dsd practice exam

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Digital System Design Sessional Exam II DE-26 Computer Engineering Department College of EME 1 DIGITAL SYSTEM DESIGN SESSIONAL II DE26 COMP Time: 1 Hrs Marks: 60 1.1 Question: Correction Vector Computation Compute the Correction Vector for computing the following equation Out = in1 + in2 + in3 in4 + in5 in6 Where in1, in2, in3, in4, in5, in6 are in Q3.5, Q7.8, Q1.6, Q8.2, Q2.4, Q3.9 format numbers respectively. 15 1.2 Question: Digital Design with Compression Tree Design and draw using dot notation an optimal logic to add twenty one-bit numbers, use Wallace Reduction Tree for the compression. 20 1.3 Question: Fixed-point Conversion Compute the floating point output for the given floating point inputs, convert the algorithm in appropriate fixed-point format using 16-bit sign arithmetic and calculate your result and compare them with the results in floating point format. Give format for yn. xn=0; xn_1=0; xn_2=0; for(i=0; i<3; i++) // compute only three iteration for the following { xn_2=xn_1; xn_1=xn; xn=input[i]; yn=0.987*xn*xn_1 + 0.7765*xn_1*xn_2+xn; } Every iteration (posedge of a clock) we get one input sample from the following stream of data. input[]={-0.234; 0.5432, 0.223,-0.889….} 25 Formatted: Bullets and Numbering Formatted: Bullets and Numbering

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Digital system design practice exam

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Page 1: dsd practice exam

Digital System Design

Sessional Exam II

DE-26 Computer Engineering Department College of EME

1DIGITAL SYSTEM DESIGN

SESSIONAL II

DE26 COMP

Time: 1 Hrs

Marks: 60

1.1 Question: Correction Vector Computation

Compute the Correction Vector for computing the following equation

Out = in1 + in2 + in3 – in4 + in5 – in6

Where in1, in2, in3, in4, in5, in6 are in Q3.5, Q7.8, Q1.6, Q8.2, Q2.4, Q3.9 format numbers respectively.

15

1.2 Question: Digital Design with Compression Tree

Design and draw using dot notation an optimal logic to add twenty one-bit numbers, use Wallace Reduction Tree for the compression.

20

1.3 Question: Fixed-point Conversion

Compute the floating point output for the given floating point inputs, convert the algorithm in appropriate fixed-point format using 16-bit sign arithmetic and

calculate your result and compare them with the results in floating point format. Give format for yn.

xn=0; xn_1=0; xn_2=0;

for(i=0; i<3; i++) // compute only three iteration for the following

{

xn_2=xn_1;

xn_1=xn;

xn=input[i];

yn=0.987*xn*xn_1 + 0.7765*xn_1*xn_2+xn;

}

Every iteration (posedge of a clock) we get one input sample from the following

stream of data.

input[]={-0.234; 0.5432, 0.223,-0.889….}

25

Formatted: Bullets and Numbering

Formatted: Bullets and Numbering