1 Crore+ students have signed up on EduRev. Have you? 
A linear delta modulator is designed to operate on speech signals limited to 3.4 kHz. The sampling rate is 10 time the Nyquist rate of the speech signal. The step size δ is 100 mV. The modulator is tested with a this test signal required to avoid slope overload is
Consider a linear DM system designed to accommodate analog message signals limited to bandwidth of 3.5 kHz. A sinusoidal test signals of amplitude A_{max} = 1 V and frequency f_{m} = 800 Hz is applied to system. The sampling rate of the system is 64 kHz.
Que: The minimum value of the step size to avoid overload is
Consider a linear DM system designed to accommodate analog message signals limited to bandwidth of 3.5 kHz. A sinusoidal test signals of amplitude A_{max} = 1 V and frequency f_{m} = 800 Hz is applied to system. The sampling rate of the system is 64 kHz.
Que: The granularnoise power would be
The output signaltoquantizationnoise ratio of a 10bit PCM was found to be 30 dB. The desired SNR is 42 dB. It can be increased by increasing the number of quantization level.In this way the fractional increase in the transmission bandwidth would be (assume log_{2}10 = 0.3)
= log C + 20 nlog2 = α + 6 ndB. This equation shows that increasing n by one bits increase the by 6 dB.
Hence an increase in the SNR by 12 dB can be accomplished by increasing 9is form 10 to 12, the transmission bandwidth would be increased by 20%
A signal has a bandwidth of 1 MHz. It is sampled at a rate 50% higher than the Nyquist rate and quantized into 256 level using a low quantizer with μ = 225.
Que: The signaltoquantizationnoise ratio is
A signal has a bandwidth of 1 MHz. It is sampled at a rate 50% higher than the Nyquist rate and quantized into 256 level using a low quantizer with μ = 225.
Que: It was found that a sampling rate 20% above therate would be adequate. So the maximum SNR, thatcan be realized without increasing the transmission bandwidth, would be
Nyquist Rate = 2 MHz
For a PCM signal the compression parameter μ = 100 and the minimum signal to quantizationnoiseratio is 50 dB. The number of bits per sample would be.
A sinusoid massage signal m(t) is transmitted by binary PCM without compression. If the signaltoquantizationnoise ratio is required to be at least 48dB, the minimum number of bits per sample will be
A speech signal has a total duration of 20 sec. It is sampled at the rate of 8 kHz and then PCM encoded. The signaltoquantization noise ratio is required to be 40dB. The minimum storage capacity needed to accommodate this signal is
The input to a linear delta modulator having f_{a }stepsize Δ = 0.628 is a sine wave with frequency f_{m} and peak amplitude E_{m}. If the sampling frequency f_{s} = 40 kHz, the combination of the sincwave frequency and the peak amplitude, where slope overload will take place is
A sinusoidal signal with peaktopeak amplitude of 1.536 V is quantized into 128 levels using a midrise uniform quantizer. The quantizationnoise power is
signal is sampled at 8 kHz and is quantized using 8 bit uniform quantizer. Assuming SNR_{q} for a sinusoidal signal, the correct statement for PCM signal with a bit rate of R is
Bit Rate = 8k x 8 = 64 kbps
A 1.0 kHz signal is flattop sampled at the rate of 180 samples sec and the samples are applied to an idealrectangular LPF with catoff frequency of 1100 Hz, then the output of the filter contains
Since the sampling rate is 1800 samples/sec the highest frequency that can be recovered is 900 Hz.
The Nyquist sampling interval, for the signal sinc(700t) + sinc(500t) is
x(t) = sinc 700t + sinc 500t
A signal x(t) = 100 cos(24π x 10^{3})t is ideally sampled with a sampling period of 50 sec and then passed through an ideal lowpass filter with cutoff frequency of 15 KHz. Which of the following frequencies is/are present at the filter output
The frequency passed through LPF are fc ,fs  fm or 12 kHz, 8 kHz
In a PCM system, if the code word length is increased form 6 to 8 bits, the signal to quantization noise ratio improves by the factor.
Four signals g_{1}(t) , g_{3}(t), g_{2}(t) and g_{4}(t) are to be multiplexed and transmitted. g_{1}(t) and g_{4}(t) have a bandwidth of 4 kHz, and the remaining two signals have bandwidth of 8 kHz,. Each sample requires 8 bit for encoding. What is the minimum transmission bit rate of the system.
signals g_{1}(t) , g_{3}(t), g_{2}(t) and g_{4}(t) will have 8 k, 8 k, 16 k and 16 k sample/sec at Nyquist rate. Thus 48000 sample/sec bit rate 48000 x 8 =384 kbps
Three analog signals, having bandwidths 1200 Hz, 600 Hz and 600 Hz, are sampled at their respective Nyquist rates, encoded with 12 bit words, and time division multiplexed. The bit rate for the multiplexed signal is
Analog signals, having bandwidth 1200 Hz, 600 Hz and 600 Hz have 2400, 1200 samples/sec at Nyquist rate. Hence 48000 sample/sec
bit rate = 48000 sample/sec x 12 = 57.6 kbps
The minimum sampling frequency (in samples/sec) required to reconstruct the following signal form its samples without distortion would be
Maximum frequency component = 3 x 1000 = 3 kHz Sampling rate = 2 f_{m} = 6kHz
19 docs177 tests

Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 
19 docs177 tests






