Question 1

Causality check for discrete-time systems: Which of the following systems is causal?

Accepted Answer

Introduction / Context:Causality is a foundational property for real-time systems: the output at time n cannot depend on future input samples. Knowing how to spot future dependencies is vital for implementability.Given Data / Assumptions:Five candidate input–output relations.Standard discrete-time indexing with n for current time, n−k for past, and n+k for future.No hidden states beyond what is shown.Concept / Approach:A causal system’s output y[n] relies only on x[n] and past values x[n−k], k ≥ 1. Any appearance of x[n+k] with k > 0 violates causality because it requires knowledge of future input.Step-by-Step Solution:Option (a): depends on x[n] and x[n−1] → causal.Option (b): includes x[n+1] → noncausal.Option (c): includes x[n+1] → noncausal.Option (d): includes x[n+1] → noncausal.Option (e): pure advance x[n+2] → noncausal.Verification / Alternative check:Physical realization: (a) can be implemented with one sample of memory (a delay). Others would need prediction of unknown future inputs, which is not causal.Why Other Options Are Wrong:Any x[n+k], k > 0, violates causality by definition.Adding current or past terms cannot fix a future dependency.Common Pitfalls:Confusing time advance (future input) with time delay (past input).Assuming linearity or time invariance alone guarantees causality—it does not.Final Answer:y[n] = 3 x[n] − 2 x[n − 1]

Question 2

Signal classification: If g(t) = A (a non-zero constant) for all t, is g(t) an energy signal, a power signal, or neither?

Accepted Answer

Introduction / Context:Signals are commonly classified as energy or power signals based on the finiteness of their total energy and average power. Constant signals are a classic test case that clarifies the definitions.Given Data / Assumptions:g(t) = A, with A ≠ 0 constant for all t.Energy: E = ∫{−∞}^{∞} |g(t)|^2 dt.Average power: P = lim{T→∞} (1/(2T)) ∫{−T}^{T} |g(t)|^2 dt.Concept / Approach:An energy signal has finite E and zero power. A power signal has finite, non-zero P and infinite energy (over infinite duration). A nonzero constant over infinite time accumulates infinite energy but has finite average power equal to A^2.Step-by-Step Solution:Compute energy: E = ∫ |A|^2 dt from −∞ to ∞ = ∞ (diverges).Compute average power: P = lim{T→∞} (1/(2T)) ∫{−T}^{T} |A|^2 dt = lim{T→∞} (1/(2T)) (2T |A|^2) = |A|^2.Hence, g(t) is a power signal (finite, non-zero average power; infinite total energy).Verification / Alternative check:For periodic or constant signals, power is finite and equal to the time average of |g(t)|^2. Here that equals A^2, consistent with power-signal classification.Why Other Options Are Wrong:Energy signal requires finite energy; here energy diverges.“Neither” is incorrect because average power is finite and well-defined.“Insufficient data” is incorrect; information given is adequate.Common Pitfalls:Assuming finite amplitude implies finite energy even over infinite duration.Confusing definitions of energy and power criteria.Final Answer:Power signal

Question 3

Basic probability without replacement: A box has 4 white and 3 red balls. Two balls are drawn successively without replacement. What is the probability that both are white?

Accepted Answer

Introduction / Context:Sampling without replacement is a fundamental probability model. Here, drawing two balls successively from a small urn illustrates conditional probability and the multiplication rule.Given Data / Assumptions:Total balls: 7 (4 white, 3 red).Two draws in succession without replacement.Event of interest: both balls are white.Concept / Approach:Use the multiplication rule: P(A ∩ B) = P(A) * P(B | A). Let A be “first ball white” and B be “second ball white given first white”.Step-by-Step Solution:P(first white) = 4/7.After removing one white, remaining white = 3, remaining total = 6.P(second white | first white) = 3/6 = 1/2.So, P(both white) = (4/7) * (3/6) = 12/42 = 2/7.Verification / Alternative check:Combinatorial approach: favorable outcomes = C(4,2) = 6; total 2-ball selections = C(7,2) = 21; so probability = 6/21 = 2/7. Matches conditional method.Why Other Options Are Wrong:4/21 and 8/21 are common arithmetic slips; 3/7 counts single white outcomes incorrectly; 1/7 is too low.Common Pitfalls:Accidentally using replacement or forgetting to update counts on the second draw.Confusing ordered draws with unordered combinations.Final Answer:2/7

Question 4

Exponential weighting property of Laplace transform: Multiplication by e^{-a t} in time corresponds to which operation in the s-domain?

Accepted Answer

Introduction / Context:The frequency-shift (exponential weighting) property of the Laplace transform is a key tool for handling signals multiplied by exponentials, as in modulation, windowing, and system responses with damping or growth.Given Data / Assumptions:Time-domain multiplication: x(t) e^{-a t}.Baseline transform: X(s) = L{x(t)}.We assume regions of convergence are adjusted accordingly.Concept / Approach:The property states L{x(t) e^{-a t}} = X(s + a). Thus multiplying by a decaying exponential shifts the transform to the right by a in the s-plane (i.e., replaces s with s + a).Step-by-Step Solution:Start with the definition: L{x(t) e^{-a t}} = ∫{0}^{∞} x(t) e^{-a t} e^{-s t} dt.Combine exponents: = ∫{0}^{∞} x(t) e^{-(s + a) t} dt.Recognize as X(s + a): a shift of +a in the s variable.Verification / Alternative check:Test with x(t) = 1 (unit step): L{e^{-a t}} = 1/(s + a), which equals X(s + a) for X(s) = 1/s, confirming the property.Why Other Options Are Wrong:Translation by −a corresponds to multiplying by e^{+a t}, not e^{−a t}.Multiplication by e^{-a s} in the s-domain is not a standard Laplace property for this operation.No change or conjugation are not applicable here.Common Pitfalls:Sign confusion: e^{-a t} → X(s + a); e^{+a t} → X(s − a).Ignoring changes in the region of convergence after the shift.Final Answer:Translation by +a in s-domain: F(s + a)

Question 5

First-order LTI realizations: Which responses cannot be realized by a first-order linear time-invariant system?

Accepted Answer

Introduction / Context:Filter order dictates the shapes a system can realize. Understanding the limitations of first-order LTI systems helps in selecting the right architecture for desired frequency responses.Given Data / Assumptions:Continuous-time or discrete-time first-order LTI systems.\u201cRealization\u201d means a proper transfer function of order one.Standard definitions: low-pass, high-pass, band-pass, band-stop (notch), and all-pass.Concept / Approach:A first-order transfer function has one pole and possibly one zero. It can realize monotonic magnitude responses such as low-pass or high-pass, and even a first-order all-pass (with a pole and a mirrored zero). To obtain a band-pass (peaked passband between two cutoffs) or a band-stop (attenuated notch around a center), you need at least two reactive elements → a second-order (or higher) system.Step-by-Step Reasoning:First-order low-pass: H(s) = K/(τ s + 1) → realizable.First-order high-pass: H(s) = K s/(τ s + 1) → realizable.First-order all-pass: H(s) = (τ s − 1)/(τ s + 1) (or discrete analog) → realizable.Band-pass or band-stop: require both a lower and an upper corner (two reactive terms) → need at least second order.Verification / Alternative check:Bode plots of first-order systems are monotonic (±20 dB/decade slopes) and cannot form a passband flanked by two roll-offs or a notch between two rises.Why Other Options Are Wrong:(a) or (b) alone are too narrow; both responses are unrealisable with first order.(d) All-pass can be implemented first-order.(e) Low-pass/high-pass are the canonical first-order forms.Common Pitfalls:Assuming any desired magnitude shape can be approximated with first order—band-selective responses need higher order.Confusing all-pass (flat magnitude, phase shaping) with band-pass.Final Answer:Both band-pass and band-stop responses

Question 6

Z-transform Region of Convergence (ROC): For the right-sided sequence x[n] = a^n · u[n], what is the correct ROC in the z-plane?

Accepted Answer

Introduction / Context:In discrete-time signal processing, the Z-transform is a powerful tool for analyzing sequences, system stability, and frequency response. A key concept is the Region of Convergence (ROC), which specifies the set of z-values for which the Z-transform series converges. For right-sided sequences like x[n] = a^n · u[n], identifying the ROC is fundamental for correct analysis and inverse transforms.Given Data / Assumptions:Sequence: x[n] = a^n · u[n], where u[n] is the unit step (u[n] = 1 for n ≥ 0, else 0).Complex z-plane variable z with |z| denoting its magnitude.a can be real or complex; ROC depends on |a|, not on the phase of a.Concept / Approach:The bilateral Z-transform is defined as X(z) = Σ from n = −∞ to ∞ of x[n] · z^(−n). For right-sided sequences (nonzero for n ≥ 0), the Z-transform typically converges outside a circle in the z-plane. Convergence is determined by comparing the sequence term a^n with the decay introduced by z^(−n).Step-by-Step Solution:Start with the series: X(z) = Σ from n = 0 to ∞ of (a^n · z^(−n)).Rewrite the term: a^n · z^(−n) = (a / z)^n.Recognize a geometric series with ratio r = a / z.A geometric series Σ r^n converges iff |r| Therefore, require |a / z| |a|.Thus, the ROC is the exterior of the circle |z| = |a| (excluding the circle itself if the pole is on it).Verification / Alternative check:Compute X(z) explicitly under the convergence condition: Σ (a/z)^n = 1 / (1 − a/z) = z / (z − a). This expression has a pole at z = a; for a right-sided sequence, the ROC must lie outside the outermost pole, confirming |z| > |a|.Why Other Options Are Wrong:|z| z > a or z Converges for all z: False; divergence occurs on or inside the pole circle for right-sided sequences.Common Pitfalls:Forgetting that ROC is magnitude-based (|z|), not a simple inequality on real numbers.Confusing left-sided and right-sided sequence ROCs.Overlooking the special case a = 0, where x[n] = 0 for n ≥ 1 and ROC becomes |z| > 0 (i.e., all z except z = 0), still consistent with |z| > |a|.Final Answer:|z| > |a|

Question 7

For an odd periodic function, what terms appear in its Fourier series expansion?

Accepted Answer

Introduction / Context:Fourier series expansion allows us to represent periodic signals as sums of sine and cosine terms. Depending on whether the function is odd, even, or general, the series simplifies in specific ways. This question tests understanding of symmetry properties and their implications on Fourier coefficients.Given Data / Assumptions:The function is periodic.The function is odd: f(−t) = −f(t).Standard Fourier series representation is assumed: f(t) = a0/2 + Σ [an cos(nω0 t) + bn sin(nω0 t)].Concept / Approach:For odd functions, integration over symmetric intervals causes cosine coefficients an to vanish because cosine is even and product of evenodd is odd, integrating to zero. The constant term a0 is also zero. The only surviving terms are sine terms (bn), which are odd functions themselves. Hence, only sine harmonics remain in the expansion.Step-by-Step Solution:Start with Fourier series: f(t) = a0/2 + Σ [an cos(nω0 t) + bn sin(nω0 t)].Odd function: f(−t) = −f(t).Check cosine terms: cos(nω0 t) is even → product with odd f(t) is odd → integral over symmetric interval cancels → an = 0.Check constant term: a0 = 0 by same logic.Check sine terms: sin(nω0 t) is odd → oddodd = even → integral is nonzero → bn survives.Therefore only sine harmonics remain.Verification / Alternative check:Take a simple odd function: f(t) = sin(t). Its Fourier series obviously has only sine terms.Why Other Options Are Wrong:Odd harmonics only: This applies to half-wave symmetry, not general odd symmetry.Even harmonics only: incorrect, cosine coefficients vanish for odd functions.Cosine harmonics only: cosine terms vanish due to odd symmetry.Both sine and cosine equally: occurs only for general non-symmetric functions.Common Pitfalls:Confusing odd symmetry with half-wave symmetry, which specifically removes even harmonics but not all cosine terms.Final Answer:Sine harmonics only

Question 8

Given the transfer function H(z) = 6 + z^(-1) + z^(-2), how should this discrete-time system be classified in terms of phase characteristics (minimum-phase, maximum-phase, or mixed-phase)?

Accepted Answer

Introduction / Context:In digital signal processing, classification of systems into minimum-phase, maximum-phase, or mixed-phase depends on the locations of zeros of the system's transfer function relative to the unit circle in the z-plane. This question tests knowledge of system classification based on pole-zero locations.Given Data / Assumptions:Transfer function: H(z) = 6 + z^(-1) + z^(-2).No poles other than at z = 0 from negative powers of z.Classification depends only on zero locations.Concept / Approach:For a system to be minimum-phase, all its zeros must lie inside the unit circle. For maximum-phase, all zeros must lie outside. For mixed-phase, some zeros lie inside and some outside. The magnitude response is the same for all three, but the phase response differs.Step-by-Step Solution:Rewrite H(z) with common denominator: H(z) = (6z^2 + z + 1)/z^2.Denominator corresponds to poles at z = 0 (stable, inside unit circle).Zeros are roots of 6z^2 + z + 1 = 0.Discriminant = 1 − 24 = −23 Magnitude of roots = sqrt(1/6) ≈ 0.408 Thus, all zeros are inside unit circle → minimum-phase system.Verification / Alternative check:Check |z| of roots explicitly: roots lie at (−1 ± j√23)/12 → magnitude less than 1.Why Other Options Are Wrong:Maximum phase: requires zeros outside unit circle, not the case here.Mixed phase: requires mix of inside and outside, not the case.None/Linear phase: irrelevant here; linear phase is a property of FIR symmetry, not applicable.Common Pitfalls:Forgetting to multiply by z^2 and solve numerator polynomial properly, or misclassifying because magnitude check is skipped.Final Answer:Minimum phase

Question 9

In an AC circuit, the fundamental component of the current lags the corresponding voltage by 20°. What is the phase relationship of the third-harmonic current component with respect to its corresponding voltage component?

Accepted Answer

Introduction / Context:In AC circuits with harmonics, phase shift between current and voltage is frequency dependent because the impedance of reactive components changes with frequency. The question explores the behavior of the third harmonic in relation to the fundamental when the fundamental lags by 20°.Given Data / Assumptions:Fundamental current lags voltage by 20°.Third harmonic frequency = 3 times fundamental frequency.Circuit contains reactive elements (inductive or capacitive).Concept / Approach:In an inductive circuit, phase angle = arctan(XL / R) increases with frequency because XL = ωL grows with frequency. Thus, higher harmonics lag more than the fundamental. In a capacitive circuit, phase lead increases with frequency, but since problem states lag, inductive behavior is implied.Step-by-Step Solution:Fundamental phase lag = 20°.At 3ω, reactance magnitude triples.Phase angle = arctan(3XL / R) > arctan(XL / R).Therefore, third harmonic current lags voltage by more than 20°.Verification / Alternative check:If R = XL, fundamental lag = 45°. At 3ω, phase lag = arctan(3) ≈ 71.6°, confirming lag increases with harmonic order.Why Other Options Are Wrong:Less than 20°: corresponds to capacitive behavior (leading), not stated here.Exactly equal to 20°: would only occur in pure resistive load.Equal to or more than 20°: vague; correct answer is specifically more than 20°.Leads by 20°: contradicts the given lag condition.Common Pitfalls:Forgetting that inductive reactance grows with frequency, thereby increasing phase lag for harmonics.Final Answer:More than 20° lag

Question 10

An ideal impulse function δ(t) in the frequency domain corresponds to what spectral property?

Accepted Answer

Introduction / Context:The Fourier transform of the unit impulse δ(t) is fundamental in signals and systems. This property explains why δ(t) is used as an excitation for LTI systems: it excites all frequencies equally.Given Data / Assumptions:Signal under consideration is δ(t).Fourier transform pair: δ(t) ↔ 1 (constant for all frequencies).Phase considerations: the transform is real and constant, hence zero phase across all frequencies.Concept / Approach:The Fourier transform of δ(t) is ∫ δ(t) e^{−jωt} dt = 1 for all ω. This means δ(t) contains all frequency components with equal weight, phase constant at zero. Thus δ(t) has infinite bandwidth and excites every frequency equally in the system frequency response.Step-by-Step Solution:Start with definition: F(ω) = ∫ δ(t) e^{−jωt} dt.Using sifting property: evaluates to e^{−jω*0} = 1.Thus F(ω) = 1 for all ω.This implies flat spectrum (all-pass) with zero phase.Verification / Alternative check:Impulse response testing in practice: every system's output to δ(t) reveals its frequency response because δ(t) excites all frequencies.Why Other Options Are Wrong:Pure DC: would correspond to constant function in time domain, not impulse.Pure AC: not defined meaningfully; impulse is broadband, not single-tone.Infinite bandwidth with linear phase variations: bandwidth is infinite, but phase is constant, not linear.Narrowband sinusoid: opposite of δ(t) properties.Common Pitfalls:Confusing δ(t) (impulse in time) with δ(ω) (impulse in frequency); they are duals.Final Answer:Entire frequency range with constant phase

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