Question 1

An AM waveform has a carrier that is 1 kV peak-to-peak. If the modulation index is 50%, what are the envelope's maximum (Emax) and minimum (Emin) voltages observed on an oscilloscope?

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Introduction / Context:For conventional amplitude modulation (AM), the oscilloscope envelope directly shows the effect of modulation index m on the carrier amplitude. Being able to compute the envelope extremes Emax and Emin is essential for avoiding overmodulation and for interpreting AM observations in lab measurements.Given Data / Assumptions:Carrier peak-to-peak voltage Vpp = 1 kV ⇒ carrier peak Ec = Vpp / 2 = 0.5 kV.Modulation index m = 0.5 (i.e., 50%).AM envelope extremes: Emax = Ec(1 + m), Emin = Ec(1 − m).Concept / Approach:In an AM signal s(t) = Ec[1 + m cos(ωm t)]cos(ωc t), the slowly varying envelope magnitudes vary between Ec(1 ± m). This directly yields the oscilloscope-measured envelope crest and trough provided the timebase is slow enough to display the envelope or a peak-detect circuit is used.Step-by-Step Solution:Compute carrier amplitude: Ec = 1 kV / 2 = 0.5 kV.Compute Emax: Emax = Ec(1 + m) = 0.5 × 1.5 = 0.75 kV.Compute Emin: Emin = Ec(1 − m) = 0.5 × 0.5 = 0.25 kV.Verification / Alternative check:Check that the envelope never goes negative (m ≤ 1). With m = 0.5, there is no overmodulation, so the trough stays positive at 0.25 kV.Why Other Options Are Wrong:2 kV, 0.5 kV and 1 kV, 0.5 kV: These misuse the peak-to-peak definition or apply m incorrectly.0.5 kV, 1.5 kV: Reverses max and min and miscomputes values.1.25 kV, 0.75 kV: Inconsistent with Ec = 0.5 kV and m = 0.5.Common Pitfalls:Confusing peak and peak-to-peak; forgetting to multiply m by Ec, not by Vpp; swapping Emax and Emin.Final Answer:0.75 kV, 0.25 kV

Question 2

An n-channel JFET has IDSS = 2 mA. At VGS = −2 V, the measured transconductance is gm = 0.5 mS. What is the pinch-off (threshold) voltage VP (use Shockley model)?

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Introduction / Context:For a JFET, the Shockley model provides the transfer characteristic and a simple relation between the transconductance and the device parameters. From a single operating point (VGS, gm) together with IDSS, one can back-calculate the pinch-off voltage VP (negative for n-channel JFETs on typical substrates).Given Data / Assumptions:IDSS = 2 mA.VGS = −2 V.gm(VGS) = 0.5 mS = 0.0005 S.Shockley relations: ID = IDSS(1 − VGS/VP)^2 and gm(VGS) = (2 IDSS / |VP|) * (1 − VGS/VP).Concept / Approach:We use the gm expression because it involves a single power of (1 − VGS/VP) and is algebraically simpler for solving VP. Remember that for an n-channel JFET, VP is negative; we often denote |VP| as the magnitude. Compute |VP|, then assign the sign.Step-by-Step Solution:Let x = |VP|. For an n-channel JFET, VP = −x.gm(VGS) = (2 IDSS / x) * (1 − VGS/VP) = (2 × 0.002 / x) * (1 − (−2)/(−x)) = (0.004/x) * (1 − 2/x).Set gm = 0.0005 S: 0.0005 = (0.004/x) * (1 − 2/x).Rearrange: 0.0005 x = 0.004 − 0.008/x ⇒ 0.0005 x^2 = 0.004 x − 0.008.Multiply by 2000: x^2 − 8x + 16 = 0 ⇒ (x − 4)^2 = 0 ⇒ x = 4 V.Thus VP = −4 V.Verification / Alternative check:Using gm0 = 2IDSS/|VP| = 4 mA/V / 4 V = 1 mS at VGS = 0; with VGS = −2 V, gm = gm0(1 − 2/4) = 1 mS × 0.5 = 0.5 mS, consistent.Why Other Options Are Wrong:−2 V or 2 V: Do not satisfy the gm relation with the given IDSS and VGS.+4 V: Sign is wrong for an n-channel device's pinch-off voltage.−1 V: Also inconsistent with the computed gm.Common Pitfalls:Dropping the negative sign of VP; mixing up mS (millisiemens) with mW; using the current equation instead of gm and creating algebraic complexity.Final Answer:-4 V

Question 3

Which expression represents a conventional double-sideband AM signal (with carrier present) using m(t) = 2 cos(2π f_m t) and carrier x_c(t) = A_c cos(2π f_c t), without exceeding 100% modulation?

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Introduction / Context:Conventional AM (double sideband with carrier) multiplies a carrier by a DC-plus-message term. Recognizing the correct mathematical form is foundational for understanding spectra, envelope detection, and constraints such as overmodulation (|μ| ≤ 1).Given Data / Assumptions:Message m(t) = 2 cos(2π f_m t).Carrier x_c(t) = A_c cos(2π f_c t).Conventional AM form: x(t) = A_c[1 + μ cos(2π f_m t)] cos(2π f_c t). Here μ is the modulation index and must satisfy |μ| ≤ 1 to avoid overmodulation.Concept / Approach:The presence of a nonzero DC term ensures the envelope never crosses zero when |μ| ≤ 1, enabling simple diode envelope detection. Expressions without the DC term correspond to DSB-SC. Quadrature combinations correspond to phase or frequency translation, not standard AM with carrier.Step-by-Step Solution:Identify conventional AM: a carrier scaled by [1 + message term].Match with options: only option (b) has a carrier multiplied by [1 + m(t)].Note about modulation index: with m(t) = 2 cos(⋯), the implicit μ would be 2, which exceeds 1. In practice, one would scale m(t) to keep |μ| ≤ 1. The form, however, is uniquely correct for conventional AM with carrier.Verification / Alternative check:Expanding x(t) = A_c cos(2π f_c t) + A_c m(t) cos(2π f_c t) produces a carrier term at f_c and symmetric sidebands at f_c ± f_m—exactly the spectrum of standard DSB-AM with carrier.Why Other Options Are Wrong:(a) A_c m(t) cos(2π f_c t) is DSB suppressed-carrier; no DC term (no carrier).(c) Uses the identity cosα cosβ + sinα sinβ = cos(α − β), which yields frequency translation, not AM with carrier.(d) Pure baseband; no modulation.(e) Linear addition of baseband and carrier is not amplitude modulation.Common Pitfalls:Confusing the correct AM form with DSB-SC; overlooking the modulation index limit and the need to scale m(t) in practice.Final Answer:x(t) = A_c [1 + m(t)] cos(2π f_c t)

Question 4

A MOSFET has threshold voltage Vth = +1 V and oxide thickness t_ox = 500 × 10^-8 cm (εr = 3.9, ε0 = 8.85 × 10^-14 F/cm). After ion implantation under the gate to tailor Vth, the required new threshold is −1 V. What implant (sheet charge) magnitude a…

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Introduction / Context:Threshold voltage engineering in MOSFETs often uses channel implants to shift Vth. The threshold shift relates to the added sheet charge via the oxide capacitance per unit area. The sign of the implant (n or p) determines whether Vth moves positive or negative for an NMOS on a p-type substrate.Given Data / Assumptions:Initial Vth = +1 V; desired Vth = −1 V ⇒ ΔVth = −2 V.Oxide thickness t_ox = 5 × 10^-6 cm.εr = 3.9, ε0 = 8.85 × 10^-14 F/cm ⇒ C_ox = εr ε0 / t_ox.Electronic charge q = 1.6 × 10^-19 C.NMOS device; n-type channel implant lowers Vth (toward negative), p-type raises Vth (toward positive).Concept / Approach:Threshold shift from a sheet charge density Qs is ΔVth = −Qs / C_ox (sign per convention). The required magnitude of the implant (in charges per cm^2) is |Qs|/q. The polarity required to move from +1 V to −1 V is that which makes Vth more negative for NMOS: n-type (donors) in the channel region.Step-by-Step Solution:Compute oxide capacitance: C_ox = (3.9 × 8.85 × 10^-14) / (5 × 10^-6) F/cm^2 ≈ 6.903 × 10^-8 F/cm^2.Required charge: |Qs| = C_ox × |ΔVth| = 6.903 × 10^-8 × 2 ≈ 1.3806 × 10^-7 C/cm^2.Convert to sheet density: N = |Qs| / q ≈ (1.3806 × 10^-7) / (1.6 × 10^-19) ≈ 8.63 × 10^11 cm^-2.Choose implant type: n-type to reduce Vth (make it more negative) in NMOS.Verification / Alternative check:Order-of-magnitude sanity: 10^12 cm^-2 level is typical for threshold trims of a few volts with thin oxides in the sub-100 nm range (scaled accordingly).Why Other Options Are Wrong:p-type implants push Vth positive (wrong direction).1.02 × 10^12 cm^-2 overestimates the required charge for a 2 V shift with the given C_ox.0.86 × 10^9 cm^-2 is far too small to create a 2 V shift.Common Pitfalls:Mixing centimeters and meters; forgetting to divide by q to convert C/cm^2 to charges/cm^2; choosing the wrong implant polarity.Final Answer:8.6 × 10^11 cm^-2, n-type implant

Question 5

Z-Transform ROC Combination: For two causal sequences x1[n] and x2[n] having Z-transforms X1(z) with ROC = R1 and X2(z) with ROC = R2, determine the region of convergence (ROC) for x1[n] + x2[n].

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Introduction / Context:In discrete-time signal processing, the Z-transform's region of convergence (ROC) determines where the transform exists and whether the associated time sequence is causal/stable. When adding two signals, understanding how their individual ROCs combine is essential for correct analysis and implementation.Given Data / Assumptions:X1(z) has ROC = R1.X2(z) has ROC = R2.x1[n] and x2[n] are specified as causal sequences.We seek the ROC for Xsum(z) = X1(z) + X2(z) corresponding to x1[n] + x2[n].Concept / Approach:For a given rational Z-transform, the ROC is the annulus in the z-plane that excludes poles and is determined by sequence sidedness (right-sided for causal). The Z-transform of a sum is the sum of transforms: Xsum(z) = X1(z) + X2(z). The ROC for the sum must include any z where both X1 and X2 individually converge; hence it must at least be the intersection of R1 and R2 (and can be larger only if exact pole-zero cancellations occur).Step-by-Step Solution:For causal (right-sided) sequences, each ROC is of the form |z| > r1 and |z| > r2 respectively.The set where both converge is |z| > max(r1, r2) = R1 ∩ R2.Therefore the guaranteed ROC for the sum is the intersection R1 ∩ R2.Only if pole–zero cancellation removes the outermost pole might the ROC expand; however, absent explicit cancellation information, the safe and correct answer is R1 ∩ R2.Verification / Alternative check:Consider examples: x1[n] = a^n u[n] ⇒ ROC |z| > |a|, x2[n] = b^n u[n] ⇒ ROC |z| > |b|. Their sum is (a^n + b^n)u[n], whose ROC is |z| > max(|a|, |b|) = R1 ∩ R2, matching the rule.Why Other Options Are Wrong:R1 ∪ R2: Union includes points where only one transform converges, but the sum then diverges—incorrect.(R1 ∪ R2) ∩ R1 and (R1 ∩ R2) ∪ R1: Both simplify to R1, which ignores R2 and is not generally valid.“Either R1 or R2 depending on cancellations”: Cancellations can enlarge the ROC, but without proof they do not replace the guaranteed intersection rule.Common Pitfalls:Confusing union with intersection when combining transforms.Forgetting sidedness: causal sequences imply a right-sided ROC.Final Answer:R1 ∩ R2

Question 6

Binary Symmetric Channel (Independent Bit Errors): For a 3-bit block with independent bit error probability p, find the probability of at most one bit error.

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Introduction / Context:In digital communications, bit errors on a memoryless channel are often modeled as independent Bernoulli events with error probability p. For small frames or blocks, exact probabilities for events like “at most one error” are calculated with the binomial distribution, guiding the design of coding and error detection.Given Data / Assumptions:Block length n = 3 bits.Each bit error is independent with probability p (0 ≤ p ≤ 1).We seek P(X ≤ 1), where X = number of bit errors in the block.Concept / Approach:Use the binomial distribution: P(X = k) = C(n, k) p^k (1 − p)^(n − k). Then sum for k = 0 and k = 1 to get “at most one error”.Step-by-Step Solution:P(0 errors) = C(3, 0) p^0 (1 − p)^3 = (1 − p)^3.P(1 error) = C(3, 1) p^1 (1 − p)^2 = 3 p (1 − p)^2.Therefore, P(X ≤ 1) = (1 − p)^3 + 3 p (1 − p)^2.Verification / Alternative check:Complement method: P(X ≥ 2) = P(2) + P(3) = 3 p^2 (1 − p) + p^3. Then 1 − P(X ≥ 2) = 1 − [3 p^2 (1 − p) + p^3] simplifies to (1 − p)^3 + 3 p (1 − p)^2, confirming the result.Why Other Options Are Wrong:3 p^2 (1 − p) + p^3: This is P(X ≥ 2), the complement of the desired probability.1 − [(1 − p)^3 + 3 p (1 − p)^2]: This equals P(X ≥ 2), not “at most one”.(1 − p)^3: Only the zero-error case; missing single error events.1 − p^3: Not a binomial sum for n = 3; incorrect.Common Pitfalls:Forgetting to include exactly one error in the “at most one error” event.Mixing complement incorrectly (using 1 − P(X ≤ 1) rather than 1 − P(X ≥ 2)).Final Answer:(1 - p)^3 + 3 p (1 - p)^2

Question 7

Routh–Hurwitz Stability Check: For the denominator polynomial s^3 + s^2 + 2 s + 24, determine the qualitative nature of the network (stable, oscillatory, or unstable).

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Introduction / Context:System stability for linear time-invariant (LTI) networks with rational transfer functions is determined solely by the denominator polynomial (the characteristic equation), provided there is no pole-zero cancellation in the right half-plane. The Routh–Hurwitz criterion gives a quick stability test without explicitly computing roots.Given Data / Assumptions:Characteristic polynomial: s^3 + s^2 + 2 s + 24.We consider the denominator only (standard stability test scenario).No special pole-zero cancellations are indicated.Concept / Approach:For a cubic, construct the Routh array and inspect the signs of the first column. A sign change indicates roots in the right half-plane (RHP), implying instability. No sign changes imply stability; a row of zeros suggests marginal cases or symmetric root pairs.Step-by-Step Solution:Arrange coefficients by descending powers: 1 (s^3), 1 (s^2), 2 (s^1), 24 (s^0).Routh array:s^3: 1    2s^2: 1    24s^1: (11 − 12)/1 = −1    (124 − 10)/1 = 24s^0: 24First column = [1, 1, −1, 24]. There is one sign change (from 1 to −1), indicating one RHP pole.Verification / Alternative check:Because a sign change exists, the system cannot be stable. Numerical root solving would show one positive-real root; however, the Routh test suffices for a qualitative conclusion.Why Other Options Are Wrong:stable: Requires all first-column entries positive—violated.oscillatory or marginally stable: Would correspond to purely imaginary roots or a zero row—none present.depends on numerator polynomial: Stability depends on poles (denominator), not zeros, assuming no cancellations.Common Pitfalls:Arithmetic slip in computing the s^1 row element.Believing zeros can “stabilize” a system; zeros do not move poles.Final Answer:unstable

Question 8

Small-Signal Input Resistance Seen by the Source: For the given common-emitter amplifier (β = 100, gm = 0.3861 A/V, rπ = 259 Ω, RB = 93 kΩ, RC = 250 Ω, RL = 1 kΩ, Rs = 1 kΩ; coupling capacitor C1 is effectively short and bypass capacitor C2 = 4.7 mF…

Accepted Answer

Introduction / Context:In small-signal BJT amplifier design, the source “sees” the Thevenin input resistance formed by the bias network and the transistor's input resistance (rπ). Correctly evaluating this resistance is essential for matching, signal attenuation estimates, and gain calculations.Given Data / Assumptions:Common-emitter stage, emitter is AC-grounded by a large bypass capacitor (C2 ≈ short at signal).Coupling capacitor C1 is large (short at signal), so it does not add reactance.Transistor parameters: β = 100, gm = 0.3861 A/V, rπ = 259 Ω (consistent with rπ ≈ β/gm).Bias resistor to base: RB = 93 kΩ.Source resistance: Rs = 1 kΩ (series with Vs).We ignore ro (given r0 = ∞).Concept / Approach:The small-signal resistance seen by Vs equals Rs in series with the amplifier's input resistance Rin. With the emitter AC-grounded, Rin at the base is RB ∥ rπ (looking into the base). The collector network does not reflect to the input significantly in this idealized case (no feedback through ro).Step-by-Step Solution:Compute Rin(base) = RB ∥ rπ = (93,000 * 259) / (93,000 + 259) Ω ≈ 259 Ω (since 93 kΩ ≫ 259 Ω).Total resistance seen by Vs = Rs + Rin(base) = 1000 Ω + 259 Ω = 1259 Ω.Rounding to provided choices gives 1258 Ω (minor rounding difference).Verification / Alternative check:Check consistency: If C2 were open (no bypass), emitter degeneration would increase input resistance; here it is bypassed, so Rin is dominated by rπ, as used.Why Other Options Are Wrong:258 Ω: Omits Rs; this is only the approximate RB ∥ rπ.93 kΩ: Ignores the strong shunting effect of rπ.∞: Would require the input to be open-circuit, which is not the case.1518 Ω: Not supported by the parallel/series combination given the numbers.Common Pitfalls:Forgetting to include source resistance Rs.Using RB alone instead of RB ∥ rπ for the base input.Final Answer:1258 Ω

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