E-jw ejw arg e-jw cos cos arg ejw arg ejw odd function let real

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IV Discrete –Time Fourier transform DTFT.

with period 2π, since a function of a
periodic function is periodic, and has the same period. Since the forward transform is a Fourier series, the inverse transform,

x(n) = 1/2π∫ X(ejw) ejwn dw

= | X(ejw) | ejφ(w)

| X(ejw) | = √ Re2 {X(ejw)} + Im2 {X(ejw)}

Then,

X(ejw) = ∑ x(n) e-jwn = ∑δ(n) e-jwn

= 1 = X(ejw)

5

1-(e-jw )N H(ejw) = ∑ 1 • (e-jw )n =
1- e-jw

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1- cos(wN) + j sin(wN) =
1- cos(w) + j sin(w)

sin(wN) φ(w) = tan-1
1- (cos(wN)

+ π u (- (1- (cos(wN) )) sin(w)
- tan-1
1-(cos(w)

Example Find the DTFT of x(n) = .5n u(n)

X(ejw) = ∑ .5n (e-jw )n = ∑ (.5 e-jw )n

X(ejw) = 1 + e-jw

x(n) = 1 / 2π∫ (1+e-jw) ejwn dw

11 Frequency Response From Difference Equation

Shift Theorem: F{x(n–no) } = e-jwn⋅X(ejw) Proof: ∑ x(n–no) e-jwn|n ← n + no = ↑

∑ ak F{y(n-k)} = ∑ bk F{x(n-k)},

Using the shift theorem,

Properties of the DTFT

(1) X(ejw) is a periodic function of w, with period 2ππππ

13 Im {X (ejw)}

= ∑ Im {x(n)[ cos(wn)-j sin(wn)] }

| X (ejw)|2 = X (ejw) • X (ejw)*

but X(ejw)* = X (e-jw)

+ πu (-∑ x(n) cos(wn) )

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(4) let x(n) be a real, even sequence,
x(n) = x(-n). Then X (ejw) is real and Im { X (ejw) } = 0.

= x(0) + 2 ∑ x(n) cos(wn) which is real and even.

16 (5) let x(n) be odd and real, x(n) = -x(-n) x(0) = 0.

= 2j ∑ x(n) (e-jwn - ejwn) / 2j

= -2j ∑ x(n) sin(wn)

= ejwn H(ejw)

Note: ∑ x(g(n)) e-jf(w)g(n) = X(ejf(w))

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Proof: Take the Fourier Transforms of both sides as
Y (ejw) = ∑∑ h(k) x(n-k) e-jwn
e-jwk e-jw(n-k)
= ∑∑ h(k) e-jwk x(n-k) e-jw(n-k)
= ∑ h(k) e-jwk∑ x(m) e-jwm
= H(ejw) • X(ejw)

= 1/ 2π∑∑ x(n) h(m) e-jnw∫ eju(n-m) du

= ∑ x(n) h(n)e-jnw

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F { x(n) ∙ h(n) } = ∑ x(n) h(n)e-jnw

= 1/2π∫∫ X(ejv) H(eju) [ δ(u + v - w) +

δ(u+v-w-2π) du dv