Errata, First Edition, Third Printing

This file contains errata, factual updates, grammatical corrections and expanded explanations for the book RF and Microwave Design: A Systems Approach by Michael Steer, First Edition, Third Printing. The third printing went to press on November 1, 2010. The printing number is the last number in the sequence above the ISBN number at the start of the book before the dedication page.

10 9 8 7 6 5 4 3 ISBN: 8791891121883

In the third edition several exercises were not numbered correctly. The correct numbers are recorded

below.

Errata History for First Edition, Third Printing

November 15, 2012 Pages 408

October 20, 2012 Pages 501

April 30, 2012 Pages 729

March 7, 2012 Pages 618, 624

March 3, 2012 Pages 677

March 1, 2012 Pages 231, 549, 571, 573

February 4, 2012 Page 558

November 19, 2011 Page 385

April 30, 2011 Pages 620, 622, 640,641, 642

March 29, 2011 Page 370

March 26, 2011 Pages 437, 445, 455

February 25, 2011 Pages 180, 288, 301

February 17, 2011 Page 190

February 2, 2011 Pages 60, 98, 239, 240, 409, 457, 594, 703, 705

November 18, 2010 Page 384

November 7, 2010 Pages 290, 502

60 MICROWAVE AND RF DESIGN: A SYSTEMS APPROACH

12. A 16QAM modulated signal has a maximumRF phasor amplitude of 4 V. If the noise on thesignal has an RMS value of 0.1 V, what is theEVM of the modulated signal?

13. A 16QAMmodulated signal has a amaximumRF phasor amplitude of 4 V. If the noise on thesignal has an RMS value of 0.1 V, what is themodulation error ratio of the modulated sig-nal in decibels?

14. A superheterodyne receiver has, in order, anantenna, a low-noise amplifier, a bandpass fil-ter, a mixer, a second bandpass filter, a secondmixer, a lowpass filter, an ADC, and a DSPwhich will implement quadrature demodula-tion. Develop the frequency plan of the re-ceiver if the input RF signal is at 2 GHz andhas a 200 kHz single-channel bandwidth. Thefinal signal applied to the ADC must be be-tween DC and 400 kHz so that I/Q demodu-lation can be done in the DSP unit. Noise con-siderations mandate that the LO of the firstmixer must be more than 10 MHz away fromthe input RF. Also, for a bandpass filter tohave minimum physical size , the center fre-quency of the filter should be as high as pos-sible. It has been determined that the appro-

priate trade-off of physical size and cost is tohave a 100 MHz bandpass filter between thetwo mixers. (Note: a 100 MHz bandpass filterhas a center frequency of 100 MHz.)

(a) Draw a block diagram of the receiver andannotate it with symbols for the frequen-cies of the LOs and the RF and IF signals.

(b) What is the LO frequency fLO1 of the firstmixer?

(c) What is the LO frequency fLO2 of the sec-ond mixer?

(d) Specify the cutoff frequency of the low-pass filter following the second mixer.

(e) Briefly discuss in less than one-half pageother design considerations as they relateto the frequency plan, filter size, and filterspecification.

15. Short answer questions. Each part requires ashort paragraph of about five lines and a fig-ure where appropriate to illustrate your un-derstanding.

(a) Explain the operation of a superhetero-dyne receiver.

(b) Compare zero-IF and low-IF receivers.

98 MICROWAVE AND RF DESIGN: A SYSTEMS APPROACH

6. On a resonant antenna, a large current is es-tablished by creating a standing wave. Thecurrent peaking that thus results establishes astrong electric field (and hence magnetic field)that radiates away from the antenna. A typicaldipole loses 15% of the power input to it as re-sistive (I2R) losses and has an antenna gainof 10 dB measured at 50 m. Consider a basestation dipole antenna that has 100 W inputto it. Also consider that the transmitted powerdensity falls off with distance d as 1/d3. Hint,calculate the power density at 50 m. [ParallelsExample 2.3 on Page 78.]

(a) What is the input power in dBm?

(b) What is the power transmitted in dBm?

(c) What is the power density at 1 km? Ex-press your answer as W/m2.

(d) What is the power captured by a receiveantenna (at 1 km) that has an effective an-tenna aperture of 6 cm2? Express your an-swer in first dBm and then watts.

(e) If the background noise level captured bythe antenna is 1 pW, what is the SNR indecibels? Ignore interference that comesfrom other transmitters.

7. Stacked dipole antennas are often found at thetop of cellphone masts, particularly for largecells and operating frequencies below 1 GHz.These antennas have an efficiency that is closeto 90%. Consider an antenna that has 40 W ofinput power, an antenna gain of 10 dB, andtransmits a signal at 900 MHz.

(a) What is the EIRP in watts?

(b) If the power density drops as 1/d3, whered is the distance from the transmit tower,what is the power density at 1 km if thepower density is 100 mW/m2 at 10 m?

8. Consider an 18 GHz point-to-point com-munication system. Parabolic antennas aremounted on masts and the LOS between theantennas is just above the tree line. As a re-sult, power falls off as 1/d3, where d is thedistance between the antennas. The gain ofthe transmit antenna is 20 dB and the gainof the receive antenna is 15 dB. The antennasare aligned so that they are in each other’s

main beam. The distance between the anten-nas is 1 km. The transmit antenna is drivenby a power amplifier with an output powerof 100 W. The amplifier drives a coaxial cablethat is connected between the amplifier andthe transmit antenna. The cable loses 75% ofits power due to resistive losses. On the re-ceive side, the receive antenna is directly con-nected to a masthead amplifier with a gain of10 dB and then a short cable with a loss of 3 dBbefore entering the receive base station.

(a) Draw the signal path.

(b) What is the loss and the gain of the trans-mitter coaxial cable in decibels?

(c) What fraction of the power input to thereceive coaxial cable is lost in the receivecable? Express your answer as a percent-age.

(d) Express the power of the transmit ampli-fier in dBW and dBm.

(e) What is the propagation loss in decibels?

(f) Determine the total power delivered tothe receive base station. Express you an-swer in watts.

9. Consider a point-to-point communication sys-tem. Parabolic antennas are mounted high ona mast so that ground effects are minimal.Thus power density falls off as 1/d2.3 where dis the distance from the transmitter. The gainof the transmit antenna is 15 dB and the gainof the receive antenna is 12 dB. These gains arenormalized to 1 m. These antenna gains arenormalized to a distance of 1 m. The distancebetween the antennas is 15 km. The outputpower of the receive antenna must be 1 pW.The RF frequency is 2 GHz; treat the receiveand transmit antennas as lossless.

(a) What is the received power in dBm?

(b) What is the path loss in decibels?

(c) What is the link loss in decibels?

(d) Using the link loss, calculate the inputpower, PT , of the transmitter. Express theanswer in dBm.

(e) What is the aperture area of the receiverin square meters?

180 MICROWAVE AND RF DESIGN: A SYSTEMS APPROACH

∆ z

z

∆z+ z,t( )Iz,t( )I ∆ zL∆ zR

∆ zGz,t( )V ∆z+ z,t( )V∆ zC __

+ +

(a) (b)

Figure 4-11 The uniform transmission line: (a) transmission line segment of

length ∆z; and (b) primary constants assigned to a lumped-element model of a

transmission line.

for narrow on-chip interconnections, as their resistance is very large.

4.6.1 Derivation of Transmission Line Properties

In this section the differential equations governing the propagation ofsignals on a transmission line are derived. Solution of the differentialequations describes how signals propagate, and leads to the extraction ofa few parameters that describe transmission line properties.From Kirchoff’s laws applied to the model of Figure 4-11(b) and taking

the limit as∆z → 0 the transmission line or telegrapher’s equations are

∂v(z, t)

∂z= −Ri(z, t)− L

∂i(z, t)

∂t(4.14)

∂i(z, t)

∂z= −Gv(z, t)− C

∂v(z, t)

∂t. (4.15)

For the sinusoidal steady-state condition with cosine-based phasors11 thesebecome

dV (z)

dz= −(R+ ωL)I(z) (4.16)

dI(z)

dz= −(G+ ωC)V (z) . (4.17)

Eliminating I(z) in Equations (4.16) and (4.17), yields the wave equation forV (z):

d2V (z)

dz2− γ2V (z) = 0 . (4.18)

Similarly,

d2I(z)

dz2− γ2I(z) = 0 , (4.19)

11 V (z) and I(z) are phasors and v(z, t) = ℜ

V (z)eωt

, i(z, t) = ℜ

I(z)eωt

. ℜwdenotes the real part of a complex numberw.

190 MICROWAVE AND RF DESIGN: A SYSTEMS APPROACH

(a) (b) (c)

Figure 4-13 Various coaxial transmission line connectors: (a) female N-type (N(f)),

coaxial connector; (b) male SMA-type (SMA(m)), coaxial connector; and (c) APC-7

coaxial connector.

(ii)

(i)

(iv)

(iii)

(a) (b) (c) (d)

Figure 4-14 Coaxial transmission line sections and tools: (a) SMA cables (from the

top): flexible cable type I, type II, type III, semirigid cable; (b) semirigid coaxial line

bender; (c) semirigid coaxial line bender with line; and (d) SMA elbow.

different types or series f connectors for high-power applications, differentfrequency ranges, low distortion, and low cost. There are also many typesof coaxial cables, as shown in Figure 4-14(a). These are cables for use withSMA connectors (with 3.5 mm outer conductor diameter). These cablesrange in cost, flexibility, and the number of times they can be reliablyflexed or bent. The semirigid cable shown at the bottom of Figure 4-14(a)must be bent using a bending tool, as shown in Figure 4-14(b) and in usein Figure 4-14(c). The controlled bending radius ensures minimal changein the characteristic impedance and propagation constant of the cable.Semirigid cables can only be bent once however. The highest precision bendis realized using an elbow bend, shown in Figure 4-14(d). Various flexiblecables have different responses to bending, with higher precision (and moreexpensive) cables having the least impact on characteristic impedance andphase variations as cables are flexed. The highest-precision flexible cablesare used in measurement systems.

204 MICROWAVE AND RF DESIGN: A SYSTEMS APPROACH

Notice that, in general, Γ may be complex, but VSWR is necessarily alwaysreal and 1 ≤ VSWR ≤ ∞. For the matched condition, Γ = 0 andVSWR = 1,and the closer VSWR is to 1, the better the load is matched to the line and themore power is delivered to the load. The reflection coefficients and standing-wave ratios for short-circuit and open-circuit terminated conditions are −1and +1, respectively, and in both cases the VSWR is infinite.To determine the position of Vmax, the voltage maxima, using Equation

(4.117) we have

Θ− 2βℓmax = 2nπ , n = 0, 1, 2, . . . .

This can be rewritten in the form

Θ− 2nπ = 22π

λglmax .

Thus the position of voltage maxima, lmax, normalized to the wavelength isgiven by

ℓmax

λg=

1

2

(

Θ

2π− n

)

, n = 0,−1,−2, . . . . (4.121)

Similarly the position of the voltage minima can be found using Equation(4.117):

Θ− 2βℓmin = (2n+ 1)π .

Rearranging the terms, we get lmin normalized to the wavelength:

ℓmin

λg=

1

2

(

Θ

2π− n+

1

2

)

, n = 0,−1,−2, . . . . (4.122)

Summarizing from Equations (4.121) and (4.122):

1. The distance between two successive maxima is λg/2.

2. The distance between two successive minima is λg/2.

3. The distance between a maximum and an adjacent minimum is λg/4.

In a similar manner to that employed above, the magnitude of the totalcurrent on the line is found to be

|I(ℓ)| =∣

∣V +0

∣

∣

Z0

∣

∣

∣1− |Γ|e(Θ−2βℓ)

∣

∣

∣. (4.123)

Hence the magnitude of the current has its maxima where the voltage isminimum and has its magnitude minima where the voltage magnitude ismaximum.

TRANSMISSION LINES 231

EXAMPLE 4. 21 Microstrip Design

Design a microstrip line to have a characteristic impedance of 75 Ω at 10 GHz. The microstripis to be constructed on a substrate that is 500 µm thick with a relative dielectric constant of5.6. What is the width of the line? Ignore the thickness of the strip. What is the effectivepermittivity of the line?

Solution:

(a) Two design formulas were introduced for microstrip: one for high impedance and onefor low. The high-impedance (or narrow-strip) formula (Equation (4.190)) is to be usedfor Zo > (44− εr) [= (44− 5.6) = 38.4] Ω.With εr = 5.6 and Z0 = 75 Ω, Equation (4.191) yields H ′ = 2.445. From Equation(4.190) w/h = 0.704, thus

w =w

h× h = 0.704× 500 µm = 352 µm.

(b) The effective permittivity formula is Equation (4.192) and so εe = 3.82 .

4.11.3 Comment on Formulas for Effective PermittivityTwo formulas have been presented that enable the effective permittivity ofa microstrip line to be calculated. Equation (4.178) provides the effectivepermittivity from the physical dimensions, the width and height of the line,and the relative permittivity of the medium. For a high-impedance line,Formula (4.192) (and Formula (4.195) for a low-impedance line) providesthe effective permittivity using an electrical characteristic, the characteristicimpedance, and the relative permittivity of the medium. Both formulasfor effective permittivity are curve fit equations, although they are basedon physical insight. So the two formulas are unlikely to provide answersthat are within only a rounding error. So which one do you really want?Most likely, once you have set the width you would really like to know theeffective permittivity as accurately as possible. Equation (4.178) is knownto be quite accurate, better than 0.2%, compared to detailed computersimulations. Can you really believe this? No, there are variations of thepermittivity from place to place as the density of the material changes.This is particularly true of composite materials such as an FR-4 circuitboard substrate, where the lower permittivity resin moves around duringmanufacture while the glass fiber stays fixed. With silicon ICs the densityof the silicon dioxide varies. Another factor is that the thickness of the stripaffects the field distribution and hence the effective permittivity.

TRANSMISSION LINES 239

7. A lossless transmission line has the followingper unit length parameters: L = 80 nH·m−1,C = 200 pF·m−1. Consider a traveling waveon the transmission line with a frequency of1 GHz.

(a) What is the attenuation constant?

(b) What is the phase constant?

(c) What is the phase velocity?

(d) What is the characteristic impedance ofthe line?

(e) Now consider that the dielectric is re-placed by a dielectric with εr = 1 (or air).The capacitance per unit length of the lineis nowC(air) = 50 pF·m−1. What is the ef-fective relative dielectric constant of theline?

8. The resonator below is constructed from a3.0 cm length of 100 Ω air-filled coaxial line,shorted at one end and terminated with a ca-pacitor at the other end.

10 kΩ

3 cm

= Ω0C 100Z

(a) What is the lowest resonant frequency ofthis circuit without the capacitor (ignorethe 10 kΩ resistor)?

(b) What is the capacitor value to achieve thelowest-order resonance at 6.0 GHz (ig-nore the 10 kΩ resistor)?

(c) Assume that loss is introduced by plac-ing a 10 kΩ resistor in parallel with thecapacitor. What is the Q of the circuit?

(d) Approximately what is the bandwidth ofthe circuit?

9. A transmission line has an attenuationof 0.2 dB/cm and a phase constant of50 radians·m−1 at 1 GHz.

(a) What is the complex propagation con-stant of the transmission line?

(b) If the capacitance of the line is 100pF·m−1 and the conductive loss is zero

(i.e., G = 0), what is the complex char-acteristic impedance of the line?

(c) If the line is driven by a source mod-eled as an ideal voltage and a seriesimpedance, what is the impedance of thesource for maximum transfer of power tothe transmission line?

(d) If 1W is delivered to the transmission lineby the generator, what is the power in theforward-travelingwave on the line at 2 mfrom the generator?

10. A transmission line is driven by a 1 GHzgenerator having a Thevenin equivalentimpedance of 50 Ω. The transmission line islossless, has a characteristic impedance of 75Ω, and is infinitely long. The maximum powerthat can be delivered to a load attached to thegenerator is 2 W.

(a) What is the total (phasor) voltage at theinput to the transmission line?

(b) What is the magnitude of the forward-traveling voltage wave at the generatorside of the line?

(c) What is the magnitude of the forward-traveling current wave at the generatorside of the line?

11. A transmission line has a characteristicimpedanceZ0 and is terminated in a loadwitha reflection coefficient of 0.86 45. A forward-traveling voltage wave on the line has a powerof 1 dBm.

(a) How much power is reflected by theload?

(b) What is the power delivered to the load?

12. A 50 Ω transmission line is terminated in aload that results in a reflection coefficient of0.5 + 0.5.

(a) What is the load impedance?

(b) What is the VSWR on the line?

(c) What is the input impedance if the line isone-half wavelength long?

240 MICROWAVE AND RF DESIGN: A SYSTEMS APPROACH

13. A transmission line has a characteristicimpedance Z0 and is terminated in a loadwith a reflection coefficient of 0.8. A forward-traveling voltagewave on the line has a powerof 1 W.

(a) How much power is reflected by theload?

(b) What is the power delivered to the load?

14. Communication filters are often constructedusing several shorted transmission line res-onators that are coupled by passive elementssuch as capacitors. Consider a coaxial line thatis short-circuited at one end. The dielectricconstant filling the coaxial line has a relativedielectric constant of 64 and the resonator is tobe designed to resonate at a center frequency,f0, of 800 MHz. [Parallels Example 4.15 onPage 209]

(a) What is the wavelength in the dielectric-filled coaxial line?

(b) What is the form of the equivalent circuit(in terms of inductors and capacitors) ofthe quarter-wavelength long resonator ifthe coaxial line is lossless?

(c) What is the length of the resonator?

(d) If the diameter of the inner conductor ofthe coaxial line is 2 mm and the inside di-ameter of the outer conductor is 5 mm,what is the characteristic impedance ofthe coaxial line?

(e) Calculate the input admittance of thedielectric-filled coaxial line at 0.99f0, f0,and 1.01f0. Determine the numericalderivative of the line admittance at f0.

(f) Derive the numeric values of the equiv-alent circuit of the resonator at the reso-nant frequency and derive the equivalentcircuit of the resonator. Hint: Match thederivative expression derived in (e) withthe actual derivative derived in Example4.15.

15. A load consists of a shunt connection of a ca-pacitor of 10 pF and a resistor of 25Ω. The loadterminates a lossless 50 Ω transmission line.The operating frequency is 1 GHz. [ParallelsExample 4.7 on Page 196]

(a) What is the impedance of the load?

(b) What is the normalized impedance ofthe load (normalized to the characteristicimpedance of the line)?

(c) What is the reflection coefficient of theload?

(d) What is the current reflection coefficientof the load? (When the term reflection co-efficient is used without a qualifier it isassumed to be the voltage reflection coef-ficient.)

(e) What is the standing wave ratio (SWR)?

(f) What is the current standing wave ratio(ISWR)? (When SWR is used on its own itis assumed to refer to the voltage stand-ing wave ratio [VSWR].)

16. The transmission line shown in the Fig-ure 4-16 consists of a source with Theveninimpedance Z1 = 40 Ω and source E =5 V (peak) connected to a quarter-wavelengthlong line of characteristic impedance Z01 =50 Ω, which in turn is connected to an in-finitely long line of characteristic impedanceZ02 = 100 Ω. The transmission lines are loss-less. Two reference planes are shown in Fig-ure 4-16. At reference plane 1 the incidentpower is PI1, the reflected power is PR1, andthe transmitted power is PT1. PI2, PR2, and(PT2) are similar quantities at reference plane2. [Parallels Examples 4.9 and 4.10 on Pages198 and 200]

(a) What is PI1?

(b) What is PT2?

17. A transmission line is driven by a 1 GHz gen-erator with a Thevenin equivalent impedanceof 50 Ω. The maximum power that can bedelivered to a load attached to the genera-tor is 2 W. The generator is connected to a10 cm long lossless transmission line with acharacteristic impedance of 75 Ω. Finally, theline is terminated in a load that has a com-plex reflection coefficient (referred to 50 Ω) of0.65 + 0.65. The effective relative permittiv-ity, ǫeff, of the transmission line medium is 2.0,and the effective relative permeability of theline is that of free space.

244 MICROWAVE AND RF DESIGN: A SYSTEMS APPROACH

• Input impedance that is a reactance of60 Ω.

(a) What is the width of the microstrip line?

(b) What is the length of the line in centime-ters?

(c) What is the effective permittivity of theline?

(d) If the line is a one-quarter wavelengthlonger than that calculated in (b), whatwill the input reactance be?

(e) Regardless of your calculations above,what is the input admittance of a one-quarter wavelength long shorted stub?

37. A load has an impedance Z = 75 + 15 Ω.

(a) What is the load reflection coefficient,ΓL, if the system reference impedance is75 Ω?

(b) Design a stub at the load that will makethe impedance of the load plus the stub,call this Z1, purely real; that is, the re-flection coefficient of the effective load,Γ1, has zero phase. Choose a stub charac-teristic impedance of 75 Ω. (Design spec-ifications require complete electrical in-formation such as whether the stub isopen- or short-circuited, and the electri-cal length of the stub.)

(c) Design a quarter-wave transformer thatwill present a matched termination

to a source with a system referenceimpedance of 50 Ω. (The design must in-clude full electrical specifications such asthe characteristic impedance of the trans-mission line and its electrical length. Thestructure is the source, a λ/4 transformer,a stub, and the load.)

(d) Now convert the electrical specificationsof the design into a physical specifica-tion. Assuming that the transmission linetechnology to be used is a microstrip lineand the substrate medium is fixed withthe following parameters: frequency f =1 GHz, substrate thickness h = 0.5 mm,substrate relative dielectric constant ǫr =10. You must design the widths andlengths of the stub and the quarter-wavetransformers.

38. Design a microstrip line to have a characteris-tic impedance of 20 Ω. The microstrip is to beconstructed on a substrate that is 1 mm thickwith a relative dielectric constant of 12. [Paral-lels Example 4.21 on Page 231]

(a) What is the width of the line? Ignore thethickness of the strip and frequency ef-fects.

(b) What is the effective permittivity of theline?

288 MICROWAVE AND RF DESIGN: A SYSTEMS APPROACH

Port 1 Port 2

21 Two−PortV1 2V

1I I 2

I2I1

Z 0Z 0

1V

V1

21

1 2

V2

2V

(a) (b)

Figure 6-1 A two-port network: (a) port voltages; and (b) with transmission lines at the ports.

and transmission. As will be seen, it is easy to convert these tomore familiar network parameters such as admittance and impedanceparameters. In this chapter S parameters will be defined and related toimpedance and admittance parameters, then it will be demonstrated thatthe use of S parameters helps in the design and interpretation of RFcircuits. S parameters have become the most important parameters forRF and microwave engineers and many design methodologies have beendeveloped around them.A graphical technique called signal flow graph analysis is introduced for

manipulating S parameter-based models of elements. Finally, polar plots ofS parameters are introduced as well as an annotated form called a Smithchart. Both are used for displaying and interpreting power flow in circuits.

6.2 Two-Port Networks

Many of the techniques employed in analyzing circuits require that thevoltage at each terminal of a circuit be referenced to a common pointsuch as ground. In microwave circuits it is generally difficult to do this.Recall that with transmission lines it is not possible to establish a commonground point. However, with transmission lines (and circuit elements thatutilize distributed effects) it was seen that for each signal current there isa signal return current. Thus at radio frequencies, and for circuits that aredistributed, ports are used, as shown in Figure 6-1,which define the voltagesand currents for what is known as a two-port network, or just two-port.1 Thenetwork in Figure 6-1(a) has four terminals and two ports. A port voltage isdefined as the voltage difference between a pair of terminals with one ofthe terminals in the pair becoming the reference terminal. Port 1 is on theleft of the diagram, where port voltage V1 is defined. The current enteringthe network at the top terminal of Port 1 is I1 and there is an equal currentleaving the reference terminal. This arrangement clearly makes sense when

1 Even when the term “two-port” is used on its own the hyphen is used, as it is referring to atwo-port network.

290 MICROWAVE AND RF DESIGN: A SYSTEMS APPROACH

z21

z11

z_21

z22

z_21

21

y21

1 211 y21y22 y21

_

y + +

(a) (b)

Figure 6-2 Circuit equivalence of the z and y parameters for a reciprocal network:

(a) z parameters; and (b) y parameters (in (b) the elements are admittances).

Impedance parameters, or z parameters, are defined as

V1 = z11I1 + z12I2 (6.1)

V2 = z21I1 + z22I2, (6.2)

or in matrix form asV = ZI . (6.3)

The double subscript on a parameter is ordered so that the first refers tothe output and the second refers to the input, so zij relates the voltageoutput at Port i to the current input at Port j. If the network is reciprocalthen z12 = z21, but this simple type of relationship does not apply to allnetwork parameters. The reciprocal circuit equivalence of the z parametersis shown in Figure 6-2(a). It will be seen that the z parameters are convenientparameters to use when an element is in series with one of the ports, as thenthe operation required in developing the z parameters of the larger networkis just addition.Figure 6-3(a) shows the series connection of a two-terminal element with

a two-port designated as network A. The z parameters of network A are

ZA =

[

z(A)11 z

(A)12

z(A)21 z

(A)22

]

, (6.4)

so that

V(A)1 = z

(A)11 I1 + z

(A)12 I2 (6.5)

V2 = z(A)21 I1 + z

(A)22 I2 . (6.6)

Now

V1 = zI1 + V(A)1 (6.7)

= zI1 + z(A)11 I1 + z

(A)12 I2 , (6.8)

so the z parameters of the whole network can be written as

Z =

[

z 00 0

]

+ ZA . (6.9)

MICROWAVE NETWORK ANALYSIS 301

0

,

Incident wave

Reflected wave

z = z =

z

ZL0Z

Figure 6-12 Transmission line of characteristic impedance Z0 and length ℓ

terminated in a load of impedance ZL.

g

g

I

V

Z

V

ZL

Figure 6-13 AThevenin equivalent sourcewith generator Vg and source impedance

Zg terminated in a load ZL.

Consider Figure 6-13, which does not have an explicit transmission line toseparate the forward- and backward-travelingwaves. Now the total voltageacross the load is

V = VgZL

Zg + ZL(6.74)

and the total current is

I =Vg

Zg + ZL. (6.75)

To develop the reflection coefficient, first define equivalent forward- andbackward-traveling waves. This can be done by imagining that betweenthe generator and the load there is a transmission line of characteristicimpedance Zg and having infinitesimal length. The incident voltage andcurrent waves (V +, I+) are the voltage and current obtained when thegenerator is conjugately matched to the load (i.e.,ZL = Z∗

g ). So the equationsof forward-traveling voltage and current become

V + = Vg

Z∗g

Zg + Z∗g

= Vg

Z∗g

2RZg(6.76)

I+ = Vg1

2RZg. (6.77)

370 MICROWAVE AND RF DESIGN: A SYSTEMS APPROACH

15. A 50 Ω lossy transmission line is shorted atone end. The line loss is 2 dB per wavelength.Note that since the line is lossy the charac-teristic impedance will be complex, but closeto 50 Ω, since it is only slightly lossy. Thereis no way to calculate the actual characteris-tic impedance with the information provided.However, this is the common situation thatmust be dealt with. That is, problems must besolved with small inconsistencies. Full accu-racy is not possible and there is often missinginformation. Microwave engineers do the bestthey can in design and always rely on mea-surements to calibrate results.

(a) What is the reflection coefficient at theload (in this case the short)?

(b) Consider the input reflection coefficient,Γin, at a distance ℓ from the load. Deter-mine Γin for ℓ going from 0.1λ to λ insteps of 0.1λ.

(c) On a Smith chart plot the locus of the in-put reflection coefficient from ℓ = 0 to λ.

(d) Calculate the input impedance, Zin,when the line is (3/8)λ long using thetelegrapher’s equation.

(e) Repeat part (d) using a Smith chart.

16. Design an open-circuited stub with an inputimpedance of +75 Ω. Use a transmission linewith a characteristic impedance of 75 Ω. [Par-allels Example 6.5 on Page 339]

17. Design a short-circuited stub with an inputimpedance of −50 Ω. Use a transmission linewith a characteristic impedance of 100 Ω. [Par-allels Example 6.5 on Page 339]

18. A load has an impedance ZL = 25− 100 Ω.

(a) What is the reflection coefficient, ΓL, ofthe load in a 50 Ω reference system?

(b) If a quarter-wavelength long 50 Ω trans-mission line is connected to the load,what is the reflection coefficient, Γin,looking into the transmission line?

(c) Describe the locus of Γin, as the length ofthe transmission line is varied from zerolength to one-half wavelength long. Usea Smith chart to illustrate your answer

19. A network consists of a source with aThevenin equivalent impedance of 50 Ω driv-ing first a series reactance of 50 Ω followed

by a one-eighth wavelength long transmissionline with a characteristic impedance of 40 Ωand a reactive element of 25 Ω in shunt with aload having an impedance ZL = 25− 100 Ω.This problem must be solved graphically andno credit will be given if this is not done.

(a) Draw the network.

(b) On a Smith chart, plot the locus of the re-flection coefficient first for the load, thenwith the element in shunt, then lookinginto the transmission line, and finally theseries element. Use letters to identify eachpoint on the Smith chart. Write down thereflection coefficient at each point.

(c) What is the impedance presented by thenetwork to the source?

20. In the circuit below, a 75 Ω lossless line is ter-minated in a 40 Ω load. On the complex planeplot the locus, with respect to the length of theline, of the reflection coefficient, looking to theline referencing it first to a 50 Ω impedance.[Parallels Example 6.6 on Page 342]

ininΓ 75 Ω40 ΩZ

LZ 0Z

21. Consider the circuit below, a 60 Ω lossless lineis terminated in a 40 Ω load.What is the centerof the reflection coeffient locus on the complexplane when it is referenced to 55 Ω. [ParallelsExample 6.7 on Page 346]

ininΓ 40 Ω60 Ω Z

LZ 0Z

22. In Section 6.11.1 on Page 350, the S parametersof a reciprocal error network were determinedby applying three loads—Z1, Z2 and Z3—andmeasuring the respective input reflection co-efficients. If Z1 is a matched load, Z2 is a shortcircuit and Z3 is an open, the S parametersof Equations (6.235), (6.236), and (6.237) arefound. Use SFG theory to derive these results.

384 MICROWAVE AND RF DESIGN: A SYSTEMS APPROACH

4

2

2

2=Z 0

+

−V

I4

Z 0

Z 2

I/ 2

+

V

+

V

−V

+

I/ 1

− −

Figure 7-10 Operation of the magnetic transformer of Figure 7-9 as a balun.

Considering Figure 7-10 and equating powers,

1

2

(2V )2

Z2

=1

2

V 2

Z0

(7.28)

2

Z2

=1

2Z0 (7.29)

Z2 = 4Z0. (7.30)

EXAMPLE 7. 2 Magnetic Transformer

Consider the magnetic transformer hybrid of Figure 7-9(a). Determine what type of hybridthis is and calculate the impedance transformations. Assume ideal coupling (k = 1).

Solution:

Since there is ideal coupling here and each coil has the same number of windings,

(V3 −V2) = V1 (7.31)

(V2 −V4) = V1. (7.32)

The current levels in the transformer depend on the attached circuitry. For this circuit tofunction as a hybrid with Port 2 isolated, the current I2 must be zero so that I3 and I4 areequal in magnitude but 180 out of phase. The loading at Ports 3 and 4 must be the same.Now V2 = 0, since I2 = 0, and so Equations (7.31) and (7.32) become

V3 = V1 (7.33)

V4 = −V1, (7.34)

so this circuit is a 180 hybrid.

To determine loading conditions at Ports 2 through 4 the following transformer equationis used

V3 = ωL2I3 + ωMI1 − ωMI4 (7.35)

where L2 is the inductance of Coil 2 and M is the mutual inductance. Now I1 = −V1/Z0,and L2 = L3 = M since coupling is ideal, and V3 = V1. So Equation (7.35) becomes

PASSIVE COMPONENTS 385

V3 = ωM(2I3 − V3/Z0) (7.36)

orV3(1 + ωM/Z0) = 2ωMI3. (7.37)

If ωM≫Z0, this reduces toZx = V3/I3 = 2Z0. (7.38)

From symmetry, Z4 = Z3. Also, I3 = −I4 = I1/2. For maximum power transfer and Z0

real, Z3 = Z∗x = 2Z0.

This above result can also be argued from maximum power transfer considerations. Theargument is as follows. An impedance Z0 is attached to coil 1 and this is an indicationthat the Thevenin equivalent circuit attached to coil 1 has a Thevenin equivalent impedanceof Z0. Maximum power transfer to the transformer through the coil requires that theinput impedance be Z0. In the ideal hybrid operation the power is split evenly betweenthe power delivered to the loads at Ports 3 and 4, since V1 = (V3 − V2) = (V2 − V4)and the power delivered to coil 1 is V 2

1 /(2Z0). The power delivered to Z3 (and Z4) is(V3 − V2)

2(2Z3) = V 21 /2Z3 = V 2

1 /(4Z0). That is, Z3 = 2Z0 = Z4.

The problem is not yet finished, as Z2 must be determined. For hybrid operation, a signalapplied to Port 2 should not have a response at Port 1. So the current at Port 2, I2, should besplit between coils 2 and 3 so that I3 = −I2/2 = I4. Thus

V4 = −I4(2Z0) = (I2/2)(2Z0) = I2Z0 = V3. (7.39)

NowV3 − V2 = V1 = 0 = V2 − V4, (7.40)

and soV2 = V4 = I2Z0 (7.41)

orZ2 = V2/I2 = Z0. (7.42)

The final 180 hybrid circuit is shown in Figure 7-9(b) with the loading conditions formatched operation as a hybrid.

In the example above it was seen that the number of windings in thecoils are the same so that the current in Coils 2 and 3 are half that in Coil1. The general rule is that with an ideal transformer the sum of the amp-turns around the magnetic circuit must be zero. The precise way the sum iscalculated depends on the direction of the windings indicated by the “dot”convention. A generalization of the rule for the transformer shown in Figure7-9 is

n1I1 − n2I2 − n3I3 = 0, (7.43)

where nj is the number of windings of coil j with current Ij . The exampleserves to illustrate the type of thinking behind the development of many RFcircuits. The emphasis on maximum power transfer provided an alternative,simpler start to the solution of the problem. Yes, this can be difficult—people write papers on ways of analyzing a new circuit element. This is

408 MICROWAVE AND RF DESIGN: A SYSTEMS APPROACH

7.16 SummaryMany passive microwave elements exploit particular physical phenomena.Many make use of the characteristics of transmission lines. Each year newvariants of microwave elements are developed and documented in patentsand publications. Microwave engineers often monitor developments inelements that are being exploited in design.

7.17 Exercises1. The Thevenin equivalent output impedance of

the amplifiers in Figure 7-24 is 5 Ω and the sys-tem impedance, R0, is 75 Ω. Choose the trans-former windings for maximum power trans-fer. [Parallels Example 7.3 on Page 400.]

2. A spiral inductor is modeled as an ideal in-ductor of 10 nH in series with a 5 Ω resistor.What is the Q of the spiral inductor at 1 GHz?

3. A 50 dB attenuator can be constructed usingthree resistors. Consider the design of a 50 dBattenuator in a 75 Ω system. [Parallels Exam-ple 7.1 on Page 380.]

(a) Draw the topology of the attenuator.(b) Write down the design equations for the

resistors.(c) Complete the design of the attenuator.

4. The circuit below is called a T attenuatorand is inserted in a system with real systemimpedance Z0; that is, a resistance of value Z0

is attached to the input and the output.

3Z 0 Z 0R

R1 R 2

Design the attenuator so that its attenuation isA dB. Show that

R1 = R2 = Z0(√N − 1√N + 1)

R3 =2Z0

√N

N − 1,

where N = log−1(A/10).

5. A 20 dB attenuator in a 17 Ω system is ideallymatched at both the input and output. Thus

there are no reflections and the power deliv-ered to the load is reduced by 20 dB from theapplied power. If a 5 W signal is applied tothe attenuator, how much power is dissipatedin the attenuator?

6. Write down the 50 Ω scattering parameters ofthe ideal transformer shown below where thenumber of windings on the secondary side(Port 2) is twice the number of windings onthe primary side (Port 1).

1 2

1:2

(a) What is S11? [Hint: Terminate Port 2 in50 Ω and determine the input reflectioncoefficient.]

(b) What is S21?(c) What is S22?(d) What is S12?

7. Derive the two-port 50 Ω scattering parame-ters of the magnetic transformer below. Theprimary (Port 1) has 10 turns, the secondary(Port 2) has 25 turns.

21

(a) What is S11?(b) What is S21?(c) What is S22?(d) What is S12?

8. Consider the hybrid shown in the figure be-low. If the number of windings of Coils 2 and3 are twice the number of windings of Coil1, show that for matched hybrid operation2Z2 = Z3 = 8Z0.

PASSIVE COMPONENTS 409

2

coil 3

coil 2

coil 1

2

2I

3V

+Z

0

+

+1

3

2

44I

Z3

Z2

Z4

4V

2V1V

3I

1I

9. A signal is applied to Ports 2 and 3 of a 180

hybrid, as shown in Figure 7-11(b). If the sig-nal consists of a differential component of 0dBm and a common mode component of 10dBm:

(a) Determine the power delivered at Port 1.

(b) Determine the power delivered at Port 4.Assume that the hybrid is lossless.

10. The balun of Figure 7-10 transforms an unbal-anced system with a system impedance of Z0

to a balanced system with an impedance of4Z0. The actual impedance transformation isdetermined by the number of windings of thecoils. Design a balun of the type shown in Fig-ure 7-10 that transforms an unbalanced 50 Ωsystem to a balanced 377Ω system. [Hint: Findthe ratio of the windings of the coils.]

11. A balun can be realized using a wire-wound transformer, and by changing thenumber of windings on the transformer itis possible to achieve impedance transfor-mation as well as balanced-to-unbalancedfunctionality. A 500 MHz balun based on amagnetic transformer is required to achieveimpedance transformation from an unbal-anced impedance of 50 Ω to a balancedimpedance of 200 Ω. If there are 20 windingson the balanced port of the balun transformer,how many windings are there on the unbal-anced port of the balun?

12. Design a lumped-element 2-way power split-ter in a 75Ω system at 1GHz. Base your designon a Wilkinson power-divider.

13. Design a 3-way power splitter in a 75 Ω sys-tem. Base your design on a Wilkinson power-divider using transmission lines and indicatelengths in terms of wavelengths.

14. Design a lumped-element 3-way power split-ter in a 75Ω system at 1GHz. Base your designon a Wilkinson power-divider.

15. Silicon RFICs use differential signal paths tominimize the introduction of substrate noise.As well, differential amplifiers are an opti-mum topology in current-biased circuits. Off-chip signals are often on microstrip linesand so the source and load, being off-chip,are not differential. The off-chip circuits arethen called single-ended. Using diagrams andexplanations, outline a system architectureaccommodating this mixed differential andsingle-ended environment.

16. A resistive power splitter is a three-port de-vice that takes power input at Port 1 and de-livers power at Ports 2 and 3 that are equal;that is, S21 = S31. However, the sum of thepower at Ports 2 and 3 will not be equal tothe input power due to loss. Design a 75 Ω re-sistive three-port power splitter with matchedinputs, S11 = 0 = S22 = S33. That is, drawthe resistive circuit and calculate its elementvalues.

17. Develop the electrical design of a rat-race hy-brid at 30 GHz in a 50 Ω system.

18. Develop the electrical design of a rat-race hy-brid at 30 GHz in a 100 Ω system.

19. Design a lumped-element hybrid at 1900MHzusing 1 nH inductors.

20. Using a schematic, show how an isolator canbe obtained from a circulator.

IMPEDANCE MATCHING 437

j5016.6j

14.933 49.5−j −j

16.6j j50

−j64.433

(a) (b)j j

−j −j16.6 50

49.514.933 j

16.6 50−j−j

64.433

(c) (d)14.933 49.5j

50−j

−j

j16.6

34.567j

j16.6 50−j

(e) (f)

50j

14.933j −j49.5

−j16.6 50j

34.567−j

−j16.6

(g) (h)

Figure 8-30 Four possible Pi matching networks: (a), (c), (e), (g) conceptual circuits;

and (b), (d), (f), (h), respectively, their final reduced Pi networks.

RL

X3

X2RS

X1

Figure 8-31 A three-element matching network.

8.6.2 Matching Network Q Revisited

To illustrate the fact that the circuitQ established by an L matching networkis the minimum circuit Q available for a network having at most threeelements, consider the design equations for RS > RL. Referring to Figure8-31,

X1 =RS

Q(8.79)

X3 = RL

(

RS/RL

Q2 + 1−RS/RL

)1/2

(8.80)

X2 =QRS +RSRL/X3

Q2 + 1. (8.81)

IMPEDANCE MATCHING 445

0.25

j 2 j

4_

j

_2 j

_j

0.5_

j

0.25_

j

4 j

2 j

4_

j

_2 j

0.5 j

4 j

0.5_

j

0.25_

j

0.25 j

_j

2

0.25j

0

j

j

G

B

10

0.5

0.5C

rLA

rS

4 ∞ ∞

Figure 8-39 Design 2: alternative design for Example 8.9.

jjX

SR

S RLX

P

XS = 66.1 Ω . XP = −75.6 Ω

Figure 8-40 Final design for Example 8.9.

EXAMPLE 8. 10 Matching Network Design With Complex Impedances

Develop a two-element matching network to match a source with an impedance of ZS =12.5 + 12.5 Ω to a load ZL = 50− 50 Ω, as shown in Figure 8-41.

Solution:

The design objective is to present conjugate matched impedances to the source and load;that is, Z1 = Z∗

S and Z2 = Z∗L. The choice here is to design for Z1; that is, elements will

be inserted in front of ZL to produce the impedance Z1. The normalized source and loadimpedances are plotted in Figure 8-42 using a normalization impedance of Z0 = 50 Ω, sozS = ZS/Z0 = 0.25 + 0.25 (Point S) and zL = ZL/Z0 = 1− (Point C).

FILTERS 549

−1

−1

L 1 L 3 C 3 CnL nC 1

Z0

Z0Vg

1n

2n

3n

2n

L 2 L n

C 2 Cn

Figure 10-33 Lumped-element odd-order (nth-order) Chebyshev bandstop filterprototypes in the type II Cauer topology.

and the transformation constant is

α =ωo

ω2 − ω1, (10.136)

where ω1 and ω2 are the band-edge radian frequencies. The resultingelement conversions are given in Figure 10-30.

Combining transformations, the element values of a lumped bandstopfilter with center radian frequency ω0 = 2πf0 and radian bandwidth ωBW =2π(f2 − f1) are as follows:

Cr =

grωBW

ω20Z0

r = odd

1

ωBWgrZ0r = even

(10.137)

Lr =

Z0

ωBWgrr = odd

grωBWZ0

ω20

r = even

. (10.138)

Similarly a lumped-element type II Cauer bandstop filter is shown inFigure 10-33. The parallel LC combination and the series LC combinationsare resonators resonant at the center frequency of the filter. The parallel LCresonator is an open circuit at the center frequency of the stopband and theseries LC resonators are short circuits.

10.9.6 Transformed Ladder PrototypesCombining the filter type transformations, and with appropriate use ofinverters, the original lowpass prototype ladder filter and its various filtertype transforms are shown in Figure 10-34.

IMPEDANCE MATCHING 455

5. A load is modeled as a 50 Ω resistance in se-ries with a reactance of 50 Ω. This load is to bematched to a source with a Thevenin equiv-alent resistance of 50 Ω. Use the Fano-Bodecriteria described in Section 8.5.2 on Page 431to determine the upper limit on the match-ing network bandwidth when the average in-band reflection coefficient is −10 dB. Notethat you must first convert the reflection coef-ficient to an absolute number.

6. A load is modeled as a 50 Ω resistance in se-ries with a reactance of 50 Ω. This load is to bematched to a source with a Thevenin equiv-alent resistance of 50 Ω. Use the Fano-Bodecriteria described in Section 8.5.2 on Page 431to determine the upper limit on the match-ing network bandwidth when the average in-band reflection coefficient is−20 dB. Note thatyoumust first convert the reflection coefficientto an absolute number.

7. The output of a transistor amplifier operatingat 1 GHz is modeled as a 100 Ω resistor inparallel with a 10 pF capacitor. The amplifiermust drive the input of a λ/2 dipole antennawith an input resistance of 73 Ω. To do this ef-ficiently a matching network is required. Con-sider that the input resistance of the antennais independent of frequency, and assume thatthe matching network is lossless. This is thesame as assuming that its bandwidth is muchgreater than the bandwidth required. If the re-quired bandwidth of the matching network is5%, and using the Fano-Bode criteria (see Sec-tion 8.5.2 on Page 431), determine the follow-ing:

(a) The lower limit on the average in-bandreflection coefficient of the matching net-work.

(b) The upper limit on the average transmis-sion coefficient of the matching network.

8. The output of a transistor amplifier is mod-eled as a current source in parallel with botha 50 Ω resistor and a 1 pF capacitor. This is tobe matched to a load consisting of a 25Ω resis-tor in series with a 0.02 nH inductor. The taskis to design a matching network that will en-able DC bias to be applied from the load to the

transistor output, thus the matching networkmust be a lowpass type. The center frequencyof the system is 10 GHz and a bandwidth of50 MHz is required.

(a) What is the fractional bandwidth of thesystem?

(b) What is the Q of the system?

(c) Indicate the form of the matching net-work if no more than four reactive ele-ments are to be used; that is, sketch thematching network.

(d) Complete the design of the amplifier pro-viding numerical element values.

9. Design a passive matching network that willachieve maximum bandwidth matching froma source with an impedance of 2 Ω (typical ofthe output impedance of a power amplifier) toa load with an impedance of 50 Ω. The match-ing network can have a maximum of three re-active elements. You need only calculate reac-tances and not the capacitor and inductor val-ues.

10. Design a passive matching network that willachieve maximum bandwidth matching froma source with an impedance of 20 Ω to a loadwith an impedance of 125 Ω. The matchingnetwork can have a maximum of four reactiveelements. You need only calculate reactancesand not the capacitor and inductor values.

(a) Will you use two, three, or four elementsin your matching network?

(b) With a diagram and perhaps equations,indicate the design procedure.

(c) Design the matching network. It is suffi-cient to use reactance values. The designmust be complete.

11. Design a passive matching network that willachieve maximum bandwidth matching froma source with an impedance of 60 Ω (typical ofthe output impedance of a power amplifier) toa load with an impedance of 5 Ω. The match-ing network can have a maximum of four re-active elements. You need only calculate reac-tances and not the capacitor and inductor val-ues.

IMPEDANCE MATCHING 457

19. A two-port matching network is shown belowwith a generator and a load. The generatorimpedance is 60 Ω and the load impedance isZL = 30 + 30 Ω. Use a Smith chart to designa lossless matching network (no credit will begiven unless a Smith chart is used to solve thisproblem). It is important that your solutioncan be followed, so youmust indicate your so-lution clearly on the chart.

M ZL

ZG

–+

ZIN

1 2E

GIN

(a) What is the condition for maximumpower transfer from the generator? Ex-press your answer using impedances.

(b) What is the condition for maximumpower transfer from the generator? Ex-press your answer using reflection coef-ficients.

(c) What system reference impedance areyou going to use to solve the problem?

(d) Plot ZL on the Smith chart and label thepoint. (Remember to use impedance nor-malization if required.)

(e) Plot ZG on the Smith chart and label thepoint.

(f) Design a lossless matching networkshowing your design process on theSmith chart. Label critical points on theSmith chart. Draw the matching networkand show the reactance values.

20. A two-port matching network is shown belowwith a generator and a load. The generatorimpedance is 30 Ω and the load impedance isZL = 90 − 30 Ω. Use a Smith chart to designa lossless matching network.

M ZL

ZG

–+

ZIN

1 2E

GIN

(a) What is the condition for maximumpower transfer from the generator? Ex-press your answer using impedances.

(b) What is the condition for maximumpower transfer from the generator? Ex-press your answer using reflection coef-ficients.

(c) What system reference impedance areyou going to use to solve the problem?

(d) Plot ZL on the Smith chart and label thepoint. (Remember to use impedance nor-malization if required.)

(e) Plot ZG on the Smith chart and label thepoint.

(f) Design a lossless matching networkshowing your design process on theSmith chart. Label critical points on theSmith chart. Draw the matching networkand show the reactance values.

21. Use Smith chart techniques to design adouble-stub matching network to match aloadwith a normalized admittance yL = 0.7−5 to a sourcewith a normalized admittance of1. The stubs are short-circuited and are sepa-rated by a length of transmission line of lengthλ/8. The load is at the position of the first stub.All transmission lines have the system charac-teristic impedance. Your design should yieldthe lengths of the two stubs.

(a) Plot the load on a Smith chart. Clearly in-dicate the load.

(b) Determine the admittances of each of thestubs. Clearly show and describe yourdesign technique so that it can be un-derstood. Label your efforts on the Smithchart and refer to them here. Note thata description is required and not simplymarkings on the Smith chart.

(c) Determine the electrical lengths of thestubs (express your answer in terms ofwavelengths or degrees).

22. Use a lossless transmission line and a seriesreactive element to match a source with aThevenin equivalent impedance of 25 + 50 Ωto a load of 100 Ω. (That is, use one transmis-sion line and one series reactance only.)

COUPLED LINES AND APPLICATIONS 501

9.13 SummaryCoupling from one transmission line to a nearby neighbor may often beundesirable. However, the effect can be exploited to realize a coupler thatdoes not have a lumped-element equivalent. This is one example of manyelements that utilize distributed transmission line effects to obtain novelfunctionality. Coupled lines are also vital components of many filters. Thesuite of elements that exploit distributed effects available to a microwavedesigner is surprisingly large.

9.14 Exercises1. Consider the crosssection of a coupled trans-

mission line, as shown in Figure 9-1.

(a) For an even mode on the coupled line wecan consider a phasor voltage of 1 V oneach of the lines above the ground plane.Sketch the electric fields in the transverseplane (the plane of the crosssection). Treatthe ground as being at 0 V.

(b) For the even mode, sketch the magneticfields in the transverse plane (the planeof the crosssection).

(c) For an odd mode on the coupled line wecan consider a phasor voltage of +1 V onthe left line and a phasor voltage of −1 Von the right line. Sketch the electric fieldsin the transverse plane (the plane of thecrosssection). Again, treat the ground asbeing at 0 V.

(d) For the odd mode, sketch the magneticfields in the transverse plane (the planeof the crosssection).

2. Electromagnetic software can be used to de-termine the even- and odd-mode parametersof a coupled line. This is usually done by set-ting phasor voltage on the coupled line andevaluating the phasor charge under each con-dition. The voltage applied to the left strip isV1 and the voltage applied to the right stripis V2. The phasor charge on the strips is Q1

and Q2, respectively. One part of the analy-sis is to redo the calculations, but this timewith the substrate removed, so that mediumis now free-space situation. In this case, thecharges are denoted by Q01 and Q02. The ma-trix of (computer-based) measurements is fol-

lows. [This problem parallels Example 9.1 onPage 473.]

Charge V1 = 1 V; V1 = 1 V;V2 = −1 V V2 = 1 V

Q1 (pC/m) 40 20Q2 (pC/m) −50 30Q01 (pC/m) 13.25 6Q02 (pC/m) −10 2.75

(a) What is the two-port capacitance matrix?(b) What is the even-mode capacitance?(c) What is the odd-mode capacitance?(d) What is the free-space (no dielectric) two-

port capacitance matrix?(e) What is the free-space even-mode capac-

itance?(f) What is the free-space odd-mode capaci-

tance?(g) What is the even-mode effective relative

permittivity?(h) What is the odd-mode effective relative

permittivity?

3. Two 50 Ω microstrip lines are to be run par-allel to each on a printed circuit board witha relative permittivity εr = 4. The signal onthe lines is 3 GHz. The thickness of the circuitboard is 1 mm. This is the distance betweenthe strips and the ground plane. The effectivepermittivity of the lines is 3.1 and it is de-termined that the approximate distance overwhich the lines will be in parallel is 1.42 cm.The coupling of the signals on the lines mustbe at least 30 dB down.

(a) What is the free-space wavelength, λ0, ofthe signal?

558 MICROWAVE AND RF DESIGN: A SYSTEMS APPROACH

(d)

n2

n21:

n2n2

n2:1

n2

n2

n2

n2 n2

n2 = 1 +

(a)

(b)

(c)

/Z1

Z 2Z 1Z 2

Z 1Z 1

Z 2Z 2

Z 1 Z 1

Z 2 Z 2

Z 1

Z 2

Z 1

Z 1Z 2

Z 2

Figure 10-42 Kuroda’s identities. Here the inverters are impedance inverters and

the designation refers to the impedance of the inverter. Recall that an inverter of

impedance Z1 can be realized by a quarter-wavelength long transmission line of

characteristic impedance Z1. (As usual, element impedances are indicated.)

10.12 Coupled Line Configurations

The coupled line model of a pair of coupled lines was presented in Figure 9-22 on Page 499. This model is repeated in Figure 10-44. The parameters of thenetwork model are related to the model impedances as follows, repeatingEquations (9.159) and (9.160):

n =1

K=

Z0e + Z0o

Z0e − Z0o(10.160)

Z0S =√

(Z0eZ0o), (10.161)

Z01 =Z0S√1−K2

, (10.162)

Z02 = Z0S

√1−K2

K2. (10.163)

Various terminating arrangements of the coupled lines result in severaluseful filter elements. One arrangement is shown in Figure 10-45. Alsoshown in this figure is the development of the network model based on

594 MICROWAVE AND RF DESIGN: A SYSTEMS APPROACH

15. Design a third-order type 2 Chebyshev high-pass filter with a corner frequency of 1 GHz,a system impedance of 50 Ω, and 0.2 dB rip-ple. There are a number of steps in the design,and to demonstrate that you understand themyou are asked to complete the table below. Foreach stage of the filter synthesis you must in-dicate whether the element is an inductance ora capacitance by writing L or C in the appro-priate cell. Other cells require a numeric valueand you must include units. TheX element isidentified in the prototype below. A Cauer 2lowpass filter prototype is shown with ωc be-ing the corner radian frequency, fc = ωc/(2π)being the corner frequency, and Z0 being thesystem impedance.

X1

X2

3

1

1

X

(a) Complete the LPF (lowpass filter) columnof the table with ωc = 1 rad/s, Z0 = 1 Ω.

(b) Complete the HPF (highpass filter) col-umn of the table with ωc = 1 rad/s, Z0 =1 Ω.

(c) Complete the second HPF column of thetable with fc = 1 GHz, Z0 = 1 Ω.

(d) Complete the third HPF column of the ta-ble with fc = 1 GHz, Z0 = 50 Ω.

ELEMENT LPF HPFωc = 1 rad/s, Z0 = 1 Ω ωc = 1 rad/s, Z0 = 1 Ω

L or C Value (units) L or C Value (units)

X1

X2

X3

ELEMENT HPF HPFfc = 1 GHz, Z0 = 1 Ω fc = 1 GHz, Z0 = 50 Ω

L or C Value (units) L or C Value (units)

X1

X2

X3

FILTERS 571

That is, (ω2rLC − 4

)2ωrL

=

(ωr

2 C0Z0 tanπ4 − 1

)Z01 tan

π4

. (10.195)

Since tan π4 = 1, (

ω2rLC − 4

)2ωrL

=

(ωr

2 C0Z01 − 1)

Z01, (10.196)

and rearranging,

C0 = C − 4

ω2rL

+2

ωrZ01. (10.197)

Another relationship comes from equating derivatives. From Equations(10.193), (10.187), and (10.191), and with ω = 1

2ωr and tan(π2ωωr

) = tan π4 = 1,

(ω2rLC + 4

)ω2rL

=1

2

(2C0Z01

ω2r

2 + π + π)

Z01ωr

2

=

(C0Z01ω

2r + 2π

)Z01ωr

. (10.198)

Substituting for C0 from Equation (10.197) and rearranging, the characteris-tic impedance of the stub is

Z01 =ωrL

4

(1 +

π

2

). (10.199)

In the passband, which is at half the resonant frequency of the stub becauseof design choice, the input impedance of the stub in the passband is

Z1 = Z01 tanβ` = Z01. (10.200)

Now consider the first resonator in Figure 10-56 with L = 0.9359 nH andC = 27.1325 pF. With ωr = 2π(2 GHz), then from Equation (10.199),

Z01 = 7.558 Ω, (10.201)

and from Equation (10.197),

C0 = 21.123 pF. (10.202)

This process is repeated for each lumped-element resonator in Figure 10-56,leading to the prototype shown in Figure 10-59.

Steps 6 and 7: Equating Characteristic Impedances of Stubs

The stubs in Figure 10-59 can be physically realized using transmissionline segments. It would be ideal if the impedances of the stubs (i.e., theimpedances looking into the stubs) are identical. To achieve this, the methodof nodal admittance matrix scaling described in Section 10.13 on Page 563is used with the impedance inverters multiplied by

√Z01/Z02 and the

FILTERS 573

j Ω Ω

ΩΩ

−j

Z01= 56.9102Zx

7.558 56.9102

= 7.588

Ω Ω−jj 7.558 56.9102

Ωj 8.7161

Ωj

ΩZ01= 8.7161

8.7161

Figure 10-62 Step 6b. Bandpass combline filter with impedance inverters. AgainZ1, Z2, and Z3 are the input impedances of the stubs.

Stub Impedance When fr = 2f0

The most commonly used design choice is to choose the resonant (orcommensurate) frequency of the resonator to be twice the center frequencyof the design. The resonant frequency fr = ωr/(2π) of the shortedtransmission line stub is the frequency at which the length of the stub is one-quarter wavelength long.3 So the common design choice is that fr = 2f0.Then Equation (10.122) becomes

Z0 =1

J tan(

π2

ω0

2ω0

) =1

J tan(π4

) =1

J= K. (10.204)

Equation (10.204) defines the characteristic impedance of the transmissionlines stubs realizing an inverter. Then the elements of the subnetwork inFigure 10-61(b) have the parameters

J = 1/K = 56.9102 Ω = 17.5715 mS (10.205)

3 Of course there are resonant frequencies at every multiple of one-quarter wavelength. Thefirst resonance occurs at fr , as then the input impedance of the short-circuited stub is anopen circuit.

618 MICROWAVE AND RF DESIGN: A SYSTEMS APPROACH

in amplitude as the signal reflects first from the source, ΓS , and then fromthe input port. That is, if

|ΓSΓIN| > 1, (11.32)

the amplifier will be unstable at the input. This situation is shown in thesignal flow graph of Figure 11-15(b), where oscillation is initiated by noise.Similarly oscillation will occur if multiple reflections between the outputand the input build in amplitude,

|ΓLΓOUT| > 1, (11.33)

with the oscillation initiated by noise as shown in Figure 11-15(c). Now

|ΓIN| =∣∣∣∣S11 +

S12S21ΓL

1− S22ΓL

∣∣∣∣ (11.34)

and

|ΓOUT| =∣∣∣∣S22 +

S12S21ΓS

1− S11ΓS

∣∣∣∣ . (11.35)

Combining Equations (11.32)–(11.35), the amplifier will be unstable if

|ΓSΓIN| =∣∣∣∣ΓSS11 +

S12S21ΓSΓL

1− S22ΓL

∣∣∣∣ > 1 (11.36)

or

|ΓLΓOUT| =∣∣∣∣ΓLS22 +

S12S21ΓSΓL

1− S11ΓS

∣∣∣∣ > 1. (11.37)

The coupling of ΓS and ΓL makes it difficult to independently designthe input and output matching networks. It is much more convenient toconsider the unconditionally stable situation whereby the input is stable nomatter what the load and output matching network present, and the outputis stable no matter what the source and input matching network present.As a first stage in design, a linear amplifier is designed for unconditionalstability. The design space is larger if the more rigorous test for stability,embodied in Equations (11.36) and (11.37), are used to determine stability.The advantage (i.e., a larger design space), however, is often small.

If the source and load are passive, then |ΓS | < 1 and |ΓL| < 1 so thatoscillations will build up if

|ΓIN| > 1 (11.38)

or|ΓOUT| > 1. (11.39)

For guaranteed stability for all passive source and load terminations (i.e.,unconditional stability), then

|ΓIN| < 1 and |ΓOUT| < 1. (11.40)

620 MICROWAVE AND RF DESIGN: A SYSTEMS APPROACH

where c is a complex number and defines the center of the circle on areflection coefficient plot, and r is a real number and is the radius of thecircle (see Section A.7). This circle defines the boundary between stable andunstable values of ΓL. Now Equation (11.45) can be rewritten as

|S11 − (S11S22 − S12S21)ΓL| = |1− S22ΓL| . (11.47)

The determinant of the scattering parameter matrix is now

∆ = S11S22 − S12S21, (11.48)

and so Equation (11.47) becomes

|S11 −∆ΓL| = |1− S22ΓL| . (11.49)

That is, removing the absolute signs bymultiplying each side by its complexconjugate

(S11 −∆ΓL) (S11 −∆ΓL)∗= (1− S22ΓL) (1− S22ΓL)

∗(11.50)

S11S∗11 +∆∆∗ΓLΓ

∗L − (∆ΓLS

∗11 +∆∗Γ∗

LS11) = 1 + S22S∗22ΓLΓ

∗L

− (S22ΓL + S∗22Γ

∗L) (11.51)

(

|S22|2 − |∆|2)

ΓLΓ∗L − (S22 −∆S∗

11) ΓL

− (S∗22 −∆∗S11) Γ

∗L = |S11|2 − 1 (11.52)

(

|S22|2 − |∆|2)

ΓLΓ∗L − (S22 −∆S∗

11) ΓL

− (S22 −∆S∗11)

∗Γ∗L = |S11|2 − 1 (11.53)

ΓLΓ∗L − (S22 −∆S∗

11) ΓL − (S∗22 −∆S∗

11) Γ∗L

|S22|2 − |∆|2=

|S11|2 − 1

|S22|2 − |∆|2(11.54)

ΓLΓ∗L − (S22 −∆S∗

11) ΓL

|S22|2 − |∆|2− (S22 −∆S∗

11)∗Γ∗L

|S22|2 − |∆|2

+(S22 −∆S∗

11) (S22 −∆S∗11)

∗

(

|S22|2 − |∆|2)2

=|S11|2 − 1

|S22|2 − |∆|2+

(S22 −∆S∗11) (S22 −∆S∗

11)∗

(

|S22|2 − |∆|2)2 (11.55)

∣

∣

∣

∣

∣

ΓL +S22 −∆S∗

11

|S22|2 − |∆|2

∣

∣

∣

∣

∣

2

=

∣

∣

∣

∣

∣

S12S21

|S22|2 − |∆|2

∣

∣

∣

∣

∣

2

. (11.56)

622 MICROWAVE AND RF DESIGN: A SYSTEMS APPROACH

|S22| < 1 |S22| > 1

STABLEout

ScrS

|Γ | < 1

UNSTABLE out|Γ | < 1

S

UNSTABLE

cS

r

STABLE

(a) (b)

Figure 11-17 Input stability circles on the ΓS plane: (a) |S22| < 1; and (b) |S22| > 1. The shaded regions

denote the values of ΓS that will result in unconditional stability at the output indicated by |Γout| < 1.

Similarly an input stability circle can be defined for ΓS , where

center : cS =(S11 −∆S∗

22)∗

|S11|2 − |∆|2(11.60)

radius : rS =

∣

∣

∣

∣

∣

S12S21

|S11|2 − |∆|2

∣

∣

∣

∣

∣

. (11.61)

The interpretation of the input stability circles, shown in Figure 11-17, is thesame as for the output stability circles.The stability criterion provided by the input and output stability circles is

very conservative. For example, the input stability circle (for ΓS) indicatesthe value of ΓS that will ensure stability no matter what passive load ispresented. Thus the stability circles here are called unconditional stabilitycircles. However, an amplifier can be stable for loads (or source impedances)other than those that ensure unconditional stability. However, the useof stability circles provides a good first pass in design of the matchingnetworks between the actual source and load and the amplifier. The stabilitycircles will change with frequency, and so ensuring stability requires a broadfrequency view. This is considered in the linear amplifier design example inSection 11.7.If an amplifier is unconditionally stable, amplifier design is considerably

simplified. A more complete stability analysis is presented by Gonzalez[183] and Suarez and Quere [184].

624 MICROWAVE AND RF DESIGN: A SYSTEMS APPROACH

combined with any one of the following auxiliary conditions [185–191]:

B1 = 1 + |S11|2 − |S22|2 − |∆|2 > 0 (11.64)

B2 = 1− |S11|2 + |S22|2 − |∆|2 > 0 (11.65)|∆| = |S11S22 − S12S21| < 1 (11.66)

C1 = 1− |S11|2 − |S12S21| > 0 (11.67)

C2 = 1− |S22|2 − |S12S21| > 0. (11.68)

The conditions denoted in Equations (11.67) and (11.68) are not indepen-dent, and it can be shown that one implies the other if the k > 1 [185].

If k > 1, then an amplifier is unconditionally stable if any one ofEquations (11.64)–(11.68) is satisfied. Rollet’s stability criteria, k and |∆|, aretabulated in Table 11-5 for the pHEMT described in Figure 11-8. The deviceis unconditionally stable at the frequencies 5–11 GHz and 22–26 GHz. Thedevice could be potentially unstable at frequencies below 5 GHz and from12 to 21 GHz. At these frequencies design needs to use stability circles indesigning matching networks.

11.5.3 Edwards-Sinsky Stability Criterion — µ-factorRollet’s stability criterion, Equations (11.63)–(11.68), assures unconditionalstability but it does not provide a relative measure of stability. That is, thek factor cannot be used to determine how close to the edge of stability aparticular design is. There is no ability to compare the relative stability ofdifferent designs. Edwards and Sinsky [185] developed a test that can beused to compare the relative stability of different designs. This is called theµ-factor stability criterion, with unconditional stability having

µ =1− |S11|2

|S22 − S∗11∆|+ |S21S12|

> 1. (11.69)

An important result is that larger values of µ indicate greater stability. The µfactor is a single quantity that provides a sufficient and necessary conditionfor unconditional stability. That is, it does not matter what passive sourceand load is presented, ΓS < 1 and ΓL < 1, the amplifier will be stable. Thiscontrasts with Rollet’s stability criterion in which two conditions must bemet.

Edwards-Sinsky stability parameters for the pHEMT transistor docu-mented in Figure 11-8 are shown in Table 11-6. The unconditionally stablefrequencies of operation are 5–11 GHz and 22–26 GHz. The same uncondi-tionally stable frequencies determined by using Rollet’s stability criterion (inTable 11-5). The additional information available with the Edwards-Sinskystability criterion is that µ indicates the relative stability. In Table 11-6, thetransistor is unconditionally stable in the 5 to 11 GHz range, and in this

640 MICROWAVE AND RF DESIGN: A SYSTEMS APPROACH

RS < RL

C

RS

RL

L

(a)

RS < RL RS > RL

RS

CR

LL

C

SR

L RL

(b) (c)

Figure 11-25 Output matching network candidates required for out-of-band stability: (a) RS < RL; (b)

RS < RL; and (c) RS > RL. The active device is on the left. (From Figure 8-13 on Page 422.)

LOADDEVICEACTIVE

C

xX

SR L R

L

C

xX

SR R

LL

xX

RS = 56.99 Ω, Xx = −75.06 Ω, RL = 50 Ω RS = 56.99 Ω, Xx = −75.06 Ω, RL = 50 Ω(a) (b)

CL

SR

SX

PX

RL o

oACTIVEDEVICE

LOAD

CL R

L

RS = 56.99 Ω, Xx = −75.06 Ω, RL = 50 Ω L1 = 1.00 nH, C1 = 1.064 pF(c) (d)

Figure 11-26 Steps in the design of the output matching network: (a) active device presents itself as

a resistance in parallel with a capacitive reactance at the output matching network; (b) with inductor

to resonate out active device reactance; (c) simplified matching network problem; and (d) final output

matching network design.

chosen, as this is closer to reality since there is a capacitance at the output ofthe transistor. Resonance, as shown in Figure 11-26(b), will be used to cancelthe effect of the active device capacitance, so that the matching problem

AMPLIFIERS 641

reduces to that shown in Figure 11-26(c). Using the procedure outlined inSection 8.4.2 on Page 424,

|QS | = |QP | =√

RS

RL− 1 =

√

56.99

50− 1 = 0.3739 (11.105)

QS =

∣

∣

∣

∣

XS

RL

∣

∣

∣

∣

=

∣

∣

∣

∣

XS

50 Ω

∣

∣

∣

∣

= 0.3739 (11.106)

QP =

∣

∣

∣

∣

RS

XP

∣

∣

∣

∣

=

∣

∣

∣

∣

56.99 Ω

XP

∣

∣

∣

∣

= 0.3739, (11.107)

so

XS = −18.70 Ω and XP = 152.4 Ω. (11.108)

Now Xx = −75.064 Ω, so 75.064 Ω must be added to XP in parallel, and thereactance of Lo is 50.29 Ω thus (at 8 GHz)

Lo = 1.00 nH and Co = 1.064 pF. (11.109)

11.7.4 Input Matching Network Design

The input stability circle at 1 GHz (Figure 11-24(d)) indicates that the inputmatching network, as seen from the transistor, could look like a short circuit,a matched load, or a capacitor at low frequencies. Figure 11-24(f), the inputstability circle at 16 GHz, indicates that the input matching network, as seenfrom the transistor, could look like a short circuit or a matched load at highfrequencies. Examining the two-element matching networks in Figure 8-13on Page 422, three candidate input matching networks are shown in Figure11-27.The reflection coefficient looking into the output matching network from

the active device is ΓL = S∗22 = 0.340 6 99.1 because of the design decision to

ignore S12 for the output matching network. Now that the output matchingnetwork has been designed, the feedback parameter need no longer beignored. So

ΓIN = S11 +S12S21ΓL

1− S22ΓL(11.110)

= (0.486 6 140.4) +(0.057 6 6.4)(3.784 6 11.2)(0.340 6 99.1)

1− (0.340 6 −99.1)(0.340 6 99.1)(11.111)

= −0.4117+ 0.3839 . (11.112)

That is, ZIN = 15.959+ 17.935 Ω. So, taking into account the bias objectivesand the output matching networks shown in Figure 11-27, the matchingnetwork topology of Figure 11-27(c) will be used (where the load is the

642 MICROWAVE AND RF DESIGN: A SYSTEMS APPROACH

RS > RL

RL

C

RS

L

(a)

RS < RL RS > RL

RS

CR

LL

C

SR

L RL

(b) (c)

Figure 11-27 Input matching network candidates required for out-of-band stability: (a) RS > RL; (b)

RS < RL; and (c) RS > RL. (The active device is on the right.)

active device). The input matching network problem is as shown in Figure11-28.

|QS | = |QP | =√

RS

RL− 1 =

√

50

15.959− 1 = 1.4605 (11.113)

QS =

∣

∣

∣

∣

XS

RL

∣

∣

∣

∣

=

∣

∣

∣

∣

XS

15.959 Ω

∣

∣

∣

∣

= 1.4605 (11.114)

QP =

∣

∣

∣

∣

RS

XP

∣

∣

∣

∣

=

∣

∣

∣

∣

50 Ω

XP

∣

∣

∣

∣

= 1.4605 (11.115)

So

XS = −23.31 Ω and XP = 34.23 Ω (11.116)

Now Xx = 17.935 Ω, so −17.935 Ω must be added to XS , and the reactanceof Ci is 41.24 Ω, thus (at 8 GHz),

Li = 681 pH and Ci = 482 fF. (11.117)

11.7.5 Summary

The final schematic of the linear amplifier design is shown in Figure 11-29.The output matching network, L2 and C2, enabled the biasing inductor tobe replaced by L2. We were not so lucky with the input matching network,L1 and C1. The gate bias network is still required. L3 should be a largeenough value for it to act as an RF choke. A value of 10 nH provides areactance of approximately 500 Ω at 8 GHz. The value of C3 = 100 pF is

AMPLIFIERS 677

and thus the noise factor is

F =SNRi

SNRo=

Si

Ni

No

So=

Si

Ni

Ni

Si/100= 100 (11.191)

and the noise figure isNF = 20 dB. (11.192)

So the noise figure of an attenuator (or filter) is just the loss of the component. This is nottrue for amplifiers of course, as there are other sources of noise, and the output impedanceof a transistor is not a thermal resistance.

11.11.4 Noise in a Cascaded SystemSection 11.11.3 developed the noise factor and noise figure measures fora two-port. This result can be generalized for a system. Considering thesecond stage of the cascade, the excess noise at the output of the secondstage, due solely to the noise generated internally in the second stage, is

Ne,2 = (F2 − 1)kT0BG2. (11.193)

Then the total noise power at the output of a two-stage cascade is

No,2 = (F2 − 1)kT0BG2 +No,1G2 (11.194)= (F2 − 1)kT0BG2 + F1kT0BG1G2. (11.195)

The second term above is the noise output from the first stage amplified bythe second gain.

Generalizing the above result yields the total noise power at the output ofthe mth stage:

No,m =

m∑n=2

[(Fn − 1) kT0B

n∏i=2

Gi

]+ F1kT0B

m∏n=1

Gn. (11.196)

Thus an m-stage cascade has a total cascaded system noise factor of FT =No,m

/(GTNi,1

), with GT being the total cascaded available gain and Ni,1

being the noise power input to the first stage. In terms of the parameters ofindividual stages, the total system noise factor is

FT = F1 +F2 − 1

G1+

F3 − 1

G1G2+

F4 − 1

G1G2G3+ · · · ; (11.197)

that is,

FT = F1 +m∑

n=2

Fn − 1∏ni=2 Gi−1

. (11.198)

This equation is known as Friis’ formula.

AMPLIFIERS 703

8. Consider the design of a 15 GHz inductively-biased Class A amplifier using the pHEMTdocumented in Figure 11-8. Use the topologyshown in Figure 11-22.

(a) If the input of the transistor is terminatedin 50 Ω, what is the impedance lookinginto the output of the transistor?

(b) Design a two-element output matchingnetwork for maximum power transferinto a 50 Ω load.

9. Design an amplifier for maximum stablegain using the discrete pHEMT transistor de-scribed in Figure 11-8. The design specifica-tions are

Gain: maximum gain at 24 GHzTopology: three two-ports (input and

output matching networks,and the active device)

Stability: broadband stabilityBandwidth: maximum that can be

achieved using two-elementmatching networks

Source Z: ZS = 10 ΩLoad Z: ZL = 50 Ω

Follow the proceduredescribed in Section 11.7on Page 636.

10. The system shown below is a receiver withband pass filters, amplifiers and a mixer. [Par-allels Example 11.4 on Page 678]

G5

= 40 dBG2

= 16 dB G3

= −1 dB G4

= 10 dBG1

=−2 dB

1= 2 dBNF

2= 3 dBNF

3= 1 dBNF

4= 5 dBNF

5= 4 dBNF

(a) What is the total gain of the system?

(b) What is the noise factor of the first band-pass filter?

(c) What is the system noise factor?

(d) What is the system noise figure?

11. An amplifier consists of three cascaded stageswith the following characteristics:

Characteristic Stage 1 Stage 2 Stage 3

Gain 10 dB 15 dB 30 dBNF 0.8 dB 2 dB 2 dB

What is the noise figure (NF) and gain of thecascade amplifier?

12. The front end of a receiver for a cellular phonehas a bandpass filter with a 25 MHz passbandand a loss in the passband of 2 dB and is fol-lowed by two amplifier stages. The first stagehas a gain of 20 dB and a noise figure of 0.5 dBand the second stage has a gain of 60 dB and anoise figure of 2 dB.

(a) Sketch the block diagram of the system asdescribed.

(b) What is the gain of the system?

(c) What is the noise figure of the bandpassfilter?

(d) What is the noise figure of the system?

(e) The system is now connected to an an-tenna with an effective noise temperatureof 30 K and which delivers a signal of10 pW to the bandpass filter. Determinethe noise temperature at the output of thesystem and hence the output noise powerin the 25 MHz bandwidth. Determine thesignal-to-noise ratio at the output of thefront-end system.

13. A 75 Ω attenuator has a loss of 16 dB and isbetween a source with a Thevenin impedanceof 75 Ω and a load of 75 Ω.

(a) What is the noise power, Ni, availablefrom the 75 Ω source resistor at standardtemperature (270 K) in a 1 MHz band-width?

(b) Now consider that the source is con-nected to the attenuator which is alsoconnected to the load. If the source gen-erates a modulated signal that is 1 MHzwide and has an available power, Si, of10 fW, what is SNRi at the input to theattenuator at standard temperature?

(c) With the attenuator connected to thesource, what is the Thevenin equivalentimpedance looking into the output of theattenuator?

(d) Calculate the noise power, No, availablefrom the attenuator with the source at-tached at standard temperature (270 K) ina 1 MHz bandwidth?

AMPLIFIERS 705

19. The distortion properties of the MOSFET cir-cuit below are captured by the nonlineartransconductance equation iDS2 = b1VGS1 +b3V

3GS2 where b1 = 0.05A/V and b3 =

−0.2A/V3.

vo

M2

C1

L 1

R2= 50 Ωvi

VDD

(a) What is the IIP3 in terms of voltage?

(b) What is the OIP3 in terms of voltage?

20. The distortion properties of the MOSFET cir-cuit below are captured by the nonlineartransconductance equations iDS1 = a1VGS1 +a3V

3GS1 and iDS2 = b1VGS1 + b3V

3GS2, where

a1 = 0.01A/V, a3 = −0.1A/V3, b1 =0.05A/V and b3 = −0.2A/V3 .

vo

M2

C1

1L= 250 ΩR1

R2= 50 Ωvi

VDD

M1

(a) What is the IIP3 in terms of voltage?

(b) What is the OIP3 in terms of voltage?

21. An amplifier consists of two cascaded stages.The first stage has a linear gain G1 =20 dB, an output 1 dB gain compression pointP1o(1 dB) = 0.1 dBm, and an output-referredthird-order intercept, P1,OIP3 = 0 dBm.The second stage has a linear gain G2 =30 dB, an output 1 dB gain compression pointP2o(1 dB) = 1 dBm, and an output-referredthird-order intercept, P2,OIP3 = 20 dBm.

(a) Develop a symbolic expression for the in-put referred 1 dB gain compression pointof the cascade amplifier.

(b) What is the input-referred 1 dB gain com-pression point of the cascade amplifier?

(c) Develop a symbolic expression for theIIP3 of the cascade amplifier.

(d) What is IIP3 of the cascade amplifier?

22. A single-stage amplifier has a linear gain of16 dB, an output 1 dB gain compression pointof 10 dBm, and an output-referred third-orderintercept point, OIP3 = 30 dBm. A commu-nication signal with a PAR of 6 dB is used.What is the maximum average power of theinput signal before the output suffers signifi-cant compression. This is defined at the pointat which the peak signal is compressed by1 dB.

MIXERS, OSCILLATORS, AND SWITCHES 729

Common Gate Common Drain Common Source

Z3

DDV

Z1

Z2

OUT

D

S

G

Z2

Z1

DDV

Z3

OUTS

DG

Z1

Z2

Z3

DDV

OUT

GS

D

(a) (b) (c)

Figure 12-18 Circuit schematics for FET feedback oscillators using Pi-type feedbacknetworks: (a) common gate; (b) common drain; and (c) common source. The currentsources provide bias.

DDV

OUT

S

D

(c)(b)(a)

Figure 12-19 Crystal oscillator: (a) schematic symbol for a crystal; (b) equivalentcircuit ; and (c) FET crystal oscillator.

counts [109, 202–206].

12.5.2 Phase Noise in OscillatorsThe performance of most RF and microwave systems is limited by oscillatornoise. The two main sources of noise in electronic devices are white noise,which has a frequency-independent power spectral density (PSD), andflicker noise, which has a PSD that varies as the inverse of frequency.2 In

2 In reality, flicker noise may not be exactly 1/f , as there are a variety of effects that canproduce flicker noise, including carrier recombination [196, 197]. However, traditionallyflicker noise is treated as having the inverse frequency dependence.