT
DTFT of Periodic Signals
Example: Find the DTFT of cos (π0π) with π0 =2π
5
π πππ = ππΏ π β2π
5β 2ππ + ππΏ(π +
2π
5β 2ππ)
β
π=ββ
β
π=ββ
= ππΏ π β2π
5+ ππΏ π +
2π
5, βπ β€ π < π
(Example 5.5 (Oppenheim)
Dr. Shadan Khattak