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1. Chapter One
Discrete-Time Signals and Systems
Lecture #4
Rediet Million
AAiT, School Of Electrical and Computer Engineering
rediet.million@aait.edu.et
March, 2018
(Rediet Million) DSP-Lecture #4 March, 2018 1 / 15

2. 1.3.LTI System and Discrete-Time Fourier Transform
1.3.2 Frequency response and Fourier Transforms
Frequency response of LTI systems
Exponential and sinusoidal sequences play a particular important role
in representing discrete time signal and system.
Complex exponential sequence are eigenfunctions of LTI systems.
Eigenfunction of LTI systems are sequences that,when input to the
system,pass through with only a change in (complex) amplitude and
phase .
If x(n) is an eigenfunction input to LTI system then the output is
y(n) = λx(n), where λ is the eigenvalue.
(Rediet Million) DSP-Lecture #4 March, 2018 2 / 15

3. Frequency response and Fourier Transforms
Frequency response of LTI systems
Signal of the form x(n) = ejwn ,for ∞ < n < ∞, are eigenfunctions of
LTI systems. This may be shown using convolution sum.
y(n) = h(n) ∗ x(n) =
∞
k=−∞
h(k)x(n − k)
y(n) =
∞
k=−∞
h(k)ejw(n−k)
= ejwn
(
∞
k=−∞
h(k)e−jwk
)
-Let us define H(ejw ) =
∞
k=−∞
h(k)e−jwk ,then the output become
y(n) = H(ejw )ejwn = λx(n)
H(ejw ) is an eigenvalue complex quantity and is called the frequency
response of the system.
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4. Frequency response and Fourier Transforms
Frequency response of LTI systems
H(ejw ) is,in general,complex-valued and depends on frequency w of
complex exponential.
It may be written in terms of its real and imaginary parts or in
terms of magnitude and phase parts.
H(ejw
) = HR(ejw
) + HI (ejw
)
H(ejw ) = |H(ejw )|ejφh(w)
|H(ejw
)| = H2
R(ejw ) + H2
I (ejw ) = H(ejw
)H∗
(ejw
)
φh(w) = tan−1
[
HI (ejw )
HR(ejw )
]
(Rediet Million) DSP-Lecture #4 March, 2018 4 / 15

5. Frequency response and Fourier Transforms
Frequency response of LTI systems
Sinusoidal response:
Let x(n) = Acos(w0n) be the input to a LTI system with a real-valued
unit sample response h(n). If x(n) is decomposed into a sum of two
complex exponential
x(n) =
A
2
ejw0n
+
A
2
e−jw0n
y(n) =
A
2
H(ejw0
)ejw0n
+
A
2
H(e−jw0
)e−jw0n
If h(n) real,then
H(e−jw0 ) = H∗(ejw0 ) = |H(ejw0 )|ejφh(w0)
y(n) = A|H(ejw0 )|[
ej(w0n+φh(w0))
2
]
y(n) = A|H(ejw0 )|cos(w0n + φh(w0))
(Rediet Million) DSP-Lecture #4 March, 2018 5 / 15

6. Frequency response and Fourier Transforms
Frequency response of LTI systems
Properties of frequency response
The frequency response is a complex-valued function of continuous
variable w and is periodic with a period 2π.
H(ej(w+2π)
) =
∞
n=−∞
h(n)e−j(w+2π)n
=
∞
n=−∞
h(n)e−jwn
e−j2πn
= H(ejw
)
- Only specified over the interval −π < w ≤ π or 0 ≤ w < 2π
Given the frequency response H(ejw ) ,the unit sample maybe recovered
by an integration:
h(n) =
1
2π
π
−π
H(ejw
)ejwn
dw
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7. Frequency response and Fourier Transforms
Frequency response of LTI systems
Example-1
Consider a simple ideal delay system defined by y(n) = x(n − N) where N
is an integer.And assume a complex sinusoidal input is x(n) = ejwn.
- The output of the delay system is :
y(n) = x(n − N) = ejw(n−N)
= e−jwNejwn
Thus,the frequency response of an ideal delay system is H(ejw ) = e−jwN
Alternatively,the frequency response maybe obtained from
H(ejw ) =
∞
n=−∞
h(n)e−jwn
note that h(n) = δ(n − N).
H(ejw ) =
∞
n=−∞
δ(n − N)e−jwn = e−jwN
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8. Frequency response and Fourier Transforms
Frequency response of LTI systems
Example-2
Consider the LTI system with unit sample response h(n) = αnu(n),where
α is a real number with |α| < 1.
- The frequency response is
H(ejw ) =
∞
n=−∞
h(n)e−jwn =
∞
n=0
αne−jwn
=
∞
n=0
(αe−jw )n =
1
1 − αe−jw
-The magnitude squared of the frequency response is
|H(ejw )|2 = H(ejw )H∗(ejw ) =
1
(1 − αe−jw )
.
1
(1 − αejw )
=
1
1 + α2 − 2α cos w
- The phase is
φh(w) = tan−1 HI (ejw )
HR(ejw )
= tan−1 −α sin w
1 − α cos w
(Rediet Million) DSP-Lecture #4 March, 2018 8 / 15

9. Frequency response and Fourier Transforms
Discrete-Time Fourier Transforms (DTFT)
The frequency domain representation of discrete-time signals and
systems may be generalized by the Fourier transform.
Many signals can be represented by a Fourier integral of the form :
x(n) =
1
2π
π
−π
X(ejw
)ejwn
dw
The integral represents x(n) as a superpostion of infinitesimally
small complex sinusoids of the form
1
2π
X(ejw )ejwndw
where X(ejw ) is the Fourier transform of x(n) , given by
X(ejw ) =
∞
n=−∞
x(n)e−jwn
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10. Frequency response and Fourier Transforms
Discrete-Time Fourier Transforms (DTFT)
The frequency response H(w) is a periodic function of w, with period
2π.
A sufficient condition for existence of the Fourier transform is that the
sequence x(n) be absolutely summable.
|X(ejw
)| = |
∞
n=−∞
x(n)e−jwn
| ≤
∞
n=−∞
|x(n)||e−jwn
| < ∞
|X(ejw
)| =
∞
n=−∞
|x(n)| < ∞
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11. Frequency response and Fourier Transforms
DTFT Properties
1. Linearity:
If
x1(n) DTFT←−−→ X1(w)
x2(n) DTFT←−−→ X2(w)
then
ax1(n) + bx2(n) DTFT←−−→ aX1(w) + bX2(w)
2.Time Shifting
If
x(n) DTFT←−−→ X(w)
then
x(n − n0) DTFT←−−→ e−jwn0
X(w)
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12. Frequency response and Fourier Transforms
DTFT Properties
3. Frequency Shifting:
If
x(n) DTFT←−−→ X(w)
then
ejw0n
x(n) DTFT←−−→ X(w − w0)
4. Time Reversal:
If
x(n) DTFT←−−→ X(w)
then
x(−n) DTFT←−−→ X(−w)
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13. Frequency response and Fourier Transforms
DTFT Properties
5. Differentiation in Frequency:
If
x(n) DTFT←−−→ X(w)
then
nx(n) DTFT←−−→ j
d
dw
X(w)
show the proof !
6. Parseval’s Theorem:
If
x(n) DTFT←−−→ X(w)
then
E =
∞
n=−∞
|x(n)|2
= x(n) =
1
2π
π
−π
|X(w)|2
dw
-The function |X(w)|2is called the energy density spectrum.
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14. Frequency response and Fourier Transforms
DTFT Properties
7.The Convolution theorem:
If
x(n) DTFT←−−→ X(w)
and
h(n) DTFT←−−→ H(w)
then
x(n) ∗ h(n) DTFT←−−→ X(w)H(w)
8.The Modulation or Windowing property:
If
x(n) DTFT←−−→ X(w)
w(n) DTFT←−−→ W (w)
then, the windowed signal would have y(n) = x(n)w(n)
Y (w) =
1
2π
π
−π
X(θ)W (w − θ)dθ
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15. Frequency response and Fourier Transforms
(#5 ) Class exercises & Assignment
1) Find the DTFT of each of the following sequences.
a. x(n) = anu(n − 5)
b. x(n) = n2nu(−n)
c. x(n) = cos(
πn
2
+
π
4
)
2) Determine the frequency & impulse response of the LTI system which
satisfy the following difference equation.
y(n) −
1
2
y(n − 1) = x(n) −
1
4
x(n − 1)
3) The input to an LTI system is
x(n) = n(
1
2
)nu(n)
and the output is
y(n) = (
1
3
)n−2u(n − 2) −
1
2
(
1
3
)n−3u(n − 3)
find the frequency response H(ejw )
(Rediet Million) DSP-Lecture #4 March, 2018 15 / 15