18EC52
Model Question Paper -1 with effect from 2020-21 (CBCS Scheme)
USN
Fifth Semester B.E. Degree Examination
Digital Signal Processing
TIME: 03 Hours Max. Marks: 100
Note: 01. Answer any FIVE full questions, choosing at least ONE question from each MODULE.
Module – 1 Marks
Why is it necessary to perform frequency domain sampling? Illustrate the relationship
(a) 6
between the sampled Fourier transform and the DFT.
Compute the N – point DFT of the following signals. (2 + 4)
(b)
Q.1 (𝑖 ) 𝑥 ( 𝑛 ) = 𝛿 (𝑛 − 𝑛0 ) ; 0 < 𝑛 0 < 𝑁 (𝑖𝑖 ) 𝑥(𝑛) = 𝑎𝑛 ; 0 ≤ 𝑛 ≤ (𝑁 − 1) 6
If 𝑥(𝑛) is real valued sequence, obtain its DFT, 𝑋 (𝑘) for the following two cases. And
(4 + 4)
(c) comment on the nature of 𝑋(𝑘).
8
(i) 𝑥(𝑛) is an even sequence. (ii) 𝑥(𝑛) is an odd sequence.
OR
Determine the 4 - point circular convolution of the sequences
2𝜋𝑛 2𝜋𝑛
(a) 𝑥1 (𝑛) = 𝑐𝑜𝑠 ( ) 𝑎𝑛𝑑 𝑥2 (𝑛) = sin ( ) ; 0 ≤𝑛≤3 4
𝑁 𝑁
Using the time – domain formula.
Q.2 Illustrate how the DFT & IDFT can be viewed as a Linear Transformations on
(b) 6
sequences {𝑥(𝑛)} and {𝑋 (𝑘)} respectively.
(c) Compute 4 – point DFT of the signal 𝑥(𝑛) = {0, 1, 2, 3} using DFT matrix. 4
Show that multiplication of the DFTs of two sequences is equivalent to the circular
(d) 6
convolution of the two sequences in the time – domain.
Module – 2
By means of DFT & IDFT, determine the response of the FIR filter with impulse
(a) 6
response ℎ(𝑛) = {5, 6, 7} to the input sequence 𝑥 (𝑛) = {1, 2, −1, 5 6}.
Using overlap & save method, compute the output of an FIR filter with impulse response
Q.3 (b) ℎ(𝑛) = {1, 2, 3} and input 𝑥(𝑛) = {2, −2, 8, −2, −2, −3, −2, 1, −1, 9, 1, 3} use only 8 – 6
point circular convolution in your approach.
(c) Develop radix – 2, DIT – FFT algorithm and write signal flow graph for 𝑁 = 8. 8
OR
(a) Develop radix – 2 decimation – in – frequency FFT algorithm. 6
Using overlap & add method, compute the output of an FIR filter with impulse response
Q.4 (b) ℎ(𝑛) = {1, − 2, 3} and input 𝑥(𝑛) = {2, −2, 8, −2, −2, −3, −2, 1, −1, 9, 1, 3} use only 8 6
– point circular convolution in your approach.
(c) Given 𝑥 (𝑛) = (𝑛 + 1) 𝑓𝑜𝑟 0 ≤ 𝑛 ≤ 7. Find 𝑋 (𝑘) using DIT – FFT algorithm. 8
Module – 3
Q.5 Determine a direct – form realization for the following linear phase filter.
(a) ℎ(𝑛) = {1, 2, 3, 4, 3, 2, 1} 4
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Prove that an FIR filter having impulse response with even length & symmetric nature has
(b) a linear – phase response. 6
A low pass filter is to be designed with the following desired frequency response:
𝜋
1, 𝑓𝑜𝑟 |𝜔| ≤
Q.5 6
𝐻𝑑 (𝜔) = { 𝜋
(c) 0, 𝑓𝑜𝑟 < |𝜔 | ≤ 𝜋 10
6
Determine the coefficients of a 25 – tap filter based on the window method with the
Hamming window.
OR
Determine the lattice coefficients corresponding to the FIR filter with system function
13 −1 5 −2 1 −3
(a) 𝐻 (𝑧 ) = 𝐴 3 (𝑧 ) = 1 + 𝑧 + 𝑧 + 𝑧 4
24 8 3
And sketch the lattice structure.
Consider a three – stage FIR lattice structure having the coefficients
(b) 𝑘1 = 0.65, 𝑘2 = −0.34 & 𝑘3 = 0.8 Realize this filter in direct form. 6
Q.6 Based on the frequency – sampling method, determine the coefficients of a linear –
phase FIR filter of length 𝑀 = 15 which has a symmetric unit sample response and
a frequency response that satisfies the conditions:
(c) 2𝜋𝑘 1, 𝑘 = 0, 1, 2, 3 10
𝐻𝑟 ( ) = { 0.4, 𝑘=4
15 0, 𝑘 = 5, 6, 7
Module – 4
Given a lowpass prototype
1
𝑯𝑷 (𝒔) =
𝑠+1
Q.7 (a) (i) Determine the high pass filter with a cut off frequency of 40 radians per second. 6
(ii) Determine the band pass filter with a center frequency of 100 rad / sec and
bandwidth of 20 rad / sec.
Realize the following digital filter using a direct form II structure.
0.7157 + 1.4314𝑧 −1 + 0.7151𝑧 −2
(b) 𝑯(𝒛) = 6
1 + 1.3490𝑧 −1 + 0.5140𝑧 −2
Using BLT design a second order digital lowpass Butterworth filter with a cut off
(c) frequency of 3.4 kHz at a sampling frequency of 8000 Hz. 8
OR
(a) Assuming that 𝑇 = 2 𝑠𝑒𝑐 in BLT, and given the following points:
(i) 𝑠 = −1 + 𝑗, on the left half of the s – plane.
(ii) 𝑠 = 1 − 𝑗, on the right half of the s – plane.
(iii) 𝑠 = 𝑗, on the positive 𝑗𝜔 on the s – plane. 8
(iv) 𝑠 = −𝑗, on the negative 𝑗𝜔 on the s – plane.
Convert each of these points in the s – plane to the z – plane, and verify the mapping
properties.
Q.8
(b) Design a digital band stop Butterworth filter with the following specifications:
Center frequency of 2.5 kHz
Passband width of 200 Hz and ripple of 3 dB 12
Stop band width of 50 Hz and attenuation of 10 dB
Sampling frequency of 8000 Hz.
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Module – 5
(a) With the block diagram explain Digital signal processors based on the Harvard
6
architecture.
Discuss briefly the following special Digital Signal Processor hardware units.
Q.9 (b)
(i) Multiplier and Accumulator (MAC) Unit (3+2+5)
(ii) Shifters 10
(iii) Address Generators
(c) Convert the following decimal numbers into Q – 15 representation.
(i) −0.1958 4
(ii) 0.560123
OR
(a) Given the FIR filter
𝑦(𝑛) = 0.9 𝑥(𝑛) + 3 𝑥(𝑛 − 1) − 0.9𝑥(𝑛 − 2)
With the passband gain of 4, and assuming that the input range occupies only ¼ of the 6
full range for a particular application. Develop the DSP implementation equations in the
Q – 15 fixed – point system.
(b) Discuss the following IEEE Floating – Point Formats.
Q.10
(i) Single Precision Format 6
(ii) Double Precision Format
(c) With the diagram explain the basic architecture of TMS320C54x family processor. 8
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Table showing the Bloom’s Taxonomy Level, Course Outcome and Programme
Outcome
Question Bloom’s Taxonomy Level Course Programme
attached Outcome Outcome
(a) L2 CO1 PO1
Q.1 (b) L3 CO2 PO1
(c) L3 CO2 PO1
(a) L3 CO1 PO1
(b) L2 CO1 PO1
Q.2
(c) L3 CO2 PO1
(d) L2 CO1 PO1
(a) L3 CO3 PO2
Q.3 (b) L3 CO3 PO2
(c) L3 CO3 PO2
(a) L3 CO3 PO2
Q.4 (b) L3 CO3 PO2
(c) L3 CO3 PO2
(a) L3 CO4 PO3
Q.5 (b) L3 CO4 PO3
(c) L3 CO4 PO3
(a) L3 CO4 PO3
Q.6 (b) L3 CO4 PO3
(c) L3 CO4 PO3
(a) L3 CO4 PO3
Q.7 (b) L3 CO4 PO3
(c) L3 CO4 PO3
(a) L3 CO4 PO3
Q.8
(b) L4 CO4 PO3
(a) L1 CO5 PO1
Q.9 (b) L2 CO5 PO1
(c) L3 CO5 PO1
(a) L4 CO5 PO1
Q.10 (b) L3 CO5 PO1
(c) L1 CO5 PO1
Lower order thinking skills
Bloom’s Remembering Understanding Applying (Application):
Taxonomy (knowledge): 𝐿1 Comprehension): 𝐿2 𝐿3
Levels Higher order thinking skills
Analyzing (Analysis): 𝐿4 Valuating (Evaluation): 𝐿5 Creating (Synthesis): 𝐿6