**Return to RF Cafe Quiz #3**

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.

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 Ω are connected between to quadrature couplers that have an intrinsic impedance of 50 + j0
Ω, then a 50 + j0 Ω 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

π
* d

^{2}) [W / m

^{2}], where d is the distance in meters from the origin.

However, the gain
of the receiving antenna, including its effective area (A

_{e}) is:

G = G

_{[receiver]}
* l

^{2} / (4

π)

Total path loss = 20 * log (4

π * 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 / T

_{a}) + 1] dB,
where T

_{a} 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:

f

_{image}
= 2 * f

_{LO} - f

_{input}.

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?

a) 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 bias).

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 H-plane 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, N

_{harmonic}
= N

_{overtone} + 1. The first harmonic of any frequency is its fundamental frequency.