Question

$\text{A signal}x\left(t\right)=sin\left(150\pi t\right)\text{is sampled at a rate of five samples per three time periods.}\text{Then the ratio of sampling frequency to the Nyquist rate is\_\_\_\_\_\_\_\_\_\_}$

• A. 0.83

Answer 1

The correct option is A 0.83
$\text{Given}x\left(t\right)=sin\left(150\pi t\right)\text{Time period,}T=\frac{2\pi }{{\omega }_0}=\frac{2\pi }{150\pi }=\frac{1}{75}sec\text{3 times periods = 3}\times T=3\times \frac{1}{75}=\frac{1}{25}sec\therefore \text{The signal sampled at a rate of five samples is}\frac{1}{25}sec\text{So, 1 sample in}\frac{1}{125}sec=T_0\text{[sampling interval]}\therefore \text{Sampling frequency = f}{}_s=\frac{1}{T_s}=125\text{samples/sec}\text{also, Nyquist rate =}{f}_N=2f_m=2\times 75[\because {\omega }_m=150\pi \Rightarrow f_m=75Hz]\text{=150 samples/sec}\therefore \text{The ratio,}\frac{f_s}{f_N}=\frac{125}{150}=\frac{5}{6}=0.83$