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1996 -
2022

Webmaster:

Kirt Blattenberger,

BSEE
- KB3UON

RF Cafe began life in 1996 as "RF Tools" in an AOL screen name web space totaling 2 MB. Its primary purpose was to provide me with ready access to commonly needed formulas and reference material while performing my work as an RF system and circuit design engineer. The Internet was still largely an unknown entity at the time and not much was available in the form of WYSIWYG ...

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Answers to RF Cafe Quiz #2

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.

**Click here for the complete list of *** RF Cafe
Quizzes*.

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

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 Ω system):

1. Left center : short circuit (0 ± j0 Ω).

2. Top center : pure inductive reactance (0 + j50 Ω).

3. Right center : open circuit (0 ± j∞ Ω).

4. Bottom center : pure capacitive reactance (0 - j50 Ω).

5. Dead center : pure 50 ohms (50 ± j0 Ω).

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 (ωL*t) * cos (ωI*t) = 1/2 * cos (ωL*t - ωI*t) + 1/2 *cos (ωL*t + ωI*t)

m2 (t) = cos (ωL*t - pi/2) * cos (ωI*t -pi/2) = sin (ωL*t) * sin (ωI*t)

= 1/2 * cos (ωL*t - ωI*t) - 1/2 *cos (ωL*t + ωI*t)

Now sum the m1 (t) and m2 (t) outputs:

f (t) = 1/2 * cos (ωL*t - ωI*t) + 1/2 *cos (ωL*t + ωI*t) + 1/2 * cos (ωL*t - ωI*t) - 1/2 *cos (ωL*t + ωI*t)

f (t) = cos (ωL*t - ωI*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 2

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.