Why Pe Is Not Proportional To Input SNR
Question
Why does the slide seem to say P_e is proportional to (S/N)_in?
Short Answer
Strictly, P_e should not be directly proportional to (S/N)_in.
If (S/N)_in means input signal-to-noise ratio, then higher input SNR normally makes the probability of error smaller:
[ (S/N)_{in} \uparrow \quad \Rightarrow \quad P_e \downarrow ]
So the slide’s last line is likely shorthand for “P_e depends on (S/N)_in”, or it is missing an inverse/decreasing relationship.
Claim status: INFERRED, based on the slide context and the standard meaning of signal-to-noise ratio.
Why Higher SNR Lowers Error Probability
Digital detection usually works by comparing a received signal value to a decision threshold.
An error happens when noise pushes the received value to the wrong side of that threshold.
- stronger signal → received levels are farther apart
- weaker noise → less random pushing
- larger SNR → less overlap between the possible received levels
- less overlap → lower probability of choosing the wrong code
That is why P_e is usually a decreasing function of SNR.
More Accurate Relationship
For many digital communication systems, P_e is not a simple linear proportion. It is more like:
[ P_e = Q(\sqrt{k \cdot SNR}) ]
or approximately decreases exponentially at high SNR:
[ P_e \approx e^{-k \cdot SNR} ]
The exact formula depends on the modulation and detection method, but the direction is the same:
[ SNR \uparrow \Rightarrow P_e \downarrow ]
How This Connects To The Slide Formula
The slide gives:
[ \left(\frac{S}{N}\right)_{out}
\frac{M^2}{1 + 4P_e(M^2 - 1)} ]
Here:
Mis the number of quantization levelsP_eis the probability of a transmission/decision error- higher
P_emakes the denominator larger - therefore higher
P_emakes output SNR worse
If:
[ P_e = 0 ]
then:
[ \left(\frac{S}{N}\right)_{out} = M^2 ]
This is the best case shown by the formula.
Practical Exam Interpretation
Read the last line as:
[ P_e \text{ is controlled by } (S/N)_{in} ]
not as:
[ P_e \propto (S/N)_{in} ]
A safer statement is:
[ P_e \propto \frac{1}{(S/N)_{in}} ]
as a rough intuition, or more accurately:
[ P_e = f((S/N)_{in}) ]
where f is a decreasing function.