Berkeley
Analysis of Memoryless Weakly Non-Lineary
Systems
Prof. Ali M. Niknejad
U.C. Berkeley
c 2014 by Ali M. Niknejad
Copyright Niknejad
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Review of Linear Systems
Z∞
y (t) =
h(t, τ )x(τ )dτ
−∞
Linear: x̂ = α x1 (t) + β x2 (t)
Z∞
Z∞
h(t, τ )x1 (τ )dτ +β
y (t) = α
−∞
−∞
|
h(t, τ )x2 (τ )dτ
{z
y1 (t)
}
|
{z
y2 (t)
}
= α y1 (t) + β y2 (t) X Linear
Complete description of a general time-varying linear system.
Note output cannot have a DC offset!
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Time-invariant Linear Systems
Time-invariant Linear Systems has h(t, τ ) = h(t − τ ).
Relative function of time rather than absolute
The transfer function is “stationary”
Z∞
y (t) =
h(t − τ ) x(τ ) dτ = h(t) ∗ x(t)
−∞
Y (jω) = H(jω)X (jω)
convolution in time is product in frequency
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Stable Systems
Linear, time invariant (LTI) system cannot generate frequency
content not present in input
Poles of H are strictly in the left
hand plane (LHP)
Y (jωk ) = 0 if X (jωk ) = 0
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Memoryless Linear System
No DC
y (t) = αx(t)
No Delay
If function is continuous at x0 , then we can do a Taylor Series
expansion about x0 :
00
f (xo )
ŷ (x) = f (xo ) + f (xo )(x − xo ) +
(x − xo )2 + ...
2!
0
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Taylor Series Expansion
This expansion has a certain radius of convergence. If we
truncate the series, we can compute a bound on the error
y − ŷ
so (x) = ŷ (x) − yo = ao + a1 x + a2 x 2 + a3 x 3 + ...
Let’s assume: x(t) =
N
P
An cos(ωn t)
n=1
Maximum excursion must be
less than radius of
convergence. Certainly the
max Ak has to be smaller
than the radius of
convergence.
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Sinusoidal Excitation
ym =
N
X
!m
An cos ωn t
=
n=1
=
N
X
An
n=1
N
X
An jωn t
e
2
=
N
X
k1 =−N k2 =−N
...
+e
−jωn t
Ao ≡ 0
ω−k = −ωk
X
()
X
|
N
X
e
!m
n=−N
=
2
!m
jωn t
X X
()
()
()
{z
}
m-times
N
X
Ak1 Ak2 ...Akm j(ωk1 +ωk2 +...+ωkm )t
e
2m
km =−N
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General Mixing Product
We have frequency components: ωk1 + ωk2 + ... + ωkm
where kp ranges over 2N values
Terms in summation: (2N)m
Example: Take m = 3, N = 2
ω1 , ω−1
(2N)m = 43 = 64
ω2 , ω−2
64 Terms in summation!
ω ∈ {−ω2 , −ω1 , ω1 , ω2 }

ω1 + ω1 + ω1 → 3ω1 ←
− HD3



ω1 + ω1 + ω2 → 2ω1 + ω2 ←
− IM3
64 Terms
ω + ω1 − ω2 → 2ω1 − ω2 ←
− IM3


 1
ω1 + ω1 − ω1 → ω1 ←
− gain expression or compression
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Vector Frequency Notation
*
Define k = (k−N , ..., k−1 , k1 , ..., kN )
2N-vector where kj denotes the number of times a particular
frequency appears in a give summation:

ω2 + ω1 + ω2  * −2 −1 0 +1 +2
ω1 + ω2 + ω2
k = 0
0 0 1
2

ω2 + ω2 + ω1
*
ω1 + ω1 − ω2 →
− k = (1
k=+N
X
0
2
0)
kj = k−N + ... + k−1 + k1 + ... + kN = m
k=−N
Sum = order of non-linearity
No DC terms (strike out middle zero frequency term from ~k
vector)
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Multinomial Coefficient
For a fixed vector k~0 , how many different sum vectors are
there?
m frequencies can be summed m! different ways, but order is
immaterial.
Each coefficient kj can be ordered kj ! ways. Therefore, we
have:
(m; k) =
m!
(k−N )! · · · (k−1 )!(k+1 )! · · · (kN )!
Multinomial coefficient
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Game of Cards (example)
3 Cards: 3! or six ways to order cards
3!
1! 2!
= 3 ways to order
Since R1 = R2
Reds not distinguished
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Making Conjugate Pairs
Usually, we only care about a particular frequency mix
generated by certain order non-linearity.
Since our signal is real, each term has a complex conjugate.
Hence, there is another:
*
k = (kN , ..., k1 , k−1 , ..., k−N )
{z
}
|
reverse order
Taking the complex conjugates in pairs:
2Re e j(ωk1 +ωk2 +...+ωkm )t = 2 cos (ωk1 + ωk2 + ... + ωkm ) t
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Amplitude of Mix
Thus the amplitude of any particular frequency component is:
*
*
2 × m; k
m; k
=
2m
2m−1
Ex: IM3 product generated by the cubic term
*
IM3 : (2ω1 − ω2 ) ⇒ k = (1 0 2 0)
*
m = 3, N = 2: m; k = 1!3!2! = 3, and 2m−1 = 22 = 4
Amplitude of IM3 relative to fundamental:
I M3 =
3 a3 si 3
3 a3 2
=
si
4 a1 si
4 a1
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Gain Compression/Expansion
How much gain compression occurs due to cubic and pentic
(x 5 ) terms?
cubic: m = 3, N = 1: ω1 + ω1 − ω1 = ω1

1
2
|{z}
|{z}
*
−ω1
+ω1
k =
appear
anywhere
This to appear
twice anywhere
*
k = (1, 2)


3!
(m; k)
3
3
2!
=
= amp. of fund: a1 si + a3 si 3
2m−1
22
4
4
3
App. Gain: = ai + a3 si 2 →
− Gain depends on signal amplitude
4
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Gain Compression/Expansion Pentic
pentic: m = 5, N = 1: ω1 − ω1 + ω1 − ω1 + ω1 = ω1
*
k = (2
*
m; k
=
3)
5·4
5!
=
= 10
3! 2!
2
2m−1 = 24 = 16
App. Gain: = a1 + 34 a3 si 2 +
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10
4
16 a5 si
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Who wins? Pentic or Cubic?
R=
Gain Reduction due to Cubic
Gain Reduction due to Pentic
=
3
2
4 a3 si
5
4
8 a5 si
=
1 a3 6
si 2 a5 5
Take an exponential transfer function and consider gain
compression:
IC = IS e VBE /VT
VBE = VQ + vs
VQ/
v
VT e s/VT
2
vs
VT
vs
+
+ ...)
= IQ (1 +
VT
2!
IC = IS e
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Compression for Exp (BJT)
a3 =
a5 =
R=
1 a3 6
1
= 2
s i 2 a5 5
si
1
VT
3
3!
5
1
VT
5!
1 3 1 5! 6
VT
3! 1 5 5
VT
=
1
si 2
1
VT
5·4·6
VT 2
=
24
2
5
si
When R = 1, pentic non-linearity contributes equally to gain
compression...
√
R = 1 ⇒ si = VT 24 ≈ 127mV
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Summary of Distortion
f (x) = a1 x + a2 x 2 + a3 x 3 + ...
Due to non-linearity, y (t) has frequency components not
present in input. For sinusoidal excitation by N tones, we
have M tones in output:
M=
(2N + m − 1) !
m! (2N − 1) !
m: Order of highest term in non-linearity (Taylor exp.)
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Amplitude of Frequency Mix
Particular frequency mix k~0 has frequency
f* =
k
N
X
kj fj = (k1 − k−1 ) f1 + ... + (kN − k−N ) fN
j=−N
j6=0
+N
X
kj = k−N + ... + k−1 + k1 + ... + kN = m
j=−N
θ* =
X
k
θj kj = (k1 − k−1 ) θ1 + ... + (kN − k−N )θN
The amplitude of any particular frequency mix


*
m; k
2m−1

|A1 |k1 +k−1 ...|AN |kN +k−N  × cos (ωk t + θk )
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Harmonic Distortion
For an input frequency ω0 , each order non-linearity (power)
produces a jth order harmonic in output j × ω0
HD3 = 41 aa31 si 2
= 14 aa133 so 2
HD2 = 12 aa21 si
= 21 aa122 so
HD2 ∝ Signal amplitude
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HD3 ∝ Signal amplitude2
2 dB increase for 1 dB signal
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Intermodulation
ω1 ≈ ω2 ≈ ωo
I M2 ≈ DC , 2ωo
I M3 ≈ ωo
For a two-tone input to a memoryless non-linearity, output
contains (2ω1 − ω2 ) and (2ω2 − ω1 ) due to cubic power and
(ω1 − ω2 ) and (ω1 − ω2 ) due to second order power.
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Filtering Intermodulation
IM2 products fall at much lower (DC) and higher frequencies
(2ω0 ). These signals appear as interference to others, but can
be attenuated by filtering.
IM3 products cannot be filtered for two close tones.
In a direct conversion receiver, IM2 is important due to DC
content of baseband signal.
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IM/Harmonic Relations
IM2
I M2 =
a2
a2
si = 2 so = HD2 (dB) + 6dB
a1
a1
I M2 = 2HD2
I M2 ∝ Signal level
IM3
I M3 =
3 a3 2 3 a3 2
si =
so = HD3 (dB) + 9.5dB
4 a1
4 a1 3
I M3 = 3HD3
I M3 ∝ Signal level2
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Triple Beat
Triple Beat: Apply three sine waves and observe effect of
cubic non-linearity
ω1 + ω2 + ω3
ω1 − ω2 + ω3
ω1 − ω2 − ω3
ω1 + ω2 − ω3
*
k TB = (0 0 0 1 1 1)
111
111
akp =
*
3; k TB
23−1
TB =
si 3 =
3!si 3
3
= si 3
4
2
3 a3 si 3
3 a3 2
=
si
2 a1 si
2 a1
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Intercept Point
Intercept Point: Apply a two tone input and plot output
power and IM powers. The intercept point in the extrapolated
signal power level which causes the distortion power to equal
the fundamental power.
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Intercept/IM Calculations
Say an amplifier has an IIP3 =-10 dBm. What is the amplifier
signal/distortion (IM3 ) ratio if we drive it with -25 dBm?
Note: IM3 = 0 dB at Pin =-10 dBm
If we back-off by 15 dB, the IM3 improves at a rate of 2:1
For Pin =-25 dBm (15 dB back-off), we have therefore IM3 =
30dBc
s 4 a1 3 a3
2
VIIP3 = 1 ⇒ VIIP3 =
I M3 =
4 a1
3 a3 |
{z
}
intercept signal level
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Gain Compression and Expansion
ω1 + ω1 + ... + ω1 − ω1 − ω1 − ... − ω1 = ω1
|
{z
} |
{z
}
k
k−1
To regenerate the fundamental for the Nth power, we need to
sum k positive frequencies with k − 1 negative frequencies, so
N = 2k − 1 implies that N must be an odd power
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P1dB Compression Point
An important specification for an amplifier is the 1dB
compression point, or the input power required to lower the
gain by 1dB
3 a3 2
3
2
s
G (si ) = a1 + a3 si = a1 1 +
4
4 a1 i
G0 = lim G (si ) = a1
si →0
si2 = 1 − 10−0.05
s
si =
−0.05
1 − 10
G0 · 10−0.05
=
G0
3 a3 2
1+
s
4 a1 i
= 10−0.05
4 a1 ⇐ Assume a3 /a1 < 0
3 a3 4 a1 q
=
1 − 10−0.05 VIIP3
3 a3
∼ 9.6dB lower than IIP3
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Dynamic Range
P−1dB is a convenient
“maximum” signal level
which sets the upper bound
on the amplifier “linear”
regime. Note that at this
power, the IM3 ∼ 20 dBc.
The lower bound is set by
the amplifier noise figure.
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Blocking (or Jamming)
Blocker: Any large interfering signal
PBL = Blocking level. Interfering signal level in dBm which
causes a +3dB drop in gain for small desired signal
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Jamming Analysis
Let: si =
s1 cos ω1 t
| {z }
small desired signal
+ S2 cos ω2 t
| {z }
Large blocker
Cubic non-linearity at ω1 : s1 k1 +k−1 , S2 k2 +k−2
s1 3 S2 0
s1 2 S2
s1 S2 2
s1 0 S2 3
Regular gain compression
Gain compression of desired signal on blocker
Gain compression of blocker on desired signal
Gain compression of blocker on blocker
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Jamming Analysis (cont)
Count the ways:
*
m; k B
*
k B = (1
0
1
1)
* s1 S2 2 3; k B
3
= s1 S2 2
2m−1
2
3 a3 2
Apparent gain = a1 1 +
S2
2 a1
Amplitude =
2
Gain reduction or expansion due to blocker ∝ Sblocker
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Blocking Power ∼ P−1dB
3 a3 2
S2
20 log 1 +
2 a1
= −3dB
s √
4 a1 0.293
S2 =
= √
3 a3
2
| {z }
VIIP3
PBL = PIIP3 − 8.3dB
= P−1dB + 1dB
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Effect of Feedback on Disto
sε = si − sfb
f <1
sfb = f so
Review from 142/242A:
so = a1 sε + a2 sε 2 + a3 sε 3 + · · ·
= a1 (si − f so ) + a2 (si − f so )2 + a3 (si − f so )3 + · · ·
= b1 si + b2 si 2 + b3 si 3 + · · ·
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New Non-Linear Coefficients
b1 =
b2 =
1
b1 ∼
= (T → ∞)
f
a1
1 + a1 f
a2
b3 =
(1 + T )3
a3 (1 + T ) − 2a2 2 f
(1 + T )5
Loop gain T = a1 f
For high loop gain, the distortion is very small. Even though
the gain drops, the distortion drops with loop gain since b2
drops with a higher power.
The cubic term has two components, the original cubic and a
second order interaction term. If an amplifier does not have
cubic, FB creates it (MOS with Rs )
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Series Inversion
sa = a1 sb + a2 sb 2 + a3 sb 3 + ...
sb = b1 sa + b2 sa 2 + b3 sa 3 + ...
sa = a1 (b1 sa + b2 sa 2 + ...) + a2 (b1 sa + b2 sa 2 + ...)2 + ...
b1 =
1
a1
b2 = −
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b3 =
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2a2 2
a3
− 4
5
a1
a1
Series Cascade
s1 = a1 si + a2 si 2 + ...
c1 = a1 b1
s0 = b1 s1 + b2 s1 2 + ...
⇒ s0 = c1 si + c2 si 2 + ...
c2 = b1 a2 + b2 a1 2
c3 = b1 a13 +
2b a a
| 2{z1 }2
+b3 a13
Second-order interaction
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IIP2 Cascade
c2
b1 a2 + b2 a12
a2 b2
=
=
+ a1
c1
b1 a1
a1 b1
1
1
a1
=
+
A
IIP2
IIP2
IIP2B
The cascade IIP2 is reduced due to the gain of the first stage:
To calculate the overall IIP2 , simply input refer the second
stage IIP2 by the voltage gain of the first stage.
The overall IIP2 is a “parallel combination” of the first and
second stage.
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IIP3 Cascade
c3
b1 a3 + b2 a1 a2 2 + b3 a13
=
=
c1
ba a1
a3 b3 2 b2
+ a + 2a2
a1 b1 1 b1
a12
1
1
+
=
2
IIP32
IIP3A IIP32B
Using the same approach, we can calculate the IIP3 of a
cascade. To simplify the result, neglect the effect of second
order interaction:
Input refer the IIP3 of the second stage by the power gain of
the first stage.
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References
UCB EECS 242 Class Notes, Robert G. Meyer, Spring 1995
Sinusoidal analysis and modeling of weakly nonlinear circuits :
with application to nonlinear interference effects, Donald D.
Weiner, John F. Spina. New York : Van Nostrand Reinhold,
c1980.
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