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Answers to RF Cafe Quiz #1

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| 29 | 30 | 31 | 32 | 33 | 34 | 35 | 36 | 37 | 38 | 39 | 40 | 41 |
| 42 | 43 | 44 | 45 | 46 | 47 | 48 | 49 | 50 | 51 | 52 | 53 | 54 |
| 55 | 56 | 57 | 58 | 59 | 60 | 61 | 62 | 63 |

All RF Cafe quizzes would make perfect 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.

Some of these books used in quizzes are available as prizes in the monthly RF Cafe Giveaway.

Note: Many answers contain passages quoted in whole or in part from the text.


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 [µ0 / ε0] = sqrt [4 π * 10-7 * 36 π * 109] = sqrt [144 * 102 * π2] = 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 [nf1 + (nf2-1)/(gain1) + (nf3-1)/(gain1*gain2) + ... + (nfM-1)/(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) 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 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 (V2 / 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
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