Final Exam Preparation

EEE238: Digital Signal Processing – Tutorial 1: Sampling and Quantization 1/8

1. Aliased signals


2. Prefiltering and reconstruction


Solution: Case (a)


Solution: Case (b)

Solution: Case (c)


Solution: Case (c) – continuation


Solution: Case (c) – continuation


3. Discrete Time Fourier Transform

Calculate frequency spectrum of the sampled signal.

Solution:


4. Quantization (specifications)

Solution:

4. Quantization (truncation)

Solution:

EEE238: Digital Signal Processing – Tutorial 2: Discrete Time Systems 1/7

1. Time invariance

Example 3.2.2: Test the time invariance of the discrete-time systems defined by

y(n)= nx(n) and the downsampler y(n)= x(2n).


2. Time invariance and linearity

Solution:


3. Impulse response of FIR filters


4. Impulse response of IIR filters


Solution:


5. Difference equation

Solution:


6. Stability

For each impulse response above determine if the LTI system is stable or not, based on

Solution:

EEE238: Digital Signal and Image Processing – Tutorial 3: FIR filtering and Convolution 1/5

1. Convolution table

Solution:

2. Sample processing (pp. 156-157)



3. Sample processing

Solution: The impulse response is read off from the coefficients of the I/O equation: h=[1,0,-1]


4. Convolution of infinite sequences


5. Index range

Solution:

EEE238: Digital Signal and Image Processing – Tutorial 4: z-Transforms 1/6

1. Transform properties

Solution:


2. Transform and Region of Convergence (ROC)

Solution:


3. Transform and Region of Convergence (ROC)

Solution:

4. Inverse transform

Solution:


5. Inverse transform (partial fraction expansions)

Solution:


Solution:


6. Transform of sinusoids

Solution:

EEE238: Digital Signal and Image Processing – Tutorial 5: Transfer Functions 1/7

1. Frequency response




2. Transient response


3. Pole/Zero Design

Solution:



3. Transfer function

Solution:

EEE238: Digital Signal and Image Processing – Tutorial 6: DFT/FFT Algorithms 1/5

1. Frequency analysis


2. Discrete Fourier Transform (DFT)

LN

LN

N

L

NN

N

NNN

NNN

WWW

WWW

WWW

A

A

)1)(1()1(0

)1(10

000

xX

with Nj

N eW

2

Solution:


Solution:


Solution:


Solution:

EEE238: Digital Signal and Image Processing – Tutorial 7: FIR & IIR Digital Filter Design 1/3

1. FIR Digital Filter Design (impulse responses)

By performing the appropriate integrations in the inverse DTFT equation, calculate the impulse response d(k) of the bandpass, lowpass and highpass ideal filters

.


2. FIR Digital Filter Design (using slides 21-23 of Lecture 7)


3. IIR Digital Filter Design

Solution:

EEE238: Digital Signal and Image Processing – Tutorial 8: Enhancement and Mathematical Morphology 1/3

1. Consider the 3-bit image A of Figure 1. For the questions below, assume that pixel values outside

the image boundaries have a zero grey level. In addition, round the output image pixel values to

the nearest integer (nint[] operator).

132

564

213

A

Figure 1: Image A.

By processing the input image A, determine the output

(a) image B resulting from a linear contrast scaling;

031

674

103

132

564

213

:Solution

14.1nint124.1nint14.1nint

:Example

14.1nint15

7nint1

16

12nint)min(

)min()max(

1nint

:Equation

1313

3

BA

AB

AAAAAAA

LB

scalingcontrastlinear

ijijijijij

(b) image C resulting from a thresholding at standard deviation of A;

011

111

101

132

564

213

:Solution

6330.13)(because1

:Example

otherwise0

if1

1.6330

)31()33()32()35()36()34()32(31)33(33

1

1,3132564213

33

11

:Equations

)(

3232

222222222

,

2

,

CA

AC

AC

AMN

AMN

AAngthresholdi

A

Aij

ij

A

A

ji

AijA

ji

ijA


(c) image C resulting from a 3x3 smoothing;

222

232

222

00000

0

0

0

132

564

213

0

0

0

00000

:Solution

21.5556nint9

64013000nint

:Example

1nint

:Equation

smoothing33

11

1

1

1

1

,

CA

C

CMN

Cm n

njmiij

(d) image D resulting from a 3x3 median filtering.

020

132

020

00000

0

0

0

132

564

213

0

0

0

00000

:Solution

3)6,5,4,3,3,2,1,,1(median

:Example

filteringmedian33

22

DA

D


2. Consider the binary image A and the structuring element B of Figure 2.

010

111

010

,

000000

000100

001110

011100

001000

000000

BA

Figure 2: Binary image and structuring element.

(a) Determine the morphological dilation of A by B;

𝐴⨁𝐵 = ⋃ (𝐵)𝑎𝑎𝜖𝐴 where (B)a is the structuring element translated by a

Solution:

000100

001110

011111

111110

011100

001000

,

010

111

010

,

000000

000100

001110

011100

001000

000000

BABA

(b) Determine the morphological erosion of A by B.

𝐴 ⊖ 𝐵 = 𝑧 |(𝐵)𝑧 ⊆ 𝐴

Solution:

000000

000000

000100

001000

000000

000000

,

010

111

010

,

000000

000100

001110

011100

001000

000000

BABA

Midterm Examination (Lecture Section 2, Fall 2015), EEE 238 DIGITAL SIGNAL AND IMAGE PROCESSING

Page 1 of 5

NAZARBAYEV UNIVERSITY

School of Engineering

Midterm Examination

(Lecture Section 2, Fall 2015)

EEE 238 DIGITAL SIGNAL AND IMAGE PROCESSING

Date: Monday, October 19, 2015 Time: 11:00 AM – 12:50 PM

Student name:

Student ID:

All questions carry indicated marks, you can obtain a maximum of 100 marks.

Only handwritten notes are permitted.

The use of calculators approved by the School of Engineering is permitted.

PLEASE WRITE YOUR ANSWERS ON THIS EXAMINATION

QUESTION PAPER. IT WILL BE COLLECTED AT THE END OF

THIS EXAMINATION.


Page 2 of 5

1. Determine the impulse response ℎ(𝑛) of the filters:

(a) 𝑦(𝑛) = 𝑥(𝑛) + 𝑥(𝑛 − 1) [5 marks]

(b) 𝑦(𝑛) = 𝑥(𝑛) + 2𝑥(𝑛 − 1) + 4𝑥(𝑛 − 2) [5 marks]

Answers:

(a)

ℎ(𝑛) = 𝛿(𝑛) + 𝛿(𝑛 − 1)

(b)

ℎ(𝑛) = 𝛿(𝑛) + 2𝛿(𝑛 − 1) + 4𝛿(𝑛 − 2)

2. Compute the convolution, 𝐲 = 𝐡 ∗ 𝐱, of the filter impulse response 𝐡 and input signal 𝐱:

𝐡 = [3 2 1], 𝐱 = [ 1 2 1 0 1 3 ]

using the convolution table method. [10 marks]

Answer:

h\x 1 2 1 0 1 3

3 3 6 3 0 3 9

2 2 4 2 0 2 6

1 1 2 1 0 1 3

𝐲 = [3, 8, 8, 4, 4, 11, 7, 3]


Page 3 of 5

3. Determine the Discrete Time Fourier Transform (DTFT) of

(a) 𝑥(𝑛) = (0.5)𝑛 𝑢(𝑛) [10 marks]

(b) 𝑥(𝑛) = 𝛿(𝑛 − 3) [5 marks]

(c) 𝑥(𝑛) = ((0.4)𝑛 𝑢(𝑛)) ∗ 𝛿(𝑛 − 2) [5 marks]

(d) 𝑥(𝑛) = ((0.3)𝑛 𝑢(𝑛)) 𝑤(𝑛) [5 marks]

where 𝑢(𝑛) and 𝛿(𝑛) are the step function and the unit impulse function, respectively. Note

that 𝑤(𝑛) is an arbitrary window. For questions (a) and (b), justify mathematically your

answer by using the DTFT equation: 𝑋(𝜔) = ∑ 𝑥(𝑛)𝑒−𝑗𝜔𝑛∞𝑛=−∞ .

Answers:

(a)

𝑋(𝜔) = ∑ 𝑥(𝑛)𝑒−𝑗𝜔𝑛

∞

𝑛=−∞

= ∑ (0.5)𝑛𝑢(𝑛)𝑒−𝑗𝜔𝑛

∞

𝑛=−∞

𝑋(𝜔) = ∑(0.5)𝑛𝑒−𝑗𝜔𝑛 =

∞

𝑛=0

∑(0.5𝑒−𝑗𝜔)𝑛

=1

1 − 0.5𝑒−𝑗𝜔

∞

𝑛=0

(b)

𝑋(𝜔) = ∑ 𝑥(𝑛)𝑒−𝑗𝜔𝑛

∞

𝑛=−∞

= ∑ 𝛿(𝑛 − 3)𝑒−𝑗𝜔𝑛

∞

𝑛=−∞

= ∑ 𝑒−𝑗𝜔𝑛 =

3

𝑛=3

𝑒−𝑗3𝜔

(c)

𝑋(𝜔) =1

1 − 0.4𝑒−𝑗𝜔𝑒−𝑗2𝜔

(d)

𝑋(𝜔) =1

1 − 0.3𝑒−𝑗𝜔∗ 𝑊(𝜔)


Page 4 of 5

4. Determine the z-transform of

(a) 𝑥(𝑛) = 𝑒𝑗𝜔0𝑛 𝑢(𝑛) [10 marks]

(b) 𝑥(𝑛) = sin(𝜔0𝑛) 𝑢(𝑛) [15 marks]

(c) 𝑦(𝑛) = ((0.5)𝑛𝑢(𝑛)) ∗ 𝛿(𝑛 − 2) [5 marks]

where 𝑢(𝑛) and 𝛿(𝑛) are the step function and the unit impulse function, respectively. For

questions (a) and (b), justify mathematically your answers by using the z-transform equation:

𝑋(𝑧) = ∑ 𝑥(𝑛)𝑧−𝑛∞𝑛=−∞ . To simplify your result for question (b), you must use Euler’s

formula for sine and cosine.

Answers:

(a)

𝑋(𝑧) = ∑ 𝑥(𝑛)𝑧−𝑛

∞

𝑛=−∞

= ∑ 𝑒𝑗𝜔0𝑛 𝑢(𝑛)𝑧−𝑛

∞

𝑛=−∞

= ∑ 𝑒𝑗𝜔0𝑛 𝑧−𝑛

∞

𝑛=0

𝑋(𝑧) = ∑(𝑒𝑗𝜔0 𝑧−1)𝑛 =1

1 − 𝑒𝑗𝜔0 𝑧−1

∞

𝑛=0

(b)

𝑋(𝑧) = ∑ 𝑥(𝑛)𝑧−𝑛

∞

𝑛=−∞

= ∑ sin (𝜔0𝑛)𝑢(𝑛)𝑧−𝑛 =

∞

𝑛=−∞

∑𝑒𝑗𝜔0𝑛 − 𝑒−𝑗𝜔0𝑛

2𝑗

∞

𝑛=0

𝑧−𝑛

𝑋(𝑧) =1

2𝑗(∑ 𝑒𝑗𝜔0𝑛𝑧−𝑛

∞

𝑛=0

− ∑ 𝑒−𝑗𝜔0𝑛𝑧−𝑛

∞

𝑛=0

) =1

2𝑗(∑(𝑒𝑗𝜔0𝑧−1)

𝑛∞

𝑛=0

− ∑(𝑒−𝑗𝜔0𝑧−1)𝑛

∞

𝑛=0

)

𝑋(𝑧) =1

2𝑗(

1

1 − 𝑒𝑗𝜔0𝑧−1−

1

1 − 𝑒−𝑗𝜔0𝑧−1) =

1

2𝑗(

1 − 𝑒−𝑗𝜔0𝑧−1 − 1 + 𝑒𝑗𝜔0𝑧−1

(1 − 𝑒𝑗𝜔0𝑧−1)(1 − 𝑒−𝑗𝜔0𝑧−1))

𝑋(𝑧) =1

2𝑗(

(𝑒𝑗𝜔0 − 𝑒−𝑗𝜔0)𝑧−1

1 − 𝑒−𝑗𝜔0𝑧−1 − 𝑒𝑗𝜔0𝑧−1 + 𝑒𝑗𝜔0𝑒−𝑗𝜔0𝑧−2) =

𝑒𝑗𝜔0 + 𝑒−𝑗𝜔0

2𝑗 𝑧−1

1 − 2𝑒𝑗𝜔0 + 𝑒−𝑗𝜔0

2 𝑧−1 + 𝑧−2

𝑋(𝑧) =sin (𝜔0)𝑧−1

1 − 2cos (𝜔0)𝑧−1 + 𝑧−2

(c)

𝑋(𝑧) = 𝑧−2

1 − 0.5 𝑧−1


Page 5 of 5

5. Consider the filter transfer function

𝐻(𝑧) =1 − 𝑧−1

1 − 0.6 𝑧−1

(a) Determine the magnitude spectrum |𝐻(𝜔)|. [10 marks]

(b) Find the Input/Output difference equation from 𝐻(𝑧). [10 marks]

(c) Draw the block diagram realization. [5 marks]

Answers:

(a)

𝐻(𝑧) =1 − 𝑧−1

1 − 0.6 𝑧−1|

𝑧 = 𝑒𝑗𝜔

, 𝐻(𝜔) =1 − 𝑒−𝑗𝜔

1 − 0.6𝑒−𝑗𝜔

|1 − 𝑎𝑒−𝑗𝜔| = |1 − acos 𝜔 − 𝑗asin 𝜔| = √(1 − acos 𝜔)2 + (− asin 𝜔)2

|1 − 𝑎𝑒−𝑗𝜔| = √1 − 2 acos 𝜔 + 𝑎2cos2 𝜔 + 𝑎2sin2 𝜔 = √1 − 2 acos 𝜔 + 𝑎2(cos2 𝜔 + sin2 𝜔)

|1 − 𝑎𝑒−𝑗𝜔| = √1 − 2 acos 𝜔 + 𝑎2

|𝐻(𝜔)| = |1 − 𝑒−𝑗𝜔

1 − 0.6𝑒−𝑗𝜔| =

√1 − 2 cos 𝜔 + 1

√1 − 1.2 cos 𝜔 + 0.36

|𝐻(𝜔)| =√2 − 2 cos 𝜔

√1.36 − 1.2 cos 𝜔

(b)

𝐻(𝑧) =𝑌(𝑧)

𝑋(𝑧)=

1 − 𝑧−1

1 − 0.6 𝑧−1

𝑌(𝑧)(1 − 0.6𝑧−1) = 𝑋(𝑧)(1 − 𝑧−1)

𝑌(𝑧) − 0.6𝑌(𝑧) 𝑧−1 = 𝑋(𝑧) − 𝑋(𝑧)𝑧−1

𝑌(𝑧) = 0.6𝑌(𝑧) 𝑧−1 + 𝑋(𝑧) − 𝑋(𝑧)𝑧−1

𝑦(𝑛) = 0.6𝑦(𝑛 − 1) + 𝑥(𝑛) − 𝑥(𝑛 − 1)

(c)

THIS IS THE END OF THE EXAMINATION

Assignment 6

Sanzhar Askaruly

October 29, 2015

Problem 1.a. Determine the Laplace transfer function Ha(s) in terms of α and Ω0, whereα = R

Lis a coefficient and Ω0 = 1/

√LC is the angular analog cutoff frequency.

Figure 1: Analog filter circuit

Proof. I = VinR+ 1

sC+sL

Ha(s) = 1RCs+s2LC+1

= 1s2LC+sRC+1

1/LC1/LC

= 1/LCs2+Rs/L+1/LC

=Ω2

0

s2+αs+Ω20

Problem 1.b. Calculate the value of the analog cutoff frequency in hertz f0.

Proof. Ω0 = 2πf0 = 1/ 2√LC

f0 = 1

2π 2√LC= 1

2π2√

10−60.253310−3= 1

s2LC+sRC+11/LC1/LC

= 1/LCs2+Rs/L+1/LC

=Ω2

0

s2+αs+Ω20

= 10000

f0=10 kHz

Problem 1.c. Determine the analog filter magnitude spectrum |Ha(Ω)| by setting s=jΩ inHa(s), where Ω is the angular frequency.

Proof. Ha(s)=Ω2

0

s2+αs+Ω20=

Ω20

−Ω2+αΩj+Ω20

G2B=|Ha(Ω)|2=

Ω40

(Ω20−Ω2)2+(αΩ)2

|Ha(Ω)|= Ω20

2√

(Ω20−Ω2)2+(αΩ)2

Problem 1.d. Apply the bi-linear transformation by setting s=K 1−z−1

1+z−1 in Ha(s) and de-termine the coefficients b0, b1, b2, a0, a1, a2 in terms of K, α and Ω0.

1

2

Problem 1. Part D

𝐻! 𝑠 =𝛺!!

𝑠! + 𝛼𝑠 + 𝛺!!=

𝛺!!

𝐾! 1− 𝑧!!1+ 𝑧!! ! + 𝛼𝐾 1− 𝑧!!

1+ 𝑧!! + 𝛺!! 1+ 𝑧!! !

1+ 𝑧!! !

=𝛺!! 1+ 𝑧!! !

𝐾! 1− 𝑧!! ! + 𝛼𝐾 1− 𝑧!! 1+ 𝑧!! + 𝛺!! 1+ 𝑧!! ! =

=𝛺!! + 2𝛺!!𝑧!! + 𝛺!!𝑧!!

𝐾! + 𝛼𝐾 + 𝛺!! + 2 𝛺!! − 𝐾! 𝑧!! + 𝐾! − 𝛼𝐾 + 𝛺!! 𝑧!!×

1𝐾! + 𝛼𝐾 + 𝛺!!

1𝐾! + 𝛼𝐾 + 𝛺!!

=

𝛺!!𝐾! + 𝛼𝐾 + 𝛺!!

+ 2𝛺!!𝑧!!𝐾! + 𝛼𝐾 + 𝛺!!

+ 𝛺!!𝑧!!𝐾! + 𝛼𝐾 + 𝛺!!

1+ 2 𝛺!! − 𝐾!

𝐾! + 𝛼𝐾 + 𝛺!!𝑧!! + 𝐾

! − 𝛼𝐾 + 𝛺!!𝐾! + 𝛼𝐾 + 𝛺!!

𝑧!!

Therefore,

𝑏! =!!!

!!!!"!!!! 𝑏! =

!!!!

!!!!"!!!!; 𝑏! =

!!!

!!!!"!!!!

𝑎! = 2 !!!!!!

!!!!"!!!!; 𝑎! =

!!!!"!!!!

!!!!"!!!!

Problem 1. Part E

𝐾 =2𝜋𝑓!𝑐𝑜𝑡 𝜋𝑓!

𝑓!= 2𝜋10! cot 𝜋10!/10! = 1.934×10!

𝛼 =𝑅𝐿 = 78.958×10!

Ω! =1𝐿𝐶

= 62,832

𝑏! =Ω!!

𝑅 = 0.0697 = 𝑏!

𝑏! = 2𝑏! = 0.1395 𝑎! = −1.1816 𝑎! = −0.4606

Problem 1. Part F

𝐻 𝑧 =0.0697𝑧! + 0.13957𝑧 + 0.0697

𝑧! − 1.1816𝑧 + 0.4606

To find poles and zeros: Numerator (zeros):

0.0697𝑧! + 0.13957𝑧 + 0.0697 = 0 0.0697 𝑧! + 2𝑧 + 1 = 0

𝑧 = −1 Denominator (poles):

𝑧! − 1.1816𝑧 + 0.4606 = 0 D = 0.6680j

𝑧!,! = 0.5908± 0.3340𝑗

3

Problem 1. Part G f_s=100e+3; %sampling frequency f_0=10e+3; %cutoff frequency f=[logspace(0,5,199)] %range of frequency %Analog filter w_a=2*pi*f;%analog frequency %H_a(f) hAnalog=(62832^2)./(62832^2-w_a.^2+1j*78958*w_a) analogDB=20.*log10(abs(hAnalog)); %magnitude in dB %Digital filter w=2*pi.*f/f_s; %digital cutoff angular frequency %H(z) shown with e^(-jw) hDigital=(0.0697+0.1395*exp(-1j*w)+0.0697*exp(-2*1j*w))./(1-1.1816*exp(-1j*w)+0.4606*exp(-1j*w)); digitalDB=20.*log10(hDigital); %magnitude in dB figure; semilogx(f, analogDB, f, digitalDB); legend('Analog filter', 'Digital filer'); title('Analog and Digital Filter Magnitude Spectrum'); xlabel('f(kHz)') ylabel('H(dB)') axis([10^0 5*10^4 -100 10]) grid on

100 101 102 103 104

f(kHz)

-100

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

10

H(d

B)

Analog and Digital Filter Magnitude Spectrum

Analog filterDigital filer

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