the minimum sampling rate (according to Nyquist frequency) for the signal [ x(t) =2 ? cos 6000 Ttt + 5 cos 10000 Ttt ] is 5000HZ 10000HZ 12000HZ 20000HZ
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- The Nyquist sampling rate required to sample the signal x(t)= [cos(6000 nt).cos(2000 nt)] is equal to. O 4000 Hz 8000 Hz O 2000 Hz O 6000 Hzthe minimum sampling rate (according to Nyquist frequency) for the signal [ x(t) =2 * ? cos 6000 Tt + 5 cos 10000 Ttt ] is 5000HZ 10000HZ 12000HZ 20000HZThe Nyquist sampling rate required to sample the signal x(t)= [sin (1000 zt)] is equal to. O 200 Hz O 1000 Hz O 2000 Hz O 100 Hz
- Qla. To digitize an analog signal m(t) whose max bandwidth is 10 kHz, the signal is sampled at Nyquist rate with a guard band of 2 kHz. The pdf of the signal is shown below: -8 1/3 fx (x) 1/6 -8 X If the PCM encoder is an 8-bit encoder Determine the resulting bit rate (b) Determine the SNR range of a uniform quantizer output (1which is trefers to time shifting of discrete time signal: a.x(n) = z(n-k) b.x(n) = z(-n-k) c.x(n) =-z(n-k) d.x(n) = z(n+k)A continuous-time signal is sampled with a sampling time interval of 0.0012 sec. then the frequency of the signal is_ Hz 1666.67 O 833.33 416.67 O 555.56
- A signal x[n] is quantized using an A/D converter (ideal uniform quantization) having a range of ±54 volts. Determine the number of bits B if the equally-sized quantization step, A, is 0.34 volts. Provide the integer number of the value of B as your answer.The analog signal x(t) = 2sin(1000nt) — sin (2000nt +) is applied to an analog-to-digital converter (ADC) module for converting the signal to digital format. The ADC has 3 bits quantizer with ± 3Vp range. i. Produce the first EIGHT (8) sampled signals when the sampling frequency is 12.5 kHZ ii. Calculate the quantized signal and encoded digital signal to represent the sampled signal obtained in (b) (i) using quantization by rounding. iii. Compute the quantization signal to noise ratio (SNR).Bandwidth of x(t) 5 cos(1000t) 500 500n cs l ow koveoeewemReepewse 57 A 2000 500/n 1000/n Ideal sampling is used Delta signal for sampling Sine wave signal for sampling Triangular signal for sampling Cosine wave signal for sampling Pulse signal for sampling
- 2000rn 1. Given sinusoid x (n) = 2 x sin obtained by the 8000 sampling analog signal x (t) = 2 x sin(2000tt) with the sampling rate of fs=80O0HZ a. Use the MATLAB DFT to compute the signal spectrum with the frequency resolution to be equal to or less than 8Hz b.Use the MATLAB FFT and zero padding to compute the signal spectrum, assuming that the data samples are available in (а).Discrete time signal x(n) for x(t) = 3 cos(500rt) when fs 1000 Hz. Your answerQ10/ --- is minimum sampling rate to avoid aliasing when a continuous time signal is given by x(t) = 5 cos 40πt. Oa) 10 Hz Ob) 20 Hz Oc) 40 Hz d) 800 Hz