example of Electromagnetic Waves, Antennas, and RF: Thermal Noise, Noise Temperature, Noise Figure, G/T, and C/N0

Electromagnetic Waves, Antennas, and RF: Thermal Noise, Noise Temperature, Noise Figure, G∕T, and C∕N₀ - Exercises and Complete Worked Solutions

This companion to EM26 turns receiver-noise theory into calculation practice. The exercises begin with Johnson–Nyquist noise and kTB, move through equivalent noise temperature and noise figure, and then build complete cascaded receiver calculations. The final problems combine system noise temperature, G∕T, C∕N₀, and bandwidth-dependent C∕N in RF and GNSS-style examples [1, 2, 3, 4, 5].

The principal relations are

(1)

(2)

(3)

(4)

(5)

and

(6)

Use

(7)

Unless otherwise stated, gains and losses used in cascade formulas are power ratios in linear units.

Part I: Exercises

Exercise 1: Johnson noise voltage and matched noise power

A 50.0 Ω resistor is at 290 K and is observed over a 1.00 MHz bandwidth. Find (a) the RMS open-circuit Johnson-noise voltage, (b) the available noise power delivered to a matched load, and (c) that noise power in dBm. Verify that the matched power is independent of the resistance value.

Exercise 2: thermal-noise power in a receiver bandwidth

At 290 K, calculate the available thermal-noise power in a 2.00 MHz bandwidth. Express the result in watts, dBW, and dBm. Compare it with the familiar −174 dBm/Hz room-temperature noise density.

Exercise 3: noise density at a colder temperature

Find N₀ = kT at T = 150 K in (a) W/Hz, (b) dBW/Hz, and (c) dBm/Hz. By how many decibels is this below the 290 K noise density?

Exercise 4: noise figure to equivalent noise temperature

An amplifier has noise figure NF = 1.20 dB. Find its linear noise factor F and equivalent input noise temperature T_e referred to T₀ = 290 K.

Exercise 5: equivalent noise temperature to noise figure

A receiver front end has equivalent input noise temperature T_e = 80.0 K. Find its noise factor and noise figure.

Exercise 6: three-stage cascaded receiver

A three-stage receiver has

G1	= 20.0 dB,	NF1	= 1.00 dB,	(8)
G2	= 15.0 dB,	NF2	= 4.00 dB,	(9)
G3	= 10.0 dB,	NF3	= 6.00 dB.	(10)

Calculate the total input-referred noise factor, total noise figure, and total equivalent input noise temperature. Quantify how strongly the first-stage gain suppresses stages 2 and 3.

Exercise 7: why loss before an LNA is expensive

A cable has 1.50 dB power loss at 290 K. An LNA has 1.00 dB noise figure and 20.0 dB gain. Compare two cases: (a) cable before LNA and (b) cable after LNA. Find the total cascade noise figure and equivalent input noise temperature in both cases.

Exercise 8: antenna temperature and system noise temperature

An antenna has noise temperature T_A = 120 K and a receiver referred to the same input plane has equivalent noise temperature T_e,rx = 80.0 K. Find (a) T_sys, (b) N₀ in W/Hz, and (c) N₀ in dBW/Hz and dBm/Hz.

Exercise 9: receiver figure of merit G∕T

A receiving antenna has directional gain 28.0 dBi and the complete receiver system has T_sys = 150 K at the same reference plane. Calculate G∕T in dB/K. Explain what physical change would improve G∕T by 3 dB.

Exercise 10: C∕N₀ from carrier power and system temperature

At the receiver reference plane, the carrier power is C = −155 dBW and T_sys = 250 K. Calculate the noise density in dBW/Hz and the resulting C∕N₀ in dB-Hz.

Exercise 11: direct C∕N₀ link budget using G∕T

An illustrative RF link has

(11)

Use the direct link-budget form to find C∕N₀. Use −10 log ₁₀k ≈ 228.60 dB-K/Hz.

Exercise 12: converting C∕N₀ to C∕N

A receiver measures C∕N₀ = 45.0 dB-Hz. What is C∕N in a noise-equivalent bandwidth of 2.00 MHz?

Exercise 13: infer receiver bandwidth from C∕N₀ and C∕N

A receiver has C∕N₀ = 48.0 dB-Hz and measured C∕N = 8.00 dB. Assuming thermal noise dominates and the stated bandwidth is a noise-equivalent bandwidth, determine B.

Exercise 14: bandwidth penalty

A fixed carrier has C∕N₀ = 50.0 dB-Hz. Compute C∕N for bandwidths of 10.0 kHz and 20.0 kHz. Show explicitly that doubling bandwidth degrades C∕N by approximately 3.01 dB.

Exercise 15: GNSS-style weak-signal example

An illustrative GNSS-like receiver has carrier power C = −158.5 dBW and system noise temperature T_sys = 400 K. Find (a) N₀ in dBW/Hz, (b) C∕N₀ in dB-Hz, and (c) C∕N in a 2.00 MHz front-end noise bandwidth. Interpret how C∕N can be negative while C∕N₀ remains a useful receiver-quality metric.

Exercise 16: reference-plane bookkeeping with feed loss and an LNA

At the antenna terminals, T_A = 100 K. A 1.00 dB feed loss at 290 K precedes an LNA having NF = 0.80 dB and 25.0 dB gain. Refer all noise to the antenna-terminal plane. Find (a) the feed equivalent noise temperature, (b) the LNA equivalent input noise temperature, (c) total system noise temperature, and (d) G∕T if the receive antenna gain is 25.0 dBi. If the carrier at the same antenna-terminal reference plane is C = −160 dBW, also find C∕N₀.

Part II: Complete Worked Solutions

Solution 1: Johnson noise voltage and matched noise power

The open-circuit mean-square voltage is

(12)

Therefore

vn,rms	=	(13)
	= 8.95 × 10−7 V.	(14)

Thus

(15)

With a matched load, only half the open-circuit voltage appears across the load. The delivered power is

N	=	(16)
	=	(17)
	= kTB.	(18)

Hence

N	= (1.380649 × 10−23)(290)(1.00 × 106)	(19)
	= 4.0039 × 10−15 W.	(20)

In dBm,

NdBm	= 10 log 10	(21)
	= −113.98 dBm.	(22)

Therefore

(23)

The resistance cancels algebraically. Changing R changes the noise voltage, but a matched resistor at the same temperature and bandwidth has the same available noise power kTB.

Solution 2: thermal-noise power in a receiver bandwidth

Using

(24)

we obtain

N	= (1.380649 × 10−23)(290)(2.00 × 106)	(25)
	= 8.0078 × 10−15 W.	(26)

Thus

(27)

The logarithmic values are

NdBW	= 10 log 10(8.0078 × 10−15) = −140.96 dBW,	(28)
NdBm	= −110.96 dBm.	(29)

At 290 K the density is approximately −173.98 dBm/Hz. A 2.00 MHz bandwidth contributes

(30)

so

(31)

in agreement with the direct calculation.

Solution 3: noise density at a colder temperature

At 150 K,

N0	= kT	(32)
	= (1.380649 × 10−23)(150)	(33)
	= 2.0710 × 10−21 W/Hz.	(34)

Therefore

N0,dBW/Hz	= 10 log 10(2.0710 × 10−21) = −206.84 dBW/Hz,	(35)
N0,dBm/Hz	= −176.84 dBm/Hz.	(36)

Thus

(37)

Relative to 290 K, the decrease is

10 log 10

= −2.86 dB.

(38)

So cooling from 290 K to 150 K lowers thermal-noise density by about 2.86 dB.

Solution 4: noise figure to equivalent noise temperature

Convert noise figure to linear noise factor:

F	= 10NF∕10	(39)
	= 101.20∕10	(40)
	= 1.3183.	(41)

Then

Te	= (F − 1)T0	(42)
	= (1.3183 − 1)(290)	(43)
	= 92.3 K.	(44)

Therefore

(45)

Solution 5: equivalent noise temperature to noise figure

Starting from

(46)

we find

F	= 1 +	(47)
	= 1.27586.	(48)

Therefore

NF	= 10 log 10(1.27586)	(49)
	= 1.058 dB.	(50)

Hence

(51)

Solution 6: three-stage cascaded receiver

First convert all gains and noise figures to linear ratios:

G1	= 1020∕10 = 100,	(52)
G2	= 1015∕10 = 31.6228,	(53)
G3	= 1010∕10 = 10,	(54)

and

F1	= 101∕10 = 1.25893,	(55)
F2	= 104∕10 = 2.51189,	(56)
F3	= 106∕10 = 3.98107.	(57)

The cascade noise factor is

Ftot	= F1 + +	(58)
	= 1.25893 + +	(59)
	= 1.27499.	(60)

Thus

NFtot	= 10 log 10(1.27499)	(61)
	= 1.055 dB.	(62)

The equivalent input temperature is

Te,tot	= (Ftot − 1)T0	(63)
	= (0.27499)(290)	(64)
	= 79.7 K.	(65)

Therefore

(66)

The stage-2 contribution is only

(67)

and stage 3 contributes only

(68)

This quantifies how the 20 dB first-stage gain suppresses later noise when referred back to the receiver input.

Solution 7: why loss before an LNA is expensive

A 1.50 dB cable loss corresponds to

(69)

so the cable power gain is

(70)

At 290 K, the cable noise factor equals its loss:

(71)

The LNA has

(72)

For cable first,

Fbefore	= Fc +	(73)
	= 1.41254 +	(74)
	= 1.77828.	(75)

Hence

(76)

The equivalent input temperature is

(77)

For LNA first,

Fafter	= FL +	(78)
	= 1.25893 +	(79)
	= 1.26305.	(80)

Thus

(81)

and

(82)

The same cable is far more damaging before the LNA because its loss directly attenuates the desired signal and its own noise is not suppressed by preceding gain.

Solution 8: antenna temperature and system noise temperature

At a common reference plane,

Tsys	= TA + Te,rx	(83)
	= 120 + 80	(84)
	= 200 K.	(85)

Therefore

N0	= kTsys	(86)
	= (1.380649 × 10−23)(200)	(87)
	= 2.7613 × 10−21 W/Hz.	(88)

The logarithmic forms are

N0,dBW/Hz	= −205.59 dBW/Hz,	(89)
N0,dBm/Hz	= −175.59 dBm/Hz.	(90)

Hence

(91)

Solution 9: receiver figure of merit G∕T

The figure of merit is

dB/K	= 28.0 − 10 log 10(150)	(92)
	= 28.0 − 21.761	(93)
	= 6.239 dB/K.	(94)

Thus

(95)

A 3 dB improvement could come from approximately doubling receive gain at unchanged system temperature, or halving system temperature at unchanged gain. Either changes the linear ratio G∕T by a factor of two.

Solution 10: C∕N₀ from carrier power and system temperature

At T_sys = 250 K,

N0	= kTsys	(96)
	= 3.4516 × 10−21 W/Hz.	(97)

In dBW/Hz,

(98)

Therefore

dB-Hz	= CdBW − N0,dBW/Hz	(99)
	= −155 − (−204.62)	(100)
	= 49.62 dB-Hz.	(101)

Thus

(102)

Solution 11: direct C∕N₀ link budget using G∕T

Use

(103)

Substitution gives

C∕N0	= 27.0 − 182.5 − 2.0 − 20.0 + 228.60	(104)
	= 51.10 dB-Hz.	(105)

Therefore

(106)

This form combines transmitter strength, path loss, miscellaneous losses, and receiver sensitivity into one bandwidth-independent link metric.

Solution 12: converting C∕N₀ to C∕N

For B = 2.00 MHz,

10 log 10B	= 10 log 10(2.00 × 106)	(107)
	= 63.010 dB-Hz.	(108)

Thus

C∕N	= 45.0 − 63.010	(109)
	= −18.01 dB.	(110)

Therefore

(111)

The negative value means that the integrated noise power over the full 2 MHz bandwidth exceeds the carrier power. It does not imply that the signal is necessarily unusable by a receiver employing correlation or other signal-processing gain.

Solution 13: infer receiver bandwidth from C∕N₀ and C∕N

From

(112)

we obtain

(113)

Therefore

B	= 1040∕10	(114)
	= 104 Hz.	(115)

Thus

(116)

Solution 14: bandwidth penalty

For B₁ = 10.0 kHz,

C∕N1	= 50.0 − 10 log 10(104)	(117)
	= 50.0 − 40.0	(118)
	= 10.0 dB.	(119)

For B₂ = 20.0 kHz,

C∕N2	= 50.0 − 10 log 10(2.00 × 104)	(120)
	= 50.0 − 43.010	(121)
	= 6.990 dB.	(122)

Hence

(123)

Doubling bandwidth doubles integrated white-noise power, and a factor of two in power is 3.010 dB.

Solution 15: GNSS-style weak-signal example

For T_sys = 400 K,

N0,dBW/Hz	= 10 log 10(kTsys)	(124)
	= −202.58 dBW/Hz.	(125)

Therefore

C∕N0	= −158.5 − (−202.58)	(126)
	= 44.08 dB-Hz.	(127)

For B = 2.00 MHz,

C∕N	= 44.08 − 63.010	(128)
	= −18.93 dB.	(129)

Thus

(130)

C∕N₀ compares a finite carrier power with noise per unit bandwidth. The wideband C∕N includes all noise integrated across 2 MHz. Spread-spectrum and correlation receivers can exploit known signal structure even when the wideband pre-correlation C∕N is below 0 dB.

Solution 16: reference-plane bookkeeping with feed loss and an LNA

The 1.00 dB feed loss is

(131)

with gain

(132)

At physical temperature 290 K, the feed equivalent input temperature is

Te,c	= (L − 1)Tp	(133)
	= (1.25893 − 1)(290)	(134)
	= 75.09 K.	(135)

The LNA noise factor is

(136)

so

Te,L	= (FL − 1)T0	(137)
	= 58.66 K.	(138)

Referred through the preceding cable to the antenna-terminal plane, the LNA contribution is

(139)

Therefore

Tsys	= TA + Te,c +	(140)
	= 100 + 75.09 + 73.85	(141)
	= 248.93 K.	(142)

Thus

(143)

For receive gain G_r = 25.0 dBi,

G∕T	= 25.0 − 10 log 10(248.93)	(144)
	= 1.039 dB/K.	(145)

Therefore

(146)

Finally,

N0,dBW/Hz	= 10 log 10(kTsys)	(147)
	≈−204.64 dBW/Hz.	(148)

With C = −160 dBW at the same reference plane,

C∕N0	= −160 − (−204.64)	(149)
	= 44.64 dB-Hz.	(150)

Hence

(151)

This problem illustrates why reference-plane discipline matters: the feed noise, the attenuated LNA noise contribution, the receive gain, and carrier power must all be referred consistently before forming T_sys, G∕T, or C∕N₀.

Julia numerical check

The following script checks the room-temperature noise floor, the three-stage cascade, bandwidth scaling, and the GNSS-style C∕N₀ example.

using Printf

k  = 1.380649e-23
T0 = 290.0

# Thermal noise density
N0 = k*T0
@printf("N0 at 290 K = %.4e W/Hz = %.3f dBm/Hz\n",
        N0, 10*log10(N0/1e-3))

# Three-stage cascade
G1, G2 = 10^(20/10), 10^(15/10)
F1, F2, F3 = 10^(1/10), 10^(4/10), 10^(6/10)
Ftot = F1 + (F2-1)/G1 + (F3-1)/(G1*G2)
@printf("Cascade NF = %.4f dB\n", 10*log10(Ftot))

# C/N from C/N0 versus bandwidth
cno = 50.0
for B in (1e4, 2e4, 1e5, 1e6)
    cn = cno - 10*log10(B)
    @printf("B=%9.0f Hz  C/N=%8.3f dB\n", B, cn)
end

# GNSS-style example
C = -158.5
Tsys = 400.0
N0_dBW_Hz = 10*log10(k*Tsys)
cno_gnss = C - N0_dBW_Hz
@printf("GNSS-style C/N0 = %.3f dB-Hz\n", cno_gnss)

The expected outputs are approximately −173.98 dBm/Hz at 290 K, 1.055 dB cascade noise figure, a 3.010 dB reduction in C∕N when bandwidth doubles from 10 to 20 kHz, and 44.08 dB-Hz for the illustrative GNSS-style case.

What EM26E1 adds to the series

EM26 introduced the thermal-noise quantities. EM26E1 makes their reference planes and logarithmic conversions operational. The calculation chain is

(152)

The next article can now add received interference power J and derive J∕S, J∕N₀, and the degradation of carrier quality in simultaneous thermal noise and interference.

References