Draw the signal constellation that used the QAM defined by the following table given below
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- 15,10 Signal 13,68 16 14 12 7.79 10 6,76g 1,37 1,89 3 4 Sample Number The given signal is digitized by using PCM using 4 bits. Which one is the correct data? O a. 000001111110001000010110 O b. 111100010010100000010110 O c. 111111011000100100010010 O d. 011111110010111001101110 e. 000101111111110100010110 Amplitude (mw) 00o4 203. 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.Calculate the offset for the physical address 00062 H and segment address FFFFH,
- Q25) For NRZ code, if Eb = 9.07 pico Joule and NO =1E-12, then the bit error rate is equal to a)1E-5 b)1E-6 c)none of the mentioned d)1E-4 This multiple choice question from DIGITAL COMMUNICATIONS course.just write for me the final answer.A 450 kHz carrier signal is modulated with a 15 kHz modulating signal. What are its sideband frequencies?(a) Draw a QPSK waveform for 11010010 (b) Draw a Multilevel FSK (M = 4) for the same bitstream as in (a).
- QPSK uses phases: 45°, 135°, 225°, 315°. If the duration of each signal is 200ns, then the data rate is a 10 Mb/s Ob 5 Mb/s Oc. 25 Mb/s Od. 100 Mb/s O e. 2.5 Mb/s CLEAR MY CHOICEQ.1: What is Quadrature Amplitude Shift Keying (QAM)? Explain with geometrically for symbol consisting of 4- bits. This means N=4. Prove that the minimum distance between the signal is d=2v0.4Eb. Your answerQ24)This multiple choice question from DIGITAL COMMUNICATIONS course.just write for me the final answer. The choice of the number of bits per PCM word depends on the a. SNR b. Bandwidth c. None of the mentioned d. SNR and bandwidth
- Question: 18 The time difference between the input signal crossing a 0.5VDD and the output signal crossing its 0.5 VDD when the output signal is changing from low to high is --- A. Propagation Delay low-to-high (Rise Propagation) B. Fall Transition Time C. Rise Transition Time D. Propagation delay high-to-low (Fall Propagation)please answer with solution Determine the “M” value and number of data bits that can be represented by a single symbol using PSK where the possible phase values per symbol can be {π/4, 3π/4, 5π/4, 7π/4}. a. M=2, N=1 bit per symbol b. M=4, N=2 bit per symbol c. M=6, N=3 bit per symbol d. M=8, N=3 bit per symbolFor RZ code, if PB = 1E-4 and NO =1 E-12, then the bit energy is equal to-------- : a. 13.7 pico joule b. 6.7 pico joule c. none of the mentioned d. 8.5 nano joule This multiple choice question from DIGITAL COMMUNICATIONS course.just write for me the final answer.