Answer to Quiz #1
1. What is the impedance of free space?
b) 120  (376.991) ohms
Impedance is defined as the square root of the ratio of the permeability (mu, ) to the
permittivity (epsilon, ) - in this case of free space.
0 = 1/(36 ) * 10-9 F/m
0 = 4  * 10-7 H/m
z0 = sqrt [0 / 0] = sqrt [4  * 10-7 * 36  * 109] = sqrt [144 * 102 * 2] = 120  = 376.991
(377) .
2. What happens to the noise figure of a receiver when a 10 dB attenuator is added at the
input?
a) Noise figure increases by 10 dB
The formula for cascaded noise figure is:
NF(M stages) = 10*log [nf1 + (nf2-1)/(gain1) + (nf3-1)/(gain1*gain2) + ... + (nfM1)/(gain1*gain2*...*gainM-1)];
where each "nf" and "gain" value is expressed as a ratio rather than in dB, and M is the total
number of stages.
The noise figure of an attenuator is equal to its insertion loss (10 dB in this case). Note that per
the equation that the noise figure of the first element in the chain is not modified by the gain of
preceding stages - as are the subsequent stages' noise figures. Therefore, any noise figure
added to the front end adds directly to the overall system noise figure - in this case an increase
of 10 dB.
3. An RF system has a linear throughput gain of +10 dB and an output 3rd-order intercept point
(OIP3) of +30 dBm. What is the input 3rd-order intercept point (IIP3)?
a) +20 dBm
A system's 3rd-order intercept points are determined by the components between its input and
output. As with the input signal power, the system's gain modifies the output intercept point
values. Simply add the gain to the IIP3 to arrive at the OIP3.
+30 dBm - 10 dB = +20 dBm.
4. Which filter type has the greatest selectivity for a given order (i.e., N=5)?
b) Chebychev (ripple=0.1 dB)
Many texts exist that list the transfer functions of the major filter types. Rather than attempt to
reiterate them all here, the following list presents them in order of increasing selectivity. The
price to be paid for increased selectivity is a greater slope in the group delay near the band
edges (bad for digital communications).
1. Bessel (Bessel-Thompson)
2. Gaussian (only slightly greater than Bessel)
3. Butterworth
4. Chebychev
5. Elliptic (a.k.a. Cauer-Chebychev, or CC)
5. Which mixer spurious product is a 5th-order product?
c) 3*LO - 2*IF
The order of any product is (±j*LO ±k*IF) simply the sum of the harmonic orders of the two
signals that create it. In this example, the 3rd harmonic of the LO (local oscillator) and the 2nd
harmonic of the IF (intermediate frequency) combine to generate the 5th-order product. The
mathematical sign of the operation does not affect the order so that Order = | j | + | k |.
Order = | 3 | + | -2 | = 5.
6. A 2.8 GHz oscillator is phase-locked to a 10 MHz reference oscillator that has a single-sided
phase noise of -100 dBc at 1 kHz offset. What is the single-sided phase noise of the 2.8 GHz
oscillator at 1 kHz offset?
a) -48.6 dBc
When an oscillator (2.8 GHz in this case) is phase-locked (PLO) to a reference source (10 MHz in
this case), the phase noise is increased in amplitude by an amount equal to 20*log(fPLO/fRef) +
2.5 dB, where the additional 2.5 dB (rule of thumb) is due to phase noise added by the phase
locking circuitry. This explains why an extremely low phase noise reference oscillator is required
when being used with a microwave frequency PLO.
-100 dBc + [20*log (2800/10) + 2.5] dB = -48.6 dBc.
7. What is the power of a 2 Vpk-pk sine wave across a 50 ohm load?
b) +10.0 dBm
Power in a sine wave is based on its rms voltage and the impedance it is imposed across. The
rms value of a sine wave's peak-to-peak value is equal to its peak value divided by the square
root of two. Rather than trying to remember whether to use 20*log or 10*log for conversion to
decibels (a real stumbling point for a lot of people), just remember to always use 10* log when
dealing with power and calculate power first using Ohm's Law (V^2 / R). Since the answer will
be in watts, you'll need to either convert to milliwatts prior to converting to dBm, or add 30 dB
to the result.
Vrms = 2 / 2 / sqrt (2) = 0.7071
P = sqr (0.7071) / 50 = 0.01 W (0.01 * 1000 = 10 mW)
10*log (0.01) = -20 dBW + 30 dB = +10 dBm
10*log (0.01 * 1000) = +10 dBm.
8. Which 2-port S-parameter is commonly referred to as "reverse isolation" in an amplifier?
c) S12
Common names for each of the four 2-port S-parameters are:
S11
S21
S12
S22
:
:
:
:
input return loss
forward gain
reverse isolation (or reverse gain)
output return loss
9. What are the minimum and maximum combined VSWR limits at an interface characterized by
a 1.25:1 VSWR and a 2.00:1 VSWR?
b) 1.60:1 (min), 2.50:1 (max)
When a signal interface is composed of two elements with differing complex impedances, part
(maybe all) of the incident signal will be reflected. Since VSWR is a scalar value, the phase
information of the reflection coefficient is lost in the conversion. Therefore a best case and a
worst case total combined VSWR is calculated as follows:
VSWR (max) = [ VSWR1 * VSWR2 ] : 1
VSWR (min) = [ VSWR1 / VSWR2 ] : 1, where VSWR1 > VSWR2.
VSWR (max) = [2.00 * 1.25] : 1 = 2.50 : 1
VSWR (min) = [2.00 / 1.25] : 1 = 1.60 : 1.
10. An ideal directional coupler has a directivity of 25 dB and an isolation of 40 dB. What is its
coupling value?
c) 15 dB
A directional coupler is characterized by five main parameters as follows:
1. Frequency band of operation.
2. Power coupling expressed as dB down from the input power level.
3. Isolation of the coupled port from the output port (essentially coupling factor from the
output port to the coupled port)
4. Directivity, which is mathematically the difference between the magnitudes of the isolation
and the coupling. If the coupler in this case had 0 dBm signals applied to both the input and
output ports, the coupled port would see -15 dBm from the input port and -40 dB from the
output port, hence, an isolation of 25 dB.
Coupling = 40 dB - 25 dB = 15 dB.
Answer to Quiz #2
1. On a Smith chart, what does a point in the bottom half of the chart represent?
b) A capacitive impedance
Points in the bottom half of the Smith chart represent capacitive impedances while points in the
top half represent inductive impedances. Both cases include a resistive component, also. Points
that lie along the center horizontal represent pure resistances.
2. While we're on the subject of Smith charts, what is the impedance of the point at the far left
edge of the center horizontal line?
b) Zero ohms (short circuit)
The Smith chart's bordering circle is the locus of points whose reflection coefficients are of
magnitude one. Here are a few of the major points on the Smith chart (50  system):
1. Left center : short circuit (0 ± j0 ).
2. Top center : pure inductive reactance (0 + j50 ).
3. Right center : open circuit (0 ± j ).
4. Bottom center : pure capacitive reactance (0 - j50 ).
5. Dead center : pure 50 ohms (50 ± j0 W).
3. A single-conversion downconverter uses a high-side local oscillator (LO) to translate the input
radio frequency (RF) to an intermediate frequency (IF). Will spectral inversion occur at IF?
a) Yes, always
Spectral inversion occurs when high frequencies within the input signal bandwidth are translated
to low frequencies in the output bandwidth, and vice versa. Since a downconversion is being
performed, the lower sideband of the mixing process is extracted, hence the difference between
the LO frequency and the RF frequency is desired. Consider the following parameters and how
spectral inversion occurs.
RF input frequency band : fc = 1250 MHz, BW = 100 MHz (1200 - 1300 MHz).
LO frequency : 1600 MHz.
IF output frequency band : fc = 350 MHz, BW = 100 MHz (300 - 400 MHz).
When the lower frequency of the input band is subtracted from the LO frequency (1600 MHz 1200 MHz = 400 MHz) a larger frequency is obtained than when the higher frequency of the
input band is subtracted from the LO frequency (1600 MHz - 1300 MHz = 300 MHz). This means
that the output spectrum is the mirror image of the input spectrum.
How to avoid spectral inversion? Always use a low-side LO (LO frequency below RF input
frequency band) for mixing, or ensure that an even number of spectral inversions are performed
in the converter (i.e., two stages of conversion with high-side LO's).
4. What happens to the noise floor of a spectrum analyzer when the input filter resolution
bandwidth is decreased by two decades?
b) 20 dB decrease
The input filter bandwidth determines the amount of power that will be present at the detector
circuitry. Since the detector performs a power integration function, it sums all of the incident
power across the band. Decreasing the bandwidth by a factor of 100 (two decades) allows one
one-hundredth of the amount of power to reach the detector, which in term of decibels is:
10*log( 1/100) = -20 dB.
5. What is a primary advantage of a quadrature modulator?
c) Single-sideband output
A quadrature modulator is comprised of two mixers, each of which receives input data and local
oscillator (LO) signals that are shifted 90 degrees relative to each other. The outputs are
summed together to generate the single-sideband signal. Deviation of the phases from the ideal
90 degrees and deviations from equal amplitudes going into the mixers will result in less than
perfect undesired sideband cancellation. Which sideband gets canceled depends on the phase
relationship of the signals entering the mixers.
The two mixer outputs are:
m1 (t) = cos (L*t) * cos (I*t) = 1/2 * cos (L*t - I*t) + 1/2 *cos (L*t + I*t)
m2 (t) = cos (L*t - pi/2) * cos (I*t -pi/2) = sin (L*t) * sin (I*t)
= 1/2 * cos (L*t - I*t) - 1/2 *cos (L*t + I*t)
Now sum the m1 (t) and m2 (t) outputs:
f (t) = 1/2 * cos (L*t - I*t) + 1/2 *cos (L*t + I*t) + 1/2 * cos (L*t - I*t)
- 1/2 *cos (L*t + I*t)
f (t) = cos (L*t - I*t)
Note that what remains is the lower sideband. Upper sideband cancellation can be achieved by
rearranging the 90 degree power splitters. If the data input is digital, the data streams can be
digitally shifted by 90 degrees and the first 90 degree power splitter can be eliminated.
6. What is meant by dBi as applied to antennas?
c) Gain relative to an isotropic radiator
An isotropic radiator (antenna) emits electromagnetic energy equally in all directions as if it
were originating from a point source. Equipotential surfaces are spheres with the isotropic
radiator at the center. If the antenna is designed to concentrate a majority of its energy in one
or more directions, it is said to be directional. Since the directional antenna radiates the same
total power as it would if it were an isotropic radiator, gain exists in the direction(s) of power
concentration. That gain is measured in decibels relative to an isotropic radiator (dBi).
7. What is the power dynamic range of an ideal 12-bit analog-to-digital converter (ADC)?
c) 72.25 dB
An ideal 12-bit ADC can assume 212 (4,096) unique voltage levels. Since power is proportional
to the square of the voltage, the maximum power sample value is 40962 (16,777,216) times the
minimum power sample value. Therefore the dynamic range is 10*log (16,777,216) = 72.25
dB.
A rule of thumb is 6 dB per bit.
8. An ideal 10 dB attenuator is added in front of a load that has a 2.00:1 VSWR. What is the
resulting VSWR of the load + attenuator?
a) 1.07:1
VSWR is related to return loss (RL) according to VSWR = [10^(RL/20) + 1] / [10^(RL/20) - 1].
It follows that increasing the return loss will result in a lower VSWR. The RL of a 2.00:1 VSWR is
9.542 dB. Add the 10 dB attenuator for a total RL of 2*10 dB + 9.542 dB = 29.542 dB. Convert
back to VSWR using the given formula for a value of 1.07:1.
Why add twice the attenuator value to the return loss? Return loss is the total decrease in signal
strength in passing through the attenuator and being reflected back through the attenuator.
Hence, the signal is decreased by twice the attenuator value.
9. What is the thermal noise power in a 1 MHz bandwidth when the system temperature is
15 °C (assume gain and noise figure are 0 dB)?
a) -114.0 dBm (in a 1 MHz BW)
Thermal noise power density is governed by the equation 10*log (k*T*B*1000) dBm, where k is
the Boltzmann constant. T is the temperature in degrees Kelvin, and B is the bandwidth in
Hertz. Multiplication by 1000 is to convert watts to milliwatts. A rule of thumb for temperatures
near 15 °C is to begin with a thermal noise density of -174 dBm/Hz, and scale accordingly (add
10 dB per decade of increased bandwidth).
10. Two equal amplitude tones have a power of +10 dBm, and generate a pair of equal
amplitude 3rd-order intermodulation products at -20 dBm. What is the 2-tone, 3rd-order
intercept point (IP3) of the system?
b) +25 dBm
2-tone, 3rd-order intermod products increase 3 dB in power for every 1 dB increase in tones
that produce them. That means the intermods increase in power at a rate of 2 dB per 1 dB
relative to the tone power. The 2-tone, 3rd-order intercept point is defined as the theoretical
point where the two original tones and the two 3-rd-order products would have equal power
(not possible in real systems due to saturation limits).
If the two original tones have a power of +10 dBm and the 3rd-order products have a power of
-20 dBm, then the intercept point will be at +10 dBm + [(+10) - (-20)]/2 dB = +10 dBm + 15
dB = +25 dBm.
Answer to Quiz #3
1. What is a primary advantage to using 90 ° (quadrature) hybrid couplers in amplifier designs?
c) Input/output impedance not dependent on devices as long as device impedances are equal.
Due to the physical construction of the quadrature coupler, as long as the two devices between
the couplers exhibit identical impedances the input and output impedances will exhibit the
intrinsic coupler impedance. For example, if matched transistors with input impedances of 12 j5  are connected between to quadrature couplers that have an intrinsic impedance of 50 + j0
, then a 50 + j0  impedance would be exhibited at the circuit input (similar for the output).
Why not always use quadrature couplers? The answer is that insertion loss, physical size and/or
cost are often intolerable.
2. Why is there a frequency term in the equation for free-space path loss?
c) Antenna geometry requires it.
Antennas are an indispensable part of all wireless systems. There is no frequency dependency in
the free-space power density equation as emitted from an isotropic radiator. Free-space power
flux density decreases with distance due to energy being spread over the surface of a sphere,
hence:
P[density] = P[transmitter] / (4  * d2) [W / m2], where d is the distance in meters from the
origin.
However, the gain of the receiving antenna, including its effective area (Ae) is:
G = G[receiver] * l2 / (4 )
Total path loss = 20 * log (4  * d / l) [dB].
3. If an amplifier has a noise temperature of 60K, what is its noise figure for an ambient
temperature of 290K?
c) 0.82 dB.
Conversion from noise temperature to noise figure is a straightforward process.
NF = 10 * log [(NT / Ta) + 1] dB, where Ta is the ambient temperature.
4. What is a primary advantage of offset-quadrature-phase-shift-keying (OQPSK) over standard
QPSK?
c) More constant envelope power.
OQPSK shifts the in-phase (I) and quadrature (Q) components of the digital data by half a bit so
that the I and Q data never change at the same moment in time. This maintains a more
constant output power.
5. A mixer has the following input frequencies: RF = 800 MHz, LO = 870 MHz. The desired
output frequency is 70 MHz. What is the image frequency?
a) 940 MHz.
By definition, the image frequency for any combination of input and LO frequencies is:
fimage = 2 * fLO - finput.
For any mixer, there are two input frequencies that, when mixed with the LO frequency, will
generate the desired output frequency. In this example, the 70 MHz output can be generated
either by taking 870 MHz - 800 MHz (desired), or by taking 940 MHz - 870 MHz (undesired).
6. What is the spurious-free dynamic range of a system with IP3 = +30 dBm and a minimum
discernible signal (MDS) level of -90 dBm?
b) 80 dB.
Spurious-free dynamic range (SFDR) is the maximum signal power above the minimum
discernible signal (MDS) power level where two tones generate 2nd-order intermodulation
products equal in power to the MDS. Input signals above that level will generate 2nd-order
products that are greater in power than the MDS power level. MDS is generally defined as the
noise power plus the minimum signal-to-noise ratio (SNR)
One form of the equation is: SFDR = 2 / 3 * (IP3 - MDS) dB.
7. A spectrum analyzer displays a component at 10 MHz @ 0 dBm, 30 MHz @ -10 dBm, 50 MHz
@ -14 dBm, 70 MHz @ -17 dBm, and all of the other odd harmonics until they disappear into
the noise. What was the most likely input signal that caused the spectrum?
a) A 10 MHz square wave (0 Vdc).
The Fourier series for a square wave with a 0 Vdc bias is the fundamental frequency and all of
its odd harmonics. Amplitudes are scaled as the reciprocal of the harmonic number; in terms of
power, the amplitudes are scaled according to 20 * log (1 / N) dB. A 10 MHz triangle wave also
contains the odd harmonics, but amplitudes fall off according to the reciprocal of the square of
the harmonic number, 40 *log (1 / N) dB.
8. On which side of a rectangular waveguide is an E-bend made?
b) The short dimension.
In a rectangular waveguide, the E-plane is in the direction of the short dimension while the Hplane is in the direction of the long dimension. The type of bend is determined by which side is
curved for the bend. A useful mnemonic is the short dimension is the [E]asy side to bend, while
the long dimension is the [H]ard side to bend.
9. During a network analyzer calibration, why are both a short circuit and an open circuit used?
b) To determine the characteristic impedance of the measurement system.
In order for the network analyzer (N/A) to make an accurate measurement, it must know what
the impedance of the measurement system is. Characteristic impedance is mathematically the
square root of the product of the short circuit impedance and the open circuit impedance. The
S/A exploits this relationship.
10. What is the first harmonic of 1 GHz?
a) 1 GHz.
Harmonic number is often mistaken for overtone number. The second harmonic of 1 GHz is 2
GHz, while the first overtone frequency of 1 GHz is 2 GHz. In other words, Nharmonic = Novertone +
1. The first harmonic of any frequency is its fundamental frequency.
Answer to Quiz #4
1. Which of the following can cause frequency intermodulation products in a system?
c) Both a) and b)
Intermodulation products are generated whenever currents of different frequencies flow in a
nonlinear junction. Dissimilar metals and impurities at connector interfaces and in cable
dielectric materials form nonlinear junctions. Ferromagnetic materials in isolators and
circulators enter into a nonlinear region when in saturation. At low powers, the products all
have powers below thermal noise, but when used in high power transmission systems, these
Passive Intermodulation (PIM) products can and often do fall within the receiver band. Read
more here.
2. What is the melting temperature of standard 60/40, tin/lead solder?
c) 186°C (386°F)
Knowing the melting temperature of the solder you use all the time is useful information.
Look here for a list of other solder alloy melting temperatures.
3. What is the frequency band for the 900 MHz GSM cellular band?
a) Tx: 880-915 MHz / Rx: 925-960 MHz
Click here for a matrix of wireless communications bands and protocols.
4. What does GSM stand for?
c) Global System for Mobile Communications
Click here for a matrix of wireless communications bands and protocols.
* GSM originally stood for Groupe Spécial Mobile - thanks to Eric for this info.
5. What does POTS stand for (in communications)?
a) Plain Old Telephone System
Maybe it's too simple to be true, but it is.
6. Which of these pairs of materials in the triboelectric series have the greatest charge
transfer potential?
c) Glass & Hard Rubber
The triboelectric series is a list of materials that, when rubbed together, transfer charge from
one to the other. The farther apart the two materials are on the series, the greater the charge
that is transferred. See here for a table of common materials.
7. Along which side of rectangular waveguide is an "E" bend made?
b) Shorter side
A common mnemonic employed to help remember is that an E-Bend is bent in the Easy
direction (along the short side). This is the direction of the E-field in the TE10 mode.
Conversely, An H-Bend is bent in the Hard direction (along the long side). This is the
direction of the H-field in the TE10 mode.
8. What is the lowest modulation index at which an FM carrier is suppressed?
a) 2.40
Depending on the modulation index (m) chosen, the carrier and certain sideband frequencies
may actually be suppressed. Zero crossings of the Bessel functions, Jn(b), occur where the
corresponding sideband, n, disappears for a given modulation index, b. The carrier is the 0th
sideband, so n=0. The next time the carrier disappears is for m=5.49. Here is more
information.
9. How much current is required through the human body to cause an onset to muscular
paralysis during electrocution?
b) 21 mA
Most non-EE's think voltage causes electrocution, but we know better (don't we?). A higher
voltage causes a higher current to flow, but ultimately it is the current that cause body
tissues to react. Click here for more current levels.
10. At what frequency is electromagnetic energy maximally absorbed due to oxygen in the
atmosphere?
c) 63 GHz
Both water and oxygen absorb electromagnetic energy in the atmosphere. Oxygen has its
first peak at around 22 GHz, but a much larger peak occurs at 63 GHz due to oxygen. Secure
communications between satellites in orbit use the 63 GHz band to prevent interception by
terrestrial receivers. Clever - eh? Check out this absorption chart.
Answer to Quiz #5
Agilent Technologies
Channel Microwave
Maury Microwave
Sage Labs
National Instruments
Synergy Microwave
Rohde & Schwarz
Piconics
Scientific Atlanta
Marki Microwave
Anadigics
Andrew Corporation
Analog Devices
Connecticut Microwave
Hittite Microwave
Datel
RF Micro Devices
Power Cube
Miteq
Power One
National Semiconductor
Vicor
Remec
Cinch
Stanford Microdevices
Delta
Texas Instruments
Huber+Suhner
Atlantic Microwave
Ansoft
Merrimac Industries
Intusoft
Trilithic
Applied Radio Labs
American Technical Ceramics
(ATC)
General Microwave
Dielectric Laboratories
Cypress
Semiconductor
Johanson Dielectrics
GHz Technology
Voltronics
Dallas Semiconductor
Answer to Quiz #6
1. Which of the following WLAN standards is on a different frequency band than the others?
a) 802.11a
802.11b/g/n are all in the 2.4 GHz ISM band, 802.11a is in the 5 GHz ISM band.
2. What does the term "ruggedness" refer to in wireless power amplifiers?
c) Ability to withstand load mismatch
A typical ruggedness spec is no damage into a 10:1 VSWR load
3. Which FCC regulation governs the unlicensed ISM band?
a) Part 15
FCC Part 15 governs the unlicensed ISM band for intentional, unintentional, and incidental
radiators.
4. In which semiconductor technology are the majority of cellphone PAs manufactured?
c) GaAs/InGaP
Many WLAN PAs are built in SiGe and Si, and one or two experimental phone PAs are being
built in Si, but the vast majority of phone PAs are built in GaAs HBT or InGaP HBT.
5. What is a major advantage of LTCC substrates?
d) All the above
Because of the high dielectric constant of the material, relatively large distributed components
can be integrated into the substrate, which decreases the number of discrete components
needed. The higher thermal conductivity also helps with heat dissipation.
6. Which phone standard supports the highest data rate?
c) EDGE
iDEN=64kbps, GPRS=21.4kbps/time slot, EDGE=384kbps, GSM=14.4kbps
7. Which component is typically not part of a front-end module (FEM)?
a) Power amplifier
Transmit modules (TxM) incorporates a FEM + PA
8. Which two systems are most likely to experience concurrent operation problems?
b) Bluetooth + WLAN
Bluetooth and WLAN (802.11b/g/n) both operate on the 2.5 GHz ISM band.
9. An isolator is typically required at the output of the PA for which transmitter system
d) CDMA/W-CDMA
CDMA/W-CDMA power amplifiers, due to their highly linear region of operation, cannot tolerate
a very high mismatch (VSWR) and must be protected with an isolator.
10. What is the commonly claimed nominal operational range for Bluetooth?
d) All the above
Sure it's stupid (and true), but a little levity in an interview can lighten the load.
Answer to Quiz #7
1. What is a “radar mile?”
c) 12.36 µs
A radar mile is the time required for a signal leaving the antenna to go out and back one
nautical mile (6076 feet). 2 * 6076 ft / [9.8356e8 ft/s] = 12.36 µs
2. Which best describes a bi-static radar?
a) Fixed transmitter and fixed receiver at different locations
Many bi-static systems use multiple receiving sites to be able to pick up weaker reflected signals
and to correlate position information with more certainty.
3. What is a radar cross-section (RCS)?
b) A target’s reflecting capacity equivalent to a perfectly reflecting surface of the same area
For radar evasion (stealth), a low RCS is desirable. To guarantee being detected (general &
commercial aviation, a high RCS is desirable. A B-52 bomber has an RCS of about 100 m2 (20
dBsm, whereas an F-117 Stealth Fighter has about a 0.003 m2 RCS (which at X-band puts it in
the realm of birds).
4. What are common units of radar cross-section (RCS)?
a) dBsm
Decibels relative to a square meter.
5. Who is known as ”The Father of Radar?”
a) Robert Watson-Watt
Sir Robert Watson-Watt, with the help of his assistant Arnold Wilkins, drafted, in February 1935,
a report titled "The Detection of Aircraft by Radio Methods." Walter Eugene O'Reilly was "Radar"
from the television show M*A*S*H.
6. Which flying (movement) condition will always result in a Doppler speed of 0 m/s?
b) Perfect concentric circle around antenna at constant altitude
Doppler speed is manifested on the boresight radial line (main radiation lobe) so, neglecting
“blind speeds,” the only way to be moving and have the Doppler speed at 0 m/s is to not have
any motion relative to the boresight radial line.
7. What does “MTI” stand for?
c) Moving Target Indication
Moving Target Indication (MTI) uses a cancellation system that blanks out stationary targets to
permit only moving targets to be displayed on the display. Doing so greatly reduces the clutter
and allows relatively weak moving targets to be seen in the midst of buildings, trees, etc.
8. Synthetic Aperture Radar (SAR) radar is mostly likely to be located on which platform?
c) Airplane
Synthetic Aperture Radar (SAR) relies on the motion of the platform to produce the “scan” that
would otherwise be provided by a rotating or steerable antenna.
9. Which feature of a “stealth” aircraft is most responsible for its low observability?
b) Multi-faceted surfaces
The angle of a signal’s reflection form a surface is equal to the angle of its incidence on the
surface. By minimizing the area presented perpendicular to any given direction (particularly
those direction most likely to be illuminated, like from below and the side), a minimum amount
of incident radar signal energy will be reflected back in the direction from which it originated
(the radar).
10. What kind of radar did the webmaster of RF Cafe work on while in the U.S.A.F?
a) MPN-14
In the U.S. government system designation standard, the "M" stands for ground Mobile, the "P"
stands for Radar, and the "N" stands for Navigational aides. The "-14" designates position in a
series.
Answer to Quiz #8
1. In which decade was the transistor invented?
a) 1940s
On December 23, 1947, William Shockley, Walter Brattain and John Bardeen, of Bell Labs,
announced their discovery of the point-contact germanium transistor to management.
2. In which decade was the telegraph invented?
b) 1840s
On May 24, 1844, Samuel Morse send his famous message, "What hath God wrought?"
3. In which decade was the Internet first implemented?
b) 1960s
The first message ever sent over what is now called the Internet took place at 10:30PM on
October 29, 1969. Back then, the Department of Defense called it ARPAnet (Advanced Research
Projects Agency network).
4. In which decade was the first solid state integrated circuit demonstrated?
a) 1950s
On September 12, 1958, Jack Kilby demonstrated the first working IC while working for Texas
Instruments, although the U.S. patent office awarded the first patent for an integrated circuit to
Robert Noyce of Fairchild.
5. In which decade were the first successful diode and triode vacuum tubes invented?
c) 1900s
In 1904, John Ambrose Fleming invented the first practical electron tube called the 'Fleming
Valve', which is a diode rectifier. In 1906, Lee de Forest invented the audion later called the
triode, which provided signal amplification.
6. In which decade was the telephone invented?
c) 1870s
Alexander Graham Bell's notebook entry of 10 March 1876 describes his successful experiment
with the telephone. Speaking through the instrument to his assistant, Thomas A. Watson, in the
next room, Bell utters these famous first words, "Mr. Watson -- come here -- I want to see
you."
7. In which decade was the AEEE (now the IEEE) founded?
a) 1880s
The IEEE (Institute of Electrical and Electronics Engineers) was formed in 1963 by the merger of
the Institute of Radio Engineers (IRE, founded 1912) and the American Institute of Electrical
Engineers (AIEE, founded 1884).
8. In which decade with the first transatlantic radio broadcast occur?
c) 1900s
On December 12, 1901, a radio transmission received by Guglielmo Marconi resulted in the first
transmission of a transatlantic wireless signal (Morse Code) from Poldhu, Cornwall, to St.
John's, Newfoundland.
9. In which decade was the SPICE simulator introduced?
b) 1970s
SPICE (Simulation Program with Integrated Circuit Emphasis) was introduced in May 1972 by
the University of Berkeley, California.
10. In which decade was the ARRL founded?
a) 1910s
On April 6, 1914, Hiram Percy Maxim proposed the formation of the American Radio Relay
League.
Answer to Quiz #9
1. Where did Bluetooth™ get its name?
b) In honor of Harald Blåtand, once king of Denmark
Bluetooth was named by Ericsson (inventor of BT) after Danish King, Harald Blåtand (Bluetooth
in English), who lived in the latter part of the 10th century. According to lore, he ate so many
blueberries that his teen turned blue (no kidding).
2. Where did ZigBee get its name?
b) From the zigzag path of a bee
According to authorities on the matter, the system was so named because it allowed networked
devices to swarm around each other and stay connected, like worker bees around a hive.
3. Who is credited with conceiving of spread spectrum radio communications?
c) Actress Hedy Lamarr
1930s actress Hedy Lamarr, “The Mother of Spread Spectrum,” is widely credited with having
introduced the concept of spread spectrum radio communications as an application for thwarting
jamming on guided torpedoes during WWII. The work of Lamarr’s (and her piano instructor
George Antheil) culminated in U.S. Patent 2,292,387, “Secret Communication System,” granted
on August 11, 1942. Those were the days when Hollywood stars were patriotic and heroic,
rather than being the cowardly traitors of today.
4. What is meant by the front-to-back ratio of a Yagi antenna?
c) Power radiated in the front main lobe vs. power in opposite direction
The series of driven (radiator) and reflector (director) elements in the Yagi design produce a
radiation pattern that is concentrated in one direction (directivity).
5. In an FM modulator with a 10 kHz deviation and a 5 kHz maximum modulating frequency,
what is the total occupied bandwidth?
d) 30 kHz
The maximum excursion on either side of carrier frequency (both above and below) is the sum
of the deviation and the modulating frequency (10 kHz + 5 kHz = 15 kHz), so, total occupied
bandwidth is twice that amount (lower + upper) of 30 kHz.
6. Which WLAN standard provides the highest data rate?
d) IEEE802.11n (2.4 GHz RF)
The spec data throughputs are as follows: 802.11a = 54 Mbits/s, 802.11b = 11 Mbits/s,
802.11g = 54 Mbits/s, 802.11n = 100+ Mbits/s (Pre-n systems delivering this rate, but yet-tobe-finalized spec calls for up to 600 Mbits/s).
7. Why might the mounting orientation of a surface mount capacitor affect frequency response?
a) The plates in the body could be either parallel to or perpendicular to the PCB, affecting
coupling
Most surface mount capacitors are of multi-layer construction with alternating conductive plates
and insulating (dielectric) layers. Edge fringing effects and the proximity to adjacent
components (including the substrate) will affect the effective capacitance depending on whether
the plates happen to be mounted parallel to or perpendicular to the those components.
8. If you were handed an unprocessed wafer of gallium arsenide (GaAs), silicon (Si), silicongermanium (SiGe), and gallium nitride (GaN), how would you know which is GaN?
a) GaN is transparent and the others are not
The wide band gap energy of GaN renders it transparent to visible light – it looks like glass. The
other wafers are all very dark in color.
9. The Smith Chart plot of a 50 ohm cable (in a 50 ohm system) spirals inward as the
impedance is plotted through multiple cycles. What is that indicative of?
a) A lossy cable
Attenuation in the cable increases the resistive component of the cable as the length increases.
10. What are the three primary JEDEC models used for ESD testing?
c) Human Body (HBM), Machine (MM), and Charged Device (CDM) Model
JEDEC (Joint Electron Device Engineering Council) specifies the Human Body Model in
EIA/JESD22-A114-x (electrically simulates the discharge RC network of the human body), the
Machine Model in EIA/JESD22-A115-x (the discharge path of a grounded machine), and the
Charged Device Model in EIA/JESD22-C101-x (the discharge path of device isolated by its
package).
Answer to Quiz #10
1. What format would a near-filed communications (13 MHz variety) antenna most likely take?
a) Inductive coil.
Most devices on the market today use an inductive coil to exploit the principle of magnetic
induction whereby a current-carrying conductor in motion relative to another conductor induces
a similar current.
2. What does SOLT stand for?
c) Short, Open, Load, Through.
In order to perform a full 2—port calibration on a vector network analyzer (VNA), it is necessary
to calibrate with both test cables using a certified set of adapters and terminations that meet
industry specifications. The VNA then calculates the set of 12 error correction terms necessary
to subtract out the effects of the system, including the test cables.
3. Which instrument would be best to use to locate a defective waveguide joint?
d) Time Domain Reflectometer (TDR).
The TDR sends a pulse of energy down the line and measures the length of time the reflected
signal takes to return. A region of poor VSWR will reflect a portion of the signal energy that is
dependent upon the degree of mismatch. If the entire length of waveguide and the termination
are properly matched, there will be no returned (reflected) signal. Many network analyzers have
this built-in capability.
4. Which entity in the U.S. determines whether an RF energy-emitting device is allowed to be
operated?
a) The Federal Communications Commission (FCC).
Whether the radiation be intentional or unintentional, all products for commercial and private
use must pass emissions testing as specified by the FCC.
5. What does 2G, 2.5G, 3G, etc., mean in reference to cellphones?
b) The “generation” of the technology.
1G was the original analog phones. 2G introduced digital technology. 3G ushered in high
bandwidth data along with voice, but it was late to arrive, so 2.5G filled the gap. 4G is now in
the works.
6. Where would you be likely to find a free wireless Internet connection?
d) All the above.
"a" and '"b" are obvious. See my Factoid for "c."
7. Who hosts the MTT-S International Microwave Symposium?
d) The Institute of Electrical and Electronics Engineers (IEEE).
Since 1958, the IEEE Microwave Theory and Techniques Society (MTT-S), has put on the show.
8. What is the “rule of thumb” for estimating RF signal propagation distance vs. time in free
space?
c) 1 foot per nanosecond.
Electromagnetic energy travels about one foot in one nanosecond in free space (actually
1.01670336 ns), and in one nanosecond, it travels about one foot (actually 0.98357106 ft). The
other three choices do not produce such close approximations (1 mm = ,3.3 ps, 1 m = 3.3 ns, 1
in = 85 ps).
9. What is the “rule of thumb” for estimating RF frequency vs. wavelength in free space?
a) 300 MHz = 1 meter.
More precisely, 300 MHz has a wavelength of 0.999308193 meters, but the error is about
0.07% - close enough. 100 MHz ≈ 3 m, 300 MHz ≈ 3.28 ft, 100 MHz ≈ 9.84 ft.
10. What is the most unique feature of a Helmholtz coil?
b) Magnetic field lines are extremely uniform within the coil.
This property makes the Helmholtz coil configuration very useful when testing magnetic
properties since a device under test can be placed within the coil to free it from influences of
outside magnetic field variations. If you chose “A,” you are thinking of a Tesla Coil.