ALL 30 CHAPTERS COVERED
SOLUTIONS MANUAL
,Radio Frequency Electronics second edition: Solutions, Chapters 1 through 10
Solution 1.1
We can almost guess that maximum power will be supplied to the load when Rload = Rsource. If
Rload becomes very large, the current goes to zero while the voltage approaches Vo, the open-
circuit voltage of the source. Thus the power dissipated by the load (current x voltage) goes to
zero. And if Rload approaches zero, the voltage across it goes to zero while the current approaches
Vo/Rsource. Again, the power goes to zero. Solving this simple max/min problem exactly, we write
an expression for Pload in terms of Rload, and set the derivative equal to zero:
I
Vo
PL
2
I ⋅RL so ( Vo) 2⋅RL
Rs + RL PL
( RS + RL) 2
( Vo) 2⋅⎡⎢
dPL 1 2 ⋅RL ⎤
− ⎥
⎣ ( RS + RL) ( RS + RL) 3 ⎦
dRL 2
Setting this derivative equal to zero, we find RL and
( Vo) 2
RS Pmax
4 ⋅RS
If we plot the normalized power, p= P/Pmax vs. the normalized load resistance,
r =Rload/Rsource, we see that Rload can differ from Rsource by a factor of
almost 6 before the power is reduced to Pmax/2.
4 ⋅r
p ( r) := Note that p(r) = p(1/r)
2
( 1 + r) ⎛ 1 ⎞ = 0.49
p ( 6) = 0.49 p⎜
r := .1 , .2 .. 10 ⎝ 6⎠
1
When the source impedance is
0.8
complex, i.e. Zsource = Rsource
+ jXsource, the load impedance
0.6 that will draw maximum power is
p ( r) Zload= Rsource - jXsource.
0.4 (The reactance of this "conjugate
load" simply cancels the source's
0.2 reactance).
0
0.1 1 10
r
,Solution 1.2
The point here is that because we're considering a lossless network (it contains L's and C's of
negligible loss and no R's), any power delivered by the generator MUST end up in the load.
With the network between the load and the source, the impedance presented to the source is
Z=R+jX (where Z, R, and X will all be functions of T). Z is the transformed impedance of the
load. The current supplied by the source will be
I = Vo/ %[Rtot2 + Xtot2 ] = Vo/ %[(rs + R)2 +X2 ]
where Vo is the open-circuit source voltage. The power delivered by the source is just the square
of the current times the resistive part of the load impedance:
P = I2R = Vo2 R / [(rs + R)2 +X2 ] .
When the network is lossless (composed, for example, of lossless inductors and capacitors), this
power must necessarily be dissipated in the load. We have seen in Problem 1.1 that the
maximum Power available from the source is
Pmax = Vo2/(4rs).
The response, P/Pmax can therefore be written as
P/Pmax = 4rsR / [(rs + R)2 +X2 ] .
,Solution 1.3 A MATLAB solution appears in the book, following the statement of the problem.
An alternative solution are presented below in Quick Basic.
'Qbasic program to calculate transmission through an rcl ladder network
RSOURCE= 1000: RLOAD = 50
FOR F=1E6 TO 2E6 STEP 5E4
OMEGA=2 * 3.14159 * F
IR=1: II=0:VR= IR*RLOAD: VI=II*RLOAD 'assume 1 amp into load.
READ TYPE$ ' "PC" is parallel capacitor, "SL" is series inductor, etc.
DO UNTIL TYPE$ = "EOF" 'EOF denotes end of circuit file.
READ VALUE
IF TYPE$ = "PC" THEN B= OMEGA * VALUE: G=0: GOSUB 400 'to update I.
IF TYPE$ = "SC" THEN X= -1/(OMEGA * VALUE):R=0: GOSUB 500 'to update V.
IF TYPE$ = "PL" THEN B= -1/( OMEGA * VALUE):G=0: GOSUB 400 'to update I.
IF TYPE$ = "SL" THEN X= OMEGA * VALUE:R=0: GOSUB 500 'to update V.
IF TYPE$ = "PR" THEN G= 1/VALUE: B=0: GOSUB 400 'to update I.
IF TYPE$ = "SR" THEN R= VALUE: X=0: GOSUB 500 'to update V.
READ TYPE$
LOOP
R=RSOURCE: X=0: GOSUB500 'to get generator voltage
'calculate fraction of maximum possible power transfer.
FRAC= (1^2*RLOAD)/((VR*VR+VI*VI)/(4*RSOURCE)
PRINT F; FRAC; 10/LOG(10)*LOG(FRAC)' freq, frac & frac in dB's.
RESTORE 600 'rewind data.
NEXT F
END
400 'subroutine to update real and imaginary parts of I.
IR=IR+(VR*G - VI*B): II=II+(VI*G + VR*B): RETURN
500 'subroutine to update real and imaginary parts of V.
VR=VR+(IR*R - II*X): VI=VI+(II*R + IR*X): RETURN
' example circuit file: a 463 pF parallel capacitor and a 23.1 microhenry series inductor
600 DATA SL,23.1E-6,PC,463E-12,EOF
Program Output
Freq. (MHz) Frac. (dB)
1.00 .4179 -3.789
1.05 .4601 -3.372
,Solution 1.4
,Solution 1.5
We will use -jejTt as the source voltage, so IR(t) will be the real part of IRcomplex..
Carrying this out, we find
,Solution 2.1
The equivalent series circuit is just a 48 ohm resistor in series with an inductor whose reactance, TL, is 39
ohms. At 100 MHz this is a .0621:H inductor. To find the equivalent parallel circuit, we note that G+jB =
1/(R+jX) or
1/Rp - j/Xp = (Rs-jXs)/(Rs2+Xs2).
Therefore Rp = (Rs2+Xs2)/Rs and Xp = (Rs2+Xs2)/Xs. Putting in the values,
Rp= (482+392)/48= 79.69 ohms and Xp = (482+392)/39 = 98.08 ohms = jTLp. At 100 MHz this is a 0.156 :H
inductor.
Remember that these equivalent circuits are strictly equivalent just at 100 MHz, the only frequency where we
have measured the impedance. But know the component is essentially a lumped resistor in series with its own
inductive leads. We can therefore be fairly confident that the equivalent series circuit will be a good model
of this component for a wide range of frequencies from dc to at least 150 MHz. The parallel circuit however,
becomes a short circuit! at low frequencies rather than just a 50 ohm resistor.
,The design recipe for "EL-networks" (Problem 2.6) involves calculating a factor QEL = %(R/r -1).
Solution 2.2
Here R = 100 and r = 50 so QEL = 1. Parallel and series reactances are then calculated according to
XP = KR/Q and XS = ±rQ.
Choosing the lower signs (circuit a), we have Xsa = - 1@ r = -50 and XPa = R/1 = 100. The C and
L values are chosen to have these reactances at the center frequency (design frequency) of 1.5 Mhz.
This yields the values Ca = 2122 pF and La = 10.61 uH.
Choosing the upper signs (circuit b), we have Xsa = 1@ r = 50 and XPa = -R/1 = -100. This yields
the values Cb = 1061 pF and Lb = 5.305 uH.
The calculated responses are shown below.
.
0
Circuit b.
Power Transfer 10. log frac
i 0.2
(dB)
0.4
1 1.25 1.5 1.75 2
f
i
6
10
Frequency (MHz
,Solution 2.3
The intermediate impedance is specified as the geometric mean of the generator and source resistances. Rint
= sqrt(50*100)=70.7. (Note: this choice will minimize the overall loss of the network).
The left-hand L-network transforms from 50 ohms to 70.7 ohms. The "design Q" , QEL, is therefore QEL= sqrt(
70.7/50 -1) = 0.644. This yields Xp1 = 70.7/.644 = 110 ohms and
Xs1 = 50(.644) = 32.2 ohms. At 1.5 MHz, these produce the values Ls = 11.67 :H and
Cp = 3296 pF.
The right-hand L-network transforms from 70.7 ohms to 100. QEL, is therefore
QEL= sqrt( 100/70.7 -1) = 0.644 (the same value as for the first network). This yields Xp2 = 100/.644 = 155
ohms and Xs2 = 70.7(.644) = 45.5 ohms. At 1.5 MHz, these produce the values Ls = 16.44 :H and Cp = 2332
pF. The calculated response is shown below.
Power transfer, 0
(dB)
10. log frac
i 0.2
0.4
1 1.25 1.5 1.75 2
f
i
6
10
Freq. (MHz)
Solution 2.4
, We have two cascaded L sections so the overall efficiency, 0TOTALis the product of the two efficiencies. Here
these individual efficiencies are equal:
Therefore
and we expect the fraction of power to be transmitted through the network at the center frequency to be 98.72%.
To simulate the lossy inductors we can add to each inductor a parallel resistor of value RP = Q XP.The lossy
network is shown below.
Analyzing this network with the program of Problem 1-3 yields the predicted result: frac = .987.
Solution 2.5
The circuit is drawn below with two equivalent circuits. In Figure b. the capacitor is divided into two
capacitors: C1+C2 = C. The total load is now seen to be two matching networks in parallel. The values of these
networks can be chosen to make the circuit equivalent to Figure c. Since the same voltage, V, say, is applied
to R1 and R2 , the powers dissipated in them are, respectively, V2/R1 and V2/R2. Since the power in R1 is to
be twice the power in R2 , we must have R2 = 2R1.
SOLUTIONS MANUAL
,Radio Frequency Electronics second edition: Solutions, Chapters 1 through 10
Solution 1.1
We can almost guess that maximum power will be supplied to the load when Rload = Rsource. If
Rload becomes very large, the current goes to zero while the voltage approaches Vo, the open-
circuit voltage of the source. Thus the power dissipated by the load (current x voltage) goes to
zero. And if Rload approaches zero, the voltage across it goes to zero while the current approaches
Vo/Rsource. Again, the power goes to zero. Solving this simple max/min problem exactly, we write
an expression for Pload in terms of Rload, and set the derivative equal to zero:
I
Vo
PL
2
I ⋅RL so ( Vo) 2⋅RL
Rs + RL PL
( RS + RL) 2
( Vo) 2⋅⎡⎢
dPL 1 2 ⋅RL ⎤
− ⎥
⎣ ( RS + RL) ( RS + RL) 3 ⎦
dRL 2
Setting this derivative equal to zero, we find RL and
( Vo) 2
RS Pmax
4 ⋅RS
If we plot the normalized power, p= P/Pmax vs. the normalized load resistance,
r =Rload/Rsource, we see that Rload can differ from Rsource by a factor of
almost 6 before the power is reduced to Pmax/2.
4 ⋅r
p ( r) := Note that p(r) = p(1/r)
2
( 1 + r) ⎛ 1 ⎞ = 0.49
p ( 6) = 0.49 p⎜
r := .1 , .2 .. 10 ⎝ 6⎠
1
When the source impedance is
0.8
complex, i.e. Zsource = Rsource
+ jXsource, the load impedance
0.6 that will draw maximum power is
p ( r) Zload= Rsource - jXsource.
0.4 (The reactance of this "conjugate
load" simply cancels the source's
0.2 reactance).
0
0.1 1 10
r
,Solution 1.2
The point here is that because we're considering a lossless network (it contains L's and C's of
negligible loss and no R's), any power delivered by the generator MUST end up in the load.
With the network between the load and the source, the impedance presented to the source is
Z=R+jX (where Z, R, and X will all be functions of T). Z is the transformed impedance of the
load. The current supplied by the source will be
I = Vo/ %[Rtot2 + Xtot2 ] = Vo/ %[(rs + R)2 +X2 ]
where Vo is the open-circuit source voltage. The power delivered by the source is just the square
of the current times the resistive part of the load impedance:
P = I2R = Vo2 R / [(rs + R)2 +X2 ] .
When the network is lossless (composed, for example, of lossless inductors and capacitors), this
power must necessarily be dissipated in the load. We have seen in Problem 1.1 that the
maximum Power available from the source is
Pmax = Vo2/(4rs).
The response, P/Pmax can therefore be written as
P/Pmax = 4rsR / [(rs + R)2 +X2 ] .
,Solution 1.3 A MATLAB solution appears in the book, following the statement of the problem.
An alternative solution are presented below in Quick Basic.
'Qbasic program to calculate transmission through an rcl ladder network
RSOURCE= 1000: RLOAD = 50
FOR F=1E6 TO 2E6 STEP 5E4
OMEGA=2 * 3.14159 * F
IR=1: II=0:VR= IR*RLOAD: VI=II*RLOAD 'assume 1 amp into load.
READ TYPE$ ' "PC" is parallel capacitor, "SL" is series inductor, etc.
DO UNTIL TYPE$ = "EOF" 'EOF denotes end of circuit file.
READ VALUE
IF TYPE$ = "PC" THEN B= OMEGA * VALUE: G=0: GOSUB 400 'to update I.
IF TYPE$ = "SC" THEN X= -1/(OMEGA * VALUE):R=0: GOSUB 500 'to update V.
IF TYPE$ = "PL" THEN B= -1/( OMEGA * VALUE):G=0: GOSUB 400 'to update I.
IF TYPE$ = "SL" THEN X= OMEGA * VALUE:R=0: GOSUB 500 'to update V.
IF TYPE$ = "PR" THEN G= 1/VALUE: B=0: GOSUB 400 'to update I.
IF TYPE$ = "SR" THEN R= VALUE: X=0: GOSUB 500 'to update V.
READ TYPE$
LOOP
R=RSOURCE: X=0: GOSUB500 'to get generator voltage
'calculate fraction of maximum possible power transfer.
FRAC= (1^2*RLOAD)/((VR*VR+VI*VI)/(4*RSOURCE)
PRINT F; FRAC; 10/LOG(10)*LOG(FRAC)' freq, frac & frac in dB's.
RESTORE 600 'rewind data.
NEXT F
END
400 'subroutine to update real and imaginary parts of I.
IR=IR+(VR*G - VI*B): II=II+(VI*G + VR*B): RETURN
500 'subroutine to update real and imaginary parts of V.
VR=VR+(IR*R - II*X): VI=VI+(II*R + IR*X): RETURN
' example circuit file: a 463 pF parallel capacitor and a 23.1 microhenry series inductor
600 DATA SL,23.1E-6,PC,463E-12,EOF
Program Output
Freq. (MHz) Frac. (dB)
1.00 .4179 -3.789
1.05 .4601 -3.372
,Solution 1.4
,Solution 1.5
We will use -jejTt as the source voltage, so IR(t) will be the real part of IRcomplex..
Carrying this out, we find
,Solution 2.1
The equivalent series circuit is just a 48 ohm resistor in series with an inductor whose reactance, TL, is 39
ohms. At 100 MHz this is a .0621:H inductor. To find the equivalent parallel circuit, we note that G+jB =
1/(R+jX) or
1/Rp - j/Xp = (Rs-jXs)/(Rs2+Xs2).
Therefore Rp = (Rs2+Xs2)/Rs and Xp = (Rs2+Xs2)/Xs. Putting in the values,
Rp= (482+392)/48= 79.69 ohms and Xp = (482+392)/39 = 98.08 ohms = jTLp. At 100 MHz this is a 0.156 :H
inductor.
Remember that these equivalent circuits are strictly equivalent just at 100 MHz, the only frequency where we
have measured the impedance. But know the component is essentially a lumped resistor in series with its own
inductive leads. We can therefore be fairly confident that the equivalent series circuit will be a good model
of this component for a wide range of frequencies from dc to at least 150 MHz. The parallel circuit however,
becomes a short circuit! at low frequencies rather than just a 50 ohm resistor.
,The design recipe for "EL-networks" (Problem 2.6) involves calculating a factor QEL = %(R/r -1).
Solution 2.2
Here R = 100 and r = 50 so QEL = 1. Parallel and series reactances are then calculated according to
XP = KR/Q and XS = ±rQ.
Choosing the lower signs (circuit a), we have Xsa = - 1@ r = -50 and XPa = R/1 = 100. The C and
L values are chosen to have these reactances at the center frequency (design frequency) of 1.5 Mhz.
This yields the values Ca = 2122 pF and La = 10.61 uH.
Choosing the upper signs (circuit b), we have Xsa = 1@ r = 50 and XPa = -R/1 = -100. This yields
the values Cb = 1061 pF and Lb = 5.305 uH.
The calculated responses are shown below.
.
0
Circuit b.
Power Transfer 10. log frac
i 0.2
(dB)
0.4
1 1.25 1.5 1.75 2
f
i
6
10
Frequency (MHz
,Solution 2.3
The intermediate impedance is specified as the geometric mean of the generator and source resistances. Rint
= sqrt(50*100)=70.7. (Note: this choice will minimize the overall loss of the network).
The left-hand L-network transforms from 50 ohms to 70.7 ohms. The "design Q" , QEL, is therefore QEL= sqrt(
70.7/50 -1) = 0.644. This yields Xp1 = 70.7/.644 = 110 ohms and
Xs1 = 50(.644) = 32.2 ohms. At 1.5 MHz, these produce the values Ls = 11.67 :H and
Cp = 3296 pF.
The right-hand L-network transforms from 70.7 ohms to 100. QEL, is therefore
QEL= sqrt( 100/70.7 -1) = 0.644 (the same value as for the first network). This yields Xp2 = 100/.644 = 155
ohms and Xs2 = 70.7(.644) = 45.5 ohms. At 1.5 MHz, these produce the values Ls = 16.44 :H and Cp = 2332
pF. The calculated response is shown below.
Power transfer, 0
(dB)
10. log frac
i 0.2
0.4
1 1.25 1.5 1.75 2
f
i
6
10
Freq. (MHz)
Solution 2.4
, We have two cascaded L sections so the overall efficiency, 0TOTALis the product of the two efficiencies. Here
these individual efficiencies are equal:
Therefore
and we expect the fraction of power to be transmitted through the network at the center frequency to be 98.72%.
To simulate the lossy inductors we can add to each inductor a parallel resistor of value RP = Q XP.The lossy
network is shown below.
Analyzing this network with the program of Problem 1-3 yields the predicted result: frac = .987.
Solution 2.5
The circuit is drawn below with two equivalent circuits. In Figure b. the capacitor is divided into two
capacitors: C1+C2 = C. The total load is now seen to be two matching networks in parallel. The values of these
networks can be chosen to make the circuit equivalent to Figure c. Since the same voltage, V, say, is applied
to R1 and R2 , the powers dissipated in them are, respectively, V2/R1 and V2/R2. Since the power in R1 is to
be twice the power in R2 , we must have R2 = 2R1.
