Q: Two signals g1(t) and g2(t) have average powers 4 and 5, respectively, and their zero-lag cross-correlation is 6. What is the average power of the sum signal g1(t) + g2(t)?

Introduction / Context: In signal analysis, the average power of a sum of signals depends on the individual powers and their correlation. This question checks understanding of how cross-correlation (at zero lag) contributes to the composite power of two simultaneous signals. Given Data / Assumptions: P1 = average power of g1(t) = 4. P2 = average power of g2(t) = 5. R12(0) = zero-lag cross-correlation between g1 and g2 = 6. Signals are wide-sense stationary or time averages are meaningful for power calculation. Concept / Approach: The time-average power of the sum is P{g1 + g2} = P1 + P2 + 2R12(0). The last term arises from the cross term 2g1(t)g2(t) when squaring and averaging. If R12(0) > 0, the sum’s power increases by twice the correlation value. Step-by-Step Solution: Start from (g1 + g2)^2 = g1^2 + g2^2 + 2g1 g2.Average both sides: P{g1 + g2} = P1 + P2 + 2 E[g1 g2].Given E[g1 g2] at zero lag = R12(0) = 6.So P{g1 + g2} = 4 + 5 + 2*6 = 21. Verification / Alternative check: If the signals were uncorrelated (R12(0) = 0), power would be 9. Positive correlation makes it larger, consistent with 21. Why Other Options Are Wrong: 9: ignores correlation term.3 or 15: not consistent with formula P1 + P2 + 2R12(0).11: would correspond to R12(0) = 1, not 6. Common Pitfalls: Confusing correlation (expected product) with covariance; omitting the factor 2. Final Answer: 21

Question 1

For a continuous-time signal with highest frequency component f, what should the sampling time T be to satisfy the Nyquist sampling criterion (i.e., avoid aliasing)?

Accepted Answer

Introduction / Context: Sampling theory is foundational in signals and systems, digital signal processing, and communications. The Nyquist–Shannon sampling theorem tells us how fast to sample a continuous-time signal so that it can be reconstructed without aliasing. This question asks for the correct relationship between sampling time T and the highest frequency component f in the analog signal. Given Data / Assumptions: Signal has a maximum (one-sided) frequency component at f hertz. Sampling is uniform with sampling period T seconds and sampling frequency fs = 1 / T. No antialias filter imperfections are considered (idealized case). Concept / Approach: For perfect reconstruction, the sampling frequency must satisfy fs ≥ 2f. This is commonly called the Nyquist rate. Since fs = 1 / T, the condition translates to 1 / T ≥ 2f, or equivalently T ≤ 1 / (2f). Choosing T strictly smaller than 1 / (2f) gives extra margin against nonidealities. Step-by-Step Solution: Let fs = 1 / T.Nyquist condition: fs ≥ 2f.Substitute fs: 1 / T ≥ 2f.Rearrange: T ≤ 1 / (2f). Verification / Alternative check: If T = 1 / (2f), then fs = 2f (Nyquist rate). Any T larger than 1 / (2f) makes fs < 2f and causes aliasing. Any T smaller than 1 / (2f) increases safety margin. Why Other Options Are Wrong: T ≥ 1 / (2f): makes fs ≤ 2f and risks aliasing.T = 1 / f or T = 2 / f: these give fs = f or fs = f/2, both below Nyquist.T = 1 / (4f): acceptable (stricter) but not the general inequality; the correct condition is T ≤ 1 / (2f). Common Pitfalls: Confusing rad/s with Hz; Nyquist uses Hz (cycles/s). Also mixing up T and fs. Final Answer: T ≤ 1 / (2f)

Question 2

Two signals g1(t) and g2(t) have average powers 4 and 5, respectively, and their zero-lag cross-correlation is 6. What is the average power of the sum signal g1(t) + g2(t)?

Accepted Answer

Introduction / Context: In signal analysis, the average power of a sum of signals depends on the individual powers and their correlation. This question checks understanding of how cross-correlation (at zero lag) contributes to the composite power of two simultaneous signals. Given Data / Assumptions: P1 = average power of g1(t) = 4. P2 = average power of g2(t) = 5. R12(0) = zero-lag cross-correlation between g1 and g2 = 6. Signals are wide-sense stationary or time averages are meaningful for power calculation. Concept / Approach: The time-average power of the sum is P{g1 + g2} = P1 + P2 + 2R12(0). The last term arises from the cross term 2g1(t)g2(t) when squaring and averaging. If R12(0) > 0, the sum’s power increases by twice the correlation value. Step-by-Step Solution: Start from (g1 + g2)^2 = g1^2 + g2^2 + 2g1 g2.Average both sides: P{g1 + g2} = P1 + P2 + 2 E[g1 g2].Given E[g1 g2] at zero lag = R12(0) = 6.So P{g1 + g2} = 4 + 5 + 2*6 = 21. Verification / Alternative check: If the signals were uncorrelated (R12(0) = 0), power would be 9. Positive correlation makes it larger, consistent with 21. Why Other Options Are Wrong: 9: ignores correlation term.3 or 15: not consistent with formula P1 + P2 + 2R12(0).11: would correspond to R12(0) = 1, not 6. Common Pitfalls: Confusing correlation (expected product) with covariance; omitting the factor 2. Final Answer: 21

Question 3

A time-limited rectangular pulse defined by f(t) = 0 for t < 0, f(t) = E for 0 ≤ t ≤ a, and f(t) = 0 for t > a is best described as which type of standard signal?

Accepted Answer

Introduction / Context: Standard test signals—step, impulse, ramp, and pulse—are frequently used in systems and signal analysis. Correct classification helps in understanding system responses and Fourier/Laplace transforms. Here, the signal is nonzero only over a finite interval [0, a] with amplitude E, which is a classic rectangular pulse. Given Data / Assumptions: f(t) = 0 for t < 0. f(t) = E for 0 ≤ t ≤ a. f(t) = 0 for t > a. Concept / Approach: A rectangular pulse can be written as the difference of two step functions: f(t) = E[u(t) − u(t − a)]. It is time-limited (finite duration) and has finite energy. It is not a single step (which stays at E forever after t ≥ 0), nor a shifted step at t = a alone (that would be zero until t ≥ a and then E thereafter). Step-by-Step Solution: Recognize finite support: f(t) ≠ 0 only on 0 ≤ t ≤ a.Represent with steps: f(t) = E[u(t) − u(t − a)].Therefore, it is a rectangular pulse (amplitude E, width a, starting at t = 0). Verification / Alternative check: Graphing shows a horizontal line at E between t = 0 and t = a, zero elsewhere—defining a pulse. Why Other Options Are Wrong: Unit step: would remain at E for all t ≥ 0.Shifted step at t = a: zero until t = a, then E forever (not finite-duration).Impulse: infinitely narrow and area-normalized; not the case here. Common Pitfalls: Confusing pulse with step or mistaking the finite duration as two impulses at boundaries. Final Answer: A pulse (rectangular) function

Question 4

A voltage with 5% fifth-harmonic content is applied to a series RC circuit. What is the fifth-harmonic content (percentage) in the circuit current?

Accepted Answer

Introduction / Context: Harmonic distortion in voltage sources translates differently into current depending on the load impedance versus frequency. In a series RC circuit, impedance magnitude decreases with frequency, so higher-frequency components produce comparatively larger currents than lower-frequency components, other things equal. Given Data / Assumptions: Applied voltage contains a fundamental and a fifth-harmonic of 5% amplitude relative to the fundamental. Load is a series RC at steady state (linear, time-invariant). No resonance conditions or nonlinearity; purely frequency-dependent impedance. Concept / Approach: Current magnitude for a frequency component at angular frequency ωk is |I(ωk)| = |V(ωk)| / |Z(ωk)|. For a series RC, |Z(ω)| = sqrt(R^2 + (1/(ωC))^2). As frequency increases, the capacitive reactance 1/(ωC) decreases, reducing |Z| and increasing current. Hence, the fifth harmonic (5ω) experiences smaller impedance than the fundamental (ω), so its current proportion increases beyond the voltage’s 5%. Step-by-Step Solution: Given voltage fifth-harmonic ratio = 5% of fundamental.For current: I1 = V1 / |Z(ω)|, I5 = V5 / |Z(5ω)|.Since |Z(5ω)| < |Z(ω)|, the ratio I5 / I1 > V5 / V1.Therefore, current fifth-harmonic content > 5%. Verification / Alternative check: Compute example with R and C: the impedance drop at higher harmonics always favors more current for RC loads. Why Other Options Are Wrong: 5% or less than 5%: would be true for RL loads (increasing impedance with frequency) but not RC.Equal to or more than 5%: overly vague; the correct insight is strictly more than 5% in an RC series.Approximately zero: not applicable. Common Pitfalls: Assuming all loads behave like resistors (frequency-neutral). Reactive impedances bias harmonic currents. Final Answer: More than 5%

Question 5

If £{f(t)} = F(s), then what is the Laplace transform of the time-shifted signal f(t − T)u(t − T)?

Accepted Answer

Introduction / Context: Time shifting is one of the most frequently used properties in Laplace-domain analysis. It enables modeling of delayed inputs and piecewise signals via step functions. The unit step u(t − T) ensures causality of the shifted segment starting at t = T. Given Data / Assumptions: £ denotes the unilateral Laplace transform. F(s) = £{f(t)} for a causal f(t) (or one for which the unilateral transform exists). We apply a shift by T > 0 with gating u(t − T). Concept / Approach: The standard time-shift property states: £{f(t − T)u(t − T)} = e^{-sT} F(s). The exponential factor in s-domain encodes the delay in time-domain. Step-by-Step Solution: Definition: £{f(t − T)u(t − T)} = ∫_0^∞ f(t − T)u(t − T) e^{-s t} dt.Let τ = t − T, with t = τ + T; lower limit becomes 0.Integral becomes e^{-sT} ∫_0^∞ f(τ) e^{-s τ} dτ = e^{-sT} F(s). Verification / Alternative check: Check with f(t) = u(t): F(s) = 1/s; delay gives e^{-sT}/s, matching tables. Why Other Options Are Wrong: e^{+sT}F(s): corresponds to advance, not delay.F(s − T): not a Laplace time-shift rule (that is a frequency shift, not time shift).sF(s), F(s)/s: relate to differentiation/integration in time, not shift. Common Pitfalls: Forgetting the gating step u(t − T); without it, the transform may not be unilateral-causal. Final Answer: e^{-sT} F(s)

Question 6

According to the Laplace time-displacement (delay) theorem, a shift of +T in the time domain corresponds to what operation in the s-domain?

Accepted Answer

Introduction / Context: Time-delay modeling is central in control systems, communications, and signal processing. In the Laplace domain, delays appear as exponential factors that modify the phase of the frequency response without changing magnitude for pure transport lags. Given Data / Assumptions: We analyze the unilateral Laplace transform. A delay of T ≥ 0 is applied to a causal signal. No growth/decay factors other than the delay are present. Concept / Approach: The property: £{f(t − T)u(t − T)} = e^{-sT}F(s) says that a time shift of T corresponds to multiplication by e^{-sT} in the s-domain. This is used to incorporate transport lags and piecewise onset times in block-diagram algebra. Step-by-Step Solution: Start from definition and substitute t = τ + T.The exponential e^{-sT} factors out of the integral.Remaining integral equals F(s), giving e^{-sT}F(s). Verification / Alternative check: For f(t) = u(t), delay T gives transform e^{-sT}/s, which matches standard tables. Why Other Options Are Wrong: Division by s or multiplication by s corresponds to integration/differentiation, not delay.Division by e^{-sT} implies time advance.Replacing s by s + T is a different frequency-shift operation. Common Pitfalls: Dropping the unit step u(t − T) in the time shift property. Final Answer: Multiplication by e^{-sT}

Question 7

An energy signal has Fourier transform magnitude G(f) = 10 (constant over the band of interest). What is its energy spectral density S(f)?

Accepted Answer

Introduction / Context: For energy signals, the energy spectral density (ESD) S(f) describes how the signal’s energy is distributed across frequencies. It relates directly to the magnitude of the Fourier transform through a simple squared-magnitude relationship. Given Data / Assumptions: G(f) denotes the Fourier transform magnitude of the energy signal (|G(f)|). |G(f)| = 10 (constant) over the relevant frequency range. Standard definition of ESD applies. Concept / Approach: By definition for energy signals, S(f) = |G(f)|^2. This ensures Parseval’s relation: total energy = ∫ S(f) df = ∫ |G(f)|^2 df. Therefore, when |G(f)| = 10, S(f) = 10^2 = 100 (with appropriate units). Step-by-Step Solution: Given |G(f)| = 10.Compute S(f): S(f) = |G(f)|^2.Therefore S(f) = 10^2 = 100. Verification / Alternative check: Parseval’s theorem: Energy E = ∫ |g(t)|^2 dt = ∫ |G(f)|^2 df = ∫ S(f) df, consistent. Why Other Options Are Wrong: 10 or 20 or 50 or 5: these fail to square the magnitude, violating the definition S(f) = |G(f)|^2. Common Pitfalls: Confusing amplitude spectrum |G(f)| with energy spectral density S(f). The former is linear, the latter is quadratic. Final Answer: 100

Question 8

Frequency translation by multiplication: If v(t) = m(t) cos(2π f_c t) and the Fourier transform of m(t) is M(f), what is the Fourier transform of v(t)?

Accepted Answer

Introduction / Context: This question checks your understanding of frequency translation (modulation) in signals and systems. Multiplying a baseband signal m(t) by a cosine of frequency f_c produces a double sideband spectrum centered at ±f_c. Recognizing how the Fourier transform behaves under multiplication by a sinusoid is fundamental in analog modulation and spectral analysis. Given Data / Assumptions: v(t) = m(t) cos(2π f_c t). Fourier transform pair: m(t) ⇄ M(f). Real-valued signals and standard engineering Fourier transform convention are assumed. Concept / Approach: Use the modulation property of the Fourier transform and the identity cos(2π f_c t) = (1/2)(e^{j 2π f_c t} + e^{-j 2π f_c t}). Multiplication in time corresponds to convolution in frequency; however, when multiplying by exponentials, the effect is simply to shift the spectrum. Step-by-Step Solution: Write the cosine as exponentials: cos(2π f_c t) = 0.5 e^{j 2π f_c t} + 0.5 e^{-j 2π f_c t}.Therefore v(t) = 0.5 m(t) e^{j 2π f_c t} + 0.5 m(t) e^{-j 2π f_c t}.In frequency: m(t) e^{j 2π f_c t} ⇄ M(f - f_c) and m(t) e^{-j 2π f_c t} ⇄ M(f + f_c).Hence V(f) = 0.5 M(f - f_c) + 0.5 M(f + f_c). Verification / Alternative check: Check symmetry: if M(f) is bandlimited around DC, the result produces two mirrored lobes around ±f_c. This is exactly the spectrum of double-sideband suppressed-carrier modulation, confirming the formula. Why Other Options Are Wrong: Only 0.5 M(f + f_c) or only 0.5 M(f - f_c): misses one of the sidebands. Duplicated shift term 0.5 M(f - f_c) + 0.5 M(f - f_c): ignores the +f_c sideband. 'Shifted by 2 f_c only': incorrect shift amount and ignores the two-sided nature. Common Pitfalls: Confusing multiplication by cos with a single frequency shift; cos always creates two symmetric shifts. Forgetting the factor 0.5 from the exponential identity. Final Answer: 0.5 M(f + f_c) + 0.5 M(f - f_c)

Question 9

Nyquist sampling for a multitone analog signal: For m(t) = 4 cos(100π t) + 8 sin(200π t) + cos(300π t), what is the Nyquist sampling interval (s)?

Accepted Answer

Introduction / Context: Nyquist sampling theory states that to avoid aliasing, a continuous-time signal with maximum frequency f_max must be sampled at a rate f_s ≥ 2 f_max. Some exams present choices as sampling intervals in seconds (T_s = 1/f_s) rather than sampling rates in hertz; you must read carefully to select the correct unit. Given Data / Assumptions: m(t) = 4 cos(100π t) + 8 sin(200π t) + cos(300π t). Frequencies present: 100π, 200π, and 300π radians/second. Nyquist criterion: f_s ≥ 2 f_max; sampling interval T_s = 1/f_s. Concept / Approach: Convert angular frequencies ω to hertz using f = ω/(2π). Then identify the highest component frequency and compute the Nyquist sampling rate f_N = 2 f_max. Finally, invert to obtain the sampling interval T_s = 1/f_N. Step-by-Step Solution: For 100π rad/s: f_1 = (100π)/(2π) = 50 Hz.For 200π rad/s: f_2 = (200π)/(2π) = 100 Hz.For 300π rad/s: f_3 = (300π)/(2π) = 150 Hz.Maximum frequency f_max = 150 Hz.Nyquist sampling rate f_N = 2 * 150 = 300 Hz.Sampling interval T_s = 1/f_N = 1/300 seconds. Verification / Alternative check: Sampling at 300 Hz ensures at least two samples per period for the 150 Hz component. Any slower rate risks aliasing of the highest frequency term into a lower frequency band. Why Other Options Are Wrong: 1/100 or 1/200: correspond to 100 Hz or 200 Hz sampling, below Nyquist for 150 Hz maximum frequency. 1/600: corresponds to 600 Hz sampling interval? No—1/600 s is faster (600 Hz), which is acceptable but the question asks the Nyquist sampling interval (minimum that satisfies criterion); 1/300 is the tightest bound. 1/150: corresponds to sampling at 150 Hz, clearly below Nyquist. Common Pitfalls: Confusing angular frequency ω with hertz f. Choosing the sampling rate instead of sampling interval; here options are in seconds. Final Answer: 1/300

Question 10

Linearity check for discrete-time systems: Which of the following systems is linear under superposition?

Accepted Answer

Introduction / Context: Determining linearity is fundamental in signals and systems. A linear system obeys the principles of superposition and homogeneity: the response to a_1 x_1[n] + a_2 x_2[n] must equal a_1 y_1[n] + a_2 y_2[n]. Given Data / Assumptions: We test three candidate input–output relations. Operations include addition, time shift, multiplication, and squaring. Zero initial conditions are assumed (standard linearity test context). Concept / Approach: Linearity is preserved by addition and time-shifting but not by multiplication or nonlinear operations like squaring. Therefore we can test each definition against linearity axioms. Step-by-Step Solution: System (a): y[n] = x[n] × x[n − 1]. Multiplication of input samples violates superposition; scaling an input by a factor does not scale the output linearly. Hence nonlinear.System (b): y[n] = x[n] + x[n − 10]. This is a linear time-invariant operator (sum of identity and a time shift). It satisfies both additivity and homogeneity.System (c): y[n] = (x[n])^2. Squaring is nonlinear; scaling the input by α scales the output by α^2, violating homogeneity. Verification / Alternative check: Apply a_1 x_1 + a_2 x_2 to (b): y = (a_1 x_1 + a_2 x_2) + (a_1 x_1[n−10] + a_2 x_2[n−10]) = a_1 (x_1 + x_1[n−10]) + a_2 (x_2 + x_2[n−10]) = a_1 y_1 + a_2 y_2, confirming linearity. Why Other Options Are Wrong: (a) Nonlinear due to multiplication of input terms. (c) Nonlinear due to squaring. (d) Incorrect because neither (a) nor (c) is linear. (e) Incorrect; (b) is linear. Common Pitfalls: Assuming any time shift or addition is nonlinear—it is linear. Overlooking that multiplication of inputs (even with delays) breaks linearity. Final Answer: y[n] = x[n] + x[n − 10]

Question 11

Time-domain integration insight: What is the result of double integrating the unit step function u(t)?

Accepted Answer

Introduction / Context: Understanding how basic signals transform under integration is essential in control and signal processing. The unit step u(t) integrates to a ramp, and integrating again produces a quadratic-in-time function—a parabola—scaled and gated by u(t). Given Data / Assumptions: Unit step u(t) = 1 for t ≥ 0, 0 for t < 0. We consider causal integration with zero initial conditions. We seek the functional form after two integrations. Concept / Approach: First integral of u(t) yields the ramp r(t) = t u(t). The second integral integrates t from 0 to t, generating a quadratic term. Multiplying by u(t) ensures causality (zero for t < 0). Step-by-Step Solution: First integration: ∫ u(τ) dτ from 0 to t = t for t ≥ 0 ⇒ r(t) = t u(t).Second integration: ∫ r(τ) dτ from 0 to t = ∫ τ dτ = (t^2)/2 for t ≥ 0.Therefore the result is (t^2 / 2) u(t), which is a parabola starting at t = 0. Verification / Alternative check: Differentiating (t^2/2) u(t) twice recovers u(t) (ignoring distributional impulses at t = 0 that depend on initial condition conventions), confirming correctness for standard engineering treatment. Why Other Options Are Wrong: Impulse or doublet: these arise from differentiation of step or ramp, not integration. Ramp: that is the result of a single integration, not double. Constant: integrating a constant once gives a ramp, twice gives a parabola, not a constant. Common Pitfalls: Confusing integration with differentiation outcomes (e.g., thinking step → impulse). Forgetting the u(t) gating in the final expression. Final Answer: A parabola

Question 12

True or False conceptual check: Is the Z-transform a non-linear operation?

Accepted Answer

Introduction / Context: The Z-transform is a cornerstone tool in discrete-time signal processing and control. A key property is linearity, which allows superposition in the Z-domain mirroring superposition in the time domain. Given Data / Assumptions: Standard bilateral or unilateral Z-transform definitions. Region of convergence (ROC) exists such that the sums converge. Zero initial conditions for simplicity when needed. Concept / Approach: Linearity means Z{a x[n] + b y[n]} = a Z{x[n]} + b Z{y[n]} provided a common ROC exists or is appropriately handled. This property underpins transfer function algebra and block-diagram manipulations. Step-by-Step Solution: Start with definition: Z{x[n]} = Σ x[n] z^{-n}.For a x[n] + b y[n], Z{a x[n] + b y[n]} = Σ (a x[n] + b y[n]) z^{-n}.Split the sum: = a Σ x[n] z^{-n} + b Σ y[n] z^{-n} = a X(z) + b Y(z).Therefore the Z-transform is linear (not non-linear). Verification / Alternative check: All standard transform pairs and properties (convolution theorem, time shifting) rely on linearity. Nonlinearity would invalidate these core identities. Why Other Options Are Wrong: True: contradicts the linearity shown above. Conditional statements tying linearity to signal length, causality, or ROC alone are incorrect; linearity is inherent to the transform definition. Common Pitfalls: Confusing linearity of the transform with linearity of systems analyzed using it. Misinterpreting ROC issues as linearity issues. Final Answer: False

Question 13

Duty cycle statement check: In a pulse waveform, the duty factor is defined as the ratio T/d (period over pulse width). Is this statement correct?

Accepted Answer

Introduction / Context: The duty factor (also called duty cycle) is a basic descriptor of periodic pulse trains, widely used in sampling, switching power converters, digital logic timing, and pulse-width modulation. Knowing its correct definition avoids design and interpretation errors. Given Data / Assumptions: A periodic pulse train with period T and on-time (pulse width) d. Duty factor commonly expressed as a fraction or percentage. Rise/fall times are not part of the ideal definition. Concept / Approach: By definition, duty factor D = (pulse width) / (period) = d / T. When expressed in percent, Duty % = 100 × (d/T). The inverse T/d is not the duty factor; it is simply the reciprocal and has different engineering meaning (average switching cycles per on-duration). Step-by-Step Solution: Identify variables: period T; on-time d.Definition: Duty factor D = d / T (0 ≤ D ≤ 1).Given statement claims D = T / d, which is the reciprocal and therefore incorrect.Hence, the correct evaluation of the statement is False. Verification / Alternative check: Check extreme cases: If the signal is always on (d = T), D = 1 (100%). The reciprocal T/d would give 1 also here, but for small on-times (e.g., d = T/10), D = 0.1 (10%) while T/d = 10, clearly not a proper duty factor. Why Other Options Are Wrong: True or conditional statements (symmetric pulses, aperiodic) do not change the definition. Dependence on rise time: ideal duty factor definition does not include edge slopes. Common Pitfalls: Confusing duty cycle with its reciprocal, particularly when computing average values in PWM converters. Mixing up frequency f = 1/T with duty factor d/T. Final Answer: False

Question 14

An analog signal m(t) = 3 sin(500π t) + 2 sin(700π t) is sampled at 600 samples per second. What is the folding (Nyquist) frequency of this sampled system?

Accepted Answer

Introduction / Context: In sampling theory for signals and systems, the folding frequency (also called the Nyquist frequency) is a crucial limit that separates positive and negative frequency images after sampling. It determines where aliasing reflections occur in the discrete-time spectrum. This question evaluates recognition of the Nyquist frequency when the sampling frequency is specified explicitly. Given Data / Assumptions: m(t) = 3 sin(500π t) + 2 sin(700π t). Sampling frequency fs = 600 samples per second (600 Hz). Standard radian–frequency to hertz conversion: ω = 2π f. Concept / Approach: The Nyquist frequency fN is one-half of the sampling frequency: fN = fs / 2. Spectral components above fN will fold (alias) back into the first Nyquist zone upon sampling. The individual analog components have ω1 = 500π rad/s and ω2 = 700π rad/s, corresponding to f1 = ω1 / (2π) = 250 Hz and f2 = ω2 / (2π) = 350 Hz; however, the folding frequency is determined solely by the sampling frequency, not by the signal content. Step-by-Step Solution: Compute Nyquist frequency: fN = fs / 2.Substitute fs = 600 Hz → fN = 600 / 2 = 300 Hz.Therefore, the folding frequency is 300 Hz. Verification / Alternative check: Component at 250 Hz lies below fN, so it will not alias. Component at 350 Hz lies above fN, so it will fold around 300 Hz to 300 − (350 − 300) = 250 Hz in the sampled spectrum, consistent with folding about 300 Hz. Why Other Options Are Wrong: 1400 Hz, 700 Hz, 500 Hz, 200 Hz: These do not equal fs / 2 for fs = 600 Hz. The Nyquist frequency only depends on fs, not on individual sinusoid frequencies. Common Pitfalls: Confusing a component frequency with the Nyquist frequency; the folding frequency is always fs / 2. Final Answer: 300 Hz

Question 15

For the time-shifted impulse signal f(t) = A·δ(t − a), what is its unilateral Laplace transform F(s)?

Accepted Answer

Introduction / Context: The Laplace transform of shifted impulses is a standard identity used widely in control systems, circuit analysis, and signals. Recognizing how δ(t − a) maps to the s-domain is crucial for modeling inputs that occur at specific times. Given Data / Assumptions: f(t) = A·δ(t − a), where a ≥ 0 and A is a constant amplitude. Unilateral Laplace transform convention is used. Concept / Approach: By definition, the Laplace transform L{δ(t − a)} = e^{-a s}, reflecting a pure delay of a seconds. A constant amplitude A scales the result linearly. Thus, L{A·δ(t − a)} = A e^{-a s}. Step-by-Step Solution: F(s) = ∫_0^∞ A·δ(t − a)·e^{-s t} dt.Impulse sampling property: ∫ δ(t − a)·g(t) dt = g(a) for a ≥ 0.Therefore, F(s) = A·e^{-s a}. Verification / Alternative check: Time shift in the time domain corresponds to multiplying by e^{-s a} in the s-domain for impulses and for general delayed signals with u(t − a). Why Other Options Are Wrong: A e^{a s}: represents a time advance, not a delay.A·a·e^{-a s} or A·a·e^{a s}: incorrectly introduces an extra factor a.A / (s + a): corresponds to L{A·e^{-a t}} for a different function, not a shifted delta. Common Pitfalls: Confusing delayed exponentials with delayed impulses; the latter transform to pure exponentials in s. Final Answer: A e^{-a s}

Question 16

A voltage V(t) is a Gaussian, ergodic random process with mean 0 and variance 4 V^2. If measured with a direct-current (DC) meter (i.e., time average reading), what value will the meter indicate?

Accepted Answer

Introduction / Context: Understanding what various meters measure for random processes is fundamental in measurement and instrumentation. A DC (average-responding) meter indicates the time-average (mean) value of the input signal. For zero-mean random processes, this has a straightforward implication. Given Data / Assumptions: V(t) is Gaussian and ergodic. Mean μ = 0 V. Variance σ^2 = 4 V^2 (so rms = √σ^2 = 2 V, but rms is not what a DC meter reads). DC meter displays time average (mean) for an ergodic process. Concept / Approach: For an ergodic process, the time average equals the ensemble average. A DC meter, which essentially performs a long-time average, therefore indicates the process mean. Since μ = 0, the meter reading is 0 V. Note this differs from an rms-responding meter, which would show 2 V for the given variance. Step-by-Step Solution: DC meter → time average → equals ensemble mean for ergodic processes.Given mean μ = 0 → meter reading = 0 V. Verification / Alternative check: If an rms meter were used, the indicated value would be √(E[V^2]) = √(σ^2) = 2 V, highlighting the difference between average and rms measurements. Why Other Options Are Wrong: 4: This is the variance, not a DC average.2 or 2√2: These relate to rms or peak equivalents, not DC average.√2: No measurement mode here yields √2 V directly. Common Pitfalls: Confusing mean (average) with rms; instruments can be average-responding or true-rms—know which is used. Final Answer: 0

Question 17

In continuous-time signal classification, the Dirac delta δ(t) is best described as which type of signal?

Accepted Answer

Introduction / Context: The Dirac delta δ(t) is a generalized function (distribution) frequently used to model impulsive events and ideal sampling. Classifying it within the conventional energy/power framework highlights important distinctions between ordinary signals and distributions used in system theory. Given Data / Assumptions: δ(t) is the unit impulse with ∫{−∞}^{∞} δ(t) dt = 1. Energy is E = ∫ |x(t)|^2 dt; average power is P = lim{T→∞} (1/(2T)) ∫_{−T}^{T} |x(t)|^2 dt. Concept / Approach: For ordinary signals, finite energy implies zero average power, and periodic power signals have infinite energy but finite average power. However, δ(t) is not a classical function; |δ(t)|^2 is not defined in the usual sense, so E and P are not finite, nor meaningfully defined. Thus δ(t) is neither an energy signal nor a power signal in the traditional sense used for LTI analysis. Step-by-Step Solution: Attempting to compute energy E = ∫ |δ(t)|^2 dt is not meaningful (distribution squared is undefined).Average power P also cannot be coherently defined for δ(t) using standard formulas.Therefore, δ(t) does not fit into energy or power categories. Verification / Alternative check: In practical engineering, δ(t) is used as an ideal input to probe LTI systems via their impulse response h(t); classification is not necessary for its utility. Why Other Options Are Wrong: Energy signal or power signal: Both require finite, well-defined integrals of |x(t)|^2, which δ(t) lacks.Both energy and power / periodic with finite energy: Contradict definitions and properties of δ(t). Common Pitfalls: Assuming δ(t) has finite energy because its area is 1; area is not energy, and squaring δ(t) is not defined. Final Answer: Neither an energy nor a power signal

Question 18

Which of the following discrete-time relations represents a dynamic and causal system most appropriately?

Accepted Answer

Introduction / Context: In discrete-time systems, “dynamic” implies memory (output depends on past values), while “causal” implies dependence only on present and past—not future—inputs/outputs. The question asks which given relation best represents a dynamic, causal behavior. Given Data / Assumptions: We assess each option for memory (dynamics) and causality. No explicit input is provided in some relations; autonomous (homogeneous) dynamics are permissible for classification. Concept / Approach: A dynamic system must involve dependence on at least one delayed term (past y or x). A causal system cannot involve future terms like y(n + 1) or x(n + 1). A static (memoryless) mapping such as y(n) = x(n) is causal but not dynamic. An equation with future terms is noncausal and not physically realizable in real time. Step-by-Step Solution: Option (a): y(n) + y(n − 1) + y(n + 1). Contains y(n + 1) (future output) → noncausal.Option (b): y(n) + y(n − 1). Only present and past output; dynamic (has memory) and causal. (Implicitly homogeneous autonomous recursion.)Option (c): y(n) = x(n). Memoryless (static) though causal; not dynamic.Option (d): y(n) + y(n − 1) + y(n + 3) = 0. Involves y(n + 3) (future output) → noncausal.Thus, (b) is the most appropriate dynamic and causal relation. Verification / Alternative check: (b) can be rearranged as y(n) = −y(n − 1) showing explicit recursion on past output—classic dynamic, causal behavior. Why Other Options Are Wrong: (a) and (d) include future terms → noncausal.(c) is memoryless → not dynamic.Added (e) y(n) = x(n + 1) would be noncausal if considered. Common Pitfalls: Equating “causal” with “has input”; causality concerns time indexing, not the presence of explicit input in the relation. Final Answer: y(n) + y(n − 1)

Question 19

Laplace transform of a rectangular pulse: For a pulse of constant magnitude E applied from t = 0 to t = a, determine L{v(t)}.

Accepted Answer

Introduction / Context: In signals and systems, the Laplace transform of time-limited rectangular pulses is a staple result used in control, communications, and circuit analysis. Recognizing how to build a pulse from step functions lets us derive the transform quickly and correctly. Given Data / Assumptions: The signal is a rectangular pulse of constant magnitude E starting at t = 0 and ending at t = a. Use unilateral Laplace transform with Re{s} large enough for convergence. Unit step u(t) indicates switching actions. Concept / Approach: A finite-duration pulse can be written as the difference of two steps: E[u(t) − u(t − a)]. The Laplace transform of u(t) is 1/s (for Re{s} > 0), and the time shift property gives L{u(t − a)} = e^{-a s}/s. Linearity then yields the pulse transform. Step-by-Step Solution: Represent pulse: v(t) = E[u(t) − u(t − a)].Apply linearity: L{v(t)} = E[L{u(t)} − L{u(t − a)}].Use step transforms: L{u(t)} = 1/s, L{u(t − a)} = e^{-a s}/s.Combine: V(s) = (E/s)(1 − e^{-a s}). Verification / Alternative check: Direct integration also works: V(s) = ∫0a E e^{-s t} dt = E(1 − e^{-a s})/s, which matches the step-based result exactly. Why Other Options Are Wrong: E(1 − e^{-a s}) lacks the essential 1/s factor. E/(s + a) is the transform of E e^{-a t} u(t), not a finite pulse. (E a)/s corresponds to the transform of a constant of area Ea, not a finite \u201cburst\u201d in time. E e^{-a s}/s is the transform of a delayed step of magnitude E, not a pulse between 0 and a. Common Pitfalls: Forgetting to subtract the delayed step, which removes the signal after t = a. Dropping the 1/s factor that always accompanies step transforms. Final Answer: (E / s) * (1 - e^{-a s})

Question 20

Final Value Theorem: Identify the correct statement for a stable system output y(t) with Laplace transform Y(s).

Accepted Answer

Introduction / Context: The Final Value Theorem (FVT) links steady-state time behavior to a limit in the Laplace domain. It is widely used to compute steady-state error in control systems without performing inverse transforms. Given Data / Assumptions: Output y(t) has Laplace transform Y(s). System stability assumptions are critical to FVT validity. No poles on the imaginary axis (except possibly a simple pole at the origin in sY(s) is not allowed). Concept / Approach: The FVT states: if all poles of sY(s) lie strictly in the left half-plane (no poles at or to the right of jω axis), then lim_{t→∞} y(t) exists and equals lim_{s→0} s Y(s). The condition on sY(s) ensures convergence and rules out oscillatory or divergent steady-state behavior. Step-by-Step Solution: Start from Y(s) = L{y(t)}.Assume poles of sY(s) are strictly in Re{s} < 0.Apply FVT: y(∞) = lim_{s→0} s Y(s).Check that conditions exclude marginally stable/unstable modes (e.g., sinusoidal terms or ramps). Verification / Alternative check: For a unit-step input through a stable first-order system G(s) = K/(τ s + 1), Y(s) = G(s)/s. Then sY(s) = K/(τ s + 1) and lim_{s→0} sY(s) = K, matching the known steady-state y(∞) = K. Why Other Options Are Wrong: Using s → ∞ is related to initial values, not final values. Replacing sY(s) by Y(s) neglects the crucial s multiplier. Claims that ignore stability constraints are unsafe and can give incorrect results for oscillatory/unstable systems. Common Pitfalls: Applying FVT to systems with poles on the imaginary axis (e.g., pure integrators or undamped oscillators) leading to wrong or undefined limits. Forgetting to include the input when forming Y(s) for closed-loop analysis. Final Answer: lim_{t→∞} y(t) = lim_{s→0} s Y(s), with all poles of sY(s) strictly in the left half-plane

Signals and Systems Questions