Chapter 2 Basic Concepts in
RF Design
1
Sections to be covered
•
•
•
•
•
2.1 General Considerations
2.2 Effects of Nonlinearity
2.3 Noise
2.4 Sensitivity and Dynamic Range
2.5 Passive Impedance Transformation
2
Chapter Outline
Nonlinearity



Harmonic Distortion
Compression
Intermodulation
Noise



Noise Spectrum
Device Noise
Noise in Circuits
Impedance
Transformation


Series-Parallel
Conversion
Matching Networks
3
The Big Picture: Generic RF
Transceiver
Overall transceiver
 Signals are upconverted/downconverted at TX/RX, by an oscillator controlled
by a Frequency Synthesizer.
4
General Considerations: Units in RF Design
Voltage gain:
rms value
Power gain:
 These two quantities are equal (in dB) only if the input and output
impedance are equal.
 Example：
 an amplifier having an input resistance of R0 (e.g., 50 Ω) and driving a load
resistance of R0 :
5
where Vout and Vin are rms value.
General Considerations: Units in RF Design
“dBm”
 The absolute signal levels are often expressed in dBm (not in
watts or volts);
 Used for power quantities, the unit dBm refers to “dB’s above
1mW”.
 To express the signal power, Psig, in dBm, we write
6
Example of Units in RF
An amplifier senses a sinusoidal signal and delivers a power of 0 dBm to a load
resistance of 50 Ω. Determine the peak-to-peak voltage swing across the load.
Solution:
 a sinusoid signal having a peak-to-peak amplitude of Vpp
 an rms value of Vpp/(2√2),
 0dBm is equivalent to 1mW,
where RL= 50 Ω
thus,
7
Example of Units in RF
A GSM receiver senses a narrowband (modulated) signal having a level of -100
dBm. 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.
Solution:
 suppose the input and output impedance are equal.
 convert the received signal level to voltage:
 -100 dBm
 is 100 dB below 632 mVpp.
 100 dB for voltage quantities is equivalent to 105.
 -100 dBm is equivalent to 6.32 μVpp.
 This input level is amplified by 15 dB (≈ 5.62),
 The output swing is 35.5 μVpp.
 Notice: For a narrowband (not sinusoid) 0-dBm signal, it is still possible to
8
approximate the (average) peak-to-peak swing as 632mV.
Voltage vs Power
 Why the output voltage of the amplifier is of interest in
this example?
– If the circuit following the amplifier does not present a 50-Ω input
impedance, the power gain and voltage gain are not equal in dB.
– Mostly, the next stage may exhibit a purely capacitive input impedance,
thereby requiring no signal “power”.
• one stage drives the gate of the transistor in the next stage.
9
dBm Used at Interfaces Without Power
Transfer
 We use “dBm” at interfaces that do not necessarily entail power transfer.
(a) LNA driving a pure-capacitive
impedance with a 632-mVpp swing,
delivering no average power.
How about the power delivery?
Assumption:
 attaching an ideal voltage buffer to node X and drive a 50-Ω load.
 the signal at node X has a level of 0 dBm,
 means that if this signal were applied to a 50-Ω load, then it would deliver 1 mW.
(b) Use of fictitious buffer to visualize the
signal level in “dBm”
voltage buffer
10
General Considerations: linearity
 A system is linear if its output can be expressed as a linear combination
(superposition) of responses to individual inputs.
For arbitrary a and b, it holds that:
 Any system that does not satisfy this condition is nonlinear. Example:
 nonzero initial conditions or dc offsets cause nonlinearity;
 However, we often relax the rule ---------- accommodate these two effects.
11
Nonlinearity: Memoryless and Static System
 Memoryless or static : if its output does not depend on the past values of its
input;
Memoryless, linear
 The input/output characteristic of a memoryless nonlinear system can be
approximated with a polynomial
Memoryless, nonlinear
If the system is time variant，αj would be general functions of time.
12
Nonlinearity: Memoryless and Static System
Example of a memoryless nonlinear circuit:
If M1 operates in the saturation region and can be
approximated as a square-law device [Lee]
then
1
𝑊𝑊
𝐼𝐼𝐷𝐷 = 𝜇𝜇𝑛𝑛 𝐶𝐶𝑜𝑜𝑜𝑜
𝑉𝑉𝑖𝑖𝑖𝑖 − 𝑉𝑉𝑇𝑇𝑇𝑇
2
𝐿𝐿
2
Common-source stage
 In this idealized case, the circuit displays only second-order nonlinearity.
13
Nonlinearity: Odd symmetry
A system has “odd symmetry” if y(t) is an odd function of x(t), i.e., if the
response to –x(t) is the negative of that to x(t).
y (t ) =+
α 0 α1 x(t ) + α 2 x 2 (t ) + α 3 x3 (t ) +
y(t) is odd symmetry if αj=0 for even j.
Such a system is sometimes called “balanced”, as exemplified by
the differential pair shown in the next page.
14
Example of Polynomial Approximation
For square-law MOS transistors operating in saturation, the characteristic
“differential pair circuit” can be expressed as
If the differential input is small, approximate the characteristic by a polynomial.
Assuming
Approximation (Taylor Expansions) gives us:
Differential pair
Input/output characteristic
15
Example of Polynomial Approximation
 Observations
 The first term represents linear operation,
 the small-signal voltage gain of the circuit (-gmRD);
 Due to symmetry, even-order nonlinear terms are absent;
 Notice: square-law devices yield a third-order characteristic in
this case.
16
General Considerations: Time Variance
 A system is time-invariant if a time shift in its input results in the same time
shift in its output.
If
y(t) = f [x(t)]
then
y(t-τ) = f [x(t-τ)]
Time Variance
Nonlinearity
Do not be confused by these two attributes.
17
Example of Time Variance
Plot the output waveform of the circuit in Fig. 1:
 vin1 = A1 cos ω1t
 vin2 = A2 cos(1.25ω1t )
Solution:
Fig.1
Switch: vout tracks vin2 if vin1 > 0 and is pulled down to zero by R1 if vin1 < 0.
vout is equal to the product of vin2 and a square wave toggling between 0 and 1.
This is an example of RF “mixers”.
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A time shift in input does not result in the same time shift in output.
Time Variance: Generation of Other Frequency
Components
S(t) denotes a square wave toggling
between 0 and 1 with a frequency of
f1=ω1/(2π)
The spectrum of square wave:
a train of impulses whose
amplitude follow a sinc
envelop.
T1= 2π /ω1
Illustration:
Multiplication in time domain
Convolution in frequency domain
 A linear system can generate frequency components that do not exist in the
19
input signal when system is time variant.
Effects of nonlinearity
•
•
•
•
•
•
? Frequency
? Amplitude
Harmonic distortion (谐波失真)
Gain compression (增益压缩)
Cross modulation (互调)
Intermodulation (交叉调制)
20
Notice
 Analog and RF circuits can be approximated with a linear model for
small-signal operation.
 In general, we have
 memoryless time-variant systems with input/output characteristic :
y (t ) ≈ α1 x(t ) + α 2 x 2 (t ) + α 3 x3 (t )
 α1 is considered as the small-signal gain.
 The nonlinearity effects primarily arise from the third-order term α3.
21
Effects of nonlinearity
•
•
•
•
Harmonic distortion (谐波失真)
Gain compression (增益压缩)
Cross modulation (互调)
Intermodulation (交叉调制)
22
Effects of Nonlinearity: Harmonic Distortion
If a sinusoid is applied to a nonlinear system:
 the output exhibits frequency components that are integer multiplies (“harmonics”)
of the input frequency.
input:
output:
x(t ) = A cos ωt
y (t ) ≈ α1 x(t ) + α 2 x 2 (t ) + α 3 x3 (t )
DC
Arising from
second-order
nonlinearity
Fundamental
Second
Harmonic
Third
Harmonic
The term with the
input frequency
23
Observations
y (t ) ≈ α1 x(t ) + α 2 x 2 (t ) + α 3 x3 (t )
 Even-order harmonics result from αj with even j and
vanish if the system has odd symmetry,
 If mismatches corrupt the symmetry, what will happen?
 The amplitude of the nth harmonic grows in proportion to?
 An .
24
Example of Harmonic Distortion in Mixer
An analog multiplier “mixes” its two inputs, producing y(t) = kx1(t)x2(t), where k is
a constant. Assume x1(t) = A1 cos ω1t and x2(t) = A2 cos ω2t.
Question:
(a) If the mixer is ideal, determine the output frequency components.
Solution:
(a)
Analog multiplier
The output contains the sum and difference frequencies.
These are “desired” components.
25
Example of Harmonic Distortion in Mixer
An analog multiplier “mixes” its two inputs below, producing y(t) = kx1(t)x2(t),
where k is a constant. Assume x1(t) = A1 cos ω1t and x2(t) = A2 cos ω2t.
(a)If the mixer is ideal, determine the output frequency components.
(b) If the input port sensing x2(t) suffers from third-order nonlinearity, determine
the output frequency components.
Solution:
(b)
Third Harmonic of x2(t)
Analog multiplier
The mixers produces two “spurious” components at ω1+3ω2 and ω1-3ω2,
 Cause other problems…
These are “undesired” components that are difficult to remove by filter.
26
Example of Harmonics on GSM Signal
The transmitter in a 900-MHz GSM cellphone delivers 1 W of power to the antenna.
Explain the effect of the harmonics of this signal.
The second harmonic?
 falls within another GSM cellphone band around 1800 MHz;
 Must be sufficiently small to impact the other users in that band.
The third, fourth, and fifth harmonics?
 do not coincide with any popular bands;
 but must still remain below a certain level imposed by regulatory organizations in
each country. (中国工信部无线电管理局/US FCC)
The sixth harmonic?
 falls in the 5-GHz band used in wireless local area networks (WLANs).
fundamental
27
Effects of nonlinearity
•
•
•
•
Harmonic distortion (谐波)
Gain compression (增益压缩)
Cross modulation (互调)
Intermodulation (交叉调制)
28
x(t ) = A cos ωt
Gain Compression– Sign of α1, α3
y (t ) ≈ α1 x(t ) + α 2 x 2 (t ) + α 3 x3 (t )
 The gain of fundamental component ω is equal to α1 + 3α3A2/4
 varied as A becomes larger.
Expansive:
 expanding the gain as the input
amplitude increases
Compressive:
 The term α3x3 “bends” the characteristic
for sufficiently large x,
 decreasing the gain as the input
amplitude increases.
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 Most RF circuit of interest are compressive, we focus on this type.
Gain Compression: 1-dB Compression Point
 With α1α3 <0, the fundamental gain is equal to α1 + 3α3A2/4 and falls as A rises.
 1-dB compression point: defined as the input signal level that causes the small
signal gain to drop by 1dB.
small signal gain
large signal gain
Plotted on a log-log scale as a function of the input level
 Output level, Aout, falls below its ideal value by 1 dB at the 1-dB compression point,
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Ain,1dB.
Calculation
At the 1-dB compression point, Ain,1dB:
3
20 log α1 + α 3 Ain2 ,1dB =20 log α1 − 1dB
4
small signal gain
Ain ,1dB
α1
= 0.145
α3
large signal gain
 Notice: The relationship of the input is
between peak value (rather than the peak-topeak value).
 Ain and Aout are voltage quantities, but
compression can also be expressed in31
terms of power quantities (dBm).
Observations
• The 1-dB compression point is typically in the range of
-20 to -25 dBm (63.2 to 35.6 mVpp in 50-Ω system) at the
input of RF receivers front-end amplifiers.
• We use the notations A1dB and P1dB interchangeably in
this book.
• Why does compression matter?
– a signal is so large as to reduce the gain of a receiver
32
Gain Compression: Effect on FM and AM
Waveforms
Are they acceptable?
 FM signal carries no information in its amplitude and hence tolerates
compression. √
 AM contains information in its amplitude, hence distorted by compression. X
33
• Gain compression on two signals?
34
Gain Compression: Desensitization
 If there are two signals: a small desired signal + a large interferer
 According to the effect of gain compression, if a signal is too large, it will
reduce the gain of a receiver.
 Desensitization: when small desired signal is superimposed on the large
interferer, the receiver gain is reduced even though the desired signal itself
is small.
 This phenomenon lowers the signal-to-noise ratio (SNR) at the receiver,35
 even if the signal contains no amplitude information.
Example of Gain Compression
A 900-MHz GSM transmitter delivers a power of 1 W 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 = -25 dBm? Assume the receiver is
1 m away and the 1.8-GHz signal is attenuated by 10 dB as it propagates across
this distance.
Solution:
 The output power at 900 MHz is +30 dBm
(1W).
 With an attenuation of 10 dB, the second
harmonic must not exceed -15 dBm at
the transmitter antenna so that it is
below P1dB of the receiver.
 Thus, the second harmonic must remain
at least 45 dB below the fundamental at
the TX output.
 In practice, this interference must be
another several dB lower to ensure the
RX does not compress.
36
How to quantify desensitization?
With the third-order characteristic,
y (t ) ≈ α1 x(t ) + α 2 x 2 (t ) + α 3 x3 (t )
the output is:
For A1 << A2
 The gain experienced by the desired signal is equal to α1 + 3α3A22/2,
 a decreasing function of A2 if α1α3 <0.
 For sufficiently large A2, the gain drops to zero, and we say the signal is
“blocked”.
37
Observations
 In RF design, the term “blocking signal” or “blocker” refers
to interferers that desensitize a circuit,
 even if they do not reduce the gain to zero.
 Some RF receivers must be able to withstand blockers that
are 60 to 70 dB greater than the desired signal.
38
Effects of nonlinearity
•
•
•
•
Harmonic distortion (谐波)
Gain compression (增益压缩)
Cross modulation (互调)
Intermodulation (交叉调制)
39
Effects of Nonlinearity: Cross Modulation
When a weak signal and a strong interferer pass through a nonlinear
system,
 “cross modulation” will happen.
Phenomenon:
 modulation (or noise) on the amplitude of the interferer will
transfer to the amplitude of the weak signal.
40
Effects of Nonlinearity: Cross Modulation
Suppose that the interferer is an amplitude-modulated signal
where m is a constant and ωm denotes the modulating frequency.
For A1 << A2
41
 Desired signal at output suffers from amplitude modulation at ωm and 2ωm.
Example of Cross Modulation
Suppose an interferer contains phase modulation but not amplitude modulation.
Does cross modulation occur in this case?
Solution:
PM interferer (A2 is constant
but Φ varies with time)
Expressing the input as:
We now note that
(1) the second-order term yields components at ω1 ± ω2 but not at ω1--------can be removed;
(2) the third-order term expansion gives 3α3A1 cos ω1t A22 cos2(ω2t+Φ),
 according to cos2x=(1+cos2x)/2,
 results in a component at ω1.
42
The desired signal at ω1 does not experience cross modulation. (constant can be removed.)
Example of Cross Modulation: observations
 Phase-modulated interferers do not cause cross
modulation in memoryless (static) nonlinear systems.
 Cross modulation commonly arises in amplifiers that
must simultaneously process many independent signal
channels.
 Examples include cable television transmitters and
system employing “orthogonal frequency division
multiplexing” (OFDM).
43
Effects of Nonlinearity: Intermodulation—
Recall Previous Discussion
 Desired Signal
 Desired Signal + one (large)
interferer
 Desired Signal + two large
interferers???
Harmonic distortion
Desensitization + Cross Modulation
Intermodulation
44
Effects of Nonlinearity: Intermodulation
 If two interferers at ω1 and ω2 are applied to a nonlinear system,
– the output exhibits components that are not harmonics of these
frequencies.
 The phenomenon of “intermodulation” (IM) arises from “mixing”
(multiplication) of the two components.
45
Effects of Nonlinearity: Intermodulation
Assume
Third-order characteristic,
Intermodulation products:
y (t ) ≈ α1 x(t ) + α 2 x 2 (t ) + α 3 x3 (t )
Expanding the right-hand side and discarding the dc terms,
harmonics, and components at ω1 ±ω2, we have:
Fundamental components:
46
Third-order IM products
 The third-order IM products at 2ω1- ω2 and 2ω2-ω1 are of
particular interest.
 if ω1 and ω2 are close to each other,
 2ω1- ω2 and 2ω2-ω1 are close to ω1 and ω2,
47
Intermodulation Product Falling on Desired
Channel
Interferer
desired
 Suppose an antenna receives:
 a small desired signal at ω0
 two large interferers at ω1 and ω2,
 Applying to a low-noise amplifier (nonlinear).
 Suppose
 The intermodulation product at 2ω1-ω2, falls onto the desired channel,
 corrupting the signal.
48
Example of Intermodulation
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;
 Users 2 and 3 transmit at 2.420 GHz and 2.430 GHz, respectively.
Explain what happens?
Solution:
 The frequencies transmitted by Users 1, 2, and 3 are equally spaced;
 The intermodulation in the LNA of RX4 corrupts the desired signal at 2.410 GHz.
49
Example of Gain Compression and Intermodulation
If the gain is not compressed, can we say that intermodulation is negligible?
A Bluetooth receiver:
a) a low-noise amplifier (LNA) having a gain of 10
b) input impedance of 50 Ω.
The LNA senses a desired signal level of -80 dBm at 2.410 GHz and two interferers
of equal levels at 2.420 GHz and 2.430 GHz.
a) For simplicity, assume the LNA drives a 50-Ω load.
a) Determine the value of α3 that yields a P1dB of -30 dBm.
b) If each interferer is 10 dB below P1dB, determine the corruption experienced by
the desired signal at the LNA output.
(a) From previous equation,
Ain ,1dB = 0.145 α1 α 3
α3 = 14,500 V-2
-80 dBm
(b) Each interferer has a level of -40 dBm (= 3.16 m Vp),
the amplitude of the IM product at 2.410 GHz as:
(c) The desired signal is amplified by a factor of α1=10=20dB,
emerging at the output at a level of -60dBm.
(d) The IM product is as large as the signal even though the LNA does not experience
significant compression.
50
Intermodulation: Third Intercept Point
 “Third Intercept Point” (IP3): if the amplitude of each tone arises, that of the output IM
products increases more sharply (∞A3). Thus, if we continue to raise A, the amplitude of
the IM products eventually becomes equal to that of the fundamental tones at the output.
 The input level at which this occurs is called the “input third interception point” (IIP3),
where the corresponding output is represented by OIP3.
A: equal amplitude of each tone
Equate the fundamental and IM amplitudes:
A1dB = 0.145
α1
α3
51
Example of Third Intercept Point
A low-noise amplifier senses a -80-dBm signal at 2.410 GHz and two -20-dBm
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 50-Ω interfaces at the
input and output.
Solution:
IM amplitude
Fundamental amplitude
At the LNA output:
>
>
Asig = 31.6μV; Aint = 31.6mV
>
> 3.16 VP
Extremely difficult to achieve
in this case.
Usually we have:
A1dB: -20 to -25dBm
+9.6dB
> +20dBm
IP3: -10 to -15dBm
52
Effects of Nonlinearity: Recall Previous
Discussion
So far we have discussed the case of:
 Single Signal
 Signal + one large interferer
 Signal + two large interferers
Harmonic distortion
Desensitization + Cross Modulation
Intermodulation
53
Effects of Nonlinearity: Cascaded Nonlinear
Stages
Consider two nonlinear
stages in cascade:
Considering only the first- and third-order terms, we have:
54
Cascaded Nonlinear Stages: Intuitive results
 divide “IP3 of the second stage” by α1.
 Effect: the higher the gain of the first stage, the more nonlinearity is contributed by
the second stage.
 α1 ↑ →
↑→
↑→ AIP3 ↓ → nonlinearity ↑
55
IM Spectra in a Cascade
 Most RF systems incorporate narrowband circuits
 the terms at ω1 ± ω2 , 2ω1, and 2ω2 are heavily attenuated at the output of
the first stage
 second-order nonlinearity is attenuated.
For more stages:
 Notice:
 The nonlinearity of the latter stages becomes increasingly more critical;
 because the IP3 of each stage is equivalently scaled down by the total gain
preceding that stage.
56
Example of Cascaded Nonlinear Stages
A low-noise amplifier having an input IP3 of -10 dBm and a gain of 20 dB is
followed by a mixer with an input IP3 of +4 dBm. Which stage limits the IP3 of the
cascade more?
Solution:
α1 = 20 dB
 The scaled IP3 of the second stage AIP3,2 is lower than the IP3 of the first stage
AIP3,1,
 The second stage limits the overall IP3 more.
57
Sections to be covered
•
•
•
•
•
2.1 General Considerations
2.2 Effects of Nonlinearity
2.3 Noise
2.4 Sensitivity
2.5 Passive Impedance Transformation
58
Noise
 The performance of RF systems is limited by noise.
 Without noise, an RF receiver would be able to detect arbitrarily
small inputs,
 allowing communication across arbitrarily long distances.
 In this section, we review basic properties of noise and
methods of calculating noise in circuits.
59
Noise: Noise as a Random Process
 “noise is random,”
 the instantaneous value of noise cannot be predicted.
 Example: a resistor tied to a battery and carrying a current.
 Due to the ambient temperature, each electron carrying the current experiences
thermal agitation,
 moving toward the positive terminal of the battery with a random path;
 Higher temperature
a) greater thermal
agitation of electrons;
b) larger fluctuations in
the current.
 The average current remains equal to VB/R;
 The instantaneous current displays random values.
60
Noise: Noise as a Random Process
 Noise cannot be characterized in terms of instantaneous voltages or
currents.
 How do we express the concept of larger random swings for a current
or voltage quantity?
 Average power
n(t) represents the
noise waveform
 Noise components in circuits have a constant average power.
 For example, Pn is known and constant for a resistor at a constant
temperature.
61
Noise Spectrum: Power Spectral Density
(PSD)
 Total area under Sx(f) represents the average power carried by x(t)
Parseval's theorem
Two-Sided
One-Sided
62
Example of Noise Spectrum
A resistor of value R1 generates a white noise voltage whose one-sided PSD is
given by
 Boltzmann constant k = 1.38 × 10-23 J/K
 The absolute temperature T.
(a) What is the total average power carried by the noise voltage?
(b) What is the dimension of Sv(f)?
(c) Calculate the noise voltage for a 50-Ω resistor in 1 Hz at room temperature.
(a) Ideally, the area under Sv(f) appears to be infinite，
 resistor noise arises from the finite ambient heat.
 In reality, Sv(f) begins to fall at f > 1 THz, exhibiting a finite total energy, i.e., thermal
noise is not quite white.
(b) The dimension of Sv(f) is voltage squared per unit bandwidth (V2/Hz), we may write the
PSD as
where
denotes the average power of Vn in 1 Hz.
(4kTRΔf) indicates the total noise in a bandwidth of Δf .
63
Example of Noise Spectrum
A resistor of value R1 generates a noise voltage whose one-sided PSD is given by
where k = 1.38 × 10-23 J/K denotes the Boltzmann constant and T the absolute
temperature.
(c) Calculate the noise voltage for a 50-Ω resistor in 1 Hz at room temperature.
(c) For a 50-Ω resistor at T = 270 C or 3000 K
a root-mean-squared (rms) quantity:
64
Noise Figure
 How does the SNR degrade as the signal travels through a given circuit?
 If the circuit contains no noise,
• the output SNR is equal to the input SNR
• (even if the circuit acts as an attenuator.)
 To quantify how noisy the circuit is, we define its noise figure (NF) as:
equal to 1 for a
noiseless stage.
in decibels:
65
Device Noise
 Model the noise by voltage and current sources:
 Thermal noise of resistors can be modeled by a series voltage source with a
PSD of
[Thevenin equivalent, 戴维南等效] or a parallel current
2
source with a PSD of
[Norton equivalent, 诺顿等效 ];
(a)
(a) Thevenin
(b)
(b) Norton models of resistor thermal noise
 Polarity of the sources is unimportant but must be kept same throughout the
calculations.
 The choice of the model sometimes simplifies the analysis.
66
Noise in MOSFETS
 Thermal noise of MOS transistors
operating in the saturation region
is approximated by a current
source tied between the source and
drain terminals;
(a) Current source
 or can be modeled by a voltage
source in series with gate.
(b) Voltage source
 γ is the “excess noise coefficient” and gm the transconductance. The value of γ：
 2/3 for long-channel transistors；
 may rise to even 2 in short-channel devices.
 The actual value of γ has other dependencies and is usually obtained by
67
measurements for each generation of CMOS technology.
Other noises in MOSFETS
 Flicker or “1/f ” noise
– Modeled by a voltage source in series with the gate,
– 1/f dependence
– For a given device size and bias current, the 1/f noise
PSD intercepts the thermal noise PSD at some frequency,
called the “1/f noise corner frequency,” fc.
– It can be negligible at high frequencies.
68
Effect of Transfer Function on Noise
 If white noise is applied to a low-pass filter, how do we determine the PSD at
the output?
 If x(t) is applied to a linear, time-invariant system with a transfer function H(s),
then the output spectrum is
where H(f)= H(s = j2πf ).
69
Example of Device Noise
Sketch the PSD of the noise voltage measured across the parallel RLC tank
depicted in figure below.
2
𝐼𝐼𝑛𝑛1
(a) RLC
ZT
(b) inclusion of resistor noise
Modeling the noise of R1 by a current source
(c) output noise spectrum due to R1
,
The transfer function Vn/In1 is equal to the impedance of the tank ZT ,
 At f0, L1 and C1 resonate, reducing the circuit to only R1. Thus, the output
noise at f0 is simply equal to 4kTR1.
 At lower or higher frequencies,
 the impedance of the tank falls ;
 the output noise falls.
70
Energy delivery of thermal noise
 What is the energy delivered by thermal noise?
 R1 at room temperature;
 R2 is an ideal noiseless resistance (at 00K) ;
 Effect:
 R1 continuous to draw thermal energy from its environment,
 converting it to noise and delivering the energy to R2.
The average power
transferred to R2 is:
Transfer of noise from one resistor to another
This quantity reaches a maximum if R2 = R1 :
PR2,max = kT (W/Hz) =-174 dBm/Hz
71
Available noise power
Observations
• If a passive circuit dissipates energy,
– it must contain a physical resistance
– must produce thermal noise.
• “lossy circuits are noisy”.
72
A Theorem about Lossy Circuit
passive
network
 If the real part of the impedance seen between two terminals of a
passive network is equal to Re{Zout},
 the PSD of the thermal noise seen between these terminals is given by
4kTRe{Zout}
73
Lossy Circuit: example
(a) Transmitting antenna
 A transmitter antenna
 dissipates energy by radiation
according to the equation
V2TX,rms/Rrad,
 where Rrad is the “radiation
resistance”.
(b) receiving antenna producing thermal noise
 As a receiving element, the
antenna generates a thermal
noise PSD of
74
Useful Tools
• Effect of Transfer Function on Noise
• A Theorem about Lossy Circuit
passive
network
75
Noise Figure
 How does the SNR degrade as the signal travels through a given circuit?
 If the circuit contains no noise,
• then the output SNR is equal to the input SNR
• even if the circuit acts as an attenuator.
 To quantify how noisy the circuit is, we define its noise figure (NF) as:
equal to 1 for a
noiseless stage.
in decibels:
76
Noise Figure
 Depends on:
 the circuit under consideration
 the SNR provided by the preceding stage.
 If SNRin=∞
 NF=∞
 even though the circuit may have finite internal noise.
77
Calculation of Noise Figure (1)
 A low-noise amplifier senses the signal received by an antenna;
 The antenna “radiation resistance,” RS, produces thermal noise.


represents the thermal noise of the antenna;
represents the output noise of the LNA.
 How to compute SNRin at the LNA input and SNRout at its output?
78
Calculation of Noise Figure (2)
If the LNA exhibits an input
impedance of Zin,
 Both Vin and VRS experience an
attenuation factor of α = Zin / (Zin +
RS)
We assume a voltage gain of Av
from the LNA input to the output;
Output signal power:
Output noise consists of two parts:
 the noise of the antenna amplified
by the LNA,
 the output noise of the LNA,
79
Calculation of Noise Figure (3)




Calculate the output noise due to the amplifier,
divide it by the gain,
normalize it to 4kTRs ,
add 1 to the result.
80
Calculation of Noise Figure (4)
 Reduce the right hand side to a simpler form:
 The numerator is the total noise measured at the output, denoted
as
,
 includes both the source impedance noise and the LNA noise;
 A0 =| α |Av is the voltage gain from Vin to Vout
 (not the gain from the LNA input to its output).
 Divide total output noise by the gain from Vin to Vout
 normalize the result to the noise of Rs
81
Noise Figure of Cascaded Stages
AP1 is the “available power gain” of the first stage,
 the “available power” at its output, Pout,av (the power that it would deliver to a
matched load) divided by
 the available source power, PS,av (the power that the source would deliver to a
matched load).
Called “Friis’ equation”,
 the noise contributed by each stage decreases as the total gain
preceding that stage increases；
 the first few stages in a cascade are the most critical.
82
Noise Figure of Lossy Circuits
 Passive circuits (filters) appear at the front end of RF transceivers;
 their loss proves critical.
 attenuate the signal
 introduce noise.
 Proposition: the noise figure of a passive circuit is equal to its “power loss,”
defined as L=Pin/Pout,
 Pin is the available source power;
 Pout is the available power at the output.
 Proof: The lossy circuit is driven by a
source impedance of RS while driving a
load impedance of RL.
83
Noise Figure of Lossy Circuits
According to the theorem of “thermal
noise model of lossy circuit”, we have
VThev
The available source power is Pin =V2in/(4RS);
According to Thevenin equivalent, the
available (matching) output power is Pout
=V2Thev/(4Rout).
 Divide total output noise by the
squared gain from Vin to Vout and
normalize the result to the noise of Rs
84
Example of Noise Figure of Lossy Circuits
A receiver
 a front-end band-pass filter (BPF) to suppress some of the interferers
that may desensitize the LNA;
 the filter has a loss of L;
 the LNA a noise figure of NFLNA;
calculate the overall noise figure.
The noise figure of the filter is NFfilt,
Friis’ equation:
According to the definition of L,
available power gain Ap1=1/L
where NFLNA can be calculated with
respect to the output resistance of
the filter
85
Example:
if L = 1.5 dB and NFLNA = 2 dB, then NFtot = 3.5 dB.
Sections to be covered
•
•
•
•
•
2.1 General Considerations
2.2 Effects of Nonlinearity
2.3 Noise
2.4 Sensitivity
2.5 Passive Impedance Transformation
86
Sensitivity
 The minimum signal level that a receiver can detect with “acceptable quality.”
 acceptable quality:
 sufficient signal-to-noise ratio,
 It depends on the type of modulation and the corruption (e.g., bit error rate) that the
system can tolerate.
All quantities
expressed in W/Hz
(or dBm/Hz)
Match of receiver and antenna
Eq. (2.91) PRS= kT= -174 dBm/Hz
Sensitivity
87
Noise Floor: the total integrated noise
Example of Sensitivity
Compare the sensitivities of these two systems if both have an NF of 7 dB.
 A GSM receiver requires a minimum SNR of 12 dB and has a channel
bandwidth of 200 kHz.
 A wireless LAN receiver specifies a minimum SNR of 23 dB and has a
channel bandwidth of 20 MHz.
Solution:
For the GSM receiver, Psen = -174+7+53+12= -102 dBm,
For the wireless LAN system, Psen = -174+7+73+23= -71 dBm.
Does this mean that the latter is inferior?
 No, the latter employs a much wider bandwidth and a more efficient
modulation to accommodate a data rate of 54 Mb/s.
 The GSM system handles a data rate of only 270 kb/s.
In other words, specifying the sensitivity of a receiver without the data rate is
not meaningful.
88
The journey of Noise
RF Receiver
Source
Digital Receiver
89
Sections to be covered
•
•
•
•
•
2.1 General Considerations
2.2 Effects of Nonlinearity
2.3 Noise
2.4 Sensitivity
2.5 Passive Impedance Transformation
– Thomas H. Lee
– Razavi
– 张肃文
90
Sections to be covered
•
•
•
•
•
2.1 General Considerations
2.2 Effects of Nonlinearity
2.3 Noise
2.4 Sensitivity
2.5 Passive Impedance Transformation
–
–
–
–
Series RLC tank
Parallel RLC tank
Series and parallel RC/RL networks
Basic Matching Networks
91
Series RLC tank circuit
The complex impedance of this circuit is given by
adding up the impedance of the components:
1
Z =R + jX =R + j (ω L −
)
ωC
Resistance reactance
1 2
Z = R + (ω L −
)
ωC
1
when
ωL =
ωC
2
V - the voltage of the power source
I - the current in the circuit
R - the resistance of the resistor
L - the inductance of the inductor
C - the capacitance of the capacitor
resonance frequency:
Z min = R
ωs =
1
LC
There is a valley in impedance at resonance.
Characteristic impedance:
1
=
ρ ω=
=
sL
ωs C
L
C
92
Attributes of impedance Z
Z
1
Z =R + jX =R + j (ω L −
)
ωC
Z =
1 2
R + (ω L −
)
ωC
2
magnitude
ϕ
ω
ωs
π
2
phase shift
ω
−
π
2
ωs
93
Quality Factor
The quality factor, Q, indicates how close to ideal an energy-storing device is.
An ideal capacitor or inductor dissipates no energy, exhibiting an infinite Q,
but a series resistance or a parallel resistance reduces its Q.
The definition of Q:
energy stored
Q =ω
average power dissipated
Q is dimensionless.
As to a series RLC tank circuit:
1
ρ ωs L ωs C 1 L
=
=
=
Q=
s
R
R
R
R C
94
Frequency response of the current: the magnitude
Z =
I
=
Imax
1 2
R + (ω L −
)
ωC
2
1
=
ωs 2
2 ω
1 + Qs ( − )
ωs
ω
1
1 + (Qs 2
ω
ωs
)2
ω= ω − ωs
Imax
ωs
is the maximum
current at resonance
frequency
95
Frequency response of the current: the phase shift
1
Z =R + jX =R + j (ω L −
)
ωC
X is the imaginary part of Z, defined as
reactance

V
I =
Z
ϕ = −tg
π
−1
X
R
ϕ
2
ωs
Q2
Q1
−
ω
Q1 > Q2
π
2
96
Bandwidth
The resonance effect can be used for filtering;
The rapid change in impedance near resonance can be used to
 Pass signals
 block signals
A key parameter in filter design is bandwidth.
 The bandwidth is measured between the 3dB-points；
 Half-power frequencies: the power passed through the circuit has
fallen to half the value passed at resonance.
B = ω2 − ω1 =
ωs
Qs
97
Q and Bandwidth B
•
Q is related to bandwidth;
– low Q circuits are wide band
– high Q circuits are narrow band
•
Q is the inverse of fractional bandwidth( B/ωs );
•
Q factor is directly proportional to selectivity, as Q factor depends
inversely on bandwidth.
Qs =
ωs
B
98
Loaded Q factor
L
C
R
RS
RL
Vs
load
Qs =
=
QL
ωs L
Unloaded Q
Rs + R
ωs L
Rs + R + RL
< Qs
99
Parallel RLC tank circuit
The complex admittance of this circuit is
given by adding up the admittances of the
components:
1
Y=
G + j (ωC −
)
ωL
Conductance
Y =
when
I in
R
C
L
Vout
susceptance
G 2 + (ωC −
1
ωL =
ωC
1 2
)
ωL
1
Ymin= G=
R
resonance frequency:
ωp =
1
LC
There is a valley in
admittance at resonance.
L
1
=
ρ ω=
=
Characteristic impedance:
pL
C
ω pC
R R
As to a parallel RLC tank circuit:
Q=
=
= Rω p C=
p
ρ ωpL
R
L/C
100
Frequency response of the voltage:
the magnitude
I in
R
Y =
V
Vout
L=
Vmax
C
G 2 + (ωC −
V
Vmax
1
=
ω ωp 2
1 + Qp ( − )
ωp ω
2
1
1 + (Q p 2
ω
ωp
ω= ω − ω p
1 2
)
ωL
Vmax
1
is the maximum
voltage at resonance
frequency
-3 dB
0.7
B = ω2 − ω1 =
ω1 ω p ω2
)2
ωp
Qp
101
Loaded Q factor
Is
RS
C
L
RP
RL
load
Qp =
1
ω p L(G p + Gs )
Unloaded Q
1
R p′ G p + Gs + GL
1
=
QL =
=
< Qp
ωpL
ωpL
ω p L(G p + Gs + GL )
102
Another example
How about the property of “not-quiteparallel RLC tank circuit”?
IL
Is
V
C
L
Please refer to “高频电子线路,张肃文”
IC
R
103
• Impedance transformation
– Matching Network
– Series/parallel
104
Series and parallel RC/RL networks
Qs =
ρ
Qp =
R
Series RC circuit
What choice of
values makes the
Equivalent parallel circuit
two networks
equivalent?
Here, ω ≈ωs
Series RL circuit
Equivalent parallel circuit
105
R
ρ
Series-to-Parallel Conversion (1)
1
Rp
jωC p
1
=
Rs +
jωCs R + 1
p
jωC p
Equating the impedance:
reforming:
R p Cs jω =
1 − R p C p Rs Csω 2 + ( R p C p + Rs Cs ) jω
Real part Equation
R p C p Rs Csω 2 = 1
(1)
1
Rp
C sω s
=
1
Rs
C pω p
Qs = Q p
106
Series-to-Parallel Conversion (2)
Real part equation
Image part equation
R p C p Rs Csω 2 = 1
0
R p C p + Rs Cs − R p Cs =
(1)
(2)
 When Cp and Cs are selected, the two impedances remain equal at all
frequencies?
 No!!!
 An approximation allows equivalence for a narrow frequency range.
For a general ω,
Substitute RpCp in (1) from (2):
=
R
Simplified but with constraints,
near ωs
p
Since Qs=1/CsωsRs, then
Substitute Qs into (1):
Qs2  1
1
+ Rs
2 2
Rs Cs ω
2
Q
( s + 1) Rs
Qs2
C p = 2 Cs
Qs + 1
=
Rp
R p ≈ Qs2 Rs
C p ≈ Cs
107
Parallel-to-Series Conversion
Observations:
 Typically, Q>4,
 Series-to-Parallel Conversion:
 retain the value of the capacitor
 but raises the resistance by a factor of Qs2.
 Parallel-to-Series Conversion:
 reduce the resistance by a factor of QP2.
 This statement applies to RL sections as well.
108
Basic Matching Networks
 a load resistance must be transformed to a lower value.
 the capacitor in parallel with RL converts this resistance to a lower series
component.
 The inductance is inserted to cancel the equivalent series capacitance.
ω≈ ωp
RL transformed
down by a factor
Setting imaginary part to zero (resonate), we have
ω≈ ωp
=
Qp
RL
=
RLC1ω p
1
C1ω p
If
109
Example of Basic Matching Networks
Design the matching network of the following figure so as to transform RL = 50 Ω
to 25 Ω at a center frequency of 5 GHz.
Solution:
 Assuming QP2 >> 1, we have C1 = 0.90 pF and L1 = 1.13 nH.
 however, QP = 1.41,
 indicating the QP2 >> 1 approximation cannot be used.
 We thus obtain C1 = 0.637 pF and L1 = 0.796 nH.
 Please fill out the detail calculation by yourself!
110
Another Example of Basic Matching Networks
Determine how the circuit shown below transforms RL.
The L1-RL branch to a parallel section produces a higher resistance.
If QS2 = (L1ω/RL)2 >> 1, then the equivalent parallel resistance is
The parallel equivalent inductance is approximately equal to L1 and is cancelled
111
by C1.
L-Section topologies
C1 transforms RL
to a smaller
series value and
L1 cancels C1
L1 transforms RL
to a larger
parallel value
and C1 cancels
the parallel
inductance
L1 transforms RL
to a smaller series
value while C1
resonates with L1
C1 transforms RL
to a larger
parallel value
and L1 cancels
the parallel
capacitance
1. In (a), the power delivered to the input port is
2. Power delivered to the load is
3. if L1 and C1 are ideal, these two powers must be equal:
 a network transforming RL to a lower value “amplifies” the voltage and attenuates the
112
current by the above factor.
Transfer a Resistance to a Higher Value
if
, the parallel combination of C1 and RL can be converted to a series network
 Viewing C2 and C1 as one capacitor, Ceq
 converting the resulting series section to a parallel circuit
RL boosted
Notice: the capacitive component must be cancelled by placing an inductor in parallel with
the input.
113
Impedance Matching by Transformers
 An ideal transformer having a turns ratio of n “amplifiers” the input voltage by
a factor of n.
 No power is lost, thus:
114
Observations
 The networks studied here operate across
 a narrow bandwidth (near resonate frequency)
 transformation ratio, e.g., 1+Q2, varies with frequency;
 The capacitance and inductance approximately resonate over a
narrow frequency range;
 Broadband matching networks can be constructed, but they typically
suffer from a high loss.
115
Loss in Matching Networks (1)
 Analyze the effect of loss.
 We define the loss as the power provided by the input divided by that delivered
to RL
The power delivered to Rin1 is entirely
absorbed by RL
Pin1=PL.
The loss of L1 modeled by a series
resistance, Rs
 This network transforms RL to a lower value,
Rin1=RL/(1+Q2P),
116
 suffering from loss even if Rs appears small.
Loss in Matching Networks (2)
The loss of L1 modeled by a parallel resistance, RP.
Pin represents the power delivered by Vin, which is entirely absorbed by RP||RL:
power delivered to the load, PL,
117
Basic matching networks
L-match
π-match
T-match
Tapped capacitor resonator as an
impedance matching network
Tapped inductor match
118
L-match (refer to P112,116)
 There are only two degrees of freedom (one can choose only L and
C);
 Once the impedance transformation ratio and resonant frequency
have been specified,
 network Q is automatically determined;
 How to get a different value of Q?
119
π-match
Cascade two L-matches
L1
L2
Original circuit
L
R in
C1
C2
RP
R in
C1
C2
RI
RP
Parallel RC
network
Two L-matches connected in cascade,
 one transforms down
 one transforms up;
The load resistance RP is transformed to a
lower resistance RI at the junction of the
two inductances;
RI is transformed up to a value Rin by a
second L-match section.
L1
R in
C1
RI
L2
C2s
Series
RI
120
equivalent
π-match: another example
C2
R in
C1
L
RP
Observations:
 We have three degrees of freedom (the two capacitance and the sum of two
inductances),
 We can independently specify
 center frequency,
 Q,
 overall impedance transformation ratio;
 In many instance, cascading several matching networks may be helpful.
121
T-match
L1
R in
Connecting up the L-sections
another way leads to the dual of
π-match, denoted T-match;
L2
RI
C1
RI
C2
RS
A single capacitor in a
practical implementation has
been decomposed explicitly into
two separate ones;
The parallel image resistance
(RI) is seen across these
capacitors, either looking to the
right or looking to the left.
122
Tapped capacitor match (refer to P107)
(a) capacitive
matching circuit
(b) simplified
circuit with
parallel-to-series
conversion
2
 Cp 
1
= 1 +
Rtot=
 Rp
2
Rs ( Ctotω )  C1 
C1C p
Ctot =
C1 + C p
(c) cascading
two capacitance
(d) simplified
circuit with
series-to-parallel
conversion
The capacitive divider
“boosts” the value of RP by a
2
factor
(1 + C
p
C1 )
123
Tapped inductor match
a) Convert the parallel combination Lp and Rp to series circuit Ls and Rs;
b) Combining L1 and Ls into Leq;
c) Converting series circuit to parallel circuit and we have:

L1
Rtot ≈ 1 +
 L
p

Ltot= L1 + Lp
2

 R p

124
Double-tapped resonator
L1
R in
L2
This circuit first boosts R2 to a
larger effective parallel resistance
across the whole tank than in a
standard tapped capacitor
network;
C1
C2
R2
It then reduces this parallel
resistance by the tapped
inductors to the desired value Rin.
125