Question

Consider a system with sampling rate 52,500 samples per second.

Answer 1

Answer:

The answer is "$22,246 \ H_z \ \ and \ \ 7.3\%$"

Step-by-step explanation:

The sampling rate: $(f_s) = 52,500 \ \frac{samples}{sec}$

In point a:

The sample signal frequency = f  

They recognize that the id $\geq$ 2f sampling frequency could be returned to its original signal unless the aliase effect is detected elsewhere,  Yet under percentage =18%

$\ sampling \ rate = 2 \times 1.18 \times f$

$\to f = \frac{52,500}{2 \times 1.18}$

       $= 22,246 \ h_z$

In point b:

Maximum 22,242 frequency is increased by $10 \ \%$

$f_a =22,246 \times 1.1$

    $= 24,470.6\\\\=24,471$

$\ sampling \ rate= x \times 2 \times 24,2471$

                    $x= \frac{52,500}{48,942} \\\\x= 1.073$

over sampling percentage $= 7.3 \ \%$