Lab 4 Hints

6. The time delay of following program is if crystal frequency is 8 MHz: LDI R15, 12 LDI R16, 14 LDI R21, 5 ADD R15, R16 ADD R15, R21 7. Write a short program that make all pins of PORTB one using RI9 register.

Decoder circuit as shown in the following Figure. if A is LSB and C is MSB, the output expression F= YO Y Y, Y, D. Y. Decoder O a. B' O b. C O'c.B O d. C

Please solve it fast

10. Subtract the following hexadecimal numbers. b) A55(16) - C5(16) = (Hex) A (Decimal Equivalent) C 5 11. Convert the following decimal numbers to 8421 BCD and Excess-3. Decimal 8421-BCD Excess-3 7910 = 3110 = 12. Complete the following table showing the 3-bit Gray codes corresponding to 010 to 810. Decimal Binary 000 Gray Code 1 001 2 010 3. 4 011 100 101 ETEC122: DIGITAL SYSTEMS 3 Wint

Design the following function F using: AB CD 00 01 11 10 00 0. 01 0. 1 11 1 10 1 1. 4:16 decoder block 2. 2:4 decoder block 3. 8:1 multiplexer block 4.4:1 multiplexer block (P.S. All empty cells are do not care) (Handwriting solution is needed: include this in your pdf file) 1.

Analyze FFT-DIT for a sequence of x(n) = {8 1 4 5 6}. Sketch the FFT structure andlabel all signal in detail in the structure.please refer to the solution(image) given and answer for my given value

An analog signal in sinus form is desired to be transmitted to the receiver with pulse code modulation. Instead of xxx given below, you write a value. Peak-to-peak amplitude = 10 volts Frequency of the signal = 100 Hz Mid-rise one-shaped quantizer: 8 levels Encode the bit strings you will create with Bipolar signaling. Encode the bit strings you will create with M=xxx multilevel pulse code modulation.

Consider the 2-bit multiplexer (MUX) shown in the figure. For OUTPUT to be the XOR of C and D, the values for A0, A₁, A2 and A3 are ExamGOAL.COM ExamCAL COM EX - ExamGOAL.COM CUM EXPTIGUALTOM COM ExamGOAL.COM Exams AL.COM ExamAL.COM A Ao 0 COM ExamGOAL.COM ExamnCOAL.COM ExamGOAL.COM ExainGOAL.COM ExamGOAL.COM Exant COM ExamGOAL.COM ExamCOAL.COM ExamGOAL.COM ExamGOALCOM ExamGOAL.COM LEA A₁ COMAL1M EXAMGOAL.COM ExamBOAL.COM ExamNGOAL.COM OM ExamGOAL.COM ExamGOAL.COM ExamGOAL.COM ExamGOAL ExamGOAL.COM ExameCIAL.C A₂ 2 OUTPUT ExamGO.COM ExamGOAL.COM ExamGOAL.COM ExamGOAL.COM ExalGOAL.COM ExaH GOAL COM ARMGOAL COM ExamGOAL.COM ExamGOAL.COM F CON EXA3 3 3 E COM ExamGOAL.COM ExamGOAL.COM M ExamGOAL.COM ExamGOAL COM S₁ SELECT ExamGOAL COM ExamGOALOM ExamGOAL.COM ExamGOAL.COM GOAL.COM MGOAL.COM L.COM Exan ExainGOAL.COM ExainGOAL.COM So COM ExamGOAL.COM ExamGOAL.COM EMIGOAL.COMmGOAL.COM ExamGOAL.COM ExamGOAL.COM EMGOAL.COMmGOAL.COM ExamGOAL.COM ExamGOALCOM EMGOAL.COM ExamGOAL.COM ExamGOAL.COM…

Consider the 2-bit multiplexer (MUX) shown in the figure. For OUTPUT to be the XOR of C and D, the values for A0, A₁, A2 and A3 are COM ExamGOAL.COM ExamGL.COM EX OAL.COM EXURIGUALTOM ExamGOAL.COM COM ExamGOAL.COM ExamGOAL.COM ExamGOAL.COM ExamGOAL.COM ExamGOAL.COM DAD M ExamGOAL.COM 0 ExamGOAL.COM ExamGOAL.COM Exani Ao COM ExamGOAL.COM ExamGOAL.COM ExamGOAL.COM ExamGOAL.COM ExamGOAL.COM Exan ExamCOAL.COM ExamGOAL.COM Exam ExamGOAL.COM A₁ 1 COM ExamGOAL.COM ExamGDAL.COM ExamGOAL.COM LEXA COM EXDALOM ExamGOAL.COM A330 ExamGOAL.COM A₂2 ExamGOAL.COM OUTPUT ExamGOA.COM ExamGOAL.COM ExamGOAL.COM ExamGOAL.COM ExamGUAL.COM Exan MGOAL.COM ExamGOAL.COM Exan COM ExamGOAL.COM ExamCOAL.COM E COM EXA3 Exam lamGOAL.COM ExamGOAL.COM SMGOAL.COM ExamGOAL.COM COM ExamGOAL.COM ExamGOAL.COM COM ExamGOAL.COM ExamGOAL.COM ExamGOAL.COM COM ExamGOAL.COM ExamGOAL.COM EsamGOAL.COM xamGOAL.COM COM ExamGOAL.COM ExamGOAL.COM S₁ SELECT GOAL.COM MGOAL.COM ExamGOAL.COM ExamGOAL COM ExamGOAL.COM Exan COM ExamGOAL.COM…

The figure below shows a multiplexer where S, and S, are the select lines. I, to I, are the input data lines, EN is the enable line, and F(P, Q, R) is the output. F is EN R F MUX R 12 13 S, So P Q

Consider the 2-bit multiplexer (MUX) shown in the figure. For OUTPUT to be the XOR of C and D, the values for A0, A1, A2 and A3 are COM ExamGOAL.COM ExamGAL.COM EXAMIGOAL.COM TEXUTIGUALTOM ExamGOAL.COM COM ExamGOAL.COM ExamGOAL.COM ExamGOAL.COM ExamGOAL.COM ExamGOAL.COM COM Exam COM ExamGOAL.COM ExamGOAL.COM ExamGOAL.COM ExamGOAL.COM ExamGOAL.COM 0 ExamGOAL.COM ExamGOAL.COM ExamGOAL.COM Exan COM ExamGOAL.COM ExamCOAL.COM ExamGOAL.COM ExamGOAL.COM ExamGOAL.COM Exam A₁ 1.COM [EBAL1M Exam GOAL.COM ExamGOAL.COM ExamGOAL.COM Exam GOAL.COM COM Exam COM COM COM ExamGOAL.COM ExamCOAL.COM E SELECT A3 3E ExamGOAL.COM ExamicAL.COM SSOAL.COM S ExamGOAL.COM ExamGOAL.COM ExamGOAL.COM ExamGOAL.COM - ExamGOAL.COM ExamGOAL.COM Exam A2COM 24 ExamGOAL.COM OUTPUT ExamGOA COM ExamGOAL.COM ExamGOAL.COM ExamGOAL.COM ExamGOAL.COM Exan MGOAL.COM ExamGOAL.COM Exan ExamGOAL.COM Exan ExamGOAL SAGI COM ExamGOAL.COM ExamGOAL.COM ExamGOAL. C ExamGOAL.COM GOAL.COMMGOAL.COM ExamGOAL.COM Exan Exan EMGOAL.COMamGOAL.COM…

Convert the Moore state diagram of the following figure into an equivalent ASM chart. Check only one variable in each decision box. Minimize the number of decision boxes in each ASM block.

No paper

Need urgent help. Will give 5 stars if correct

3. Some GPS receivers use 1 bit ADCS (Analog-to-digital Converters) to sample and quantize the signal i.e. the output of the ADC is simply 2 quantization levels: -1 or +1. This is typically done to save on power. (a) What is the quantization SNR? How does it compare to typical thermal noise SNR of a GPS signal before correlating with the CDMA code? (b) Justify, how come it is possible to sample the signal using a single bit and still get good performance.

We want to design a digital circuit that converts Gray code (ABC) to Binary code (xyz). Set up a 8-to- 1 line multiplexer so that the output gives y as a function of ABC. "A" is given to the most significant bit of the control inputs of the multiplexer. "C" is given to the least significant bit of the control inputs of the multiplexer.

A major advantage of PRTS is that they are very linear. A disadvantage is that they have poor sensitivity. Select one: True False

1- A music signal has frequency content from 0 Hz up to 8.75 kHz. (a) What sampling frequency must be exceeded for successful A/D conversion? (b) What is another name for the minimum sampling frequency?

2) In an A-to-D process, what should be the sampling rate for a signal with maximum frequency of 16kHz ? 8000 samples per second. 16000 samples per second. 32000 samples per second. 64000 samples per second. 2) 6) A 6 kHz channel transmits a signal experiencing SNR of 15. What is the maximum Bit Rate of the channel?? 24000bps. 12000bps. 6Kbps. 6000bps

Determine the binary values for which the following SOP expression is equal to 1: XYZ + XYZ + XYZ + XYZ + XYZ Is this a standard SOP expression?

Section 3: Pulse Code Modulation (PCM) A standard encoding format for digital telephony is Pulse Code Modulation. Assume that the number of code levels used for PCM in a telephone system is 16. The following sequence of bits are the PCM encoded digital data received by a destination telephone (assume no errors). Question 9: 0101001100110010010001111010101010111100 On the figure below draw the output analog audio signal at the destination telephone. The horizontal dotted lines should be used as the levels, and the vertical dashed lines as the sample points.

Generate binary data from the given graph with the process of delta modulation. Writing the output bit stream is enough. No need to draw the signal.

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. X Y Figure_6 Part 4:The data-input and data-select waveforms in Figure 7 are applied to the multiplexer. Determine the output waveform F in relation to the inputs. DO D1 D2 D3 1. SO s1 Figure_7

For the above given signal each bit is separated by a vertical line. If total time is given as 12 ms, which answer is correct? O a. Data: 0 10001101011 Modulation Rate: 2 KBaud Encoding: Manchester O b. Data: 011001011110 Data Rate: 0,08 Kbps Encoding: Differantial Manchester c. Data: 011001011110 Modulation Rate: 1 Kbps Encoding: Manchester O d. Data: 010001101011Modulation Rate: 0,17 MBaud Encoding: Manchester O e. Data: 110010111101 Modulation Rate: 2 MBaud Encoding: Differantial Manchester

For the signal constellation shown below, Answer the following questions: 1. What type of modulation does this represent? 32-QAM 2. How many symbols are represented?! 3. How many bits per symbol are used? 4. If the baud rate is 50000 symbols per second, what is the bit duration? Rb = baude N baud=b Rb- 180 90 00000 00001 10001 10000 ● 00100 01100 01000 11000 11100 10100 00101 01101 01001 11001 11101 10101 00111 01111 01011 10111 11111 11011 ● · 00110 01110 01010 10110 11110 11010 • . . 00010 00011 10010 10011 270 ты Tb =(RD). ть Rb = Baud * N 0 (175) 2 2 AM = A Fe 200 Fm