Wireless 101
M09 · L01
Module 9 · Lesson 1

The Floor You Have Been Standing On

Every link budget since M2-L4, Shannon’s capacity, every BER curve — all of it assumed a noise floor nobody derived. One constant and one temperature is all it takes.

01 / 11
Wireless 101
M09 · L01
Johnson & Nyquist, 1928

Warm Things Hiss

Charge carriers in any resistor above absolute zero are in thermal motion, so a resistor makes noise. What Nyquist proved is what the answer leaves out: not resistance, not material — only temperature and bandwidth.

Available thermal noise power
N = kTB \;\;\Big|\;\; N_0 = \tfrac{N}{B} = kT
02 / 11
Wireless 101
M09 · L01
T₀ = 290 K, the IEEE reference

Memorise −174

The noise floor in dBm
N_{\text{dBm}} = -174 + 10\log_{10} B_{\text{Hz}}

kT₀ = 1.380649×10⁻²³ × 290 = 4.0039×10⁻²¹ W/Hz = 4.0039×10⁻¹⁸ mW/Hz, and 10log₁₀ of that is −173.98 dBm/Hz. At 300 K you would get −173.8 — pick one and stay with it.

03 / 11
Wireless 101
M09 · L01
−174 + 10log₁₀B

Bandwidth Is Noise

Widening 20 MHz to 100 MHz is a factor of five, which is 7 dB — and the floor rises by exactly 7 dB. Wide channels buy capacity with noise, which is the trade Shannon priced in M6-L1.

1 MHz · +60.0
−114 dBm
20 MHz · +73.0
−101 dBm
100 MHz · +80.0
−94 dBm
04 / 11
Wireless 101
M09 · L01
SNR and its relatives

Say Which Ratio You Mean

SNR, Eb/N₀ and spectral efficiency
\mathrm{SNR} = \tfrac{E_b R_b}{N_0 B} = \tfrac{E_b}{N_0}\cdot\tfrac{R_b}{B}

One radio, three numbers: 20 dB of SNR in 20 MHz is C/N₀ = 93 dB-Hz and, at 4 bit/s/Hz, Eb/N₀ = 14 dB. Quote the wrong one and the budget is out by 73 dB.

05 / 11
Wireless 101
M09 · L01
Try it — build a sensitivity

Floor, Noise Figure, Sensitivity

Three sliders, four numbers. Watch the ladder climb from thermal floor to the weakest signal this receiver can use.

Bandwidth 20.0 MHz
Noise fig NF 5.0 dB
Req SNR SNR 20 dB
thermal −101.0 dBm with NF −96.0 dBm sensitivity −76.0 dBm Shannon 133 Mb/s
06 / 11
Wireless 101
M09 · L01
Noise figure

What the Receiver Adds

Noise factor and equivalent noise temperature
F = \tfrac{\mathrm{SNR}_{in}}{\mathrm{SNR}_{out}} \;\Big|\; T_e = T_0(F-1)

NF 1 dB is F = 1.259, so Te = 290 × 0.259 = 75 K. NF 3 dB is F = 1.995, so Te = 289 K — the receiver doubles the noise. F says nothing about gain.

07 / 11
Wireless 101
M09 · L01
Friis, 1944 — linear ratios only

The First Stage Decides

Cascaded noise factor
F_{tot} = F_1 + \tfrac{F_2-1}{G_1} + \tfrac{F_3-1}{G_1G_2} + \cdots

LNA (1 dB, G 20 dB) → mixer (8 dB, G 10 dB) → IF (15 dB): 1.259 + 5.31/100 + 30.6/1000 = 1.3427, i.e. NF 1.28 dB. Put the mixer first and the same parts give 6.31 + 0.0259 + 0.0306 = 6.366, i.e. 8.04 dB.

08 / 11
Wireless 101
M09 · L01
Worked — 20 MHz, NF 5 dB, 64-QAM

Sensitivity in Four Terms

Receiver sensitivity
S_{\text{dBm}} = -174 + 10\log_{10}B + \mathrm{NF} + \mathrm{SNR}_{req}

−174 + 73.0 + 5 + 20 = −76 dBm. Real radios miss that by an implementation margin — and in a crowded band the floor they actually see is set by neighbours, not by kTB (M9-L3).

09 / 11
Wireless 101
Knowledge Check

Check what stuck

Four questions from this lesson. Answer to see why — the explanation appears whether you were right or wrong. Nothing is scored or saved.

Question 1 of 0
Score 0/0

10 / 11
Wireless 101
M09 · L01
Recap

What you learned

  • N = kTB, and kT₀ at 290 K is −174 dBm/Hz
  • Any floor is −174 + 10log₁₀B: −101 dBm in 20 MHz
  • SNR = (Eb/N₀)(Rb/B); C/N₀ adds 10log₁₀B
  • NF = 10log₁₀F and Te = T₀(F−1): 3 dB is 289 K
  • Friis: 1.28 dB LNA-first, 8.04 dB mixer-first
Up next in Module 9
Bit Error Rate →
All lessons Read article
11 / 11