Q3] A discrete memoryless source emits the following alphabet symbols: M1 1/2 M2 1/8 M3 1/8 M4 1/16 M5 M6 M7 M8 1/16 1/16 1/32 1/32 (a) Write the fixed length code for every symbol (no source coding). (b) Compute the Huffman code for this source. (c) Compute the code efficiency: with and without source encoding. Compare between the results.
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- USE DIGITAL LOGIC AND DESIGN Part 1: In Figure_4; we have 4-bit Comparator using 2-bit Comparators block. You have to satisfy given condition by applying all data on figure 4. At the end, given condition should produce HIGH output and other two should be LOW. A3 A2 A1 A0 = 1101 and B3 B2 B1 B0 = 1110 Figure_4 Part 2: The serial data-input waveform (Data in) and data-select inputs (S0 and S1) are shown in Figure_5. Determine the data-output waveforms from D0 through D3. Figure_5 Part 3: Decoder can be useful when we have to decode some specific numbers from their equivalent code. Figure 6 has a concept of 3 to 8 line decoder from which you have to generate output waveform from D0 to D7 with proper relationship to input. Figure_6 Part 4: The data-input and…Adocs.google.com is a logic switch that select information from one of the input line and directs it to a single line Encoder Decoder Adder Multiplexer Binary to octal can be implemented used 3-8 De-Multiplexer Multiplexer Decoder Encoder System base 12 include numbers from ... to .... 1-B 1-C 1-12 1-11 ۹:۰۳ F(c) Figure Q3(c)(i) shows a register and Figure Q3(c)(ii) shows the input waveforms (CLOCK and Data in) to the circuit. A1 A9 A10 A2 Function generator A3 A11 A12 AS A13 A6 A14 A7 A15 Data in Bop.7) ip.r 82p.7) Logic analyser U1 U2 U3 U4 UO 6. 1. 6 1 6 INVERTER 3 CLK 3 CLK oCLK CLK 5 K K 5 K K 4027 Clock Function generator Figure Q3(c)(i) (i) Determine the type of register as shown in Figure Q3(c)(i).
- H.W :- 1) A four logic-signal A,B,C,D are being used to represent a 4-bit binary number with A as the LSB and D as the MSB. The binary inputs are fed to a logic circuit that produces a logic 1 (HIGH) output only when the binary number is greater than 01102-610. Design this circuit. 2) repeat problem 1 for the output will be 0 (LOW) when the binary input is less than 01112-710- Saleem LateefDesign a code converter that converts a decimal digit from BCD to excess-3 code, the input variables are organized as (A BC D) respectively with A is the MSB, the output variables are organized as (W XY Z) respectively with W is the MSB, put the invalid decimal numbers as don't care. X= BCD'+B'D+B'C X= BC'D'+B'D+BC X= BC'D'+B'D+B'C X= BC'D'+BD+B'CWrite an assembly 8051 code to count a hexadecimal digit every second and display it on the 7-segment.
- Design a code converter that converts a decimal digit from BCD to excess-3 code, the input variables are organized as (A BC D) respectively with A is the MSB, the output variables are organized as (W X Y Z) respectively with W is the MSB, put the invalid decimal numbers as don't care. X= BCD'+B'D+B'C X= BC'D'+B'D+BC X= BC'D'+B'D+B'C X= BC'D'+BD+B'CGiven A = 40 and B = 60 in base 10. %3D %3D convert A and B to binary using 7 digits find – B in binary using 2's complement Find S = A+ (-B)in binary, Convert the result using 2' scomplement method to find the S in decimal representation. (you should get – 20).The numbers from 0-9 and a no characters is the Basic 1 digit seven segment display * .can show False True In a (CA) method of 7 segments, the anodes of all the LED segments are * "connected to the logic "O False True Some times may run out of pins on your Arduino board and need to not extend it * .with shift registers True False
- bit numbers and store lower byte of the result in accumulator A and higher byte in accumulator B.. Similarly lower byte of 2nd number is stored in R2 and higher byte in R3. Now subtract these two 16 numbers in the RAM such that lower byte of 1t number is stored in R0 and higher byte in RI. manipulate 16 bits numbers as well. Write an assembly language program that stores two 16 bits 8051 is an 8-bit microcontroller that performs arithmetic on 8-bit numbers. However, we can Question#3: 8051 is an 8-bit microcontroller that performs arithmetic on 8-bit numbers. HoweveLogic effort and parasitic delayin Intel 8086 microprocessor, the memory address for the code segment 771Fh and offset 3584 h is O 7A774 3CF5F О АСАЗО OACA3 in Intel 8086 microprocessor, the memory address for the code segment 3021 h and offset 21D7h is 323E7 51F80 051F8 24D91