16 posts.
In standard AM the sideband share is μ² ÷ (2 + μ²), and μ can't pass 1 without distortion, so it tops out at 1/3. The derivation and a worked 10 V example.
A diode's small-signal resistance is rd = nVT ÷ ID: 258.5 Ω at 0.1 mA, 25.85 Ω at 1 mA, 2.585 Ω at 10 mA. The derivation, a worked example and when it fails.
Raising VCE narrows the base, so IC creeps up and the BJT's output resistance is ro ≈ VA ÷ IC. The physics, a worked 50 V example and the effect on gain.
The Gibbs overshoot: a square wave's partial sums peak near 1.179, about 9% of the jump, at any number of terms. Computed sums, a worked peak and the traps.
Each Hamming parity bit checks the positions whose binary index has that bit set, so the syndrome spells out the flipped position. A worked (7,4) example.
Don't-cares in a K-map are optional 1s, not compulsory ones. Four common mistakes, a worked example checked by brute force, and why minimal answers can differ.
A 2 pF base-collector capacitance can act like 195 pF at a CE amplifier's input. A worked Miller effect example, the bandwidth it costs and how exact it is.
Indent the Nyquist contour around the pole, then map the detour: it becomes an infinite arc closing clockwise. Worked K/(s(s+1)(s+2)) example, stable for K < 6.
An ideal integrator has infinite DC gain, so any DC input ramps it into a rail: 0.5 V into 10 kΩ and 100 nF hits −12 V in 24 ms. The offset trap and the Rf fix.
A lone zero in the Routh first column gets an ε; a whole zero row needs an auxiliary polynomial. Worked examples checked against the real roots, and the traps.
Series Q = ω0L/R falls as R rises; parallel Q = R/(ω0L) rises. Why, with one worked tank: L = 1 mH, C = 10 nF, Q = 31.6 at 10 Ω series or 10 kΩ in parallel.
Setup slack uses the slowest path and the clock period; hold slack uses the fastest path and no period. A worked 5 ns example with skew gives 0.9 and 0.05 ns.
Skin depth δ = 1/√(πfμσ) shrinks as 1/√f: copper's is 9.35 mm at 50 Hz, 66.1 µm at 1 MHz and 2.09 µm at 1 GHz. Where it comes from and what it does to a wire.
With only dependent sources, Voc and Isc are both zero, so Voc ÷ Isc gives 0 ÷ 0. Use a test source: the method, a worked GATE EC example and the usual traps.
VSWR = Vmax ÷ Vmin gives |Γ|; the first minimum's position gives its phase. Worked 50 Ω line: VSWR 3, |Γ| = 0.5, return loss 6.02 dB, ZL = 24.05 − j30.50 Ω.
A z-transform means nothing without its region of convergence: 1/(1 − az⁻¹) is aⁿu[n] or −aⁿu[−n−1]. Worked poles at 0.5 and 2, three ROCs, which is stable.