Advanced Electronics & Communication interview questions

125 questions with answers (ECE-301 to ECE-425). Try each one aloud before reading the answer.

Signals & Systems

ECE-301 · A bandpass signal occupies 20–25 MHz. What is the minimum sampling rate without aliasing?

Use bandpass sampling: fs_min = 2fH/⌊fH/B⌋ = 50/⌊25/5⌋ = 10 MHz, well below the 50 MHz lowpass Nyquist rate. Because 10 MHz sits exactly at the edge of the valid range, a practical design uses guard bands and a slightly different fs.

ECE-302 · Classify the modulator y(t) = x(t)·cos(ω₀t).

It is linear (superposition holds), time-varying (shifting x does not just shift y because the cosine is fixed in time), causal and BIBO stable. It is therefore not LTI and cannot be described by an impulse response alone.

ECE-303 · An RC low-pass filter has R = 1 kΩ and C = 1 µF. Find its cutoff and 10–90% rise time.

f₃dB = 1/(2πRC) = 1/(2π × 10⁻³) ≈ 159 Hz. Rise time tr ≈ 2.2RC = 2.2 ms, which matches tr ≈ 0.35/f₃dB.

ECE-304 · An oscilloscope has 100 MHz bandwidth. What rise time can it show, and how does this affect a measurement?

Its own rise time is about 0.35/100 MHz = 3.5 ns. The displayed rise time is approximately √(t_signal² + t_scope²), so measuring a 3.5 ns edge would show about 4.9 ns. A scope 3–5 times faster than the signal is needed for accuracy.

ECE-305 · What is the PSD of a random polar NRZ data stream?

Its autocorrelation is triangular over ±Tb, so the PSD is A²Tb·sinc²(fTb). The first null is at Rb, which is why NRZ is said to need about Rb of bandwidth. It also has strong low-frequency content.

ECE-306 · How is narrowband noise represented, and what are its envelope statistics?

n(t) = nI(t)cos ω_ct − nQ(t)sin ω_ct, where nI and nQ are independent low-pass Gaussian processes. The envelope √(nI² + nQ²) is Rayleigh-distributed and the phase is uniform; with a sinusoid added, the envelope becomes Rician.

ECE-307 · White noise of PSD N₀/2 passes through an ideal low-pass filter of bandwidth B. What is the output noise power?

Output PSD is |H(f)|²·N₀/2 over −B to B, so power = (N₀/2)·2B = N₀B. This is why receiver noise power scales directly with filter bandwidth.

ECE-308 · What is the noise-equivalent bandwidth of an RC low-pass filter?

Bn = ∫|H(f)|²df/|H(0)|² = 1/(4RC) = (π/2)·f₃dB. It is about 57% larger than the 3 dB bandwidth, which matters when computing kTB noise.

ECE-309 · What amplitude droop does a zero-order hold DAC cause at the Nyquist frequency?

A ZOH has response sinc(f/fs). At f = fs/2 this is sin(π/2)/(π/2) ≈ 0.637, about −3.9 dB. Designers use an inverse-sinc compensation filter or oversampling.

ECE-310 · Two tones at 1.000 kHz and 1.100 kHz must be resolved by spectral analysis. What minimum observation time is needed?

Resolution is roughly the reciprocal of the record length, so T ≥ 1/Δf = 1/100 Hz = 10 ms with a rectangular window. Windows with lower side lobes widen the main lobe, so in practice 15–20 ms or more is needed.

Analog Communication

ECE-311 · How do you choose the IF in a superheterodyne receiver?

A high IF puts the image far from the wanted signal, making preselector image rejection easy. A low IF makes narrow, sharp channel filtering easier. Dual conversion uses a high first IF for image rejection and a low second IF for selectivity.

ECE-312 · An AM receiver tuned to 1000 kHz has IF = 455 kHz and a preselector with Q = 100. Find the image rejection.

Image = 1910 kHz. ρ = 1910/1000 − 1000/1910 = 1.91 − 0.524 = 1.386. Rejection α = √(1 + Q²ρ²) ≈ 138.6, or about 42.8 dB.

ECE-313 · An FM signal has fm = 5 kHz and Δf = 25 kHz. Find β and bandwidth.

β = Δf/fm = 5. Carson's rule gives B = 2(25 + 5) = 60 kHz. The Bessel-function view agrees: about β + 1 = 6 significant sideband pairs, so 2 × 6 × 5 kHz = 60 kHz.

ECE-314 · How can FM deviation be calibrated using the carrier-null method?

The carrier amplitude is J₀(β), which first goes to zero at β ≈ 2.405. Increase the modulating level with a known fm until the carrier vanishes on a spectrum analyzer; then Δf = 2.405·fm, for example 2.405 kHz at fm = 1 kHz.

ECE-315 · How does the Armstrong (indirect) FM transmitter work?

It generates narrowband FM with a stable crystal oscillator and a phase modulator fed by the integrated message. Frequency multipliers then raise both carrier and deviation to the required Δf, and a mixer shifts the carrier to the final frequency without changing deviation.

ECE-316 · An AM carrier is modulated by two tones with μ₁ = 0.6 and μ₂ = 0.8. What is the effective modulation index?

μt = √(μ₁² + μ₂²) = √(0.36 + 0.64) = 1.0. The signal is at 100% modulation, and total power is Pc(1 + μt²/2) = 1.5·Pc.

ECE-317 · How does a PLL demodulate FM?

The VCO tracks the incoming instantaneous frequency. Because VCO frequency is proportional to its control voltage, that voltage reproduces the message. The loop bandwidth must cover the message bandwidth while rejecting noise.

ECE-318 · For equal transmitted power, how does output SNR compare for DSB-SC and SSB?

Both give output SNR equal to the baseband reference SNR. SSB halves the bandwidth and so halves the noise, but DSB-SC coherent detection adds both sidebands coherently, giving a 3 dB gain that compensates. SSB's advantage is bandwidth, not SNR.

ECE-319 · Design the RC of an envelope detector for fc = 1 MHz, fm = 5 kHz and μ = 0.8.

To avoid ripple, RC ≫ 1/fc = 1 µs. To avoid diagonal clipping, RC ≤ √(1 − μ²)/(ωm·μ) = 0.6/(2π × 5000 × 0.8) ≈ 24 µs. A choice of about 10–20 µs satisfies both.

ECE-320 · What happens if an SSB receiver's carrier reinsertion is off by 50 Hz, compared with DSB-SC?

In SSB, every audio component shifts by 50 Hz, so speech sounds pitch-shifted but stays intelligible. In DSB-SC, the two sidebands produce a 50 Hz beating that makes the output fade periodically, which is far more damaging.

Digital Communication

ECE-321 · What is a matched filter?

A filter whose impulse response is the time-reversed (conjugate) copy of the expected pulse, h(t) = s*(T − t). It maximizes output SNR at the sampling instant, reaching 2E/N₀ in white noise, and is equivalent to a correlator.

ECE-322 · How do BPSK, QPSK and 16-QAM compare?

BPSK carries 1 bit/symbol, QPSK 2 and 16-QAM 4. BPSK and QPSK have the same BER versus Eb/N0; 16-QAM needs about 4 dB more Eb/N0 for the same BER. Higher orders give more throughput per hertz but need higher SNR and linearity.

ECE-323 · A link must carry 20 Mbps with 16-QAM and raised-cosine roll-off 0.25. What bandwidth is needed?

Symbol rate Rs = 20/4 = 5 Mbaud. Passband bandwidth = Rs(1 + α) = 5 × 1.25 = 6.25 MHz.

ECE-324 · Why is QAM preferred over PSK for M ≥ 16?

QAM spreads points over the whole I–Q plane, giving a larger minimum distance for the same average power. 16-QAM is about 4 dB better than 16-PSK. PSK's constant envelope matters only when the power amplifier must run saturated.

ECE-325 · A receiver has C/N₀ = 70 dB-Hz and data rate 1 Mbps with BPSK. Is a 10⁻⁵ BER achievable?

Eb/N₀ = C/N₀ − 10·log₁₀(Rb) = 70 − 60 = 10 dB. BPSK needs about 9.6 dB for 10⁻⁵, leaving only 0.4 dB margin. Adding FEC coding gain or reducing the rate would make the link robust.

ECE-326 · What is adaptive modulation and coding (AMC)?

The transmitter selects modulation order and code rate according to reported channel quality, using high-order QAM and high code rates in good conditions and robust settings at the cell edge. It maximizes throughput while meeting a target block error rate.

ECE-327 · How is symbol timing recovered in a digital receiver?

A timing error detector such as early-late gate, Gardner or Mueller–Müller estimates the offset of the sampling instant. A loop filter then adjusts an interpolator or sampling clock, much like a PLL.

ECE-328 · What is the processing gain of a DSSS system with 3.84 Mcps chips and 12.2 kbps voice?

Gp = Rc/Rb = 3.84 × 10⁶ / 12.2 × 10³ ≈ 315, or about 25 dB. This is the factor by which narrowband interference and other users' signals are suppressed after despreading.

ECE-329 · How do DSSS and FHSS compare?

DSSS spreads with a fast chip sequence, giving processing gain, multipath resolution (RAKE) and CDMA capability, but needs tight power control. FHSS hops a narrowband carrier among channels, avoiding interference and jamming; it is simpler but lacks fine multipath resolution.

ECE-330 · Why is Gray coding used in constellation mapping?

Adjacent constellation points differ by only one bit, and most symbol errors go to a neighbour. A symbol error therefore usually causes just one bit error, so BER ≈ SER/log₂M.

ECE-331 · What do common distortions look like on a QAM constellation?

Uniform fuzzy clouds indicate AWGN. A spinning constellation shows carrier frequency offset, and arcs show phase noise. Compressed outer points indicate PA compression, and a skewed or elliptical shape indicates I/Q gain or phase imbalance.

Information Theory & Coding

ECE-332 · Which channel codes are used in LTE and 5G NR?

LTE uses turbo codes for data and tail-biting convolutional codes for control. 5G NR uses LDPC codes for data channels, which give high throughput with parallel decoding, and polar codes for control channels.

ECE-333 · What is the capacity of a 20 MHz channel at 20 dB SNR?

SNR = 100, so C = 20 × 10⁶ × log₂(101) ≈ 20 × 10⁶ × 6.66 ≈ 133 Mbps.

ECE-334 · What are the bandwidth-limited and power-limited regimes?

At high SNR, C ≈ B·log₂(SNR): capacity grows linearly with bandwidth but only logarithmically with power, so systems use high-order modulation. At low SNR, C ≈ 1.44·P/N₀, independent of bandwidth, so power and coding gain dominate, as in deep-space links.

ECE-335 · A source has probabilities 0.4, 0.2, 0.2, 0.1, 0.1. Find the Huffman average length and efficiency.

Huffman coding gives an average length L = 2.2 bits/symbol. Entropy H = 0.4(1.32) + 2 × 0.2(2.32) + 2 × 0.1(3.32) ≈ 2.12 bits. Efficiency η = H/L ≈ 96.5%.

ECE-336 · How many parity bits are needed for a single-error-correcting Hamming code with 11 data bits?

Choose the smallest r with 2ʳ ≥ k + r + 1. For k = 11, r = 4 gives 16 ≥ 16. The code is (15,11) with rate 0.73.

ECE-337 · What errors can an r-bit CRC reliably detect?

With a well-chosen generator of degree r (including a constant term), it detects all single-bit errors, all burst errors of length ≤ r, and any odd number of errors if (x + 1) is a factor. CRC-32 thus catches every burst up to 32 bits.

ECE-338 · What is HARQ, and how do Chase combining and incremental redundancy differ?

Hybrid ARQ combines FEC with retransmission, keeping soft information from failed attempts. Chase combining resends the same coded bits and adds them. Incremental redundancy sends new parity bits, lowering the effective code rate, and generally performs better.

ECE-339 · What is the benefit of soft-decision decoding?

The decoder uses reliability information (quantized or real-valued samples) instead of hard 0/1 decisions. In AWGN this gives about 2 dB coding gain over hard decisions for typical convolutional codes.

ECE-340 · How complex is Viterbi decoding of a constraint-length 7 code?

The trellis has 2^(K−1) = 64 states, each needing an add-compare-select per bit. Complexity grows exponentially with K, which limits practical convolutional codes to K of about 7–9.

ECE-341 · How does MIMO change channel capacity?

In rich scattering with Nt transmit and Nr receive antennas, capacity at high SNR grows roughly linearly with min(Nt, Nr), giving multiple parallel spatial streams without extra bandwidth or power.

ECE-342 · What is water-filling power allocation?

For parallel channels with different gains (such as OFDM subcarriers or MIMO eigenmodes), capacity is maximized by giving more power to stronger channels, so that noise-to-gain level plus power reaches a common "water level". Very weak channels may get no power.

EM Theory & Transmission Lines

ECE-343 · Design a quarter-wave transformer to match a 100 Ω load to a 50 Ω line at 1 GHz on a microstrip with εeff = 3.3.

Z₁ = √(50 × 100) ≈ 70.7 Ω. Guide wavelength λg = 300 mm/√3.3 ≈ 165 mm, so the section is about 41 mm long. Matching is exact only at 1 GHz; cascaded sections widen bandwidth.

ECE-344 · Find the input impedance of a λ/8 short-circuited 50 Ω line.

Zin = jZ₀·tan(βl) = j50·tan(π/4) = j50 Ω, an inductive reactance. Short stubs shorter than λ/4 act as inductors; between λ/4 and λ/2 they act as capacitors.

ECE-345 · An antenna has VSWR 1.5. How much power is lost to reflection?

|Γ| = (1.5 − 1)/(1.5 + 1) = 0.2, so |Γ|² = 4% is reflected. Mismatch loss is −10·log(1 − 0.04) ≈ 0.18 dB, which is usually acceptable.

ECE-346 · A TDR shows a reflection 100 ns after the incident step on a cable with velocity factor 0.66. Where is the fault?

Distance = v·t/2 = 0.66 × 3 × 10⁸ × 100 × 10⁻⁹/2 ≈ 9.9 m. An upward step indicates higher impedance, such as an open; a downward step indicates lower impedance, such as a short or crushed cable.

ECE-347 · How do you get a 50 Ω microstrip line on FR-4?

Characteristic impedance depends on trace width W, dielectric height h and εr. On FR-4 (εr ≈ 4.3–4.6), 50 Ω needs W roughly twice h. A field solver should confirm the exact width, accounting for copper thickness and solder mask.

ECE-348 · The load has |Γ| = 0.5, but the cable has 3 dB one-way loss. What reflection is seen at the transmitter?

The reflected wave suffers the loss twice (6 dB, an amplitude factor of 0.5), so |Γin| = 0.5 × 0.5 = 0.25 and VSWR appears to be 1.67 instead of 3. Lossy cables mask a bad antenna match.

ECE-349 · What is the loss tangent of a dielectric?

tan δ = σ/(ωε′), or ε″/ε′, the ratio of conduction (loss) current to displacement current. Low-loss microwave substrates have tan δ around 0.001 or less, while FR-4 is about 0.02.

ECE-350 · What fraction of light reflects at an air–glass interface (n = 1.5), and how is it reduced?

Γ = (1 − 1.5)/(1 + 1.5) = −0.2, so |Γ|² = 4% of power is reflected at normal incidence. A quarter-wave anti-reflection coating with n = √(1 × 1.5) ≈ 1.22 cancels the reflection at the design wavelength.

ECE-351 · How much absorption loss does a shield of thickness 3δ give?

Absorption loss A = 8.686·(t/δ) dB, so t = 3δ gives about 26 dB. Total shielding effectiveness also includes reflection loss, which dominates for electric and plane-wave fields.

ECE-352 · What termination schemes are used for high-speed digital transmission lines?

Series (source) termination with R ≈ Z₀ − R_driver uses little power and suits point-to-point links. Parallel termination to ground or Vtt at the load works for multidrop but draws DC. AC (RC) and Thevenin terminations trade power against complexity.

ECE-353 · How does a waveguide below cutoff work as a shield or attenuator?

Below cutoff the mode is evanescent and decays as e^(−αz), with α = √(kc² − k²) ≈ 2π/λc when f ≪ fc. Honeycomb ventilation panels use this: long, narrow tubes give high attenuation of EMI while letting air through.

Antennas & Wave Propagation

ECE-354 · A 2.4 GHz link has Pt = 30 dBm, Gt = Gr = 10 dBi and range 1 km. Find received power.

FSPL = 20·log(1) + 20·log(2400) + 32.44 ≈ 100 dB. Pr = 30 + 10 + 10 − 100 = −50 dBm, ignoring cable and fade losses.

ECE-355 · How is a phased array steered, and how are grating lobes avoided?

Each element gets a progressive phase β = −k·d·sin θ₀ to point the beam at θ₀. Grating lobes stay out of visible space if d ≤ λ/(1 + |sin θmax|), so λ/2 spacing allows steering to ±90°.

ECE-356 · What is the free-space path loss to a geostationary satellite at 12 GHz?

With d ≈ 36,000 km, FSPL = 20·log(36,000) + 20·log(12,000) + 32.44 ≈ 91.1 + 81.6 + 32.4 ≈ 205 dB. High-gain dishes and low-noise receivers make up for it.

ECE-357 · What is G/T for a satellite earth station?

G/T (dB/K) is antenna gain minus 10·log of system noise temperature. It is the receive figure of merit because C/N₀ = EIRP − path loss + G/T + 228.6 dB, where 228.6 is −10·log(k).

ECE-358 · How is a rectangular microstrip patch antenna designed?

Its resonant length is about λ₀/(2√εeff), slightly shortened for fringing. Width sets radiation resistance and efficiency. Bandwidth is only a few percent; thicker, lower-εr substrates widen it but raise surface waves and feed inductance.

ECE-359 · How does rain affect links above about 10 GHz, and how is it handled?

Raindrops absorb and scatter energy, causing attenuation that rises steeply with frequency and rain rate. Designers use statistical rain models to size a fade margin for the target availability, plus adaptive coding, power control or site diversity.

ECE-360 · What does the two-ray ground-reflection model predict?

Beyond a crossover distance, the direct and ground-reflected waves nearly cancel, so Pr ≈ Pt·Gt·Gr·(ht·hr)²/d⁴. Power falls 40 dB per decade instead of 20 dB, and higher antennas help significantly.

ECE-361 · When is the Okumura–Hata model used?

It is an empirical path-loss model for macrocells, roughly 150–1500 MHz, 1–20 km distances and base antennas 30–200 m high. It has corrections for urban, suburban and open areas. COST-231 Hata extends it to about 2 GHz.

ECE-362 · How do you choose an operating frequency for an HF sky-wave link?

Predict the MUF for the path and time of day, then operate near the optimum working frequency, about 85% of MUF. This leaves margin for ionospheric variation while staying well above the D-layer absorption region.

ECE-363 · What loss does a knife-edge obstacle cause when it just touches the line of sight?

At the grazing condition (Fresnel–Kirchhoff parameter ν = 0) the diffraction loss is about 6 dB. Loss rises as the obstacle blocks more of the first Fresnel zone.

ECE-364 · Why does antenna noise temperature matter in a satellite receiver?

A dish pointed at cold sky may have antenna temperature of only tens of kelvin, so a low-noise LNB greatly improves G/T. If the dish sees warm ground through side lobes, or rain, its noise temperature rises and degrades C/N.

Microwave Engineering

ECE-365 · How is an equal-split Wilkinson divider designed?

Two λ/4 arms of impedance √2·Z₀ (70.7 Ω for a 50 Ω system) split the input, and a 2Z₀ (100 Ω) resistor joins the outputs. All ports are matched, outputs are isolated and each gets −3 dB, in phase.

ECE-366 · How is a 90° branch-line hybrid designed?

Four λ/4 lines form a square: two series arms of Z₀/√2 (35.4 Ω) and two shunt arms of Z₀ (50 Ω). Power splits equally to two outputs 90° apart, with the fourth port isolated. Bandwidth is about 10–20%.

ECE-367 · Choose rectangular waveguide dimensions for single-mode operation from 9 to 10 GHz.

TE₁₀ cutoff c/(2a) must be below 9 GHz, so a > 16.7 mm. TE₂₀ cutoff c/a must exceed 10 GHz, so a < 30 mm, and TE₀₁ needs b < 15 mm. Standard WR-90 (22.86 × 10.16 mm) fits.

ECE-368 · How is unconditional stability of a microwave transistor checked?

Using its S-parameters, Rollett's factor K = (1 − |S₁₁|² − |S₂₂|² + |Δ|²)/(2|S₁₂S₂₁|) must exceed 1 with |Δ| = |S₁₁S₂₂ − S₁₂S₂₁| < 1. Otherwise some source or load impedances cause oscillation, and stability circles show which.

ECE-369 · How are lumped filter prototypes converted to distributed microwave filters?

Richards' transformation replaces inductors and capacitors with short- and open-circuited stubs of length λ/8 at the cutoff. Kuroda identities then separate the stubs with unit elements and convert series stubs to shunt stubs that are practical to build.

ECE-370 · Why must a VNA be calibrated, and what methods are used?

Calibration removes systematic errors from directivity, source and load match, tracking and cables. SOLT (short-open-load-thru) is common for coax; TRL (thru-reflect-line) suits on-wafer or fixtured measurements where good standards are hard to make.

ECE-371 · How does radar range scale with transmit power?

From the radar equation, R_max⁴ ∝ Pt·G²·λ²·σ/((4π)³·P_min). Doubling range needs 16 times (12 dB) more power, or the equivalent in gain, sensitivity or integration.

ECE-372 · What bandwidth gives 1 m radar range resolution?

ΔR = c/(2B), so B = 3 × 10⁸/(2 × 1) = 150 MHz. Pulse compression with chirps achieves this bandwidth with long pulses, keeping average power high.

ECE-373 · What is the maximum unambiguous range of a radar with PRF 1 kHz?

Ru = c/(2·PRF) = 3 × 10⁸/2000 = 150 km. Echoes from beyond 150 km arrive after the next pulse and appear at false short ranges.

ECE-374 · What Doppler shift does a 10 GHz radar see from a target moving radially at 30 m/s?

fd = 2v/λ = 2 × 30/0.03 = 2000 Hz. The factor 2 arises because the wave travels to the target and back.

Digital Signal Processing

ECE-375 · How do FIR and IIR filters compare?

FIR filters are always stable and can have exactly linear phase, but need high order for sharp transitions. IIR filters achieve sharp responses with low order and less computation, but can be unstable, are sensitive to quantization and have nonlinear phase.

ECE-376 · Estimate the FIR order for 60 dB stop-band attenuation and a transition width of 0.1π rad/sample.

Kaiser's formula gives N ≈ (A − 8)/(2.285·Δω) = 52/(2.285 × 0.314) ≈ 73 taps. Halving the transition width roughly doubles the order.

ECE-377 · Why are polyphase structures used for decimation filters?

A polyphase decimator splits the FIR into M sub-filters running at the low output rate. It avoids computing outputs that downsampling would discard, cutting computation by a factor of M with identical results.

ECE-378 · What is a CIC filter and when is it used?

A cascaded integrator-comb filter performs large-ratio decimation or interpolation using only adders and delays, with no multipliers. Its sinc-shaped passband droops, so a short compensating FIR usually follows it. It is common in digital down-converters.

ECE-379 · How do you convert audio from 44.1 kHz to 48 kHz?

The ratio is 48,000/44,100 = 160/147. Upsample by L = 160, low-pass filter at the lower Nyquist band edge, then downsample by M = 147. Practical designs use polyphase or multistage filters, or an arbitrary-ratio resampler.

ECE-380 · How much does oversampling improve ADC SNR?

With white quantization noise, SNR improves by 10·log₁₀(OSR) dB, so each 4× oversampling adds about 6 dB (one bit). Sigma-delta modulators add noise shaping, giving much larger gains, such as about 15 dB per doubling for a second-order loop.

ECE-381 · An ADC measures SINAD = 62 dB. What is its ENOB?

ENOB = (SINAD − 1.76)/6.02 = (62 − 1.76)/6.02 ≈ 10 bits. A 12-bit converter with this SINAD loses two bits to noise and distortion.

ECE-382 · Why is the Goertzel algorithm used for DTMF detection?

DTMF needs only 8 frequency bins. Goertzel computes a single DFT bin with a second-order IIR recursion, using far fewer operations than a full FFT when only a few bins are needed.

ECE-383 · How does an LMS adaptive filter work?

It updates coefficients each sample as w(n+1) = w(n) + μ·e(n)·x(n) to minimize mean-square error. Larger μ converges faster but adds misadjustment noise and can go unstable. It is used for echo cancellation, equalization and noise cancelling.

ECE-384 · How do Butterworth, Chebyshev and elliptic filters compare?

Butterworth is maximally flat with a gentle roll-off. Chebyshev type I trades passband ripple for steeper roll-off. Elliptic has ripple in both bands and the sharpest transition for a given order, but the worst phase response.

ECE-385 · A 10th-order IIR filter becomes unstable after fixed-point coefficient quantization. How do you fix it?

High-order direct forms have very sensitive pole positions. Implement the filter as cascaded second-order sections (biquads) with pole-zero pairing and ordering, scale each stage to prevent overflow, and use wider coefficient and accumulator words.

Wireless & Cellular

ECE-386 · What is OFDM, and why is a cyclic prefix used?

OFDM sends data on many closely spaced orthogonal subcarriers, each narrow enough to see flat fading. The cyclic prefix copies the symbol's end to its start; if it exceeds the delay spread, it absorbs ISI and turns linear convolution into circular, so one-tap equalization per subcarrier works after the FFT.

ECE-387 · What are the key technologies of 5G?

mmWave spectrum, massive MIMO with beamforming, flexible numerology, LDPC and polar coding, a service-based core with network slicing, and support for eMBB, URLLC (low latency) and mMTC (massive IoT).

ECE-388 · How much spectrum does a 20 MHz LTE carrier actually occupy?

It has 100 resource blocks × 12 subcarriers × 15 kHz = 18 MHz. The remaining 2 MHz acts as guard bands, giving 90% spectral occupancy.

ECE-389 · How is the OFDM cyclic prefix length chosen?

It must exceed the channel's maximum excess delay spread plus timing error, while staying small relative to the symbol to limit overhead. LTE's normal CP is about 4.7 µs (about 7% overhead), with an extended CP for large cells.

ECE-390 · What is 5G NR numerology?

Subcarrier spacing is 15 × 2^µ kHz (15, 30, 60, 120 or 240 kHz) and slot duration is 1 ms/2^µ. Wider spacing tolerates phase noise and Doppler at mmWave and cuts latency; narrower spacing suits large cells with long delay spreads.

ECE-391 · What is PAPR in OFDM and how is it reduced?

Adding many subcarriers can produce peaks far above the average power, forcing PA back-off and lowering efficiency. Mitigations include clipping and filtering, selected mapping, partial transmit sequences, tone reservation and DFT-spreading (SC-FDMA).

ECE-392 · A cell handles 300 calls per hour with average holding time 3 minutes. What is the offered traffic?

A = λ·h = 300 × (3/60) = 15 Erlangs. The number of channels needed for a given blocking probability is then found from the Erlang B table.

ECE-393 · Why is massive MIMO effective, and why does it favour TDD?

Many antennas form narrow beams toward each user, increasing SNR and serving multiple users in the same resources with low interference. TDD channel reciprocity lets the base station estimate the downlink channel from uplink pilots, avoiding huge feedback overhead.

ECE-394 · Estimate the co-channel S/I for N = 7, path-loss exponent 4 and six first-tier interferers.

S/I = (√(3N))ⁿ/i₀ = (4.58)⁴/6 ≈ 440/6 ≈ 73, or about 18.7 dB. This meets the roughly 18 dB target of analog AMPS, which is why N = 7 was used.

ECE-395 · How is the cell radius found from a link budget?

Compute the maximum allowable path loss (MAPL) = EIRP − receiver sensitivity + antenna gains − losses − fade and interference margins. Then invert a propagation model such as Okumura–Hata to get the distance; the uplink is usually the limiting direction.

ECE-396 · Why does mmWave 5G need beamforming and dense small cells?

At mmWave, the small wavelength means isotropic path loss is high and penetration and diffraction are poor. High-gain beamforming from large antenna arrays recovers the link budget, while dense small cells keep links short and limit blockage.

ECE-397 · Users see ping-pong handovers between two cells. How do you fix it?

Increase the handover hysteresis (offset) and time-to-trigger so a neighbour must be stronger for longer before handover. Also check for coverage overlap, antenna tilt and neighbour list problems; too large settings cause late handovers and drops.

Optical Communication

ECE-398 · Estimate the chromatic-dispersion limit for 10 Gb/s at 1550 nm with D = 17 ps/(nm·km) and source width 0.1 nm.

Pulse spread Δτ = D·L·Δλ. Requiring Δτ ≤ T/4 = 25 ps gives L = 25/(17 × 0.1) ≈ 15 km. Longer spans need narrow-linewidth external modulation, dispersion compensation or coherent DSP.

ECE-399 · What is the modal dispersion of a step-index fiber with n₁ = 1.48 and n₂ = 1.46?

Δτ/L ≈ (n₁ − n₂)/c × (n₁/n₂) ≈ 0.02/(3 × 10⁸) ≈ 67 ns/km. This limits bandwidth-distance to only a few Mb/s·km, which is why graded-index or single-mode fiber is used.

ECE-400 · How is an optical rise-time budget calculated?

System rise time is t_sys = √(t_tx² + t_modal² + t_chrom² + t_rx²). For NRZ data, t_sys should be at most about 0.7/B, where B is the bit rate.

ECE-401 · What does coherent optical detection with DSP achieve?

A local-oscillator laser and 90° hybrid recover amplitude, phase and both polarizations, enabling formats such as DP-QPSK and DP-16QAM. DSP then compensates chromatic dispersion and PMD electronically, so long links need no optical dispersion compensation.

ECE-402 · What fiber nonlinearities limit WDM systems?

Self- and cross-phase modulation, four-wave mixing, and stimulated Raman and Brillouin scattering. FWM is worst near zero dispersion, so non-zero dispersion-shifted fiber is used. Nonlinearities cap launch power per channel.

ECE-403 · How does a passive optical network (PON) work?

An OLT at the central office feeds a single fiber that a passive splitter divides among many ONUs at subscribers. Downstream is broadcast, and upstream uses TDMA. A 1:32 split costs about 15 dB ideal loss (10·log 32), plus excess loss.

ECE-404 · What limits the noise figure of an EDFA?

Amplified spontaneous emission. The quantum limit of a high-gain optical amplifier's noise figure is 3 dB. In a chain of amplifiers, ASE accumulates and sets the OSNR at the receiver.

ECE-405 · Why is PMD a problem at high bit rates?

Fiber birefringence gives the two polarization modes slightly different group delays, randomly varying with time and temperature. The mean differential group delay grows with √L and must stay a small fraction of the bit period, so it matters most at 10 Gb/s and above.

ECE-406 · A 10 Gb/s campus link must span 2 km. Multimode or single-mode fiber?

Single-mode. 10GBASE-SR over OM3 multimode reaches only about 300 m (400 m on OM4), while single-mode optics such as 10GBASE-LR reach 10 km.

ECE-407 · How do you interpret an OTDR trace?

A sloped line shows fiber attenuation. A sharp spike indicates a reflective event such as a connector or break, and a step drop without a spike indicates a fusion splice or bend. Distance = c·t/(2n), and dead zones hide events right after strong reflections.

Data Communication & Networking

ECE-408 · What throughput does TCP get with a 64 KB window and 100 ms RTT?

Throughput ≤ window/RTT = 65,535 × 8/0.1 ≈ 5.2 Mbps, regardless of link speed. The TCP window scale option allows larger windows to fill high bandwidth-delay paths.

ECE-409 · Divide 172.16.0.0/16 into at least 500 subnets. What mask and host count result?

500 subnets need 9 subnet bits (2⁹ = 512), giving /25. Each subnet has 7 host bits: 128 addresses, 126 usable hosts.

ECE-410 · A router link serves 1000 packets/s and receives 800 packets/s. Using M/M/1, what is the average delay?

T = 1/(µ − λ) = 1/(1000 − 800) = 5 ms. At 95% load the delay rises to 20 ms, showing how queueing delay grows sharply near saturation.

ECE-411 · What mechanisms provide QoS in IP networks?

Classification and marking (DiffServ DSCP), policing and shaping (token bucket), queue scheduling (priority, WFQ), active queue management (RED) and admission control. Voice is typically given expedited forwarding with strict priority.

ECE-412 · Why does classic Ethernet have a 64-byte minimum frame?

In shared CSMA/CD, a sender must still be transmitting when a collision from the farthest node returns. The 512-bit slot time at 10 Mbps set the minimum frame at 64 bytes and constrained maximum network diameter.

ECE-413 · What do VLANs and the Spanning Tree Protocol do?

VLANs split one physical switch network into separate broadcast domains using 802.1Q tags. STP blocks redundant links to make a loop-free tree, preventing broadcast storms while keeping backup paths ready.

ECE-414 · Estimate TCP throughput with MSS = 1460 B, RTT = 100 ms and loss rate 10⁻⁴.

The Mathis model gives throughput ≈ (MSS/RTT)·(1.22/√p) = (11,680 bits/0.1 s) × 122 ≈ 14 Mbps. Even small loss severely limits long-RTT TCP flows.

ECE-415 · What is software-defined networking (SDN)?

SDN separates the control plane from the data plane. A centralized controller programs switch forwarding tables via protocols such as OpenFlow, enabling network-wide policy, automation and traffic engineering.

ECE-416 · Why is Wi-Fi throughput well below the advertised PHY rate?

Overheads include preambles, MAC headers, ACKs, inter-frame spacing and contention backoff, plus half-duplex shared medium, retransmissions, rate adaptation and interference. Aggregation in newer standards reduces, but does not remove, these losses.

RF Circuits

ECE-417 · Find the cascaded noise figure: LNA (NF 1 dB, G 15 dB), mixer (NF 10 dB, G −7 dB), IF amp (NF 3 dB).

Linear: F₁ = 1.26, G₁ = 31.6, F₂ = 10, G₂ = 0.2, F₃ = 2. F = 1.26 + 9/31.6 + 1/(31.6 × 0.2) ≈ 1.26 + 0.28 + 0.16 = 1.70, or about 2.3 dB.

ECE-418 · How do you calculate cascaded IIP3?

In linear power units, 1/IIP3_total = 1/IIP3₁ + G₁/IIP3₂ + G₁G₂/IIP3₃ + …. Later stages dominate linearity because they see amplified signals, the opposite of noise figure.

ECE-419 · What LO phase noise is needed to tolerate a −40 dBm blocker 1 MHz away when receiving a −100 dBm signal needing 10 dB SNR in 200 kHz?

Reciprocal-mixing noise must stay below −110 dBm. PN ≤ −110 − (−40) − 10·log(200 kHz) = −70 − 53 = −123 dBc/Hz at 1 MHz offset.

ECE-420 · How are RF power amplifiers linearized?

Digital predistortion (DPD) applies the inverse PA nonlinearity in baseband. Feedforward cancels distortion with an error amplifier. Envelope tracking and Doherty architectures improve efficiency at back-off, which high-PAPR signals require.

ECE-421 · How does a Doherty amplifier work?

A carrier amplifier (class AB) and peaking amplifier (class C) are combined through a λ/4 impedance inverter. At low power only the carrier amplifier runs near saturation; at high power the peaking amplifier load-modulates it, keeping efficiency high over about 6 dB back-off.

ECE-422 · How much image rejection does an image-reject mixer give with 0.1 dB gain and 1° phase mismatch?

IRR ≈ 4/(ε² + θ²) with ε ≈ 0.0115 and θ ≈ 0.0175 rad, giving about 9000 or roughly 40 dB. Higher rejection needs calibration of I/Q imbalance.

ECE-423 · A 2.4 GHz radio uses a 20 ppm crystal. What carrier offset can occur?

Δf = 20 × 10⁻⁶ × 2.4 × 10⁹ = 48 kHz per device, and up to 96 kHz between two devices with opposite errors. The receiver must estimate and correct this frequency offset.

ECE-424 · Why is LNA matching a trade-off between noise and gain?

Minimum noise figure needs the source reflection to equal Γopt, while maximum gain needs a conjugate match. These usually differ, so designers compromise using noise and gain circles. Inductive source degeneration brings them closer and provides a real input impedance.

ECE-425 · What trade-offs exist in PLL frequency synthesizer design?

Integer-N synthesizers have channel spacing equal to the reference frequency, forcing narrow loop bandwidth. Fractional-N allows a high reference and wide bandwidth but needs sigma-delta modulation to suppress fractional spurs. Wide loop bandwidth gives fast lock and suppresses VCO noise; narrow bandwidth filters reference noise.

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