Dung Trinh, PhD
HCMUT / 2019
Chapter 2
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Noise, Nonlinear Distortion and
System Parameters
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Trinh Xuan Dung, PhD
dung.trinh@hcmut.edu.vn
Department of Telecommunications
Faculty of Electrical and Electronics Engineering
Ho Chi Minh city University of Technology
Dept. of Telecoms Engineering
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Dung Trinh, PhD
HCMUT / 2019
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1. General Considerations
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Contents
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2. Nonlinear Distortion in Microwave System
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3. Noise in RF/Microwave System
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1. Units in RF Design
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 Units in RF Design:
The voltage gain and power gain are expressed in decibels (dB):
𝑉𝑜𝑢𝑡
𝐴𝑣 𝑑𝐵 = 20𝑙𝑜𝑔
𝑉𝑖𝑛
𝑃𝑜𝑢𝑡
𝐴𝑃 𝑑𝐵 = 10𝑙𝑜𝑔
𝑃𝑖𝑛
•
These two quantities are equal (in dB) only if the input and output voltages appear
across equal impedances. For example, an amplifier having an input resistance of
R0 (50Ω) and driving a load resistance of R0 satisfies the following equations:
2
𝑉𝑜𝑢𝑡
𝑃𝑜𝑢𝑡
𝑉𝑜𝑢𝑡
𝑅0
𝐴𝑃 𝑑𝐵 = 10𝑙𝑜𝑔
= 10𝑙𝑜𝑔 2 = 20𝑙𝑜𝑔
= 𝐴𝑣 𝑑𝐵
𝑃𝑖𝑛
𝑉𝑖𝑛
𝑉𝑖𝑛
𝑅0
•
The absolute signal levels are often expressed in dBm rather than watts or volts.
The unit dBm refers to “dBs” above “1mW”. To express the signal power, Psig, in
dBm, we write:
𝑃𝑠𝑖𝑔
𝑃𝑠𝑖𝑔 𝑑𝐵𝑚 = 10𝑙𝑜𝑔
1 𝑚𝑊
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1. Units in RF Design
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 Solution: 0dBm is equivalent to 1mW, we have:
2
𝑉𝑃𝑃
= 1𝑚𝑊
8𝑅𝐿
𝑉𝑃𝑃 = 632𝑚𝑉.
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 Example 1: An amplifier senses a sinusoidal signal and delivers a power of 0 dBm to a
load resistance of 50Ω. Determine the peak-to-peak voltage swing across the load.
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 Output voltage of the amplifier is of interest in the above example. This may occur if
the circuit following the amplifier does not present a 50Ω input impedance, and hence
the power gain and voltage gain are not equal in dB.
 Only for a sinusoid can we assume that the rms value is equal to the peak-to-peak value
divided by 2 2. Fortunately, for a narrowband 0-dBm signal, it is still possible to
approximate the (average) peak-to-peak swing as 632mV.
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1. General Considerations
 Solution:
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• -100dBm is 100 dB below 632 mVpp.
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 Example 2: A GSM receiver senses a narrowband (modulated) signal having a level of
-100dBm. If the front-end amplifier provides a voltage gain of 15 dB, calculate the
peak-to-peak voltage swing at the output of the amplifier.
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• 100 dB for voltage quantities is equivalent to 105.
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• Thus, -100 dBm is equivalent to 6.32𝜇𝑉𝑃𝑃 .
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• This input level is amplified by 15𝑑𝐵 ≈ 5.62, resulting in an output swing of
35.5 𝜇𝑉𝑃𝑃 .
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1. Nonlinearity
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 A system is called “memoryless” or “static” if its output does not depend on the past
values of its input.
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 For a memoryless linear system, the input/output characteristic is given by:
𝑦 𝑡 = 𝛼𝑥(𝑡)
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 For a memoryless nonlinear system, the input/output characteristic can be approximated
with a polynomial:
𝑦 𝑡 = 𝛼0 + 𝛼1 𝑥 𝑡 + 𝛼2 𝑥 2 𝑡 + +𝛼3 𝑥 3 𝑡 + ⋯
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 The following figure shows a common-source stage as an example of a memoryless
nonlinear circuit (at low frequencies). If M1 operates in the saturation region and can be
approximated as a square-law device, then
1
𝑉𝑜𝑢𝑡 = 𝑉𝐷𝐷 − 𝐼𝐷 𝑅𝐷 = 𝑉𝐷𝐷 − 𝑘𝑛 𝑉𝑖𝑛 − 𝑉𝑇𝐻 2 𝑅𝐷
2
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 The system has “odd symmetry” if y(t) is an odd function of
x(t). This occurs if 𝛼𝑗 = 0 for even j. Such system is called
“balanced” as exemplified by the differential pair.
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1. Nonlinearity
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 Example 3: The MOS transistors operating in saturation, characteristic can be
expressed as
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1
4𝐼𝑆𝑆
2
𝑉𝑜𝑢𝑡 = − 𝑘𝑛 𝑉𝑖𝑛
− 𝑉𝑖𝑛
𝑅𝐷
2
𝑘𝑛
If the differential input is small, approximate the characteristic by a polynomial.
4𝐼𝐷𝑆𝑆
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𝑉𝑜𝑢𝑡
and applying the approximation 1 − 𝜀 ≈ 1 − 𝜀/2, we have
𝑘𝑛 2
≈ − 𝑘𝑛 𝐼𝐷𝑆𝑆 𝑉𝑖𝑛 1 −
𝑉 𝑅
8𝐼𝐷𝑆𝑆 𝑖𝑛 𝐷
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• Assuming 𝑉𝑖𝑛
≪
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 Solution:
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𝑉𝑜𝑢𝑡 ≈ − 𝑘𝑛 𝐼𝐷𝑆𝑆 𝑅𝐷 𝑉𝑖𝑛 +
3
𝑘𝑛 2
8 𝐼𝐷𝑆
3
𝑅𝐷 𝑉𝑖𝑛
• The first term on the right-hand side represents linear
operation, revealing the small signal voltage gain of the
circuit (−𝑔𝑚 𝑅𝐷 ).
• Due to symmetry, even-order nonlinear terms are absent.
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2. Nonlinear Distortion in Microwave Systems
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 A system is linear if its output can be expressed as a linear combination
(superposition) of responses to individual inputs:
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𝑦1 𝑡 = 𝑓 𝑥1 𝑡
𝑎𝑦1 𝑡 + 𝑏𝑦2 𝑡 = 𝑓 𝑎𝑥1 𝑡 + 𝑏𝑥2 𝑡
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𝑣𝑜 = 𝑎𝑣𝑖
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 For a linear system:
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𝑦2 𝑡 = 𝑓 𝑥2 𝑡
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 For a nonlinear system: The
input/output characteristic of a
memory-less nonlinear system can be
approximated with a polynomial:
𝑣𝑜 = 𝑎0 + 𝑎1 𝑣𝑖 + 𝑎2 𝑣𝑖2 + 𝑎3 𝑣𝑖3 + ⋯
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Dung Trinh, PhD
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2. Nonlinear Distortion in Microwave Systems
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 The nonlinearity of the system causes the following issues:
Harmonic generation (multiples of a fundamental signal)

Gain Compression (gain reduction in an amplifier)

Inter-modulation Distortion (products of a two-tone input signal)

Cross-modulation (modulation transfer from one signal to another)

AM-PM conversion (amplitude variation causes phase shift)

Spectral regrowth (intermodulation with many closely spaced signals)
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
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2A. Nonlinear – Harmonic Distortion
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 Input signal of the system:
𝑣𝑖 𝑡 = 𝑉0 𝑐𝑜𝑠𝜔0 𝑡
2
+ 𝑎3 𝑉0 𝑐𝑜𝑠𝜔0 𝑡
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𝑣0 𝑡 = 𝑎0 + 𝑎1 𝑉0 𝑐𝑜𝑠𝜔0 𝑡 + 𝑎2 𝑉0 𝑐𝑜𝑠𝜔0 𝑡
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 Output signal of the system:
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+⋯
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Fundamental
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1
3
1
1
= 𝑎0 + 𝑎2 𝑉02 + 𝑎1 𝑉0 + 𝑎3 𝑉03 𝑐𝑜𝑠𝜔0 𝑡 + 𝑎2 𝑉02 𝑐𝑜𝑠2𝜔0 𝑡 + 𝑎3 𝑉03 𝑐𝑜𝑠3𝜔0 𝑡 + ⋯
2
4
2
4
Second
Harmonic
Third
Harmonic
 Even-order harmonics result from αj with even j
 nth harmonic grows in proportion to An.
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2A. Nonlinear – Harmonic Distortion
 Problems:
More signal loss and distortion.

Interference to other systems.

Sometimes can be used to create
frequency multipliers.
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GSM900 Band
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
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Dung Trinh, PhD
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2A. Nonlinear – Harmonic Distortion
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 Example 4: An analog multiplier “mixes” its two inputs, ideally producing 𝑦 𝑡 =
𝑘𝑥1 (𝑡)𝑥2 (𝑡) where k is a constant. Assume 𝑥1 𝑡 = 𝐴1 𝑐𝑜𝑠 𝜔1 𝑡 and 𝑥2 𝑡 =
𝐴2 𝑐𝑜𝑠 𝜔2 𝑡 .
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a. If the mixer is ideal, determine the output frequency components.
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b. If the input port sensing 𝑥2 (𝑡) suffers from third-order nonlinearity, determine the
output frequency components
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 Solution:
𝑘𝐴1 𝐴2
𝑐𝑜𝑠
2
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=
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𝑦 𝑡 = 𝑘𝐴1 𝑐𝑜𝑠 𝜔1 𝑡 𝐴2 𝑐𝑜𝑠 𝜔2 𝑡
a. We have:
𝜔1 + 𝜔2 𝑡 +
𝑘𝐴1 𝐴2
𝑐𝑜𝑠
2
𝜔1 − 𝜔2 𝑡
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b. The third harmonic of 𝑥2 𝑡 is 𝛼3 𝐴32 /4, we have:
𝑦 𝑡 = 𝑘𝐴1 𝑐𝑜𝑠 𝜔1 𝑡
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=
𝑘𝐴1 𝐴2
𝑐𝑜𝑠
2
+
𝑘𝛼3 𝐴1 𝐴32
𝑐𝑜𝑠
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𝛼3 𝐴32
𝐴2 𝑐𝑜𝑠 𝜔2 𝑡 +
𝑐𝑜𝑠 3𝜔2 𝑡
4
𝜔1 + 𝜔2 𝑡 +
𝑘𝐴1 𝐴2
𝑐𝑜𝑠
2
𝜔1 + 3𝜔2 𝑡 +
𝜔1 − 𝜔2 𝑡
𝑘𝛼3 𝐴1 𝐴32
𝑐𝑜𝑠
8
𝜔1 − 3𝜔2 𝑡
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2A. Nonlinear – Harmonic Distortion
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The mixer now produces two “spurious” component at 𝜔1 + 3𝜔2 and 𝜔1 − 3𝜔2 ,
one or both of which often prove problematic. For example, if 𝜔1 = 2𝜋 × 850𝑀𝐻𝑧 and
𝜔2 = 2𝜋 × 900𝑀𝐻𝑧, then 𝜔1 − 3𝜔2 = 2𝜋 × 1850𝑀𝐻𝑧, an undesired component that is
difficult to filter out because it lies close to the desired component at 𝜔1 + 3𝜔2 =
2𝜋 ×1750MHz.
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2B. Nonlinear – Gain Compression
GAIN: 𝑮 =
𝟑
𝒂𝟏 𝑽𝟎 + 𝟒 𝒂𝟑 𝑽𝟑𝟎
𝑽𝟎
𝟑
= 𝒂𝟏 + 𝒂𝟑 𝑽𝟐𝟎
𝟒
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Fundamental
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1
3
1
1
𝑣0 = 𝑎0 + 𝑎2 𝑉02 + 𝑎1 𝑉0 + 𝑎3 𝑉03 𝑐𝑜𝑠𝜔0 𝑡 + 𝑎2 𝑉02 𝑐𝑜𝑠2𝜔0 𝑡 + 𝑎3 𝑉03 𝑐𝑜𝑠3𝜔0 𝑡 + ⋯
2
4
2
4
14
Expansive
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Most RF/microwave
component/circuits
are compressive !
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2B. Nonlinear – Gain Compression
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 P1dB: power at 1-dB compression.
 Psat: power at saturation.
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 If we need very good linearity, we should operate a few dBs below the P1dB point;
this is called power back-off.
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𝟑
𝟐𝟎𝒍𝒐𝒈 𝒂𝟏 + 𝒂𝟑 𝑽𝟐𝒊𝒏,𝟏𝒅𝑩 = 𝟐𝟎𝒍𝒐𝒈 𝒂𝟏 − 𝟏
𝟒
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𝑽𝒊𝒏,𝟏𝒅𝑩 =
𝒂𝟏
𝟎. 𝟏𝟒𝟓
𝒂𝟑
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2B. Nonlinear – Gain Compression

Amplitude Modulation:
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Frequency Modulation:
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
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 Effect on modulated signal:
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 FM signal carries no information in its amplitude and hence tolerates compression.
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 AM contains information in its amplitude, hence distorted by compression.
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 Desensitization: the receiver gain is reduced by the large excursions produced by the
interferer even though the desired signal itself is small.
vi  V1 cos 1t  V2 cos 2t
3
3


vo  ...   a1V1  a3V13  a3V23  cos 0t  ...
4
2


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2B. Nonlinear – Gain Compression
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 Example 5: A 900-MHz GSM transmitter delivers a power of 1W to the antenna. By
how much must the second harmonic of the signal be suppressed (filtered) so that it
does not desensitize a 1.8-GHz receiver having 𝑃1𝑑𝐵 = −25dBm? Assume the receiver
is 1m away and the 1.8-GHz signal is attenuated by 10 dB as it propagates across this
distance.
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 Solution:
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The output power at 900MHz is equal to 30dBm. With an attenuation of 10dB, the
second harmonic must not exceed -15dBm at the transmitter antenna so that it is below
𝑃1𝑑𝐵 of the receiver. Thus, the second harmonic must remain at least 45dB below the
fundamental at the TX output. In practice, this interference must be another several dB
lower to ensure the RX does not compress.
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2C. Nonlinear – Inter-modulation Distortion
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 Single Signal:
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Harmonic distortion
Desensitization
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 Single and one
large interferer:
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 Single and two
large interferers:
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Intermodulation
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2C. Nonlinear – Inter-modulation Distortion
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 A received small desired signal along with two large interferers
 Intermodulation product falls onto the desired channel, corrupts signal.
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interferers
𝟐𝝎𝟐 − 𝝎𝟏 = 𝝎𝟎
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desired
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2C. Nonlinear – Inter-modulation Distortion
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 The two interferers at frequency 𝜔1 and 𝜔2 go into the system:
𝑣𝑖 𝑡 = 𝑉0 𝑐𝑜𝑠𝜔1 𝑡 + 𝑐𝑜𝑠𝜔2 𝑡
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 The signal at the output of the system:
2
+ 𝑎3 𝑉03 𝑐𝑜𝑠𝜔1 𝑡 + 𝑐𝑜𝑠𝜔2 𝑡
3
+⋯
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𝑣0 𝑡 = 𝑎0 + 𝑎1 𝑉0 𝑐𝑜𝑠𝜔1 𝑡 + 𝑐𝑜𝑠𝜔2 𝑡 + 𝑎2 𝑉02 𝑐𝑜𝑠𝜔1 𝑡 + 𝑐𝑜𝑠𝜔2 𝑡
ng
th
1
1
= 𝑎0 + 𝑎1 𝑉0 𝑐𝑜𝑠𝜔1 𝑡 + 𝑎1 𝑉0 𝑐𝑜𝑠𝜔2 𝑡 + 𝑎2 𝑉02 1 + 𝑐𝑜𝑠2𝜔1 𝑡 + 𝑎2 𝑉02 1 + 𝑐𝑜𝑠2𝜔2 𝑡
2
2
+ 𝑎3 𝑉03
+ 𝑎3 𝑉03
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3
1
3
1
𝑐𝑜𝑠𝜔1 𝑡 + 𝑐𝑜𝑠3𝜔1 𝑡 + 𝑎3 𝑉03
𝑐𝑜𝑠𝜔2 𝑡 + 𝑐𝑜𝑠3𝜔2 𝑡
4
4
4
4
cu
+ 𝑎3 𝑉03
du
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+ 𝑎2 𝑉02 𝑐𝑜𝑠 𝜔1 − 𝜔2 𝑡 + 𝑎2 𝑉02 𝑐𝑜𝑠 𝜔1 + 𝜔2 𝑡
3
3
3
𝑐𝑜𝑠𝜔1 𝑡 + 𝑐𝑜𝑠 𝟐𝝎𝟏 − 𝝎𝟐 𝑡 + 𝑐𝑜𝑠 2𝜔1 + 𝜔2 𝑡
2
4
4
3
3
3
𝑐𝑜𝑠𝜔2 𝑡 + 𝑐𝑜𝑠 𝟐𝝎𝟐 − 𝝎𝟏 𝑡 + 𝑐𝑜𝑠 2𝜔2 + 𝜔1 𝑡 + ⋯
2
4
4
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2C. Nonlinear – Inter-modulation Distortion
Harmonics
Formulation
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𝑎0 + 𝑎2 𝑉02
DC
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3
3
𝑎1 𝑉0 + 𝑎3 𝑉0 𝑐𝑜𝑠𝜔1 𝑡 + 𝑎1 𝑉0 + 𝑎3 𝑉03 𝑐𝑜𝑠𝜔2 𝑡
4
4
an
Fundamental
𝑎2 𝑉02 𝑐𝑜𝑠 𝜔1 − 𝜔2 𝑡 + 𝑎2 𝑉02 𝑐𝑜𝑠 𝜔1 + 𝜔2 𝑡
2nd harmonic
1
1
2
𝑎2 𝑉0 𝑐𝑜𝑠2𝜔1 𝑡 + 𝑎2 𝑉02 𝑐𝑜𝑠2𝜔2 𝑡
2
2
du
o
ng
th
2nd order IMD
3
3
3
𝑎3 𝑉0 𝑐𝑜𝑠 2𝜔1 − 𝜔2 𝑡 + 𝑎3 𝑉03 𝑐𝑜𝑠 2𝜔2 − 𝜔1 𝑡
4
4
3rd order IMD (IM3)
3
3
𝑎3 𝑉03 𝑐𝑜𝑠 2𝜔1 + 𝜔2 𝑡 + 𝑎3 𝑉03 𝑐𝑜𝑠 2𝜔2 + 𝜔1 𝑡
4
4
cu
u
3rd order IMD (IM3)
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2C. Nonlinear – Inter-modulation Distortion
cu
u
du
o
ng
th
an
co
ng
.c
om
Example 6: Suppose four Bluetooth users operate in a room as shown in figure below.
User 4 is in the receive mode and attempts to sense a weak signal transmitted by User 1
at 2.410 GHz. At the same time, Users 2 and 3 transmit at 2.420 GHz and 2.430 GHz,
respectively. Explain what happens.
 Since the frequencies transmitted by Users 1, 2, and 3 happen to be equally spaced,
the intermodulation in the LNA of RX4 corrupts the desired signal at 2.410 GHz.
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2C. Nonlinear – Inter-modulation Distortion
.c
om
 IM3 products degrades system performance because they are very close to the
passband of the system
th
du
o
ng
3
𝑎3 𝑽𝟑𝟎 𝑐𝑜𝑠 2𝜔1 − 𝜔2 𝑡
4
an
co
ng
 The power of IM3 products increases more rapidly than the fundamental output
when the input power is increased
cu
u
3
𝑎3 𝑽𝟑𝟎 𝑐𝑜𝑠 2𝜔2 − 𝜔1 𝑡
4
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HCMUT / 2019
2C. Nonlinear – Inter-modulation Distortion
.c
om
 Third-order intercept point designates the point where the third-order
intermodulation product has the same power as the fundamental output power.
The input power at the IP3 point is call input-referred IP3: IIP3.

The output power at the IP3 point is call output-referred IP3: OIP3.
an
th
ng
 In practice, the amplifier never
works close to the IP3 point
because it would have
saturated way before that.
co
ng

cu
u
du
o
 IP3 is extrapolated from
measurements
of
the
fundamental output and IM3
products in the linear region of
the amplifier.

IP3 is really a measure of
linearity, rather than
power handling.
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2C. Nonlinear – Inter-modulation Distortion
.c
om
 Third Intercept Point: is the point where the output power at 𝜔1 equals to output
power at 2𝜔1 − 𝜔2 .
th
an
co
ng
3
𝑎1 𝑉𝐼𝐼𝑃3 = 𝑎3 𝑽𝟑𝑰𝑰𝑷𝟑 → 𝑉𝐼𝐼𝑃3 =
4
𝑎1
𝑉
→ 𝐼𝐼𝑃3
=
0.145
𝑉1𝑑𝐵
𝑎3
4
≃ 9.6𝑑𝐵
0.435
cu
u
du
o
ng
𝑉1𝑑𝐵 =
4 𝑎1
3 𝑎3
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HCMUT / 2019
2C. Nonlinear – Inter-modulation Distortion
.c
om
 Example 7: A Bluetooth receiver employs a low-noise amplifier having a gain of 10
and an input impedance of 50Ω. The LNA senses a desired signal level of -80dBm at
2.410 GHz and two interferers of equal levels at 2.420 GHz and 2.430 GHz. For
simplicity, assume the LNA drives a 50Ω load.
ng
a. Determine the value of 𝛼3 that yields a 𝑃1𝑑𝐵 of -30dBm.
an
co
b. If each interferer is 10 dB below 𝑃1𝑑𝐵 , determine the corruption experienced by the
desired signal at the LNA output.
th
 Solution:
ng
a. Note that −30𝑑𝐵𝑚 = 20𝑚𝑉𝑃𝑃 = 10𝑚𝑉𝑃 . We have:
du
o
𝑉1𝑑𝐵 =
𝑎1
0.145
= 10𝑚𝑉𝑃
𝑎3
cu
u
• Since 𝛼1 = 10 then 𝛼3 = 14.500 𝑉 −2
b. Each interferer has a level of -40dBm (6.32 m𝑉𝑃𝑃 ). Setting 𝐴1 = 𝐴2 = 6.32 m𝑉𝑃𝑃 ,
we determine the amplitude of the IM product
3𝛼3 𝐴12 𝐴2
= 0.343𝑚𝑉𝑃 = −59.3𝑑𝐵𝑚
4
• The desired signal is amplified by 0 (20dB), produce an output at level -60dBm.
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cu
u
du
o
ng
th
an
co
ng
 Input IP is the point where the output power
at 1 equals to output power at (21 - 2 )
.c
om
2C. Nonlinear – Inter-modulation Distortion
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2C. Nonlinear – Inter-modulation Distortion
ng
.c
om
Example 9: A low-noise amplifier senses a -80dBm signal at 2.410 GHz and two 20dBm interferers at 2.420 GHz and 2.430 GHz. What IIP3 is required if the IM
products must remain 20 dB below the signal? For simplicity, assume 50Ω interfaces at
the input and output.
co
Solution:
an
 At the LNA output:
ng
th
20𝑙𝑜𝑔10 𝑎1 𝑉𝑠𝑖𝑔 − 20 = 20𝑙𝑜𝑔10
du
o
 Then:
30
𝑎3 𝑽𝟑𝒊𝒏𝒕
4
 Thus:
cu
u
𝑎1 𝑉𝑠𝑖𝑔 =
3
𝑎3 𝑽𝟑𝒊𝒏𝒕
4
𝐼𝐼𝑃3 =
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4 𝑎1
= 3.65𝑉𝑃 = 15.2(𝑑𝐵𝑚)
30 𝑎3
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2D. Nonlinear – Cross Modulation
.c
om
 Suppose that the interferer is an amplitude-modulated signal:
ng
𝑣𝑖𝑛𝑡𝑒𝑟𝑓𝑒𝑟𝑒𝑟 𝑡 = 𝑉0 1 + 𝑚𝑐𝑜𝑠𝜔𝑚 𝑡 𝑐𝑜𝑠𝜔2 𝑡
co
 Then:
cu
u
du
o
ng
th
an
𝑣𝑜 𝑡 = 𝑉0 1 + 𝑚𝑐𝑜𝑠𝜔𝑚 𝑡 𝑐𝑜𝑠𝜔2 𝑡
3
𝑎1 𝑉0 + 𝑎3 𝑉03 𝑐𝑜𝑠𝜔0 𝑡
4
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HCMUT / 2019
3. Noise in RF/Microwave System
.c
om
 The effect of noise is critical to the performance of most RF and microwave
communications, radar, and remote sensing systems.
cu
u
du
o
ng
th
an
co
ng
 Noise ultimately determines the threshold for the minimum signal that can be reliably
detected by a receiver.
 Noise power in a receiver will be introduced from:

The external environment through the receiving antenna – External Noise.

Internal circuits of receivers – Internal Noise.
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Dung Trinh, PhD
HCMUT / 2019
3. Noise in RF/Microwave System
Thermal noise from the ground

Cosmic background noise from the sky

Noise from stars (including the sun)

Lightning

Gas discharge lamps

Radio, TV, and cellular stations

Wireless devices

Microwave ovens

Deliberate jamming devices
cu
u
du
o
ng
th
an
co

ng
.c
om
 External noise may be introduced into a system either by a receiving antenna or by
electromagnetic coupling. Some sources of external RF noise include the following:
Noise vs. Interferer
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Dung Trinh, PhD
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3. Noise in RF/Microwave System
.c
om
 Internal Noise generated in a device or component is usually caused by random
motions of charges or charge carriers in devices and materials. Such motions may be
due to any of several mechanisms, leading to various types of noise:
Thermal noise is the most basic type of noise, being caused by thermal vibration
of bound charges. It is also known as Johnson or Nyquist noise.

Shot noise is due to random fluctuations of charge carriers in an electron tube or
solid-state device.

Flicker noise occurs in solid-state components and vacuum tubes. Flicker noise
power varies inversely with frequency, and so is often called 1/ f -noise.
cu
u
du
o
ng
th
an
co
ng

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3. Thermal Noise

.c
om
 A resistor not connected to anything can have a voltage across it.
The average voltage is of course 0 and its amplitude is random and very small
ng
 Random motion of charge carriers produce a random variation of voltage (or current)
with respect to time  noise voltage and/or noise current.
co
 A conductor (equivalently a resistor) has lots of charge carriers.
cu
u
du
o
ng
th
an
 The higher the temperature, the more random motion, and therefore the higher the
noise.
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Higher temperature
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3. Thermal Noise
ℎ𝑓
𝑒 𝑘𝑇
ng
−1
Noise voltage: 𝑽𝒏 = 𝟒𝒌𝑻𝑩𝑹
h = 6.626 x 10-34 J∙s, Planck’s constant
k = 1.38 x 10-23 J/K, Boltzmann’s constant
B, bandwidth in Hz
ℎ𝑓
:
𝑘𝑇
f, center frequency of the bandwidth
R, resistance in Ω
T, temperature in Kelvin
𝑽𝒏
Maximum Noise Power: 𝑷𝒏 =
𝟐𝑹
𝟐
𝑹 = 𝒌𝑻𝑩
cu
u

−1≃
du
o

ℎ𝑓
𝑘𝑇
ng
 At low frequencies (f < 5 THz), 𝑒
th
an
 It can be shown that *:
4ℎ𝑓𝐵𝑅
co
𝑉𝑛2 =
.c
om
 The noisy resistor can be replaced
with a Thevenin equivalent circuit
of a noiseless resistor and a random
noise voltage generator Vn.
𝑷𝒏 𝒅𝑩𝒎 = 𝟏𝟎𝒍𝒐𝒈𝟏𝟎 𝟏. 𝟑𝟖 × 𝟏𝟎−𝟐𝟑 𝑻𝑩 = 𝟏𝟎𝒍𝒐𝒈𝟏𝟎 𝑻𝑩 − 𝟏𝟑𝟖. 𝟓
When:
𝑇 = 290𝐾, 𝐵 = 1𝐻𝑧
𝑇 = 290𝐾, 𝐵 = 1𝐾𝐻𝑧
𝑇 = 290𝐾, 𝐵 = 1𝑀𝐻𝑧
→ 𝑷𝒏 = −𝟏𝟕𝟒𝒅𝑩𝒎
→ 𝑷𝒏 = −𝟏𝟒𝟒𝒅𝑩𝒎
→ 𝑷𝒏 = −𝟏𝟏𝟒𝒅𝑩𝒎
* For derivation, refer to: http://www.pas.rochester.edu/~dmw/ast203/Lectures/Lect_20.pdf
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3. Thermal Noise
.c
om
 At most RF and microwave frequencies, the thermal noise from a resistor appear
uniform with respect to frequency, as is evident from the absence of a frequency
variable in the Vn expression.
cu
u
du
o
ng
th
an
co
ng
 We call this type of frequency independent noise “white noise”.
 We also note that the noise power is independent of the resistance!
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3. Noise Figure
𝒅𝒆𝒔𝒊𝒓𝒆𝒅 𝒔𝒊𝒈𝒏𝒂𝒍 𝒍𝒆𝒗𝒆𝒍
𝒖𝒏𝒅𝒆𝒔𝒊𝒓𝒆𝒅 𝒏𝒐𝒊𝒔𝒆 𝒍𝒆𝒗𝒆𝒍
du
o
ng
th
an
co
ng
Signal to Noise Ratio (SNR): 𝑆𝑁𝑅 =
.c
om
 Noise Figure: measures the degradation in the signal to noise ratio between the input
and output of the component.
u
 It is important to understand that while amplifiers are great at making your signal
stronger, they also make noise stronger!
cu
 Noise Factor (linear scale):
𝐹=
 Noise Figure (dB scale):
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𝑺𝑵𝑹𝒊𝒏
𝑺𝑵𝑹𝒐𝒖𝒕
𝑁𝐹 = 𝟏𝟎𝒍𝒐𝒈𝟏𝟎 𝑭
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3. Noise Figure
.c
om
Example 10: Consider a device of bandwidth 𝐵 = 20𝑀𝐻𝑧 and having a gain 𝐺 =
60𝑑𝐵 with noise figure of 𝑁𝐹 = 5𝑑𝐵. What is the noise power seen at the output of the
device?
co
ng
Solution:
cu
u
du
o
ng
th
an
Nout = -35.82dBm.
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3. Noise Figure
.c
om
 The noise contributions from subsequent stages are suppressed by the preceding
stages. It follows that the first stage in a cascade system needs to have low noise and
high gain.
ng
 Amplifiers specifically designed to do this are called low noise amplifiers (LNA).
LNAs are often designed for the lowest noise, sacrificing efficiency, linearity,
and even gain.

To a large extent, LNAs determine the system sensitivity.
th
an
co

cu
u
du
o
ng
𝑁1 = 𝐺1 𝑘𝑇0 𝐵 + 𝐺1 𝑘𝑇𝑒1 𝐵
𝑁0 = 𝐺2 𝑁1 + 𝐺2 𝑘𝑇𝑒2 𝐵
= 𝐺1 𝐺2 𝑘𝐵 𝑇0 + 𝑇𝑒1 +
𝑇𝐶𝐴𝑆 = 𝑇𝑒1 +
1
𝑇
𝐺1 𝑒2
𝑇𝑒2
𝑇𝑒3
+
+⋯
𝐺1 𝐺1 𝐺2
𝑇𝑒 = (𝐹 − 1)𝑇0
𝐹2 − 1 𝐹3 − 1
𝐹𝐶𝐴𝑆 = 𝐹1 +
+
+⋯
𝐺1
𝐺1 𝐺2
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HCMUT / 2019
3. Noise Figure
an
co
ng
.c
om
Example 11: The block diagram of a wireless receiver front-end is shown in following
figure. Compute the overall noise figure of this subsystem. If the input noise power
from a feeding antenna is 𝑁𝑖 = 𝑘𝑇𝐴 𝐵, where 𝑇𝐴 = 150𝐾, find the output noise power in
dBm. If we require a minimum signal-to-noise ratio (SNR) of 20𝑑𝐵 at the output of the
receiver, what is the minimum signal voltage that should be applied at the receiver
input? Assume the system is at temperature 𝑇𝑜 , with a characteristic impedance of 50Ω
and an IF bandwidth of 10MHz.
cu
u
du
o
ng
th
Solution:
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3. Noise Figure
.c
om
 We have:
𝐺𝑓 = −1𝑑𝐵 = 0.79
𝐹𝑓 = 1𝑑𝐵 = 1.26
ng
𝐺𝑎 = 10𝑑𝐵 = 10
𝐹𝑎 = 2𝑑𝐵 = 1.58
𝐹𝑓 −1
𝐺𝑎
+
𝐹𝑚 −1
𝐺𝑎 𝐺𝑓
= 1.58 +
1.26−1
2.51−1
+
10
10×0.79
= 1.8 = 2.55𝑑𝐵
an
𝐹 = 𝐹𝑎 +
co
 The overall noise figure of the system:
𝐺𝑚 = −3𝑑𝐵 = 0.5
𝐹𝑚 = 4𝑑𝐵 = 2.51
th
 The equivalent noise temperature of the overall system:
du
o
 The output noise power:
ng
𝑇𝑒 = 𝐹 − 1 𝑇0 = 1.8 − 1 × 290 = 232𝐾
𝑁𝑜 = 𝑘 𝑇𝐴 + 𝑇𝑒 𝐵𝐺 = 1.38 × 10−23 150 + 232 × 107 × 3.95 = −96.8𝑑𝐵𝑚
𝑆𝑖 =
𝑆𝑜
𝐺
=
cu
u
 For an output SNR of 20dB, the input signal power must be:
𝑆𝑜 𝑁𝑜
𝑁𝑜 𝐺
=
2.08×10−13
100
3.95
= −82.8𝑑𝐵𝑚
 For a 50Ω system impedance, this corresponds to an input signal voltage of
𝑉𝑖 =
𝑍0 𝑆𝑖 = 1.62 × 10−5 𝑉 = 16.2𝜇𝑉
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3. Noise Figure of a Lossy Element
ng
.c
om
 Let’s assume that a resistor at the same physical temperature is connected to the input of
the network. Since they in thermal equilibrium, the resistor network can be treated as
one noise source at T, the output noise power should be
co
𝑁0 = 𝑘𝑇𝐵
ng
th
an
 We can also treat the input resistor as the signal source, then the output noise power
should be:
1
1
𝑁0 = 𝑘𝑇𝐵 + 𝑘𝑇𝑒 𝐵
𝐴
𝐴
du
o
 Equating the two equation gives: 𝑇𝑒 = 𝐴 − 1 𝑇
cu
u
 Also we have: 𝑇𝑒 = 𝐹 − 1 𝑇. Finally, 𝐹 = 𝐴.
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A = attenuation, defined as Pin/Pout
B, bandwidth in Hz
T, physical temperature (in Kelvin) of the
network
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3. Noise Figure Measurement
.c
om
 Cold source method: a termination impedance at room temperature, also called “cold
source”, is placed at the DUT input.
co
ng
 If the gain of the amplifier is well known over the frequency of interest it can be
subtracted from the output power noise resulting in the amplifier’s excess noise factor F.
an
𝑁𝑜𝑢𝑡 = 𝑁𝑖𝑛 × 𝐺 × 𝐹
cu
u
du
o
ng
th
 The gain G of the amplifier can be accurately measured over a bandwidth B by a
calibrated two-port measurement with a VNA.
https://www.youtube.com/watch?v=IgXKS0yJ1-E
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3. Noise Figure Measurement
ng
co
𝑁1 = 𝐺𝑘𝑇1 𝐵 + 𝐺𝑘𝑇𝑒 𝐵
.c
om
 Y factor method: the idea is
simple: measure the noise from
two known noise sources and
compare
the
noise
power
difference at the output.
th
an
𝑁1 𝑇1 + 𝑇𝑒
𝑌=
=
𝑁2 𝑇2 + 𝑇𝑒
ng
𝑁2 = 𝐺𝑘𝑇2 𝐵 + 𝐺𝑘𝑇𝑒 𝐵
𝑇1 − 𝑌𝑇2
𝑇𝑒 =
𝑌−1
du
o
 It seems that taking one measurement is enough to
calculate Te but:
That would be less accurate; you would have to
know GB precisely.

Taking two measurements allows you to get rid of
measurement errors that are common to both
measurements.

You need two or more measurement points to
https://www.youtube.com/watch?v=bbdTRX4_2DE
accurately extrapolate the zero crossing value.
cu
u

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Dung Trinh, PhD
HCMUT / 2019
3. Noise Figure
.c
om
Example 12: An X-band amplifier has a gain of 20 dB and a 1 GHz bandwidth. Its
equivalent noise temperature is to be measured via the Y-factor method. The following
data are obtained:
𝑁1 = −62𝑑𝐵𝑚
For 𝑇2 = 77𝐾,
𝑁2 = −64.7𝑑𝐵𝑚
co
ng
For 𝑇1 = 290𝐾,
ng
th
an
Determine the equivalent noise temperature of the amplifier. If the amplifier is used
with a source having an equivalent noise temperature of 𝑇𝑆 = 450𝐾, what is the output
noise power from the amplifier, in dBm?
 The Y factor in dB is:
du
o
Solution:
cu
u
𝑌 = 𝑁1 − 𝑁2 = −62 + 64.7 = 2.7𝑑𝐵 = 1.86
 This gives the equivalent noise temperature as
𝑇𝑒 =
𝑇1 −𝑌𝑇2
𝑌−1
=
270−1.86×77
1.86−1
= 170𝐾
 If 𝑇𝑆 = 450𝐾, the output noise power from the amplifier is
𝑁𝑜 = 𝐺𝑘𝑇𝑠 𝐵 + 𝐺𝑘𝑇𝑒 𝐵 = 100 × 1.38 × 10−23 × 109 × 450 + 170 = −60.7𝑑𝐵𝑚
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Dung Trinh, PhD
HCMUT / 2019
.c
om
3. Noise Figure Measurement
co
th
an
𝑃𝑜𝑓𝑓 = 𝑘𝑇𝐶 𝐵𝐺𝐹
ng
 In the OFF or cold state, the noise
source
is
switched
off,
corresponding to a temperature TC
and the noise power at the output is
then:
u
du
o
ng
 With the noise source ON or in hot
state, the excess noise PN of the
noise source is added to the output
noise power:
cu
𝑃𝑜𝑛 = 𝑘𝑇𝐶 𝐵𝐺𝐹 + 𝑃𝑁 𝐺𝐵 = 𝑘𝑇ℎ 𝐵𝐺𝐹
 The Excess Noise Ratio of the used
noise figure is calculated as:
𝑇ℎ − 𝑇𝑜
𝐸𝑁𝑅 =
𝑇𝑜
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𝐸𝑁𝑅
𝐹=
𝑌−1
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Dung Trinh, PhD
HCMUT / 2019
3. Noise Figure Measurement
.c
om
 Would it help if we take more than two
measurements?
Good accurate noise sources can be
expensive

Getting two noise sources with
drastically different noise temperature
can also be difficult

Remember that temperature is specified
in Kelvin; 0°C and 100°C are only about
30% different.
u
du
o
ng
th
an
co

ng
 That would require more noise sources:
cu
 Example: Calculate the Noise Figure NF of your Spectrum Analyzer:
1. Connect a 50Ω termination at the analyzer’s input
2. Select a resolution bandwidth RBW, e.g. RBW=10KHz
3. Place a marker at the center frequency and read out the power level 𝑃𝑁 in
dBm, e.g. 𝑃𝑁 = −110𝑑𝐵𝑚.
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Dung Trinh, PhD
HCMUT / 2019
3. Sensitivity
.c
om
Sensitivity: Minimum detectable signal by the receiver according to a fixed SNR
determined by the BER.

Path loss.

Receiver sensitivity – govern by
standards and applications.

Required SNR – depends on BER
requirement
and
modulation
scheme.
du
o
ng
th
Path Loss
Receiver Sensitivity
Input Noise Floor (No/G)
Required SNRo
u
Noise floor – thermal noise or
circuit noise limited depending on
the modulation schemes.
cu

co
Transmit Power – FCC regulated.
Transmit Power
an

ng
 To find Receiver NF:
Noise Figure
Input Noise (Ni)
 Example: Mobile network: -100dBm; GPS: -127.5dBm.
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HCMUT / 2019
co
ng
.c
om
Dung Trinh, PhD
cu
u
du
o
ng
th
an
Q&A
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