Cascaded microwave stages use a 10 dB first-stage gain, 2 dB first-stage noise figure, and 6 dB second-stage noise figure. What is total noise figure?
A 2.0 dB
B 2.8 dB
C 3.0 dB
D 6.0 dB
Answer: B. 2.8 dB. Convert to noise factors: F1≈1.585 and F2≈3.981. Friis gives Ftotal=F1+(F2−1)/10≈1.883, or about 2.8 dB.
Measured at an amplifier output, the carrier is 0 dBm and a third-order product is −30 dBm. What is OIP3?
A +10 dBm
B +30 dBm
C +15 dBm
D +45 dBm
Answer: C. +15 dBm. The carrier-to-IM3 separation is 30 dB. The third-order intercept lies half that distance above the carrier: 0+30/2=15 dBm.
Mixing tones at 10.000 GHz and 10.010 GHz produces a lower third-order spur at what frequency?
A 9.980 GHz
B 9.990 GHz
C 10.000 GHz
D 10.020 GHz
Answer: B. 9.990 GHz. The lower third-order product is 2f1−f2. Here it equals 2(10.000)−10.010=9.990 GHz.
For a 1.2 m waveguide with group velocity 0.8c, what is its group delay?
A 2 ns
B 5 ns
C 4 ns
D 8 ns
Answer: B. 5 ns. Group delay is length divided by group velocity: 1.2/(0.8×3×10⁸)=5 ns.
When a resonator has Q0=600 and external Qe=1200, what is its loaded Q?
A 200
B 600
C 900
D 400
Answer: D. 400. The reciprocal relation is 1/QL=1/Q0+1/Qe. Thus QL=(1/600+1/1200)⁻¹=400.
With Q0=800 and Qe=1600, how is the resonator coupled?
A Critically coupled
B Overcoupled
C Undercoupled
D Uncoupled
Answer: C. Undercoupled. The coupling factor β=Q0/Qe=0.5. Since β<1, the resonator is undercoupled.
At critical coupling in a one-port resonator, what happens at resonance?
A Reflection reaches zero
B Power doubles
C Q becomes infinite
D Phase stays constant
Answer: A. Reflection reaches zero. At β=1, the reflected wave cancels at resonance in the ideal one-port case, giving zero input reflection.
For a 25 Ω load on a 50 Ω line, what is the voltage reflection coefficient?
A −1/2
B +1/3
C +1/2
D −1/3
Answer: D. −1/3. Γ=(ZL−Z0)/(ZL+Z0)=(25−50)/(25+50)=−1/3.
Reciprocity in a linear two-port network requires which S-parameter relation?
A S11 = S22
B S21 = S12
C S11 = S21
D S12 = S22
Answer: B. S21 = S12. For a reciprocal two-port with the same reference convention, the forward and reverse transmission coefficients are equal.
In TRL calibration, which reflect-standard property need not be known exactly?
A Reflection magnitude
B Reflection phase
C Port identity
D Line impedance
Answer: A. Reflection magnitude. The reflect standard must be the same at both ports and highly reflective, but its exact magnitude need not be known. Its phase must be known within one-quarter wavelength.
A valid TRL line standard must differ from the thru in what way?
A Connector color
B Electrical length
C Cable jacket
D Port number
Answer: B. Electrical length. TRL requires the line and thru to have different electrical lengths; an identical length would not provide the needed phase distinction.
Group delay is defined from a network’s transmission phase as what?
A Positive amplitude slope
B Input return loss
C Power ratio
D Negative phase slope
Answer: D. Negative phase slope. Group delay is the negative derivative of phase with respect to angular frequency: τg=−dφ/dω.
YIG filter center frequency is tuned primarily by changing what?
A Junction capacitance
B Guide width
C Magnetic bias field
D LO power
Answer: C. Magnetic bias field. A YIG resonator’s ferrimagnetic resonance frequency shifts with applied magnetic bias, enabling electronic tuning.
In a magnetron operating in the π-mode, adjacent resonant cavities have what phase difference?
A 0 degrees
B 90 degrees
C 180 degrees
D 270 degrees
Answer: C. 180 degrees. The π-mode means neighboring anode cavities oscillate with a phase difference of π radians, or 180 degrees.
Wideband gain in a TWT requires synchronism between the electron beam and which wave property?
A RF phase velocity
B RF group delay
C Guide cutoff
D Cavity Q
Answer: A. RF phase velocity. Efficient traveling-wave interaction requires the RF wave’s phase velocity to be close to the electron-beam velocity.
Directional ferrite devices become nonreciprocal when subjected to what?
A Acoustic drive
B Optical pumping
C Static magnetic bias
D Thermal cycling
Answer: C. Static magnetic bias. A DC magnetic bias magnetizes the ferrite and produces different wave behavior in opposite directions.
Among circular-waveguide modes, which has the lowest cutoff frequency?
A TE11
B TM01
C TE21
D TM11
Answer: A. TE11. For a hollow circular guide, TE11 has the lowest cutoff and is the dominant mode.
For an air guide with a=40 mm and b=20 mm, what is the single-mode frequency range?
A 1.88–3.75 GHz
B 3.75–7.5 GHz
C 5–10 GHz
D 7.5–15 GHz
Answer: B. 3.75–7.5 GHz. TE10 cuts on at c/(2a)=3.75 GHz. The next modes TE20 and TE01 cut on at 7.5 GHz.
Near a waveguide mode’s cutoff, what happens to its group velocity?
A It approaches c
B It doubles
C It approaches zero
D It becomes negative
Answer: C. It approaches zero. For a propagating waveguide mode, group velocity falls toward zero as frequency approaches cutoff from above.
An RF input is −10 dBm and its IF output is −17 dBm. What is mixer conversion loss?
A 3 dB
B 10 dB
C 17 dB
D 7 dB
Answer: D. 7 dB. Conversion loss is input RF power minus output IF power in dB: −10−(−17)=7 dB.
Double-balanced mixers are often selected to suppress which unwanted signal?
A Guide cutoff
B Cavity ringing
C Thermal drift
D LO feedthrough
Answer: D. LO feedthrough. A double-balanced topology cancels common feedthrough terms, including much of the LO and RF leakage.
The 1 dB compression point marks where amplifier gain has fallen by how much?
A 1 dB
B 0.1 dB
C 3 dB
D 10 dB
Answer: A. 1 dB. At P1dB, the measured gain is 1 dB below its small-signal value because the amplifier is entering compression.
Time-domain gating on a VNA can isolate which feature?
A Noise figure
B LO frequency
C DC offset
D A delayed reflection
Answer: D. A delayed reflection. A frequency sweep can be transformed to time domain; gating then retains or removes responses from selected delay regions.
When fixture S-parameters are known, de-embedding chiefly removes what?
A Fixture effects
B Source harmonics
C Thermal noise
D Port mismatch
Answer: A. Fixture effects. De-embedding mathematically removes the characterized fixture response so the DUT behavior can be estimated at its terminals.
Reducing a VNA’s IF bandwidth usually trades sweep speed for what?
A Lower measurement noise
B Wider frequency span
C Higher output power
D More harmonics
Answer: A. Lower measurement noise. A narrower IF bandwidth reduces integrated receiver noise, but it takes longer to settle at each frequency point.