Tag Info

1

Under appropriate assumptions, some consider $$\frac{\mathbb{E}[P]}{\mathbb{E}[N]}$$ and $$\mathbb{E}\left[\frac{P}{N}\right]$$ can be related through limits or approximations. One typical assumption can be that the variable at the denominator takes only positive values, with a non-vanishing expectation. This could suit to the case of looking at energies. ...

1

I have understood that SNR is often defined as ... That's the only sensible definition of SNR (assuming $P$ is signal power, and $N$ is noise power)! (other forms are just equivalent) Would the definition E[P/N] be equally intuitively valid? No. As you note yourself, it's mathematically simply not true. Long story short: if $P$ and $N$ are lognormal, then ...

5

Would the definition be equally intuitively valid? No. Using the expectation operator $\mathbb{E}$ implies that both $P$ and $N$ are functions of time. The problem with the second definition is that $P(t)$ can be occasionally very small or zero in which case the quotient gets very large or infinite. These "low noise" instances will dominate the ...

1

UPDATE**** Attempt # 2: After thinking about this overnight, I do not believe that looking at the peaks does anything useful. SNR is a measure of a signal power to signal noise, so I think I need to look at the power of these signals. With that in mind, here is my thought: We start with: S = clean signal power N = noise power Ps = observer S + n power (I can ...

6

Give me an A! Give me a D! Give me a converter! What have we got? An A/D converter! Go Team! Let $X$ denote a standard Gaussian random variable with pdf $\phi(x)$ and complementary CDF $Q(x)$. Let $Y$ denote a quantized version of $X$. First, for simplicity (and to flex our analytical muscles), suppose that Y = \begin{cases}+\alpha, & X \geq 0,\\ -\...

5

Thus, for a partial statistical characterization of the quantizer in terms of output signal-to-(quantization) noise ratio, we need only find the mean-square value of the quantization error Q. All that stuff does not imply that you need to find the mean-square value. This may just be an English a usage thing -- in most contexts, "we need only" ...

3

With the input M having zero mean and the quantizer assumed to be symmetric, it follows that the quantizer output V and, therefore, the quantization error Q will also have zero mean. That doesn't seem to be correct unless there are additional assumptions. It's easy enough to construct a counter example: Let's assume the quantization step is $1$ and the ...

Top 50 recent answers are included