All RF Cafe Quizzes make great fodder for
employment interviews for technicians or engineers - particularly those who are
fresh out of school or are relatively new to the work world. Come to think of it,
they would make equally excellent study material for the same persons who are going
to be interviewed for a job. Bonne chance, Viel Glück, がんばろう,
buena suerte, удачи, in bocca al lupo, 행운을 빕니다,
ádh mór, בהצלחה, lykke til, 祝你好運.
Well, you know what I mean: Good luck!
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**RF Cafe Quizzes**.
Note: Some material based on books have quoted passages.
Return to RF Cafe Quiz #1 1. What is the impedance of free space? b) 120
π (376.991) ohms Impedance is defined as the square
root of the ratio of the permeability (mu, µ) to the permittivity (epsilon, ε) - in this case of free space.
ε0 = 1/(36 π) * 10^{-9} F/m µ0 = 4
π * 10^{-7} H/m z0 = sqrt [µ_{0}
/ ε_{0}] = sqrt [4 π * 10^{-7} * 36
π * 10^{9}] = sqrt [144 * 10^{2} *
π^{2}] = 120
π = 376.991 (377) Ω 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 [nf_{1}
+ (nf_{2}-1)/(gain_{1}) + (nf_{3}-1)/(gain_{1}*gain_{2}) + ... + (nf_{M}-1)/(gain_{1}*gain_{2}*...*gain_{M-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) Chebyshev
(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. Chebyshev 5. Elliptic (a.k.a. Cauer-Chebyshev, 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 V_{pk-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 : input return loss
S21 : forward gain S12 : reverse isolation (or reverse gain) S22 : 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 |