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Solution Manual for Microwave Circuit Design A Practical Approach Using ADS, 1st Edition

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    Note: When Figures in the text are cited, it is denoted by Figure and figures in this End-of chapter solutions are denoted by Figure. CHAPTER 1 PROBLEMS 1.1 Waveguide generally has lower line loss than microstrip. SIW(Substrate Integrated Waveguide) can be regarded as the planar version of waveguide. How SIW is configured using substrate? Solution A post-wall waveguide (also known as substrate integrated waveguide (SIW) or a laminated waveguide) is a synthetic rectangular electromagnetic waveguide formed in a dielectric substrate by densely arraying metallized posts or via-holes which connect the upper and lower metal plates of the substrate. The waveguide can be easily fabricated with low-cost and mass-production using through-hole techniques where the post walls consists of via fences. The post-wall waveguide is known to have similar guided wave and mode characteristics to the conventional rectangular waveguide with equivalent guided wavelength. The structure is shown below. Since SIW provides low loss, filters using SIW have studied by many researchers. Figure 1.1 The structure of SIW 1.2 Find the microwave integrated circuit example built using LTCC in the web-site. Solution In the web site http://www. barryltcc.com/, there is a TR (Transmit and Receive) module for radar. 1.3 How the ALC in Figure 1.7 is constructed? Solution Figure 1.2 shows a block diagram for ALC. Usually PA module has the terminal for output power control. If not, a driver amplifier should be designed to control the output power level. A small portion of output power is sampled through a directional coupler as shown in Fig. 1.2 and the RF power level is detected through a diode detector. The reference signal is a variable DC voltage

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    source. Sometimes when the variable voltage source is difficult, it is generated using a lowpass filter PWM (Pulse width modulation) signal generated from the controller. The average DC voltage corresponds to the desired power level. Since it is a negative feedback loop, the detector voltage is equal to the reference voltage at the steady state. Thus the output power level can be controlled by changing the reference voltage. Figure 1.2 ALC block diagram 1.4 Refer to SA605 which is used to demodulate the FM signal. Explain its block diagram. Solution FM demodulation in SA605 is based on quadrature FM demodulator shown in Fig. 1.3. In coming FM signal is split in two paths; one is directly applied to the mixer and the other is coupled to the resonant circuit centered at the FM carrier frequency by a high reactance capacitor. As a result, the signal coupled to the high reactance capacitor suffers a 90 phase shift. The FM modulated signal suffers additional phase shift proportional to the resonator slope and AM modulated waveform proportional to FM modulation appears at V2. As a result of mixing demodulated FM waveform appears. Figure 1.3 Quadrature FM demodulator 1.5 Refer the web-site of PLL IC. Explain the synthesizer data bus shown in Fig. 1.7. Solution Refer to Analog Devices, RF PLL Frequency Synthesizers ADF4110/4111/4112/4113, August, 2012.

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    1.6 How the PLL can be modulated? Explain how to set the PLL loop bandwidth. Solution For FM signal generation at the VCO output, the modulation signal may be directly injected to VCO tuning input. The fast varying signal than PLL loop bandwidth can modulate VCO successfully, however the slow varying one will be compensated by the PLL and does not present the modulated output. Figure 1.4 FM modulation using PLL CHAPTER 2 PROBLEMS 2.1 A company A fabricates a thin film resistor whose sheet resistivity is 50 ohm/square and in the case of a company B, an identical thin film resistor has a sheet resistivity of 100 ohm/square. What is the difference between the processes of the two companies? Furthermore, given that the material’s volume resistivity is ρ and its thickness is t , find its sheet resistivity. Solution The thickness of the resistor material of the company A is two times thicker than that of the company B. The resistance R with a volume resistivity of ρ and a thickness of t is S L L R R tW W ρ = = Thus S R t ρ =

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    2.2 Given that the sheet resistivity is 50 ohm/(square) and we want to design a 100 ohm resistor. If the current flowing in this resistor is 2 mA, find its minimum width. Its rated current per width is 0.5 mA/μm. Solution The minimum width W min theta can sustain 2mA is min 2(mA) 4( μ m) 0.5(mA/ μ m) W = = Taking W min into consideration, the width is determined as 10 m m. Thus the length should be 20 m m to yield 100 ohm resistor. 2.3 The dielectric material of a MIM capacitor is Silicon Nitride and its dielectric constant is 7.2. If the area is 50 μm 2 find the thickness of a 0.53 pF capacitor. In addition, find the capacitance per unit area in (F/ μm 2 ). Solution 0.53( pF C = 8.854( pF ) o r A t ε ε = = 2 /m) 7.2 50( μ m × × ) ( μ m t 6 3 ) 8.854 10 7.2 50 6.0 10 ( μ m) 6.0(nm) 0.53 t − − × × × = = × = The sheet capacitance C S is 2 2 0.53(pF) 10.6 (fF / μ m ) 50( μ m ) S C = = 2.4 In the equivalent circuit of an inductor such as that in Fig . 2.10, show that the approximate magnitude of the impedance at resonance is ( ) 2 max o L Z R ω = 1 o LC ω = Solution The impedance of Z ( ω ) of the inductor equivalent circuit ( ) ( ) 1 1 R j L j C Z R j L C ω ω ω ω ω + =   + −     Near the resonance Z ( ω ) can be approximated as ( ) ( ) ( ) ( ) ( ) 2 1 1 1 1 o o o o o R j L R j L j L L j C j C j C Z R R R R j L C ω ω ω ω ω ω ω ω ω ω + + = ≅ ≅ =   + −    

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    2.5 In this text, we have covered the method of extracting the equivalent circuit of a capacitor from | Z ( ω )| characteristics obtained from impedance analyzer such as that shown in Fig . 2.8. The impedance of the equivalent circuit of a capacitor is ( ) 1 Z R jX R j L C ω ω ω   = + = + −     Using this, the real and imaginary parts for frequency can be obtained by graphical representation. Then show that the approximate impedance near the resonant frequency ω o is ( ) ( ) 2 o Z R j L ω ω ω ≅ + − Solution ( ) ( )( ) ( )( ) ( ) 2 2 2 2 2 2 1 1 1 1 2 o o o o o o o Z R jX R j L R j L R j L C LC R j L R jL R jL R j L ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω       = + = + − = + − = + −             − +   − = + = +     − + = + ≅ + − 2.6 Similar to Problem 2.5 above, for a parallel resonant circuit, show that the admittance near the resonant frequency, Y ( ω ) is given by ( ) ( ) 2 o Y G j C ω ω ω ≅ + − Solution The admittance of a parallel resonant circuit can be expressed as ( ) 1 Y G jB R j C L ω ω ω   = + = + −     This is similar to Z ( ω ) in problem 2.5. Thus, ( ) ( ) 2 o Y G j C ω ω ω ≅ + − 2.7 (ADS Problem) In Example 2.2, the equivalent circuit of the 10 pF capacitor from Murata are obtained using the slope of the reactance versus frequency and resonance frequency such as 4 o f X L f π ∂ = ∂ and 1 o LC ω = . In another way, the equivalent circuit can be obtained using the impedance plot in log-log scale. Compute the equivalent circuit values using ADS. Solution Fig. 2.1 shows the simulation schematic for impedance plot.

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    Figure 2.1 Simulation schematic setup for impedance plot. Here log sweep is set to 30 points/decade The simulated impedance is shown in Fig. 2.2. Using the impedance values at markers, the series equivalent circuit values can be obtained. From the minimum value, the value of R can be obtained and R =0.363 Ω . The following equations for C and L are written in the Display window. Eqn C=1/(2*pi*m2*indep(m2)) Eqn L=m3/(2*pi*indep(m3)) The computed C = 9.9 pF and L =0.96 nH. The values of R , C , and L are close to those computed in Example 2.2 but shows little differences. The error of R is mainly due to the sparse sweep points. The differences of the other values are believed to come from different approximations in the computation of element values. Figure 2.2 Simulated impedance for the Murata 10 pF capacitor.

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    CHAPTER 3 PROBLEMS 3.1 In Example 3.1, a traveling sine wave propagating along z direction is given by cos( ω t– β z ). For a constant C , if ω t– β z = C is chosen, what is the meaning? Also, when this constant is taken as the peak of the sine wave, prove that its velocity is v p . Solution When ω t– β z = C is chosen, the wave propagating along z direction becomes cos(C). This means that time and space which yields constant phase is chosen. For example, let C =0. Then it represents a wave at its peak value. This will propagate along z direction with time, which one usually observes the wave velocity. Thus the phase velocity is { } 0 p d dz t z v dt dt ω β ω β β − = → = = p v ω β = 3.2 In the special case of the transmission line when θ is small, find the values of L and C using the results in Fig . 3.25. In addition, show that the following equivalent circuits yield the same wave equation as θ → 0. (a) (b) Figure 3.1 Equivalent circuit of a small length transmission line: (a) Tee and (b) Pi shape Solution For the circuit in Fig. 3.1(a), from Tee-equivalent circuit in section 3.6.3.2, 1 tan 2 2 2 a c o o o p z j L Z Z jZ jZ jZ v θ θ ω ω ∆   = = = ≅ =     Replacing Z o and v p in the above equation by L and C given by (3.6) and (3.14), 1 2 L z L ∆ = This is phaysically obvious. Similar procedure leads to 1 C C z = ∆ For the circuit in Fig. 3.1(b), by inspection, one can obtain the following results as

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    2 L L z = ∆ 2 1 2 C C z = ∆ For the circuit in Fig. 3.1(a), define v ( z , t ) and i ( z , t ) as shown in Fig. 3.2 then ( ) ( ) { } ( ) ( ) , , 1 1 , , 2 2 i z t i z z t v z z t v z t L z L z t t ∂ ∂ + ∆ − + ∆ − = ∆ + ∆ ∂ ∂ Figure 3.2 Definition of voltages and currents Dividing both sides by ∆ z , ( ) ( ) ( ) ( ) , , , , 1 1 2 2 v z z t v z t i z t i z z t L L z t t + ∆ − ∂ ∂ + ∆ − = + ∆ ∂ ∂ In the limit ∆ z → 0, the above equation becomes ( ) ( ) , , v z t i z t L z t ∂ ∂ = − ∂ ∂ This is equal to the result in (3.2). The other equation can be similarly derived. Thus, the wave equation in (3.4) is obtained irrespective of the equivalent circuit forms. 3.3 Considering the 2-port circuit in Fig. 3.3 as a transmission line, determine its equivalent characteristics impedance using open and short-circuit method presented in Section 3.2.4. Figure 3.3 LC circuit Solution The open circuit impedance is

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    1 and open short Z j L Z j L C ω ω ω   = − =     The equivalent characteristic impedance is 2 2 1 o open short o L Z Z Z C ω ω   = ⋅ = −     Where 2 1 o LC ω = 3.4 (1) If the electrical length at a frequency of 1 GHz is 90 ° , what is the electrical length at a frequency of 3 GHz? Solution The electrical length is 1GHz 1GHz 90 f p f l v ω θ = = = = ° . Since the electrical length is proportional to frequency, the electrical length at 3 GHz is 270°. (2) Also, if the phase velocity is equal to the speed of light, what is its length? Solution Since v p = c , 1GHz 1GHz 1GHz 90 2 f f p f l l v c ω ω π θ = = = = = = = ° Hence, the length is 2 c l π π ω = = 8 3 10 2 2 π × 9 3 0.1(m) 7.5(cm) 4 10 = × = × (3) Finally, what is its propagation constant? Solution Since the propagation constant is defined as the phase delay per length (rad/m), it can be computed as 2 20.9(rad / m) 7.5(cm) 2 0.075 c l π ω θ π β = = = = = × 3.5 In the definition of the reflection coefficient shown below, when the real part of Z L is positive, prove that | Γ L | ≤ 1. L o L L o V Z Z V Z Z − + − Γ = = +

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    Solution Define the normalized impedance z L by Z o as for > 0 L z r jx r = + ∞ ≥ Then proving the following relation is the problem. 1 1 1 1 1 L L L L L z z z z − Γ = ≤ → − ≤ + + Squaring both sides and subtracting, one can find that { } 2 2 2 2 ( 1) ( 1) 4 0 r x r x r + + − − + = ≥ Thus | Γ L | ≤ 1 and the equality is satisfied for r =0. 3.6 Figure 3.4 shows a slot line measurement. Voltage standing wave minimum points are shown by the numbers for two measurements. When the load is replaced by a short, the first minimum occurs at 90cm. The distance d min from the first minimum to the short is 15 cm (=90 – 75cm in Fig. 3.4) at a frequency of 375 MHz. Find the load impedance. Figure 3.4 Slot line measurements Solution This can be solved using Smith chart. Since VSWR=5, one can draw a circle in the Smith chart corresponding to VSWR=5. Furthermore the half wavelength is found to be 40 cm (=75 – 35) from Fig. P3.3. Thus, the wavelength is 80cm. The d min normalized by wavelength is 0.1875 (=15/80). Thus the load impedance Z L is found by reading the impedance at point B shown in Fig. 3.5. Z L reads as 56.5 94.5 L Z j = −

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    Figure 3.5 Slot line measurements 3.7 In Fig. 3.6, V + and V – represent incident and reflected phasors at load plane. Write voltage V 1 using V + and V – . When the VSWR at the input is 2, find the magnitude of Γ in and the corresponding return loss. In above condition, setting Z o =50 ohms, find the value of resistor Z L . Figure 3.6 Transmission line terminated by load. Solution V 1 is composed of two voltage waves V f and V b . V f is propagating to the right and resulting in V + at the load plane. The phase of the voltage V f advances that of V + by θ . This can be expressed as j f V V e θ + = The other propagating to the left from the load whose value is V – at the load plane. The phase of V b is delayed by θ compared with V – . Thus j b V V e θ − − =

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    As a result, 1 j j f b V V V V e V e θ θ + − − = + = + Denote | Γ in |= ρ, then 1 1 1 2 20 log 9.54(dB) 1 3 3 VSWR RL ρ ρ ρ +   = = → = → = − =   −   Since ρ=1/3, 1 1 1 2 or 1 3 2 L o L L L L o L Z Z z z z Z Z z ρ − − = = = → = = + + 3.8 Using the scale at the bottom of Smith chart, find VSWR and return loss corresponding to Γ =0.6 using compass. Solution The scale in the bottom of the Smith chart is shown in Fig. 3.7. From Fig. 3.7, the return loss and VSWR corresponding to Γ =0.6 are about 4.4 dB and 4, respectively. Figure 3.7 The scale in the bottom of the Smith chart. 3.9 For a transmission line with Z o and θ , derive ABCD parameters defined as, 1 2 1 2 V V A B I I C D       =             Solution The voltage and current of a transmission line at θ can be expressed as 2 j j V V e V e θ θ + − − + = + ( ) 2 1 j j o I V e V e Z θ θ + − − + = − .(1) From 1 V V V + − = +

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    ( ) 1 1 o I V V Z + − = − , V + and V – can be obtained as 1 1 1 1 and 2 2 o o V Z I V Z I V V + − + − = = . Substituting the above V + and V – into (1), 1 1 1 1 2 1 1 cos sin 2 2 j j o o o V Z I V Z I V e e V jZ I θ θ θ θ − + + − = + = + 1 1 1 1 1 2 1 1 sin cos 2 2 j j o o o o V Z I V Z I jV I e e I Z Z θ θ θ θ − + + −   = + = +     Thus cos , sin , and sin o o A D B jZ C jY θ θ θ = = = = . 3.10 Figure 3.8 shows an equivalent circuit of a distribution amplifier similar to that in Fig . 3.1 in the text. Find V 1 and V 2 . Figure 3.8 Equivalent circuit of the output of a distribution amplifier. Solution The voltage V 1 and V 2 can be computed using superposition principle. The voltage V c1 across the current source I is computed to be 1 2 o c Z I V = This will propagates to both ends of the transmission lines and the resulting voltages V o1 and V o2 are 2 1, 1 2, 1 and 2 2 j j o o c c Z I Z I V e V e θ θ − − = = Similarly for the current source Ie -j θ 2 3 1, 2 2, 2 = and 2 2 2 j j j j o o o c c Z Ie Z I Z I V e e V e θ θ θ θ − − − − = = Thus 2 2 1 1, 1 1, 2 2 = 2 j j o c c o Z I V V V e Z Ie θ θ − − = + = ×

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    ( ) 3 2 2, 1 2, 2 2 j j o c c Z I V V V e e θ θ − − = + = + 3.11 If a 50 ohm transmission line has a 1/4 wavelength length at 1 GHz and its end is short- circuited, can this be considered a parallel resonant circuit near the 1 GHz frequency? What then is the capacitance of the parallel resonant circuit? Solution The transmission line of a quarter wavelength whose end is short-circuited can be approximated by parallel resonant circuit according to section 3.5.2 and the capacitance of the parallel resonant circuit is 9 1 1 1 = = (nF) 8 8 10 50 400 o o C f Z = × × 3.12 In Fig. 3.9, answer the following. Find the characteristic impedance Z x for maximum power transfer to load 2 Z o . For this, find the time domain voltage waveform v 1 ( t ). Find also the time domain voltage waveform v 2 ( t ). Figure 3.9 Problem 3.12 circuit. Solution For the maximum power transfer, the impedance Z in should be 2 2 2 2 x x in o x o L o Z Z Z Z Z Z Z Z = = = → = Since Z in =Z o , 1 1 ( ) cos 2 o o v t E t ω = This voltage propagates the transmission line and is delivered to the load 2Z o .whichis equal to the power delivered to Z o . Thus 2 2 1 2 4 2 1 2 2 o o o o V E V E Z Z = → = The phase of V 2 should be – 90° because the voltage propagates the quarter wavelength transmission line. Thus

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    2 2 1 1 2 2 j o o V E e j E π − = = − And 2 1 1 ( ) Re sin 2 2 o j t o o o v t jE e E t ω ω   = − =     3,13 (ADS Problem) The characteristic impedance of a transmission line can be found applying a pulse signal to the transmission line. Fig. 3.10 shows a simulation set up to compute the characteristic impedance of the transmission line. By computing the current and voltage pulses and the ratio of the pulse, show that the ratio is the characteristic impedance given by 50 ohms. Figure 3.10 Simulation setup to compute the characteristic impedance using a pulse Solution In Fig. 3.10, the time of arrival t d is 8 150 0.5 sec 3 10 [ / sec] d l m t c m m = = = × Fig. 3.11 shows the simulated pulse for Vin and Iin.i . From Fig. 3.11, it can be found that the pulse appears after 0.5 m sec. Also from the values of current and voltage, the characteristic impedance is 50 Ω .

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    Figure 3.11 The simulated voltage and current pulses. 3,14 (ADS Problem) The standing wave can be simulated by sweeping the length of the transmission line as shown in Fig. 3.12. Plot of mag( v 1) vs l 1 results in the standing wave pattern. Also plot the standing waveform with time as a parameter. It can be found that the minimum point does not move despite of time change, which is why it is called “standing wave.” Figure 3.12 Standing waveform simulation. Solution After the simulation given in Fig. 3.12, v1 is plot for the length variable l1. Fig. 3.13 shows the simulated standing waveform. Fig. 3.14 shows the time variation of the standing waveform. This can be obtained using ts() function in ADS. As can be seen from Fig. 3.14, the minimum point called “node” does not change with respect to time. This is the reason why we naming “standing wave.”

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    Figure 3.13 Standing waveform Figure 3.14 Standing waveform change for time 3.15 (ADS Problem) Using ADS, extract the equivalent circuit of corner-discontinuity MCORN which has a width of 50 ohm line microstrip width. Note that it can be represented by Tee equivalent circuit because the corner is a passive circuit. Here the substrate is a 10 mil thick alumina with ε r =9.6. Solution Fig. 3.15 shows the schematic setup for the simulation of the corner discontinuity. Also the microstrip line length which has the corner discontinuity along the center line is also included for the comparison. The variable w 50 represents the 50 ohm microstrip line width which is computed using LineCalc in ADS.

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    Figure 3.15 Schematic setup for the corner discontinuity simulation After the simulation, the transmission characteristics are compared in Fig. 3.16. From Fig. 3.16, one can find that the length of the corner discontinuity has a shorter than that of the microstrip line. Figure 3.16 The S 21 of the corner discontinuity and microstrip line Generally, the corner discontinuity shows a shorter length than the microstrip line with the length of the corner discontinuity along the center line. Since the corner discontinuity can be represented by a Tee-equivalent circuit shown in Fig. 3.12.

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    Figure 3.17 T-shape equivalent circuit The values of Z A and Z B can be obtained from Z-parameters as Eqn ZA=Z11-Z12 Eqn ZB=Z12 Eqn YB=1/ZB Writing the above equations in the Display window, their imaginary parts can be plotted as shown in Fig. 3.18. Here Y B =1/ Z B is plotted. (a) (b) Figure 3.18 Imaginary parts of Z A and Z B vs frequency.(a) YB (=1/ Z B ) and (b) Z A From Fig. 3.18(a), Z B is close to a capacitor. In Fig. 3.18(b) Z A is plotted with the microstrip line Z A . Since the microstrip line Z A is tan 2 A o Z jZ θ   =     , it will increase linearly with frequency. On the contrary, Z A of the corner discontinuity decrease with frequency and its value is negative. This phenomenon occurs because the length of Z A is negative. Using the marker values, the inductance and capacitance corresponding to Z A and Z B can be computed inserting equations as

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    Eqn LA=(m4-m3)/(indep(m4)-indep(m3))/(2*pi) Eqn CB=(m2-m1)/(indep(m2)-indep(m1))/(2*pi) The value of C B and L A are 33 fF and 7.3 pH B A C L = = − . CHAPTER 4 PROBLEMS 4.1 For a circuit in Fig. 4.1, calculate 2-port S-parameters normalized by the reference impedance Z o . Figure 4.1 A two-port circuit Using the previous results, calculate two-port S-parameters for the circuit where two transmission lines of electrical lengths θ 1 and θ 2 are added to the circuit in Fig. 4.1 as shown in Fig. 4.2. Figure 4.2 A two-port circuit with transmission lines Solution Due to symmetry, S 12 = S 21 and S 11 = S 22 . 1 1 2 2 11 21 1 1 2 2 1 2 , and 3 3 o o o o o o o Z E Z Z Z Z S S Z Z E − + = = − = = + Therefore 1 2 3 3 2 1 3 3 −   =   −   S For the circuit in Fig. 4.2, S can be computed as

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    1 1 2 1 2 2 2 ( ) 1 2 3 3 ( ) 2 2 1 3 3 j j j j e e e e θ θ θ θ θ θ − − + − + −   − =   −   S , using the shift of reference plane explained in Section 4.1.5. 4.2 Calculate the S-parameters of the following parallel and serial resonance circuits. (a) (b) Figure 4.3 (a) Parallel resonator and (b) the serial resonator Solution In the case of Fig. 4.3(b), the S-parameters are presented in Example 4.5 in Section 4.1.5 and are 11 22 1 1 2 a S S jQ δ − = = + and 21 12 2 1 2 a a jQ S S jQ δ δ = = + . Here Q a and δ are defined as 2 o a o L Q Z ω = and = o o ω ω δ ω − In the case of Fig. 4.3(a), the S-parameters are ( ) 21 12 2 || || p o o o p Z Z S S Z Z Z = = + and 11 22 || || p o o o o p Z Z Z S S Z Z Z − = = + . Z o || Z p is computed as 1 || and , = 1 4 2 o o o p b o o b o Z Z Z Q Z C jQ ω ω ω δ δ ω − ≅ = + Substituting Z p into S 21 ( ) ( ) 21 11 2 || || 1 2 , and || 1 2 || 1 2 p o p o o b o o p b o p o b Z Z Z Z Z jQ S S Z Z Z jQ Z Z Z jQ δ δ δ − − = = = = + + + + 4.3 Explain how to obtain the circuit values of the above parallel resonant or series resonant circuits using | S 21 | and | S 11 | (Refer Example 4.5). Solution

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    In the case of Fig. 4.3(b), 11 21 0 0 and 1 o S S ω ω ω ω → → ∞  = =  =  Thus both | S 21 | and | S 11 | have the bell-shaped frequency response. From the peak value of | S 21 | and | S 11 |, one can find the resonance frequencies, which are equal to ω o =2 π f o as 2 1 o LC ω = In addition, Q a and Q b of the circuits in Fig. 4.3(a) and (b) can be found from the 3-dB bandwidth BW as 2 o o o a f CZ Q BW ω = = and 2 o o b o f L Q BW Z ω = = Using Q s and ω o , the values of L and C in Fig 4.3 can be determined. 4.4 Find the maximum value of | S 21 | for the circuit shown in Fig. 4.4, and sketch the frequency response ( ω o =2 π f o =2 π ×10 GHz). Figure 4.4 A series resonant circuit Solution The transmission lines at the input and output have no effects on | S 21 |. Excluding the transmission lines, one can form a new two port which yield the same | S 21 |. Since S 21 can be expressed as 2 21 2 V S E = | S 21 | approaches to 0 as frequency away from the resonance frequency 10 GHz. Thus, | S 21 | has a bell-shaped frequency response centered at 10 GHz. The maximum value of | S 21 | appears at the resonance and the value is 2 21 2 100 2 0.953 2 105 o o Z V S E Z r = = = = + 4.5 Find voltage V out for the circuit in Fig. 4.5.

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    Figure 4.5 Circuit for problem 4.5 Solution P=polar ( dbmtow (0),0) in P_1Tone port means that, when a conjugate matched load is connected to this port, the power of 0 dBm is delivered the conjugate matched load. Thus the power 0 dBm is delivered to 50 ohm load. The voltage V out is 2 3 2 1 1mW 10 W 10 0.1(V) 2 out out out o V V V Z − − = = → = → = Note that the value V out represents the peak value. 4.6 Given that the insertion loss in a passive lossless 2-port network is 3 dB, calculate the return loss. Solution Since lossless, 2 2 11 21 1 1 1 1 2 2 S S = − = − = . Thus the return loss= 3dB. 4.7 Calculate the S-parameters of the attenuator circuit shown below. Determine whether | S 21 | 2 + | S 11 | 2 =1 is satisfied. Figure 4.6 T-type 3 dB attenuator Solution S 11 and S 22 can be manipulated as 11 22 {(50 8.56) ||141.8 8.56} 50 0 {(50 8.56) ||141.8 8.56} 50 S S + + − = = = + + +

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    21 12 141.8 || (50 8.56) 50 2 50 8.56 141.8 || (50 8.56) 50 8.56 2 S S + = = × + + + + = 41.44 2 × 100 50 × 1 0.707 58.56 2 = = 2 2 11 21 1 1 2 S S + = < Since the circuit is lossy, | S 11 | 2 +| S 21 | 2 <1. 4.8 Prove equation (4.83) Solution The noise figure in (4.82) is ( ) ( ) { } 2 2 2 1 1 u n u n s s s s s s s G R G R F Y Y G G B B G G G G γ γ γ = + + + = + + + + + For B s change, minimum occurs at opt B B γ = The resulting noise figure is ( ) ( ) ( ) 2 2 2 1 1 2 1 2 2 u n s s s u n n n s n u n n s G R F G G G G G R G R G R G R G G R R G G γ γ γ γ γ = + + + + = + + + ≥ + + + The minimum occurs at ( ) ( ) 2 2 u n u n s opt s n G R G G R G G G G R γ γ + = → = + Then F min is ( ) ( ) 2 min 1 2 2 1 2 n u n n n opt F R G G R R G R G G γ γ γ = + + + = + + Using F min , G opt , and B opt , the noise figure relation can be re-arranged as ( ) ( ) { } 2 2 min n s opt s opt s R F F G G B B G = + − + − 4.9 Prove equation (4.84) Solution From problem 4.8, the relation can be transformed into the equation using normalized admittances as ( ) ( ) { } 2 2 2 min min 2 1 n o n s opt s opt s opt s o s o R Y r F F G G B B F y y G Y g Y = + − + − = + − Substituting the following

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    1 1 , and 1 1 opt s s opt s opt y y − Γ − Γ = = + Γ + Γ , Into F , the following equation can be obtained after manipulation. ( ) 2 min 2 2 4 1 1 n s opt s opt r F F Γ − Γ = + − Γ + Γ 4.10 For the circuit in Fig. 4.7, find F min and Γ opt . Use the definition in equation (4.68) after transforming the circuit into Norton equivalent circuit at port 2 Figure 4.7 A two-port resistor circuit. Solution From Fig. 4.8, the Norton current i T at the port2 2 1 o s T s s o s o Y G i i i i G Y G Y = + − + + Figure 4.8 Equivalent Norton current computation. The contribution of the source noise to Norton current is , o T s s s o Y i i G Y = + Here

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    ( ) ( ) ( ) 2 2 2 1 2 4 , and 4 s o s o o E i kT G E i E i kT Y = = = Thus ( ) ( ) 2 2 2 2 2 2 2 1 1 1 2 2 T o s o s o s s s o o s o s o o o s s s o o s E i Y G Y G Y G G G Y Y F G Y G Y Y Y G E i G Y Y G     + + + = = + + = + +         = + + + Hence min 3 2 2 F = + and 2 1 50 2 0.172 2 1 opt opt R s − = → = = + 4.11 The noise temperature T e is defined in terms of the noise factor as T e =( F –1) T o . Using this relationship, with the noise temperature of each stage known, derive the expression below by expressing the noise temperature in terms of Frii’s formula. 2 3 1 1 1 2 e e e e T T T T G G G = + + + ⋅⋅⋅ Solution The noise temperature of a linear two-port network is often used in place of noise figure. Given that the linear two-port network has the transducer power gain G (which is defined as the ratio of the available power from the source to delivered power to a load and will be explained in detail in Chapter 8). When the input termination with the temperature T s . is connected to the linear two port network, the delivered output power P L can be expressed as ( ) L s e P kG T T f = + ∆ . In the above equation, kT e ∆ f represents the added equivalent thermal noise power density at the input due to the noises of the two-port network. Thus the noisy linear two-port network can be modeled as shown in Fig. 4.9. Figure 4.9 Two port’s noise temperature concept One can find that from section 4.2.3, ( 1) ( 1) e o e o kT f k F T f T F T ∆ = − ∆ → = − Now consider the following cascaded chains shown in Fig. 4.10.

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    . (a) (b) Figure 4.10 Cascaded block diagram (a) cascaded block and (b) its equivalent representation Excluding thermal noise source contribution to P L , the total output noise power excluding the contribution of the source noise power can be related as { } 1 2 1 2 1 2 2 L T e n e n e n e n en P G T G G G T G G G T G G T G T k f = = = + + + ∆     Dividing both sides with total gain G T 2 3 1 1 1 2 e e e e T T T T G G G = + + + ⋅⋅⋅ 4.12 The attenuation with an attenuator is L ( L is not in decibel scale), given that its temperature is T o , determine the noise factor of the attenuator. Solution When one end (source port) of the attenuator is terminated by 50 ohm termination, the impedance seen from the other end (load port) is 50 ohm. Since the available power of a 50 ohm resistor is kT o , the delivered power to the 50 ohm load connected to the load port is kT o . On the other hand, the delivered power from 50 ohm termination connected to the source port of the attenuator is kT o / L . Thus, from the definition of the noise figure 1 o o kT F L kT L = = 4.13 When the attenuator in Problem 4.12 is connected next to a noise source having ENR 1 , what is the resulting ENR 2 ? Assume the cold temperature of the noise source is T o .

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    Solution Defining the hot and cold state noise temperatures T h and T c , the ENR of the noise source is 1 h c c T T ENR T − = Normally, T c = T o . After the attenuator, the corresponding output noise powers are 1 c o N kT kT = = 2 1 1 1 h h o N kT kT kT L L   = = + −     Thus 2 1 2 1 1 1 1 1 1 h o o h o o o T T T T T kT kT L L ENR ENR kT T LT L   + − −   − −   = = = = 4.14 (ADS problem) It is explained that icor (1,1) and icor (2,2) appearing after S-parameter noise analysis represent the short circuit noise current squares at ports 1 and 2 respectively. The short circuit noise currents can be obtained through AC simulation in ADS. The following figure shows AC noise simulation to obtain the short circuit noise currents. Since the noise currents cannot be obtained directly, the short circuit noise currents are computed using the node voltages across small valued resistors. Compute the short circuit noise currents through the AC noise simulation and compare the results with the icor (1,1) and icor (2,2) obtained through the S-parameter noise simulation. Figure 4.11 AC noise simulation to compute the short circuit noise currents Solution After the AC noise simulation, the short circuit noise current squares are computed in the display window using the equations as Eqn Ii_sq=(Prob_15b..vi.noise/1e-3)**2

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    Eqn Io_sq=(Prob_15b..vo.noise/1e-3)**2 Figure 4.12 AC noise simulation to compute the short circuit noise currents In the above equations, Prob_15b..vi.noise and Prob_15b..vo.noise represent the simulated noise voltages in the dataset name Prob_15b. To represent the noise voltage, Ads uses the extension .noise after the node voltage name. The division factor 1e-3 appears tocompute the current across 1m Ω resistors. In addition, the S-parameter noise simulation for the same device is performed. The computed two values are compared in Fig. 4.12. As can be seen in Fig. 4.12, the exact match can be found. CHAPTER 5 PROBLEMS 5.1 Figure 5.1 shows a simplified equivalent circuit of GaAs MESFET at low frequency. Figure 5.1 A simplified FET equivalent circuit

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    Figure 5.2 A simplified FET equivalent circuit with a parallel feedback resistor (1) Calculate S-parameters setting the reference impedance Z o . Solution Since input is open, S 11 =1. Also S 12 =0 by inspection. In the case of S 21 , 2 21 1 1 2 2 m o m o g Z E V S g Z E E − = = = − When the gate is termed by Z o , and source is applied to the drain, g m V gs =0. Thus the impedance seen from the drain is open, which means that S 22 =1. Therefore 1 0 2 1 m o g Z   =   −   S (2) Setting Z o =50 ohms, compute g m when S 21 =5 ∠ 180°. Solution 21 2 5 50 (mA/V) m o m S g Z g = − = − → = (3) When resistor f R is connected in parallel as shown in Fig. 5P.2, find the value of R f that makes S 11 =0. Solution For S 11 =0, Z in =50. Applying KVL at the input 1 1 1 1 ( ) f m o V R I I g V Z = + − Thus the following condition can be obtained. 2 1 1 1 f o in o f m o m o R Z V Z Z R g Z I g Z + = = = → = + 5.2 Figure 5.3 shows a simplified equivalent circuit of FET with source resistance R s at low frequency. Due to R s , the equivalent trans-conductance g me is lowered from g m as explained in Example 5.3. Defining the equivalent trans-conductance g me as Drain shorted d me gs i g v =

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