advanced39 min studyLesson 747 in Mathematics

Gaussian channels and water filling

Fix real-use and power conventions before allocating signal power across independent Gaussian noise modes.

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What you’ll understand

  • Derive real scalar AWGN capacity from the Gaussian maximum-entropy property.
  • Allocate a total per-vector power budget by water filling over independent real modes.
  • Separate differential-entropy scale changes from invariant mutual information.

Gaussian noise turns a power budget into an information constraint. The capacity formula depends on whether one use is a real scalar, a complex symbol or a vector of parallel modes, so a factor of two can encode a modeling choice rather than a numerical slip. Water filling then allocates power to modes whose noise floors justify it. A noisier mode can receive no signal power even though the total budget is positive; equal power is not automatically the best allocation.

One real use fixes the factor one half

Consider Y=X+Z with independent zero-mean Gaussian noise of variance N>0. Uses are memoryless and the block codewords satisfy squared norm at most nP, where P≥0 is the average power per real use. The corresponding single-use information maximization allows E X²≤P. Differential entropy of the output is bounded by that of a Gaussian with the same variance, while the conditional output entropy equals the entropy of the independent noise.

A zero-mean Gaussian input of variance P attains the information maximum, yielding half log₂ of one plus signal-to-noise ratio. The operational coding theorem can enforce the block power constraint through suitable long code constructions rather than assuming every iid Gaussian draw already obeys it. A complex use contains two real dimensions under its own total noise and power convention; importing its formula without translating units can silently double a claimed real-use rate.

h(Z)=12log⁡2(2πeN),Creal=max⁡EX2≤PI(X;X+Z)=12log⁡2(1+PN)h(Z)=\frac12\log_2(2\pi eN),\qquad C_{\mathrm{real}}=\max_{\mathbb E X^2\le P}I(X;X+Z)=\frac12\log_2\left(1+\frac PN\right)
The noise variance is positive and the noise is independent of the input. Differential entropies use consistent coordinate units; capacity is bits per real scalar use.↔ Scroll wide equations horizontally.

Equalize marginal gains above the noise floors

Now one vector use contains m independent real Gaussian noise modes with strictly positive variances N_i. Allocate powers P_i≥0 with their sum at most one total budget P. Independent Gaussian input components maximize information for a chosen allocation, and the rates add across modes. The resulting total is bits per vector use, not per scalar component. Dividing by m would be a separate normalization, while a time or bandwidth rate needs an additional physical sampling convention.

The objective is concave in the powers. For active modes, the derivative is proportional to one divided by N_i+P_i, so the constrained optimum equalizes those total levels. Inactive modes have noise already above the common level. The level is chosen so the allocated powers sum to P when P is positive. This Karush–Kuhn–Tucker argument also explains why a mode begins receiving power only when the water level reaches its noise floor.

C=max⁡Pi≥0∑iPi≤P12∑ilog⁡2(1+PiNi),Pi=(μ−Ni)+,∑i(μ−Ni)+=PC=\max_{\substack{P_i\ge0\\\sum_iP_i\le P}}\frac12\sum_i\log_2\left(1+\frac{P_i}{N_i}\right),\qquad P_i=(\mu-N_i)_+,\qquad\sum_i(\mu-N_i)_+=P
The noise modes are independent, the budget is per vector use, and (a)_+=max(a,0). For P=0 every allocated power is zero; a convenient water level can be chosen below all noise floors.↔ Scroll wide equations horizontally.

Entropy changes with units while information survives invertible changes

Differential entropy can be negative and changes under a nonsingular linear coordinate map by the logarithm of its absolute determinant. That does not contradict information invariance: applying a bijective measurable change to one observed variable preserves its mutual information with another. In the Gaussian proof, coordinate-dependent differential entropies occur in a difference whose scale terms cancel. Treating differential entropy alone as a count of distinguishable messages loses that cancellation.

The water-filling result assumes the stated independent additive Gaussian modes and quadratic sum constraint. Correlated noise can be diagonalized when the available transformations and input constraint justify doing so. Peak-amplitude restrictions, non-Gaussian noise or a different cost can change the optimizing distribution and allocation. A channel designer must therefore identify the allowed signal set and noise law before interpreting a logarithmic expression as an achievable capacity.

Fill two different noise floors

Two real parallel modes have noise variances one and four. A vector use has total power five. Find the optimum allocation and total capacity.

  1. If both modes are active, (mu−1)+(mu−4)=5, so mu=5. This is above both noise floors, validating the active-set assumption.

  2. The powers are four and one. Their signal-to-noise ratios are four and one quarter, respectively.

  3. Capacity is one half log₂5 plus one half log₂1.25, or one half log₂6.25=log₂2.5≈1.321928 bits per vector use. The allocation spends more power on the quieter mode while equalizing total levels.

The result

The optimum is (4,1). Its total rate should not be reported as bits per scalar mode without applying the extra normalization.

Explore the model

Interactive lab

Interactive lab · matrix

Track a coordinate-entropy change without inventing information

Pause and reset, then set both controls to a = 1 and b = 0, predicting determinant one. For a fresh comparison, pause and reset, then set a = 2 and b = 0, predicting determinant two. For a two-dimensional density with finite differential entropy, this selected transformation adds log₂2=one bit to differential entropy, yet preserves mutual information with another variable because it is invertible. The instrument displays geometry and determinant, not Gaussian-channel capacity; readouts describe the selected final matrix during playback.

Track a coordinate-entropy change without inventing informationThe selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Current readings: Area scale · det A: 1.4; Singular values · σmax, σmin: 1.597, 0.876; Condition number · κ₂(A): 1.823. All parameters are adjustable with the labeled controls below.Matrix A with first row 1.4, 0.6 and second row zero, one; display interpolates between identity and A while final determinant remains 1.4.−4−2024−2−1012x · equal geometric scaleyA = [ 1.4 0.6 ; 0 1 ]Display blend: 0% · final values below
Interpolated geometryOriginal unit squareSelected final transform
Area scale · det A
1.4
Singular values · σmax, σmin
1.597, 0.876
Condition number · κ₂(A)
1.823
t = 0 s

The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forCoordinate-dependent differential entropy and invariant mutual information answer different questions.

Check your understanding

Why can an optimum parallel Gaussian allocation leave a noisy mode unused?

Try it yourself

Two real modes have noise variances one and three and total power one per vector use. Find the allocation and capacity. At what total power does the second mode first become eligible for positive allocation?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Test an active set containing only the quieter mode, then compare the resulting water level with three.

Compare with the solution

Putting the full unit into the first mode sets mu=2, below the second noise floor. Hence the optimum is (1,0) and capacity is half log₂2=0.5 bits per vector use. The level reaches three when the first allocation reaches two, so the threshold total power is two. At exactly that threshold the second allocation is still zero; it becomes positive for larger budgets. This active-set check is necessary before solving a formula that assumes both modes receive power.

Take it with you

  • Real, complex and vector uses require consistent noise, power and rate units.
  • Water filling equalizes active signal-plus-noise levels under a quadratic sum budget.
  • Invertible coordinate changes alter differential entropy without altering mutual information.

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