Answer:
The answer is "[tex]22,246 \ H_z \ \ and \ \ 7.3\%[/tex]"
Step-by-step explanation:
The sampling rate: [tex](f_s) = 52,500 \ \frac{samples}{sec}[/tex]
In point a:
The sample signal frequency = f
They recognize that the id [tex]\geq[/tex] 2f sampling frequency could be returned to its original signal unless the aliase effect is detected elsewhere, Yet under percentage =18%
[tex]\ sampling \ rate = 2 \times 1.18 \times f[/tex]
[tex]\to f = \frac{52,500}{2 \times 1.18}\\\\[/tex]
[tex]= 22,246 \ h_z[/tex]
In point b:
Maximum 22,242 frequency is increased by [tex]10 \ \%[/tex]
[tex]f_a =22,246 \times 1.1[/tex]
[tex]= 24,470.6\\\\=24,471[/tex]
[tex]\ sampling \ rate= x \times 2 \times 24,2471 \\[/tex]
[tex]x= \frac{52,500}{48,942} \\\\x= 1.073\\\\[/tex]
over sampling percentage [tex]= 7.3 \ \%[/tex]
