SET2 AK

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PARISUTHAM INSTITUTE OF TECHNOLOGY AND SCIENCE
Department of ECE
AY 2015-16
IV Year-CSE /VII Sem.
CAT 1
CS2403 DIGITAL SIGNAL PROCESSING
Duration: 3 Hours
Max. Marks: 100
Answer all the questions
PART-A (10x2 marks =20 marks)
1. State the sampling theorem.
ο‚· A continuous time signal can be completely represented in its samples and recovered
back if the sampling frequency 𝑓𝑠 ≥ 2πœ”.
ο‚· Here 𝑓𝑠 is the sampling frequency and πœ” is the maximum frequency.
2. State any properties of LTI systems.
Linearity =>
𝑇[π‘Ž1 π‘₯1 (𝑛) + π‘Ž2 π‘₯2 (𝑛)] = π‘Ž1 𝑇[π‘₯1 (𝑛)] + π‘Ž2 𝑇[π‘₯2 (𝑛)]
Time Shifting =>
π‘Œ(𝑛) = 𝑇[π‘₯(𝑛)], then
π‘Œ(𝑛 − 𝐾) = 𝑇[π‘₯(𝑛 − 𝐾)]
= 𝑍 −𝐾 𝑇[π‘₯(𝑁)]
𝑍 −𝐾 = π‘‘π‘’π‘™π‘Žπ‘¦
3. Determine the Z transform for 𝒙(𝒏) = −𝒏𝒂𝒏 𝒖(−𝒏 − 𝟏).
The Z-Transform for given values is
π‘₯(𝑛) = −π‘›π‘Žπ‘› 𝑒(−𝑛 − 1)
π‘Žπ‘§ −1
π‘₯(𝑧) =
(1 − π‘Žπ‘§ −1 )2
π‘Žπ‘§
π‘₯(𝑧) =
(∴ |𝑧| < |π‘Ž|)
(𝑧 − π‘Ž)2
4. Find whether the signal π’š = π’πŸ 𝒙(𝒏) is linear.
𝑦(𝑛) = 𝑛2 π‘₯(𝑛)
we have π‘Œ1 (𝑛) = 𝑇[π‘₯1 (𝑛)] = 𝑛2 π‘₯1 (𝑛)
π‘Œ2 (𝑛) = 𝑇[π‘₯2 (𝑛)] = 𝑛2 π‘₯2 (𝑛)
The weighted sum of output is
π‘Ž1 𝑇[π‘₯1 (𝑛)] + π‘Ž2 𝑇[π‘₯2 (𝑛)] = π‘Ž1 𝑛2 π‘₯1 (𝑛) + π‘Ž2 𝑛2 π‘₯2 (𝑛)
The output to the weighted sum of input is
π‘Œ3 (𝑛) = 𝑇[π‘Ž1 π‘₯1 (𝑛) + π‘Ž2 π‘₯2 (𝑛)] = 𝑛2 π‘Ž1 π‘₯1 (𝑛) + 𝑛2 π‘Ž2 π‘₯2 (𝑛)
Hence the system is NON-LINEAR
5. State and prove parseval’s theorem.
If DFT [x(n)] = X(k) and
DFT [y(n)] = Y(k) then
𝑁−1
𝑁−1
∑ π‘₯(𝑛) 𝑦
∗ (𝑛)
𝑛=0
𝑛=0
𝑁−1
∑ π‘₯(𝑛) 𝑦
𝑛=0
1
= ∑ 𝑋(π‘˜)π‘Œ ∗ (π‘˜)
𝑁
∗ (𝑛)
𝑁−1
𝑁−1
𝑛=0
𝑛=0
1
1
= ∑ π‘₯(𝑛) [ ∑ π‘Œ ∗ (π‘˜)𝑒 −𝑗2πœ‹π‘˜π‘›⁄𝑁 ]
𝑁
𝑁
𝑁−1
𝑁−1
𝑛=0
𝑛=0
1
1
= ∑ π‘Œ ∗ (π‘˜) [ ∑ π‘₯(𝑛)𝑒 −𝑗2πœ‹π‘˜π‘›⁄𝑁 ]
𝑁
𝑁
𝑁−1
1
= ∑ π‘Œ ∗ (π‘˜) 𝑋(π‘˜)
𝑁
𝑛=0
Hence Proved
6. Compute the DFT of the four point sequence 𝒙(𝒏) = {𝟎, 𝟏, 𝟐, πŸ‘ … }.
𝑁−1
𝑋(π‘˜) = ∑ π‘₯(𝑛)𝑒 −𝑗2πœ‹π‘˜π‘›⁄𝑁
𝑛=0
𝑋(π‘˜) = {6, −2 + 2𝑗, −2, −2 − 2𝑗}
7. Draw the basic butterfly of the Radix-4 DIT algorithm.
8. What is phase factor or twiddle factor?
The complex number π‘Šπ‘ is called phase factor or twiddle factor.
π‘Šπ‘ = 𝑒
−𝑗2πœ‹
𝑁
π‘Š0 = 1
π‘Š1 = 0.707 − 𝑗0.707
π‘Š2 = −𝑗
π‘Š3 = −0.707 − 𝑗707
9. Find the DFT for 𝒙(𝒏) = {𝟏, −𝟏, 𝟏, −𝟏}.
The 4-point DFT of x(n) is given by
3
𝑋(𝐾) = ∑ π‘₯(𝑛) 𝑒 −𝑗2πœ‹π‘›π‘˜⁄4
𝑛=0
= π‘₯(0) + π‘₯(1)𝑒 −π‘—πœ‹π‘˜⁄2 + π‘₯(2)𝑒 −π‘—πœ‹π‘˜ + π‘₯(3)𝑒 −𝑗3πœ‹π‘˜⁄2
∴ π‘₯(0) = 1, π‘₯(1) = −1, π‘₯(2) = 1, π‘₯(3) = −1
= 1 − 𝑒 −π‘—πœ‹π‘˜⁄2 + 𝑒 −π‘—πœ‹π‘˜ − 𝑒 −𝑗3πœ‹π‘˜⁄2 ; π‘“π‘œπ‘Ÿ π‘˜ = 0, 1, 2, 3.
10. What are the requirements for converting a stable analog filter into a stable digital filter?
ο‚·
Mapping of desired digital filter specifications into equivalent analog filter.
ο‚·
Analog transfer function is derived for the analog filter.
ο‚·
Then, analog prototype filter is transformed into equivalent digital filter transfer function.
PART –B (5x 16 marks = 80 marks)
11. (a) (i) Determine the Z-transform and compute the ROC of the following signal
𝟏 𝒏
πŸ‘
𝟏 𝒏
πŸ‘
𝒙(𝒏) = ( ) 𝒖(−𝒏) + ( ) 𝒖(𝒏).
(16)
∞
𝑋(𝑧) = ∑ π‘₯(𝑛)𝑍 −𝑛
𝑛=−∞
∞
𝑋(𝑧) = ∑ π‘₯(𝑛)𝑍
𝑛=−∞
∞
−𝑛
𝑋(𝑧) = [− ∑(1⁄3)
𝑛=0
−𝑛
∞
+ ∑ π‘₯(𝑛)𝑍 −𝑛
𝑛=−∞
∞
𝑛
𝑍 +𝑛 ] + [− ∑(1⁄3) 𝑍 −𝑛 ]
𝑛=0
1
𝑍
+
(1
3𝑧 − 1 𝑍 − ⁄3)
(OR)
(b) (i) Compute the linear convolution of the signals x(n)={1,2,3,4,5,3,-1,-2}&
h(n)={3,2,1,4} using Tabulation method & Matrix method
(8)
ο‚· Tabulation method: y(n)={3,8,14,24,34,35,24,15,7,-6,-8} (4 marks)
ο‚· Matrix Method: y(n)={3,8,14,24,34,35,24,15,7,-6,-8} (4 marks)
(ii) Find the output of the linear convolution using Circular Convolution
x(n)={1,-3,5,-7,9,-11} & h(n)={-4,8,-16} [ Matrix Method]
.
(8)
y(n)={-4,20,-60,116,-172,228,-232,176}
ANS:
12. (a) (i) State and explain sampling theorems.
(8)
Definition – 2marks
Explanation – 2marks
π‘₯(𝑑) = 𝑠𝑖𝑛 𝑒𝑑
π‘₯(𝑛𝑇) = 𝑠𝑖𝑛 Ω𝑛𝑇 = 𝑠𝑖𝑛 πœ”π‘› → 2π‘šπ‘Žπ‘Ÿπ‘˜π‘ 
𝑓𝑠 ≥ 2π‘“π‘š → 2π‘šπ‘Žπ‘Ÿπ‘˜π‘ 
(ii) Find the Z-transform auto correlation function.
Definition – 2marks
(8)
∞
𝛾π‘₯π‘₯ (𝑙) = ∑ π‘₯(𝑛)π‘₯(𝑛 − 𝑙) − > 2π‘šπ‘Žπ‘Ÿπ‘˜π‘ 
𝑛=−∞
∞
∞
π‘Œ(𝑍) = ∑ [ ∑ π‘₯(π‘˜)π‘₯(𝑛 − π‘˜)] 𝑧 −𝑛 → 2π‘šπ‘Žπ‘Ÿπ‘˜π‘ 
𝑛=−∞ π‘˜=−∞
𝛾π‘₯π‘₯ (𝑙) = 𝑋(𝑍)𝑋(𝑍 −1 ) → 2π‘šπ‘Žπ‘Ÿπ‘˜π‘ 
(OR)
(b) (i) Suppose a LTI system with input x(n) & output y(n) is characterized by its
unit sample response h(n)= (0.8)n u(n). Find the response y(n) of the system having
the input x(n)= u(n).
(10)
ANSWER:
h(n)= (0.8)n u(n) => an u(n)=
Y(z)=H(z).X(z)=
𝒛
.
𝑧
𝒛−𝟎.πŸ– 𝑧−1
𝒛
; H(z)=
𝒛−𝒂
𝒛
𝒛−𝟎.πŸ–
; X(z)=
𝑧
𝑧−1
(taking inverse Z-Transform using partial fraction method)
Y(n)= (-4. (0.8)n +5) u(n)
(ii) A causal system represented by the difference equation,
𝟏
π’š(𝒏) + πŸ’ π’š(𝒏 − 𝟏) = 𝒙(𝒏) +
𝟏
𝟐
𝒙(𝒏 − 𝟏). Compute its transfer function H(z).
ANSWER: Taking inverse transform and compute the H(z)= Y(z)/ X(z)
𝟏
H(z)=
(6)
𝟏+(𝟐)𝒛^−𝟏
𝟏
𝟏+(πŸ’)𝒛^−𝟏
13. (a) Given 𝒙(𝒏) = {𝟎, 𝟏, 𝟐, πŸ‘, πŸ’, πŸ“, πŸ”, πŸ•} find 𝑿(π’Œ) using DIT FFT algorithm.
(16)
X(K)= {28, -4+j 9.656, -4+j4, -4+j 1.656, -4 ,-4-j 1.656, -4-j 4, -4-j 9.656}
(OR)
(b) Given 𝒙(𝒏) = πŸπ’ compute 𝑿(π’Œ) using DIF FFT algorithm for N = 8 and 𝒏 ≥ 𝟎
(16)
14. (a) (i) Derive the key equations of radix-2 DIF FFT algorithm and draw the relevant flow
graph taking the computation of an 8 point DFT for your illustration.
(8)
ο‚· Explanation – 2marks
ο‚· Graph diagram – 6marks
(ii)Compute the FFT for the sequence 𝒙(𝒏) = 𝒏 + 𝟏 where N = 8 using the inplace radix 2
decimation frequency algorithm.
(8)
(OR)
(b) (i) List out all the properties of DFT.
Definition – 2marks
(8)
Radix 2 flow graph – 2marks
Properties and explanation – 4marks
(ii) Determine the circular convolution of the sequences
(8)
π’™πŸ (𝒏) = {𝟏
⏟ , 𝟏, 𝟐, 𝟏}
↑
π’™πŸ (𝒏) = {𝟏
⏟ , 𝟐, πŸ‘, πŸ’}
↑
1
[1
2
1
1
1
1
2
2
1
1
1
𝑦(0)
1 1
2 ] [2] = [𝑦(1)]
1 3
𝑦(2)
1 4
𝑦(3)
π’š(𝒏) = {πŸπŸ‘, πŸπŸ’, 𝟏𝟏, 𝟏𝟐}
15. (a) Determine the cascade and parallel realization for the system, described by the system
function
𝑯(𝒛) =
(16)
𝟏𝟎(𝟏 − (𝟏⁄𝟐)𝒛−𝟏 )(𝟏 − (𝟐⁄πŸ‘)𝒛−𝟏 )(𝟏 + πŸπ’›−𝟏 )
[(𝟏 − (πŸ‘⁄πŸ’)𝒛−𝟏 )(𝟏 − (𝟏⁄πŸ–)𝒛−𝟏 )(𝟏 − (𝟏⁄𝟐 + 𝒋 𝟏⁄𝟐)𝒛−𝟏 )(𝟏 − (𝟏⁄𝟐 − 𝒋 𝟏⁄𝟐)𝒛−𝟏 )]
π‘ͺ𝒂𝒔𝒄𝒂𝒅𝒆:
Parallel:
(OR)
(b) Determine H(z) for a butterworth filter satisfying the following constraints
𝝅
√𝟎. πŸ“ ≤ |𝑯(π’†π’‹πŽ )| ≤ 𝟏
𝟎≤𝝎≤
𝟐
π’‹πŽ
πŸ‘π…
⁄πŸ’ ≤ 𝝎 ≤ 𝝅
|𝑯(𝒆 )| ≤ 𝟎. 𝟐
with T = 1s. Apply impulse invariant transformation.
(16)
Ω𝑝 = πœ”π‘ ; Ω𝑠 = πœ”π‘  → 2π‘šπ‘Žπ‘Ÿπ‘˜π‘ 
πœ† = 4.989; πœ– = 1 → 2π‘šπ‘Žπ‘Ÿπ‘˜π‘ 
𝑁≥
πœ†
log πœ–
Ω
log Ω 𝑠
= 4 → .4π‘šπ‘Žπ‘Ÿπ‘˜π‘ 
𝑝
𝐻(𝑆) =
𝐻(𝑧) =
1
→ 2π‘šπ‘Žπ‘Ÿπ‘˜π‘ 
(𝑠 2 + 0.76537𝑠 + 1)(𝑠 2 + 1.8477𝑠 + 1)
1.454 + 0.1839𝑧 −1
−1.454 + 0.2307𝑧 −1
+
→ 6π‘šπ‘Žπ‘Ÿπ‘˜π‘ 
1 − 0,387𝑧 −1 + 0.055𝑧 −2 1 − 0.1322𝑧 −1 + 0.301𝑧 −2
Faculty Signature:
Faculty Name: S.SHANMUGA PRIYA
HOD Signature
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