By combining $\Delta=\frac{R}{2^{b+1}}$ with $P_n=\sigma_e^2=\Delta^2/12$ and substituting the result into SQNR = 10 $log_{10}⁡ \frac{P_x}{P_n}$, what is the final expression for SQNR = ?

Question

TuteeHUB · Accepted Answer

NA

TuteeHUB · Answer

6.02b + 16.81 + $20log_{10}\frac{R}{σ_x}$

TuteeHUB · Answer

6.02b + 16.81 – $20log_{10}⁡ \frac{R}{σ_x}$

TuteeHUB · Answer

6.02b – 16.81 – $20log_{10}⁡ \frac{R}{σ_x}$

TuteeHUB · Answer

6.02b – 16.81 – $20log_{10}⁡ \frac{R}{σ_x}$

1.	By combining \(\Delta=\frac{R}{2^{b+1}}\) with \(P_n=\sigma_e^2=\Delta^2/12\) and substituting the result into SQNR = 10 \(log_{10}⁡ \frac{P_x}{P_n}\), what is the final expression for SQNR = ?
A.	6.02b + 16.81 + \(20log_{10}\frac{R}{σ_x}\)
B.	6.02b + 16.81 – \(20log_{10}⁡ \frac{R}{σ_x}\)
C.	6.02b – 16.81 – \(20log_{10}⁡ \frac{R}{σ_x}\)
D.	6.02b – 16.81 – \(20log_{10}⁡ \frac{R}{σ_x}\)
Answer» C.

By combining \(\Delta=\frac{R}{2^{b+1}}\) with \(P_n=\sigma_e^2=\Delta^2/12\) and substituting the result into SQNR = 10 \(log_{10}⁡ \frac{P_x}{P_n}\), what is the final expression for SQNR = ?

Discussion

No Comment Found

Related MCQs

Reply to Comment