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Digital Fundamentals

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|QUESTION: A1 |
|a. |
|CLK |
|Data Input |
|Q0 |
|Q1 |
|Q2 |
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|0 |
|x |
|1 |
|1 |
|0 |
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|1 |
|1 |
|1 |
|1 |
|1 |
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|2 |
|0 |
|0 |
|1 |
|1 |
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|3 |
|1 |
|1 |
|0 |
|1 |
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|4 |
|1 |
|1 |
|1 |
|0 |
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|5 |
|0 |
|0 |
|1 |
|1 |
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|b. |
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|[pic] |

|QUESTION: A2 |
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|a. |
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|X |
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|b. |
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|If Preset of J-K flip flop is tied to ground, A will be always ‘1’ |
|B will follows the input Z all the time |

|QUESTION: A3 |
|a. |
|Q2Q1Q0 |
|1 1 0 |
|1 0 1 |
|1 0 0 |
|0 1 1 |
|0 1 0 Reset Count |
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|Thus Mod-4 |
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|Max. Clock Freq = /(N x TPD) =1/(3 x 20ns) = 16.67 MHz |
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|Freq of Q0 = 20k / 2 = 10 KHz. |

|QUESTION: A4 SOLUTIONS CONFIDENTIAL |
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|a) Memory Type: ROM |
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|b)Word Length = 8 bits |
|Word Capacity = 212 = 22 x 210 = 4 K words |
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|c) |
|32K÷4K ==8 |
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|32Kx4 module will need 8 of the memory modules. |
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|QUESTION: A5 |
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|a. Binary output of this ADC for an input voltage of 2.5V: |
|2.5 / 0.15 = 16.67 |
|For Digital Ramp ADC, round up to 17. |
|Hence, binary output for 2.5V input is 0100012 (= 17) |
| |
|b. Conversion time needed for part (c) if the input clock rate is 40 kHz. |
|Period of Clock is TCLK = 1 / 40k |
|= 0.025 ms |
|Conversion time = No.of Steps x TCLK |
|= 17 x 0.025 ms |
|= 0.425 ms |
| |
|c. For 6-bit Successive Approximation ADC (SAC), the new conversion time needed for part (c) is fixed. |
|New Conversion time= No.of Bits of SAC x TCLK |
|= 6 x 0.025 ms |
|= 0.15 ms |
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|QUESTION: B1 |
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|a. |
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|1K x 4 |
|2K x 4 |
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|Word capacity |
|1K (210) |
|2K (211) |
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|No data lines |
|4 |
|4 |
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|No of ICs |
|1 x 2 = 2 |
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|b |
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|Memory#3: 00010000 to 00011111. |

|QUESTION: B2 |
|a. |
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|QA |
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|JA = 1, KA =1 |
|QB |
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|JB = QA, KB = QA |
|QC |
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|JC = QB, KC = QB |
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|QA= 0 |
|JA = 1, KA =1 |
|QB = 0 |
|JB = 1, KB = 1 |
|QC = 0 |
|JC =0, KC=0 |
| |
|1st clk |
|QA= 1 |
|JA = 1, KA =1 |
|QB = 1 |
|JB = 0, KB = 0 |
|QC = 0 |
|JC =1, KC=1 |
| |
|2nd clk |
|QA= 0 |
|JA = 1, KA =1 |
|QB = 1 |
|JB = 1, KB = 1 |
|QC = 1 |
|JC =1, KC=1 |
| |
|3rd clk |
|QA= 1 |
|JA = 1, KA =1 |
|QB = 0 |
|JB = 0, KB = 0 |
|QC = 0 |
|JC =0, KC=0 |
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|4th clk |
|QA= 0 |
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|QB = 0 |
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|QC = 0 |
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|QUESTION: B2 |
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|Count Sequence 2: |
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|QA |
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|JA = 1, KA =1 |
|QB |
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|JB = QA, KB = QA |
|QC |
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|JC = QB, KC = QB |
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|QA= 0 |
|JA = 1, KA =1 |
|QB = 0 |
|JB = 1, KB = 1 |
|QC = 1 |
|JC =0, KC=0 |
| |
|1st clk |
|QA= 1 |
|JA = 1, KA =1 |
|QB = 1 |
|JB = 0, KB = 0 |
|QC = 1 |
|JC =1, KC=1 |
| |
|2nd clk |
|QA= 0 |
|JA = 1, KA =1 |
|QB = 1 |
|JB = 1, KB = 1 |
|QC = 0 |
|JC =1, KC=1 |
| |
|3rd clk |
|QA= 1 |
|JA = 1, KA =1 |
|QB = 0 |
|JB = 0, KB = 0 |
|QC = 1 |
|JC =0, KC=0 |
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|4th clk |
|QA= 0 |
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|QB = 0 |
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|QC = 1 |
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|[pic] |
| |

|QUESTION: C1 |
| |
|a. Select the 4081 IC. ---4 marks |
| |
|Using 7408: |
|Voh(min) – 1.3V noise swing= 2.4-1.3=1.1V which is less than the Vih(min) required vg of 2V |
|Vol(max)+0.3V noise swing = 0.4+0.3 =0.7V which is within the Vil(max) limit of 0.8V |
|Using 4081: |
|Voh(min) – 1.3V noise swing= 4.95-1.3=3.65V which is more than the Vih(min) required voltage of 3.5V |
|Vol(max)+0.3V noise swing = 0.05+0.3 =0.35V which is within the Vil(max) limit of 1.5V |
| |
|7408 noise margin in HIGH state is only 0.4V whereas 4081 has a noise margin of 1.45V. thus select 4081. |
| |
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|b. |
|[pic] |

----------------------- AB0

AB9

AB10

C

DB0

DB3

I/O0

I/O3

A0

A9

A0

A9

I/O0

I/O3

CS

CS

R/W

R/W

R/W

000 [0]

110 [6]

011 [3]

100 [4]

QAQBQC

CLK

FF2

QA

JA

KA

QA

CLK0

JB

KB

QB

QB

CLK1

KC

QC

JC

CLK2

FF0

FF1

PRE

PRE

PRE

CLR

CLR

CLR

5V

5V

QC

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Cmgt 554 Week 2 Assignment

...Media communication is the process in which data is transferred from one computer to another. This involves transmission of digital of information to different devices through wireless or cabled connections. The data transmitted over networks could be either digital or analog. * Analog signals are continuous signals that vary in strength. Sound is an example of an analog signal. Telephones have transmitters that encode sound waves into electromagnetic waves, which then travel over wires toward their destination. The receiving telephone decodes the electromagnetic waves back into sound waves. Our brains then decode the sound waves into the words we hear. Computer modems use the same principle. Analog signals can be represented digitally. For instance, a high electromagnetic voltage could be interpreted as 1 and low voltage as 0. * Digital signals are discrete rather than continuous. Either there is a signal or there isn't a signal. Telegraphs transmit data with discrete signals. You either hear a tap or you do not hear a tap. Discrete signals can be represented by on and off pulses. The duration of a discrete signal can be varied, as with dots and dashes in Morse Code. To explain how this data is transmitted over the network, first I had to explain the mediums. Mediums are ways, which the data use to travel from one place to another. These mediums may vary depending the environment or type of network. The most commonly used data communication media include: *...

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