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Homework & Tutorial Problems

TutorialI Property of DTFT: 5.24I Difference equation of LTI systems: 5.36

HomeworkI 5.23, 5.29, 5.33

Signals & Systems DT Fourier Transform P1

Convolution Property & LTI Systems (1/2)

Let h[n] be the impulse response of certain LTI system

The output y [n] of input x [n] is given by y [n] = x [n] ∗ h[n]

For input signal x [n] = e jωn,

y [n] =∞∑

k=−∞

h[k]e jω(n−k) = e jωn∞∑

k=−∞

h[k]e−jωk︸︷︷︸Frequency Response H(e jω)

For periodic input signal x [n] =∑

k=<N>

akejk( 2π

N )n:

y [n] =∑

k=<N>

akH(e jk(2πN ))e jk(

2πN )n

bk = akH(e jk(2πN ))


Convolution Property & LTI Systems (2/2)

Convolution Property

If y [n] = x1[n] ∗ x2[n], then

Y (e jω) = X1(e jω)X2(e jω)

For general input signal x [n]:

y [n] = x [n] ∗ h[n]←→ Y (e jω) = X (e jω)H(e jω)

Observation: It’s easier to evaluate LTI systems in frequency domain

Drawback: Not every LTI system has frequency responseI h[n] = anu[n] (a > 1)I Stable LTI system has frequency response, because

∞∑n=−∞

|h[n]| < ∞


Example: Ideal Low-Pass Filter (1/2)

What’s ideal low-pass filter?

Perfectly maintain the low-frequency component

Perfectly cancel the high-frequency component


Example: Ideal Low-Pass Filter (2/2)

Impulse response:

h[n] =1

2π

π∫−π

H(e jω)e jωndω =1

2π

ωc∫−ωc

e jωndω =sinωcn

πn

Pros: no distortion in frequency domain

Cons: non-causal

See textbook, Example 5.12


Example: Band-Stop Filter (1/2)

Given low-pass filter Hlp(e jω), how to generate band-stop effect fromlow-pass filter?


Example: Band-Stop Filter (2/2)

Two branches: low-pass + high-pass

(−1)n = e jπn ⇒W1(e jω) = X (e j(ω−π))

⇒W2(e jω) = Hlp(e jω)X (e j(ω−π))

⇒W3(e jω) = Hlp(e j(ω−π))X (e j(ω−2π)) = Hlp(e j(ω−π))X (e jω)

See textbook, Example 5.14


Multiplication Property

Multiplication Property

Let y [n] = x1[n]x2[n], then

Y (e jω) =1

2π

∫2π

X1(e jθ)X2(e j(ω−θ))dθ,

which is periodic convolution of X1(e jω) and X2(e jω)

Comparison: Multiplication of periodic signals

x1[n]←→ ak and x2[n]←→ bk

⇒ x1[n]x2[n]←→∑

k=<N>

akbn−k discrete-time periodic convolution


Example: Sampling on Discrete-Time Signals

In digital signal processing, we may sample the discrete-time signals

Example: image, audio, video compression

Its mathematical model is give below:


Review: Discrete-Time Impulse Chain

What’s the Fourier transform of x [n] =+∞∑

k=−∞δ[n − kN] ?

First of all, we calculate the Fourier series:

ak =1

N

∑n=<N>

x [n]e−jk2πN n

=1

N

∑n=<N>

+∞∑k=−∞

δ[n − kN]e−jk2πN n

=1

N

∑n=<N>

δ[n]e−jk2πN n =

1

N

Then, we have

X (e jω) =∞∑

l=−∞

N−1∑k=0

2π

Nδ(ω − k(2π/N)− 2πl) =

2π

N

∞∑k=−∞

δ(ω − 2πk

N)

Time domain period × Frequency domain period = ?


Example: Frequency Analysis

P(e jω) =2π

N

∞∑k=−∞

δ(ω − kωs) where ωs = 2π/N

Xp(e jω) =1

2π

∫2π

P(e jθ)X (e j(ω−θ))dθ =1

N

N−1∑k=0

X (e j(ω−kωs ))


Duality in DTFS

Analysis and synthesis equations of DTFS share a similar form:

x [n] =∑

k=<N>

akejk(2π/N)n

ak =1

N

∑k=<N>

x [n]e−jk(2π/N)n (1)

Every of DTFS has a dual (See Table 3.2)I Time shift v.s. Frequency shiftI Time multiplication v.s. Frequency multiplication

If g [n]←→ f [k], then f [n]←→ 1N g [−k]


Dualtiy Between CTFS and DTFT

DTFT transform pair is given by

x [n] = 12π

∫2π

X (e jω)e jωndω

X (e jω) =∑∞

n=−∞ x [n]e−jωn

CTFS transform pair is given by

x(t) =∞∑

k=−∞

akejkω0t

ak =1

T

∫T

x(t)e−jkω0tdt

Frequency Domain of DTFT ⇔ Time Domain of CTFS

Frequency Domain of CTFS ⇔ Time Domain of DTFT

Discrete-Time Low-pass Filter v.s. Continuous-Time Square Wave


Summary of Duality

See textbook, Table 5.3


LTI by Difference Equation

A number of DT LTI systems can be written as the following linearconstant-coefficient difference equation

N∑k=0

aky [n − k] =M∑k=0

bkx [n − k]

What’s the frequency response?

Taking Fourier transform on both side, we have∑Nk=0 ake

−jkωY (e jω) =∑M

k=0 bke−jkωX (e jω)

⇒ H(e jω) = Y (e jω)X (e jω) =

∑Mk=0 bke

−jkω∑Nk=0 ake

−jkω

What’s the impulse response?


Example: Difference Equation

Please calculate the frequency and impulse response of the following LTI systems

y [n]− 3

4y [n − 1] +

1

8y [n − 2] = 2x [n].

SolutionAccording to the last slide,

H(e jω) =2

1− 34e−jω + 1

8e−2jω .

Moreover, by partial fraction expansion

H(e jω) =4

1− 12e−jω −

2

1− 14e−jω ⇒ h[n] = 4(

1

2)nu[n]− 2(

1

4)nu[n]


Discrete Fourier Transform (DFT)In practice, there is a huge demand on processing finite duration signals

Given a finite duration signal {x [0], x [1], ..., x [N − 1]}, its Fourier transform is

X (e jω) =N−1∑n=0

x [n]e−jωn

Drawback: The spectrum of DTFT is continuous ⇒ Cannot be handle bycomputer.

Discrete Fourier Transfrom (DFT) is developed for digital processing of finiteduration signals

X [k] =1

N

N−1∑n=0

x [n]e−jk(2π/N)n k =< N >

DFT: frequency sampling of DTFT

X [k] =1

NX (e j2kπ/N) k =< N >


Inverse DFT

Equation system of DFT:(X [0]

X [1]...

X [N − 1]

)= 1

N

e−j0 e−j0 ... e−j0

e−j0 e−j2π/N ... e−j2(N−1)π/N

... ... ... ...

e−j0 e−j2(N−1)π/N ... e−j2(N−1)(N−1)π/N

︸︷︷︸

F

(x[0]x[1]...

x[N − 1]

)

Transform matrix F is full rank.

Observation: {X [k]|k =< N >} maintains all the information of{x [0], x [1], ..., x [N − 1]}Inverse DFT is feasible:

x [n] =N−1∑k=0

X [k]e jk(2π/N)n n = 0, 1, ...,N − 1


DFT and DTFS

DFT is for finite duration signals; DTFS is for periodic signals

Define x [n] as the periodic extension of x [n]: Repeat x [n] with period N

Fourier series of x [n]:

1

N

N−1∑n=0

x [n]e−jk(2π/N)n =1

N

N−1∑n=0

x [n]e−jk(2π/N)n ⇒ DFT of x [n]

DFT of a finite-duration signal = DTFS of its periodic extension

Reference on DFT:I Textbook: Problem 5.53, 5.54I http://en.wikipedia.org/wiki/Discrete Fourier transform


Comparison


Periodic Convolution of Finite Duration Signals

Periodic ConvolutionLet x and y be two finite duration signals with duration N, x and y be theassociated periodic extension, then the periodic convolution of finite durationsignals is defined as

x [n] ~ y [n] := x [n] ~ y [n] =∑

k=<N>

x [k]y [n − k] n = 0, 1, 2...,N − 1

Convoluation Property of DFT

Time domain periodic convolution is equivalent to frequency domainmultiplication, thus,

x [n] ~ y [n]←→ NX [k]Y [k]


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