Given an analog signal x(t) = 10 cos(11000 pit) + 5 sin(15000 pi t), for t>=0 sa

Question

Given an analog signal x(t) = 10 cos(11000 pit) + 5 sin(15000 pi t), for t>=0 sampled at a rate of 8000 Hz. Sketch the spectrum of the original signal and the sampled signal from 0 kHz and 25 kHz. Sketch the recovered analog signal spectrum if an ideal lowpass filter with a cutoff frequency of 4 kHz is used to filter the sampled signal. Determine the frequency/frequencies of aliasing noise. > Assume that a 3-bit ADC channel accepts analog input ranging from -2.5 to 2.5 volts, determine the following: Number of quantization levels; Step size of the quantizer or resolution; Quantization level when the analog voltage is 0.75 Volts; Binary code produced by the ADC; Quantization error. Work out the following problem: If the analog signal to be quantized is a sinusoidal waveform, that is, x(t) = 6 sin (6283t), and if the bipolar quantizer uses 8 bits, determine Number of quantization levels; Step size of the quantizer: The SNR. Given an analog signal x(t) = 5 cos(5000 pi t) + 2 cos(6400 pi t), for t>=0 sampledat a rate of 8000 Hz. Sketch the spectrum of the original signal and the sampled signal from 0 kHz and 25 kHz. Sketch the recovered analog signal spectrum if an ideal lowpass filter with a cutoff frequency of 4 kHz is used to filter the sampled signal.

Explanation / Answer

Q. assume a 3 bit adc ....

a) no. of levels = 2^n = 2^3 = 8 levels

b) resolution = Vrange/(2^n - 1) = (2.5 - (-2.5))/(2^3 - 1) = 5/7 V

c) 0.75 volt

range -2.5 to 2.5

level = (0.75 - (-2.5))/(5/7) + 1= 4.55 + 1 = 5.55 = 6 (rounding to integer)

d) 6th level = 101 ( starting from 000 which is frist)

e) quantization errror = V- + resolution*(level - 1) - actual value = -2.5 + (5/7)*5 - 0.75 = 9/28 V

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