Exam text content

ELT-47206 Basics of RF Engineering, TUT — 12.9.2017 10of5

 

Exam
Examiner: Olli-Pekka Lunden

Students may use own calculators (including programmable types) and any hand
written or printed materials, including books, lecture materials, and own notes, but
not mobile phones, laptops, nor any other communication devices. Students may
keep the problem sheet. Evaluation shall be based on the evidence student demon-
strates about the learning outcome statements of this course.

Problem 1: The following elaims may contain inaccuracies or falsities. First you must
identify them and explain what is wrong. Then you must provide a corrected!
claim. If the claim is already correct, you must state this to be the case, and
not comment or edit it.

(a) The wavelength of an RF signal (voltage or current) at 3 GHz is 10
cm in a coaxial cable, such as the one shown in Figure 1.

 

Figure 1: Coaxial cable Huber+Suhner Sucoflex 104. The insulator of cables is
foamed polytetrafluoroethylene, that has effective relative permittivity of €, = 1.7.
The cable loss is less than 1.1 dB/m up to about 18 GHz.

(b) Consider two microstrip lines on a 1-mm thick alumina substrate (A1,03,
€, = 9.6) with different widths: wj, = | mm and w, =2 mm. Suppose
their electrical length is the same, 0, = 0, =/4. The line with greater
width is longer when measured in millimeters.

(c) A two-port (Figure 2) is described by its S-parameters that are refer-
enced to 50 0. With the given component values, Sj, = 0at | GHz.
The $>> -parameter of this circuit is also zero, moreover, its trans-
ducer power gain and reverse isolation are very close to 0 dB at 1
GHz. Hint: You will not have to calculate anything.

"General instructions: Corrections should be made with minimum editing. However just negating
a sentence is not appreciated. If the claim is "Bats are birds”, you should not correct it as "Bats are
not birds” but rather as "Bats are mammals”. The actual claim is in bold-font. The normal-font text
should always be correct.
ELT-47206 Basics of RF Engineering, TUT — 12.9.2017 20fsS

 

100 k

 

Figure 2: A two-port (within the dashed line) is connected to a high-resistance load.
Each capacitor is 3.2 pF.

(d) Consider now the input reflection coefficient (T;n) of the two-port in
Figure 2. Now, because the load resistance is high, |T',| = 1, at all
freguencies.

(e

A particular transistor (BFG520) is going to be used for a single-transistor,
common-emitter RF amplifier. The amplifier is supposed to be uncondi-
tionally stable. Figure 3 shows a typical Edwards-Sinsky stability factor
Wof this transistor. There is no need for stabilization as long as the
operating freguency is anything from 1 to 2 GHz and the DC oper-
ating point is V,, =6 V/1. = 30 mA.

 

Mu1

 

 

 

aa mihmäiktea? 11 Sith
0.0 0.2 04 06 08 10 12 14 16 18 2.0

freg, GHz

Figure 3: Typical stability factor u of BFG520 transistor from 40 MHz to 2 GHz in
the DC operating point of V,, = 6 V and 1, = 30 mA.

(f) The reflection coefficient I of a load is indicated by a red circle in
Figure 4. This load can be matched to 50 ohms using two capacitors,
one series-connected and one parallel-connected, in either order, as
suggested in Figure 5.

Each subproblem yields 1 point at max.
ELT-47206 Basics of RF Engineering, TUT — 12.9.2017 30f5

 

 

Figure 4: A simplified Smith chart. T7 indicates a load reflection coefficient. As-
sume the chart uses 50 ohms as the reference impedance.

Figure 5: Suggested matching circuits for I7.

Problem 2: Still considering the load reflection coefficient I'7, in Figure 4:

(a) What is the respective load impedance? (7 point)

(b) Which one of the generators shown in Figure 6 is the one that provides
more power to this load? Consider this problem without any impedance
matching networks inserted, just having a generator directly connected
to the load. (7 point)

(c) How many microwatts is that power? (2 points)

(d) How high will the peak-to-peak voltage across the load be in that case?
(2 points)

(50 +550) 2 (50 — j50) 2

E, = 40 mV pp E; = 40 mV pp

Figure 6: Two sinewave generators. Note that the electromotive force is a peak-to-
peak value, 2 times the r.m.s. value.
ELT-47206 Basics of RF Engineering, TUT — 12.9.2017 40f5

 

Problem 3: A simple impedance matching topology of Figure 7 uses a segment of trans-
mission line and a shunt capacitor. Suppose the characteristic impedance of
the transmission line is 50 O.

Plot, on the accompanying Smith chart, either the locus or the area of nor-
malized impedances zz = Z7 /Zo that this impedance matching topology can
match to 50 2 if C were arbitrary but the transmission line length were fixed
to 42/8.

79 =500,1=/8

Figure 7: A simple impedance matching circuit with load (Zz). Suppose the trans-
mission line length / is fixed but C is arbitrary.

Problem 4: An oscilloscope and a spectrum analyzer are used to simultaneously monitor
fairly low-freguency signals generated by a 50-ohm source as illustrated in
Figure 8.

Give a reasonable estimate for the upper freguency limit of this measurement
setup.

The input impedance of the spectrum analyzer is 50 & while that of the oscil-
loscope is high. Oscilloscope is connected directly, without a probe head. All
cables are of the Huber+Suhner Sucoflex 104 type, each 120 cm long, also
shown in Figure 1.

 

 

 

 

 

 

 

spectrum -
ralyzer oscilloscope
(50 0) (high impedance)
cable B
cable C
signal
source
(50 0) cable A

 

 

 

Figure 8: The test setup uses three 50-92 cables each 120 cm long and a T-junction.
RADIALLY SCALED PARAMETERS

x

 

 

 

 

 

 

TOWARD LOAD —: < TOWARD GENERATOR
4 3,35 2 18, 1614 1211115, 10) 7 3.4 3 3 i
—
15 0 08 6 84 32 104f 10712 1314 16 18.2 3 45 Pm -
ISKI JI W 2 W 20 300 01 02 — 04 06 08 1.15. 2 3.4 3,6, , 10 15=
05 04 03 = 02 01 005 001' ojo 11 12 13 14 15 1617 18192 253 45 To
07,06 108, 0, 04 0,03 002,01 [1 0 095. 09 08 — 07 — 06 05,04 03 02 01.0
x
CENTER
00 8 LO L, 03 OA 1,08, L, 08 |, 07 08 0,09 1 12 118 1, 819 2
ORIGIN

 
 

RADIALLY SCALED PARAMETERS

 

 

 

 

ERE

 

 

TOWARD LOAD. < TOWARD GENERATOR. '
=10040, 0,10 5.4 328 2 NS 1614 12 LT N 110 7 4 3 3 1
an ,
40 3 » 15 v 8 es aj m TR 13:14 1618 2 5 s
o ! ASSI JIM R 20 30e 01 02 — 04 05 08 1.15. 2 3 3.6, , 0150
1 0908 07 06 05 04 03 — 02 01 0us 0" ojan 1 TT 015 14 15 1617 18192 25 3 715 To
1 0908 07, 506,00 08 0,04, 0003 0,02 00, ]jI 0 098, 09 08 , 07 06,05, 04 03 02 01,0
x
CENTER
PAO J OAK KAN JNKU 1 A NI 2
ORIGIN