
Your RF Cafe

Authors: William Domino, Nooshin Vakilian, & Darioush Agahi
(all currently work for Skyworks Solutions, Inc.)
For a digital cellular system such as GSM (Global System for Mobile Communications), there are wellspecified blocking tests, that include both inband and outofband interferers. In these tests, the receiver front end must be able to reject the blocker while amplifying the desired signal, without violating the maximum allowed bit error rate. For outofband blockers, much of the rejection comes from a receiveband filter placed in front of the lownoise amplifier (LNA). Some architectures also place a similar filter after the LNA and preceding the mixer, while others utilize an imagereject mixer. In the latter case, the LNA/mixer combination is often implemented as a single IC, and it must exhibit particularly good blocker resistance, as there is only one receiveband filter placed ahead of it.
V_{o} = k_{1} V_{in} + k_{3} V_{in}^{3} (1)
But gain compression is generally a saturationtype function that is not well modeled by the 3^{rd}order component alone, which causes a downward turn of the gain curve. Figure 1 is an illustration of how the addition of higher order terms can produce a curve that saturates. Of course, to truly model a saturation condition, one would require an infinite number of terms, however a few terms
Figure 1 Polynomial Compression Models
are enough to produce an adequate model. In all cases the use of the model must be stopped above the point where it strays from true saturation.
With 1^{st}, 3^{rd}, 5^{th}, and 7^{th} terms:
V_{o} = k_{1} V_{in} + k_{3} V_{in}^{3} + k_{5} V_{in}^{5} + k_{7} V_{in}^{7 } (2)
To obtain the nonlinear gain, the above equation is divided by the input voltage, which yields
Gain = k_{1} + k_{3} V_{in}^{2} + k_{5} V_{in}^{4} + k_{7} V_{in}^{6 } (3)
Note that the gain equation has only evenorder terms. An equation of this type is particularly useful when V_{in} is a large blocker, and the gain is the smallsignal voltage gain in the presence of this blocker. Of course, with V_{in }always representing the blocker, the coefficients will be different for the blocker gain and the smallsignal gain, since the small signal always is compressed faster than the large signal. (Refer to appendix A.)
If the transfer function model is limited to 3^{rd}order and 5^{th}order nonlinearity, and the coefficients are defined as positive quantities, then one can write the following equation for the large signal gain (LSG):
LSG = k_{1}  3/4 k_{3} (V_{Large})^{2} + 5/8 k_{5} (V_{Large})^{4 } (4)
Where V_{Large} is peak amplitude of the large signal blocker. Similarly the small signal gain (SSG) is:
SSG = k_{1}  3/2 k_{3} (V_{Large})^{ 2} + 15/8 k_{5} (V_{Large}^{ })^{4 } (5)
Note that the 2^{nd} order term is doubled for the small signal gain, while the 4^{th} order term is tripled. The k_{3} term causes the small signal to be compressed faster than the large signal. The k_{5} term is the one that pulls the gain curve back up and keeps it from turning downward. Even though this term is tripled for the small signal gain, it is generally not enough to keep the k_{3} term from causing the faster compression.
Below in Figure 2 and Figure 3 are given some example results of compressions that occur according to these equations, where k_{1} is unity.
Figure 2 Small Signal Gain Compression with Coefficient K3 = .0025
Figure 3 Small Signal Gain Compression with Coefficient K3 = .0050
In the method to be described, the coefficients of the SSG compression are directly determined by measurement and curve fit, rather than assuming they keep the precise relationship with the LSG coefficients that is seen in eq. (4) and (5). The noise figure increase that occurs is also measured and related to the small signal gain.
Figure 4 is a diagram of our model showing how the blocker affects the gain and noise of the system.
Figure 4 Model for Blocker Effects
SSG = a_{0}  a_{2}(Blocker)^{2} + a_{4}(Blocker)^{4} (6)
The three noise contributors are the basic noise figure of the system, the noise figure degradation due to the blocker, and the LO phase noise mixed onto the blocker. It can be seen that the blocker amplitude affects the last two of these. In our modeling exercise, it is most convenient to relate the noise figure degradation to the smallsignal gain, which is compressed by the blocker. This relationship is best represented using 0^{th}, 1^{st}, and 2^{nd} order terms:
NoiseFactor = n_{0}  n_{1}(SSG) + n_{2}(SSG)^{2} (7)
Actually the blocker causes the small signal gain to be reduced while the system noise increases, causing a composite degradation in the noise figure of the system. This is taken into account when the above equations are fit to actual measurements.
Finally, after the blocker itself sees some compression, the LO noise is reciprocally mixed onto it (refer to appendix B), and the noise at the blockerdesired frequency offset falls into the desired band. This noise is then summed in with the other contributors.
Figure 5 Test Setup
the HP8970B noise figure meter. The blocker generator is filtered so as not to emit noise in the desired band, and the LO generator is filtered so as not to emit noise that would mix with the blocker. Also, the IF output is filtered so the blocker does not hit the NF meter with excessive power. The combiner has isolators on either side of it. On the noise source side, the isolator protects the noise source from the blocker. On the blocker side, the isolator insures that the port sees 50W both inside and outside the passband of the blocker filter.
The gain and NF with no blocker is measured, then a blocker is applied and the measurement is repeated for different blocker amplitudes. The measurements are stopped once the blocker reaches the highest value expected in the GSM system, or once the noise figure degrades by more than about 68dB.
It should also be noted that the frequencies of the measured noise band and of the blocker need to be chosen to avoid mixer spurs. If the blocker is located on or near a loworder mixer spur, then the NF meter will register an incorrectly high measurement.
Figure 6 Small Signal Gain vs. Blocker Voltage
Figure 7 is a plot of the noise factor vs. small signal gain for the Conexant RF212 LNA/Mixer. The equation has terms of order 0^{th}, 1^{st} and 2^{nd}. This equation defines (indirectly) the blocker dependence of the “noise figure degradation” box in Figure 4.
Figure 7 Noise Factor vs. Small Signal Gain
Once these polynomial equations are generated, they are used in a process where all of the noise contributions are summed, to find the carrier/noise ratio (C/N) at the LNA/mixer output. An example case is shown in Figure 8, where a desired signal of –102dBm and a blocker of –16dBm are incident on the LNA/mixer input.
To find the composite C/N, the basic quantities of gain, noise figure, and P_{1dB} are required, as are the coefficients of the smallsignal gain polynomial (a0, a2, a4), and the noise factor polynomial (n0, n1, n2). We find the small signal gain and the noise figure based on the level of the blocker, and then add the reciprocally mixed LO phase noise at the end.
We must apply the large signal compression to the blocker itself, since in our model this compression occurs before the LO noise reciprocal mixing is applied. A polynomial with higher orders can be applied for the LSG just as for the SSG, but in our experience it is adequate to apply 3^{rd}order distortion only for the LSG, as long as the large signal does not go far beyond the P1dB point of the system. Therefore the LSG coefficients can be derived either in the same manner as the SSG coefficients, or they can be taken from the measured P_{1dB} and the relationship in eq. (4) where k5=0. Appendix C details the P_{1dB} derivation.
Figure 8 Example of Model's Prediction of C/N (click to enlarge)
For our example, the –16dBm blocker at the input of the Conexant RF212 LNA/Mixer causes the noise figure to degrade from 2.5dB (the typical value with no blockers present) to 6.5dB. The small signal gain drops in the process from 21.6dB to 16.75dB. Then the cumulative effects of the compressed SSG and NF result in a 12.5dB C/N before the LO noise is added, reciprocally mixed onto the blocker which has been amplified by the LSG. The sum total C/N is 7.8dB.
What makes the approach powerful is that once the model is derived, in can be plugged into the chain calculations for a complete receiver. Then, various whatif analyses can be done with accurate results for any blocker level hitting the LNA/mixer. For example, the effect of different frontend SAW filters with different levels of stopband attenuation for outofband blockers can be checked, and an accurate tradeoff can be made between the SAW’s passband insertion loss and its stopband selectivity.
Furthermore, the approach can be used to accurately estimate the sensitivity level for a receiver in the presence of a blocker. In the typical GSM receiver, the sensitivity level is where the C/N drops to about 6.0dB. For our example with a –16dBm blocker, the sensitivity level estimate is –103.8dBm.
The calculation can also show which contributors are the most important. In this example, it turns out that even with the seemingly high noise figure of 6.5dB, the LO phase noise floor of –150dBc/Hz is still the most significant contributor to the system noise, due to the blocker’s presence. If the LO phase noise improves by 1dB, then the sensitivity improves by 0.7dB.
1) W. E. Sabin and E. O. Schoenike, “Single Sideband Systems and Circuits”, second edition, McGraw Hill, 1995
2) Behzad Razavi, “RF Microelectronics”, Prentice Hall, 1998
Nooshin Vakilian is Systems Engineer for GSM RF Systems Engineering at Conexant Systems. She may be reached via email at nooshin.vakilian@conexant.com.
Darioush Agahi is Director of GSM RF Systems Engineering at Conexant Systems. He may be reached via email at darioush.agahi@conexant.com.
V_{o}(t) = k_{1} v_{i}(t) + k_{2} v_{i}^{2}(t) + k_{3} v_{i}^{3}(t) + k_{4} v_{i}^{4}(t) + k_{5} v_{i}^{5}(t) + ……………………. (a1)
Note that if the gain were perfectly linear, then k_{1} would be the only nonzero coefficient, and the gain would be identical to k_{1}.
For a twotone input, V_{i}(t) is:
V_{i}(t) = A Cos(w_{1}t) + B Cos(w_{2}t) (a2)
Inserting (a2) into (a1) and using the wellknown trigonometric equalities, one can expand the expression for V_{o}(t) to the following:
{k_{1} B + 3/4 k_{3} B^{3} + 3/2 k_{3} A^{2}B + 5/8 k_{5} B^{5} + 15/4 k_{5} A^{2}B^{3} + 15/8 k_{5} A^{4}B} Cos(w_{2}t) +
{1/2 k_{2} A^{2} + 1/2 k_{4} A^{4} + 3/2 k_{4} A^{2} B^{2}} Cos(2w_{1}t) + { 1/2 k_{2} B^{2} + 1/2 k_{4} B^{4} + 3/2 k_{4} A^{2} B^{2}} Cos(2w_{2}t) +
{k_{2} AB + 3/2 k_{4} A^{3}B + 3/2 k_{4} AB^{3}} Cos((w_{1 }+_{ }w_{2}) t ) + {k_{2} AB + 3/2 k_{4} A^{3}B + 3/2 k_{4} AB^{3}} Cos((w_{1 } _{ }w_{2}) t ) +
{1/4 k_{3}A^{3} + 5/16 k_{5}A^{5} + 5/4 k_{5}A^{3}B^{2}} Cos(3w_{1}t) + {1/4 k_{3}B^{3} + 5/16 k_{5}B^{5} + 5/4 k_{5}A^{2}B^{3}} Cos(3w_{2}t) +
{3/4 k_{3} A^{2}B + 5/4 k_{5}A^{4}B + 15/8 k_{5} A^{2}B^{3}} Cos((2w_{1 }±_{ }w_{2}) t ) +
{3/4 k_{3} AB^{2} + 5/4 k_{5}AB^{4} + 15/8 k_{5} A^{3}B^{2}} Cos((w_{1 }±_{ }2w_{2}) t ) +
1/2 k_{4} A^{3}B Cos((3w_{1 } ±_{ } w_{2}) t ) + 1/2 k_{4} AB^{3} Cos((w_{1 }±_{ }3w_{2}) t ) + 3/4 k_{4} A^{2}B^{2} Cos((2w_{1 }+_{ }2w_{2}) t ) +
1/8 k_{4}A^{4} Cos(4w_{1}t) + 1/8 k_{4}B^{4} Cos(4w_{2}t) ……………………. (a3)
The two terms that are of interest are the firstorder gain terms. At frequency w_{1} the output voltage is:
V_{o}(w_{1}) = k_{1} A + 3/4 k_{3} A^{3} + 3/2 k_{3} AB^{2} + 5/8 k_{5} A^{5} + 15/4 k_{5} A^{3}B^{2} + 15/8 k_{5} AB^{4} (a4)
And similarly at w_{2} the output voltage is:
V_{o}(w_{2}) = k_{1} B + 3/4 k_{3} B^{3} + 3/2 k_{3} A^{2}B + 5/8 k_{5} B^{5} + 15/4 k_{5} A^{2}B^{3} + 15/8 k_{5} A^{4}B (a5)
Dividing both sides of equations (a4) and (a5) by their respective inputs would yield gain at the corresponding frequencies:
G(w_{1}) = k_{1} + 3/4 k_{3} A^{2} + 3/2 k_{3} B^{2} + 5/8 k_{5} A^{4} + 15/4 k_{5} A^{2}B^{2} + 15/8 k_{5} B^{4} (a6)
G(w_{2}) = k_{1} + 3/4 k_{3} B^{2} + 3/2 k_{3} A^{2} + 5/8 k_{5} B^{4} + 15/4 k_{5} A^{2}B^{2} + 15/8 k_{5} A^{4} (a7)
Let’s assume A represents the large signal blocker and B the small desired signal. This means A>>B, therefore we can approximate the above gain terms by letting B go to zero:
G(w_{1}) = k_{1} + 3/4 k_{3} A^{2} +
G(w_{1}) = k_{1} + 3/4 k_{3} A^{2} + 5/8 k_{5} A^{4} (a9)
Similarly we have;
G(w_{2}) = k_{1} +
G(w_{2}) = k_{1} + 3/2 k_{3} A^{2} + 15/8 k_{5} A^{4} (a11)
Equations (a9) and (a11) relate the large signal gain and small signal gain respectively of an amplifier. The interesting point is that both gains depend on the large signal amplitude, and further that under large signal interference, the small signal gain suffers faster. This is apparent by comparing the coefficients of A^{2} and A^{4} in the above equations.
Figure B1 RX Input Signals, and LO with Phase Noise Spectrum
Figure B2 Signals Downconverted to IF, with Overlap of ReciprocallyMixed Noise
V_{o}(w_{1}) = k_{1} A + 3/4 k_{3} A^{3} + 3/2 k_{3} AB^{2} + 5/8 k_{5} A^{5} + 15/4 k_{5} A^{3}B^{2} + 15/8 k_{5} AB^{4} (c1)
Setting the higher order terms equal to zero^{ }yields
V_{o} = { k_{1} A + 3/4 k_{3} A^{3} + 3/2 k_{3} AB^{2} } (c2)
Setting B = 0 and dividing both sides by A yields
V_{o} = { k_{1} A + 3/4 k_{3} A^{3} } (c3)
G = V_{o}/A= k_{1} + 3/4 k_{3} A^{2} (c4)
Equation (c4) is the gain with thirdorder nonlinearity.
At the P_{1dB} point, the overall gain is reduced by 1 dB from the linear gain, that is, the voltage gain becomes k_{1}*(10^{1/20}). To find the amplitude at the P_{1dB} point occurs, one needs to solve the following:
k_{1} + 3/4 k_{3} A^{2} = 10^{1/20} k_{1} Þ A^{2} = 4/3 { 10^{1/20 }  1 } (k_{1}/k_{3}) (c5)
Note that at this point, “A” represents amplitude at which P_{1dB} occurs.
For ease of manipulation, we set
a = 4/3 { 10^{1/20 }  1 } (c6)
Then,
A = √(a k_{1}/k_{3}) (c7)
Equation (c7) represents the amplitude at which gain is compressed by 1dB.
To find 1dB compression point in terms of power, we can start with equation (c3) again:
V_{o} = { k_{1} A + 3/4 k_{3} A^{3} }
Next raise both sides to power 2, which yields
V_{o}^{2 } = { k_{1} A + 3/4 k_{3} A^{3} }^{2} (c8)
After some routine arithmetic and replacing A with its equivalent a,
V_{o}^{2 } = { a + 3/2 a^{2} + 9/16 a^{3}} (k_{1}^{3}/k_{3}) (c9)
However a can be numerically evaluated as
a = 4/3 { 10^{1/20 }  1 } =  0.145
And V_{o}^{2} becomes
V_{o}^{2 } =  0.11518 (k_{1}^{3}/k_{3}) (c10)
Now V_{o} is a peak voltage (call it V_{p}). Converting to dBm yields
P_{dBm} = 10 Log { (V_{p}^{2}/2R) 1000 } = 10 Log {V_{p}^{2} 50/R 10} (c11)
where R is the system source resistance .
Inserting (c10) into (c11) gives P_{1dB} in terms of the amplifier’s coefficients:
P_{1dB} = 10 Log { 0.11518 (k_{1}^{3}/k_{3}) 50/R 10 } (c12)
Since an amplifier’s k_{3} is a negative term, its sign can be absorbed and the term can be presented in the absolute form, which then results in:
P_{1dB} = 10 Log { (k_{1}^{3}/k_{3}) 50/R } + 0.614 (c13)
In a 50 W system the R term vanishes and P_{1dB} reduces to:
P_{1dB} = 10 Log {k_{1}^{3}/k_{3}} + 0.614 (c14)