Question

Consider the signal $x\left(t\right)=\cos \left(6\pi t\right)+\sin \left(8\pi t\right)$, where t is in seconds. The Nyquist sampling rate (in samples/second) for the signal y(t)=x(2t+5) is

• A. 16
• B. 12
• C. 32
• D. 8

Answer 1

The correct option is A 16

$x\left(t\right)=\cos \left(6\pi t\right)+\sin \left(8\pi t\right)$
$Y\left(t\right)=x(2t+5)$
$Y\left(t\right)=\cos \left[6\pi \right(2t+5)+\sin (8\pi (2t+5)\left)\right]$
$=\cos (12\pi t+30\pi )+\sin (16\pi t+40\pi )$
$f_{m_1}=6Hz;f_{m_2}=8Hz$

$\text{Nyquist sampling rate,}$
$f_s=2f_{max}$
$\text{= 16 samples/second}$