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JTO MCQs

Practice JTO-oriented Electronics & Telecommunication Engineering MCQs with answers and explanations for technical exam preparation. Questions are arranged in pages of up to 100.

Question 501

Signals & SystemsConvolution

If x(t)=e^(-t)u(t) and h(t)=delta(t), the output is?

Ae^(-t)u(t)
Bdelta(t)
Cu(t)
De^t u(t)

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Correct Answer: A. e^(-t)u(t)

Explanation: Convolution with delta leaves the signal unchanged.

Reference: Standard signals and systems: convolution

Question 502

Signals & SystemsDiscrete-Time LTI

Discrete-time convolution uses?

AA summation over an integer index
BOnly a continuous integral
CMatrix inversion always
DDifferentiation only

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Correct Answer: A. A summation over an integer index

Explanation: y[n]=sum x[k]h[n-k].

Reference: Standard signals and systems: discrete-time lti

Question 503

Signals & SystemsDiscrete-Time LTI

Discrete-time LTI BIBO stability requires?

ASum of |h[n]| over all n is finite
Bh[n]=n
Ch[n] periodic
Dh[0]=0

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Correct Answer: A. Sum of |h[n]| over all n is finite

Explanation: The impulse response must be absolutely summable.

Reference: Standard signals and systems: discrete-time lti

Question 504

Signals & SystemsDiscrete-Time LTI

For a causal discrete-time LTI system, h[n] must be zero for?

An<0
Bn>0
Cn=0
DAll positive even n

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Correct Answer: A. n<0

Explanation: Causality requires h[n]=0 for negative n.

Reference: Standard signals and systems: discrete-time lti

Question 505

Signals & SystemsSystem Properties

A system is causal if its output at any time depends only on?

AFuture input values
BPresent and past input values
CFuture output values only
DThe input at t=infinity

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Correct Answer: B. Present and past input values

Explanation: A causal system cannot depend on future input values; its output may depend on present and past inputs.

Reference: Standard signals and systems: system properties

Question 506

Signals & SystemsFourier Series

Fourier series is primarily used to represent?

APeriodic signals
BOnly nonperiodic impulses
COnly random noise
DDC circuits only

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Correct Answer: A. Periodic signals

Explanation: Fourier series represents periodic signals using harmonically related components.

Reference: Standard signals and systems: fourier series

Question 507

Signals & SystemsSystem Properties

Which system is memoryless?

Ay(t)=x(t-2)
By(t)=x(t)+x(t-1)
Cy(t)=3x(t)
Dy(t)=integral from -infinity to t of x(tau)d tau

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Correct Answer: C. y(t)=3x(t)

Explanation: A memoryless system output at time t depends only on the input at that same time.

Reference: Standard signals and systems: system properties

Question 508

Signals & SystemsFourier Series

For period T0, fundamental angular frequency is?

AT0/(2pi)
B2pi/T0
C1/T0^2
Dpi T0

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Correct Answer: B. 2pi/T0

Explanation: omega0=2pi/T0.

Reference: Standard signals and systems: fourier series

Question 509

Signals & SystemsSystem Properties

The continuous-time system y(t)=x(t-1) is?

ANoncausal
BCausal with memory
CMemoryless and causal
DUnstable for every input

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Correct Answer: B. Causal with memory

Explanation: It depends on a past input value, so it is causal and has memory.

Reference: Standard signals and systems: system properties

Question 510

Signals & SystemsFourier Series

A periodic signal has period 2 ms. Its fundamental frequency is?

A50 Hz
B500 Hz
C2 kHz
D5 kHz

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Correct Answer: B. 500 Hz

Explanation: f0=1/0.002=500 Hz.

Reference: Standard signals and systems: fourier series

Question 511

Signals & SystemsSystem Properties

A system is BIBO stable if every bounded input produces?

AA bounded output
BAn unbounded output
CZero output only
DA periodic output only

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Correct Answer: A. A bounded output

Explanation: BIBO means bounded-input bounded-output stability.

Reference: Standard signals and systems: system properties

Question 512

Signals & SystemsFourier Series

Fourier-series harmonic frequencies are?

AInteger multiples of the fundamental
BAlways irrational multiples
CAlways lower than fundamental
DIndependent of fundamental

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Correct Answer: A. Integer multiples of the fundamental

Explanation: The kth harmonic is k times the fundamental frequency.

Reference: Standard signals and systems: fourier series

Question 513

Signals & SystemsLTI Systems

For a continuous-time LTI system, BIBO stability requires?

AIntegral of |h(t)| over all time is finite
Bh(t) must be periodic
Ch(t) must be odd
DIntegral of h(t) must always be zero

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Correct Answer: A. Integral of |h(t)| over all time is finite

Explanation: A continuous-time LTI system is BIBO stable if its impulse response is absolutely integrable.

Reference: Standard signals and systems: lti systems

Question 514

Signals & SystemsLTI Systems

For a causal continuous-time LTI system, h(t) must be zero for?

At>0
Bt<0
Ct=1 only
DAll t

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Correct Answer: B. t<0

Explanation: Causality requires h(t)=0 for t<0.

Reference: Standard signals and systems: lti systems

Question 515

Signals & SystemsConvolution

The output of an LTI system is obtained from x(t) and h(t) by?

AMultiplication only
BConvolution
CDifferentiation only
DTime reversal only

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Correct Answer: B. Convolution

Explanation: For an LTI system, y(t)=x(t)*h(t).

Reference: Standard signals and systems: convolution

Question 516

Signals & SystemsConvolution

Continuous-time convolution is?

AIntegral x(tau)h(t-tau)d tau over all tau
BIntegral x(t)h(t)dt only
Cx(t)+h(t)
Dx(t)/h(t)

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Correct Answer: A. Integral x(tau)h(t-tau)d tau over all tau

Explanation: Continuous-time convolution integrates x(tau)h(t-tau) over all tau.

Reference: Standard signals and systems: convolution

Question 517

Signals & SystemsConvolution

Which is a property of convolution?

Ax*h=h*x
Bx*h=x-h
CIt is never associative
DIt requires x=h

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Correct Answer: A. x*h=h*x

Explanation: Convolution is commutative.

Reference: Standard signals and systems: convolution

Question 518

Signals & SystemsConvolution

Convolution with the unit impulse gives?

A0
Bx(t)
C1
Ddx(t)/dt

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Correct Answer: B. x(t)

Explanation: The unit impulse is the identity for convolution.

Reference: Standard signals and systems: convolution

Question 519

Signals & SystemsConvolution

x(t)*delta(t-t0) equals?

Ax(t+t0)
Bx(t-t0)
Cdelta(t)
Dx(t0) only

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Correct Answer: B. x(t-t0)

Explanation: A shifted impulse shifts the convolved signal by the same amount.

Reference: Standard signals and systems: convolution

Question 520

Signals & SystemsConvolution

If x(t)=u(t) and h(t)=u(t), their convolution is?

Au(t)
Bt u(t)
Cdelta(t)
De^(-t)u(t)

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Correct Answer: B. t u(t)

Explanation: The overlap integral has length t for t>=0, giving t u(t).

Reference: Standard signals and systems: convolution

Question 521

Signals & SystemsFourier Series

For a real even periodic signal, the trigonometric Fourier series contains primarily?

ACosine terms
BSine terms only
CNo harmonics
DComplex frequencies only

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Correct Answer: A. Cosine terms

Explanation: Even symmetry eliminates sine coefficients, apart from a possible DC term.

Reference: Standard signals and systems: fourier series

Question 522

Signals & SystemsFourier Series

For a real odd periodic signal, the trigonometric Fourier series contains primarily?

ACosine terms and DC
BSine terms
COnly DC
DNo frequency components

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Correct Answer: B. Sine terms

Explanation: Odd symmetry eliminates cosine and DC terms, leaving sine terms.

Reference: Standard signals and systems: fourier series

Question 523

Signals & SystemsFourier Series

Parseval’s relation connects average power with?

ASquared magnitudes of Fourier-series coefficients
BOnly maximum amplitude
COnly fundamental period
DDerivative at t=0

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Correct Answer: A. Squared magnitudes of Fourier-series coefficients

Explanation: Parseval expresses signal power through the squared Fourier coefficient magnitudes.

Reference: Standard signals and systems: fourier series

Question 524

Signals & SystemsFourier Series

A standard symmetric 50% duty-cycle square wave contains?

AOnly odd harmonics
BOnly even harmonics
CNo harmonics
DOnly second harmonic

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Correct Answer: A. Only odd harmonics

Explanation: Its even harmonics vanish due to symmetry.

Reference: Standard signals and systems: fourier series

Question 525

Signals & SystemsLTI Frequency Response

If an LTI system is excited by e^(j omega t), the response is the same exponential multiplied by?

AH(j omega)
BSampling period only
CZero always
Domega squared always

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Correct Answer: A. H(j omega)

Explanation: Complex exponentials are eigenfunctions of LTI systems; output is H(jomega)e^(jomega t).

Reference: Standard signals and systems: lti frequency response

Question 526

Signals & SystemsFourier Transform

If x(t) ↔ X(jω), then x(t-t0) has Fourier transform?

AX(jω)e^(-jωt0)
BX(jω)e^(jωt0)
CX(j(ω-t0))
Dt0X(jω)

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Correct Answer: A. X(jω)e^(-jωt0)

Explanation: A time delay t0 introduces the linear phase factor e^(-jωt0).

Reference: Standard signals and systems: fourier transform

Question 527

Signals & SystemsFourier Transform

If x(t) ↔ X(jω), multiplication of x(t) by e^(jω0t) produces?

AX(j(ω-ω0))
BX(j(ω+ω0))
Cω0X(jω)
DX(jω)e^(-jω0)

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Correct Answer: A. X(j(ω-ω0))

Explanation: Time-domain modulation by e^(jω0t) shifts the spectrum by +ω0: X(j(ω-ω0)).

Reference: Standard signals and systems: fourier transform

Question 528

Signals & SystemsFourier Transform

Differentiation in time corresponds in the Fourier domain to?

ADivision by jω
BMultiplication by jω
CTime reversal only
DMultiplication by t

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Correct Answer: B. Multiplication by jω

Explanation: If x(t) ↔ X(jω), then dx(t)/dt ↔ jωX(jω), under standard conditions.

Reference: Standard signals and systems: fourier transform

Question 529

Signals & SystemsFourier Transform

Convolution in the time domain corresponds to what operation in the Fourier domain?

AAddition
BMultiplication
CDifferentiation only
DTime reversal

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Correct Answer: B. Multiplication

Explanation: The convolution theorem states x(t)*h(t) ↔ X(jω)H(jω).

Reference: Standard signals and systems: fourier transform

Question 530

Signals & SystemsFourier Transform

Parseval’s relation for the Fourier transform connects signal energy with?

AIntegral of |X(jω)|² in frequency
BOnly X(0)
COnly signal period
DMaximum signal amplitude only

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Correct Answer: A. Integral of |X(jω)|² in frequency

Explanation: Parseval's relation expresses time-domain energy equivalently in terms of the integral of spectral magnitude squared, with the appropriate 1/(2π) factor.

Reference: Standard signals and systems: fourier transform

Question 531

Signals & SystemsLaplace Transform

The bilateral Laplace transform of x(t) is?

A∫x(t)e^(-st)dt over all time
B∫x(t)e^(st)dt from 0 to 1
Cx(t)/s
Ddx(t)/dt only

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Correct Answer: A. ∫x(t)e^(-st)dt over all time

Explanation: The bilateral Laplace transform is X(s)=∫ from -∞ to ∞ x(t)e^(-st)dt.

Reference: Standard signals and systems: laplace transform

Question 532

Signals & SystemsLaplace Transform

The Laplace transform of e^(-at)u(t), with the appropriate ROC, is?

A1/(s-a)
B1/(s+a)
Cs/(s+a)
Da/(s²+a²)

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Correct Answer: B. 1/(s+a)

Explanation: For a right-sided exponential e^(-at)u(t), X(s)=1/(s+a) with ROC Re{s}>-a.

Reference: Standard signals and systems: laplace transform

Question 533

Signals & SystemsLaplace Transform

For x(t)=e^(-2t)u(t), the ROC of its bilateral Laplace transform is?

ARe{s}>-2
BRe{s}<-2
CRe{s}=0 only
DAll s including the pole

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Correct Answer: A. Re{s}>-2

Explanation: A right-sided exponential has ROC to the right of its pole at s=-2.

Reference: Standard signals and systems: laplace transform

Question 534

Signals & SystemsLaplace Transform

For x(t)=-e^(-2t)u(-t), whose transform is 1/(s+2), the ROC is?

ARe{s}>-2
BRe{s}<-2
CRe{s}>2
DAll s

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Correct Answer: B. Re{s}<-2

Explanation: The left-sided version has the same algebraic transform but an ROC to the left of the pole: Re{s}<-2.

Reference: Standard signals and systems: laplace transform

Question 535

Signals & SystemsLaplace Transform

The ROC of a bilateral Laplace transform can contain a pole of X(s)?

AAlways
BNever
COnly for causal signals
DOnly at s=0

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Correct Answer: B. Never

Explanation: The Laplace-transform integral does not converge at a pole, so poles are excluded from the ROC.

Reference: Standard signals and systems: laplace transform

Question 536

Signals & SystemsLaplace Transform

For a causal rational LTI system, the ROC of H(s) lies?

ATo the right of the rightmost pole
BTo the left of the leftmost pole
COnly between every pair of poles
DExactly on the jω-axis

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Correct Answer: A. To the right of the rightmost pole

Explanation: A causal rational impulse response is right-sided, so its ROC is to the right of the rightmost pole.

Reference: Standard signals and systems: laplace transform

Question 537

Signals & SystemsLaplace Transform

A causal rational continuous-time LTI system is BIBO stable if all its poles lie?

AStrictly in the left half of the s-plane
BStrictly in the right half-plane
COnly on the jω-axis
DAt the origin only

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Correct Answer: A. Strictly in the left half of the s-plane

Explanation: For a causal rational system, BIBO stability requires the jω-axis to lie in the ROC, which means all poles must be strictly in the left half-plane.

Reference: Standard signals and systems: laplace transform

Question 538

Signals & SystemsZ-Transform

The bilateral Z-transform of x[n] is defined as?

AΣ x[n]z^(-n) over all integer n
BΣ x[n]z^n for n=0 only
Cx[n]/z
Ddx[n]/dn

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Correct Answer: A. Σ x[n]z^(-n) over all integer n

Explanation: The bilateral Z-transform is X(z)=Σ from n=-∞ to ∞ x[n]z^(-n).

Reference: Standard signals and systems: z-transform

Question 539

Signals & SystemsFourier Transform

The continuous-time Fourier transform X(jω) of x(t) is defined as?

A∫x(t)e^(-jωt)dt over all time
B∫x(t)e^(jωt)dt from 0 to 1
Cx(t)e^(-jωt)
Ddx(t)/dt

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Correct Answer: A. ∫x(t)e^(-jωt)dt over all time

Explanation: The CTFT analysis equation is X(jω)=∫ from -∞ to ∞ x(t)e^(-jωt)dt.

Reference: Standard signals and systems: fourier transform

Question 540

Signals & SystemsFourier Transform

The Fourier transform of the unit impulse delta(t) is?

A0
B1
Cjω
D2πdelta(ω) only

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Correct Answer: B. 1

Explanation: Using the CTFT definition, ∫delta(t)e^(-jωt)dt=1.

Reference: Standard signals and systems: fourier transform

Question 541

Signals & SystemsFourier Transform

The Fourier transform of delta(t-t0) is?

Ae^(-jωt0)
Be^(jωt0)
Cdelta(ω-t0)
D1/(jω)

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Correct Answer: A. e^(-jωt0)

Explanation: The shifted impulse property gives delta(t-t0) ↔ e^(-jωt0).

Reference: Standard signals and systems: fourier transform

Question 542

Signals & SystemsZ-Transform

A causal rational discrete-time LTI system is BIBO stable when all its poles are?

AStrictly inside the unit circle
BStrictly outside the unit circle
CExactly on the unit circle
DAt z=1 only

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Correct Answer: A. Strictly inside the unit circle

Explanation: For a causal rational system to be stable, the unit circle must lie in the ROC; therefore all poles must lie strictly inside it.

Reference: Standard signals and systems: z-transform

Question 543

Signals & SystemsZ-Transform

The frequency response of a stable discrete-time LTI system can be obtained from H(z) by evaluating it on?

AThe unit circle z=e^(jω)
BThe real axis only
Cz=0 only
Dz=infinity only

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Correct Answer: A. The unit circle z=e^(jω)

Explanation: When the unit circle is in the ROC, H(e^(jω)) is obtained by evaluating H(z) at z=e^(jω).

Reference: Standard signals and systems: z-transform

Question 544

Signals & SystemsSampling

According to the sampling theorem, a signal band-limited to 6 kHz should be sampled at at least?

A3 kHz
B6 kHz
C12 kHz
D36 kHz

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Correct Answer: C. 12 kHz

Explanation: The Nyquist rate is twice the highest signal frequency: fs≥2×6 kHz=12 kHz.

Reference: Standard signals and systems: sampling

Question 545

Signals & SystemsSampling

A sinusoid of frequency 3 kHz is sampled at 8 kHz. Does the sampling rate satisfy the basic Nyquist criterion?

AYes
BNo
COnly if amplitude is zero
DOnly for a DC signal

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Correct Answer: A. Yes

Explanation: The Nyquist rate is 2×3=6 kHz, and 8 kHz exceeds it.

Reference: Standard signals and systems: sampling

Question 546

Signals & SystemsSampling

Ideal reconstruction of a properly sampled band-limited signal is theoretically performed using a?

AIdeal low-pass reconstruction filter
BHigh-pass differentiator only
CRectifier
DLimiter only

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Correct Answer: A. Ideal low-pass reconstruction filter

Explanation: After ideal impulse-train sampling above the Nyquist rate, an ideal low-pass filter can select the original baseband spectral replica.

Reference: Standard signals and systems: sampling

Question 547

Signals & SystemsSampling

A continuous-time signal band-limited to 4 kHz is sampled at 6 kHz. Which phenomenon can occur?

AAliasing
BPerfect reconstruction guaranteed
CNo spectral repetition
DOnly amplitude scaling

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Correct Answer: A. Aliasing

Explanation: Since 6 kHz is below the 8 kHz Nyquist rate, shifted spectra can overlap, causing aliasing.

Reference: Standard signals and systems: sampling

Question 548

Signals & SystemsZ-Transform

The Z-transform of a^n u[n] is?

A1/(1-az^-1), ROC |z|>|a|
B1/(1+az^-1), ROC |z|<|a|
Cz-a, all z
Da/(z+1), all z

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Correct Answer: A. 1/(1-az^-1), ROC |z|>|a|

Explanation: For the right-sided sequence a^n u[n], the geometric series gives X(z)=1/(1-az^-1) with ROC |z|>|a|.

Reference: Standard signals and systems: z-transform

Question 549

Signals & SystemsZ-Transform

For a right-sided sequence a^n u[n], the ROC is?

A|z|>|a|
B|z|<|a|
C|z|=|a| only
Dz=a only

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Correct Answer: A. |z|>|a|

Explanation: Convergence of the geometric series requires |az^-1||a|.

Reference: Standard signals and systems: z-transform

Question 550

Signals & SystemsZ-Transform

For a causal rational discrete-time LTI system, the ROC of H(z) is generally?

AOutside the outermost pole
BInside the innermost pole
COnly on the unit circle
DBetween zero and every pole

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Correct Answer: A. Outside the outermost pole

Explanation: A causal rational impulse response is right-sided, so its ROC extends outward from the outermost pole to infinity.

Reference: Standard signals and systems: z-transform

Question 551

Signals & SystemsFrequency Response

For an LTI system, if x(t)=e^(j3t) and H(j3)=2e^(-jπ/4), the output is?

A2e^(j(3t-π/4))
Be^(j(6t-π/4))
C2e^(-j3t)
De^(j3t)/2

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Correct Answer: A. 2e^(j(3t-π/4))

Explanation: Complex exponentials are eigenfunctions of LTI systems, so y(t)=H(j3)e^(j3t)=2e^(j(3t-π/4)).

Reference: Standard signals and systems: frequency response

Question 552

Signals & SystemsFrequency Response

If an LTI system has H(jω)=0 at ω=100 rad/s, a sinusoidal component exactly at 100 rad/s is ideally?

APassed unchanged
BCompletely rejected
CFrequency doubled
DConverted to DC

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Correct Answer: B. Completely rejected

Explanation: A zero frequency-response magnitude at that frequency means the corresponding sinusoidal steady-state component is suppressed.

Reference: Standard signals and systems: frequency response

Question 553

Signals & SystemsFilters

A continuous-time ideal low-pass filter has cutoff 1 kHz. Which sinusoid is in its ideal passband?

A500 Hz
B2 kHz
C5 kHz
D10 kHz

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Correct Answer: A. 500 Hz

Explanation: An ideal low-pass filter passes frequency components below its 1 kHz cutoff.

Reference: Standard signals and systems: filters

Question 554

Signals & SystemsFilters

An ideal high-pass filter with cutoff 2 kHz ideally rejects which component?

A500 Hz
B3 kHz
C5 kHz
D10 kHz

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Correct Answer: A. 500 Hz

Explanation: An ideal high-pass filter rejects components below the cutoff frequency, including 500 Hz.

Reference: Standard signals and systems: filters

Question 555

Signals & SystemsFilters

A band-pass filter is designed primarily to pass?

AA selected range of frequencies
BOnly DC
CAll frequencies equally
DOnly frequencies below zero

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Correct Answer: A. A selected range of frequencies

Explanation: A band-pass filter passes frequencies within a specified band while attenuating frequencies outside it.

Reference: Standard signals and systems: filters

Question 556

Signals & SystemsSampling

If x(t) has bandwidth 5 kHz, the theoretical Nyquist sampling rate is?

A2.5 kHz
B5 kHz
C10 kHz
D20 kHz

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Correct Answer: C. 10 kHz

Explanation: For a baseband signal band-limited to 5 kHz, the Nyquist rate is 2×5=10 kHz.

Reference: Standard signals and systems: sampling

Question 557

Signals & SystemsSystem Function

For an LTI system with impulse response h(t)=e^(-2t)u(t), the transfer function H(s) is?

A1/(s+2)
B1/(s-2)
Cs/(s+2)
D2/(s^2+4)

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Correct Answer: A. 1/(s+2)

Explanation: The Laplace transform of e^(-2t)u(t) is 1/(s+2), with ROC Re{s}>-2.

Reference: Standard signals and systems: system function

Question 558

Signals & SystemsSampling

A 7 kHz sinusoid sampled at 10 kHz produces an aliased frequency in the baseband of?

A1 kHz
B3 kHz
C5 kHz
D7 kHz

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Correct Answer: B. 3 kHz

Explanation: The sampled sinusoid aliases to |7-10|=3 kHz, which lies below fs/2=5 kHz.

Reference: Standard signals and systems: sampling

Question 559

Signals & SystemsSystem Function

For H(s)=1/(s+4), the impulse response of the causal system is?

Ae^(-4t)u(t)
Be^(4t)u(t)
C4e^(-t)u(t)
Ddelta(t-4)

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Correct Answer: A. e^(-4t)u(t)

Explanation: The inverse Laplace transform of 1/(s+4) with a right-sided ROC is e^(-4t)u(t).

Reference: Standard signals and systems: system function

Question 560

Signals & SystemsSampling

A 9 kHz sinusoid is sampled at 12 kHz. Its principal aliased frequency is?

A2 kHz
B3 kHz
C6 kHz
D9 kHz

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Correct Answer: B. 3 kHz

Explanation: The equivalent baseband frequency is |9-12|=3 kHz, within the 0 to fs/2 range.

Reference: Standard signals and systems: sampling

Question 561

Signals & SystemsSystem Function

A causal system has H(s)=(s+2)/[(s+1)(s+5)]. Its poles are?

A-2 only
B-1 and -5
C1 and 5
D-2 and -5

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Correct Answer: B. -1 and -5

Explanation: Poles are roots of the denominator, hence s=-1 and s=-5.

Reference: Standard signals and systems: system function

Question 562

Signals & SystemsSampling

If a band-limited signal is sampled above its Nyquist rate, ideal reconstruction uses sinc interpolation because the ideal reconstruction filter has a?

ARectangular frequency response
BTriangular time response only
CConstant impulse at every sample
DHigh-pass response

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Correct Answer: A. Rectangular frequency response

Explanation: An ideal brick-wall low-pass filter has a rectangular frequency response and a sinc-shaped impulse response, leading to sinc interpolation.

Reference: Standard signals and systems: sampling

Question 563

Signals & SystemsSystem Function

For H(s)=(s+2)/[(s+1)(s+5)], the finite zero is at?

As=-2
Bs=-1
Cs=-5
Ds=2

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Correct Answer: A. s=-2

Explanation: A finite zero is a root of the numerator, so s+2=0 gives s=-2.

Reference: Standard signals and systems: system function

Question 564

Signals & SystemsDTFT

The discrete-time Fourier transform X(e^(jω)) is periodic in ω with period?

Aπ
B2π
C4π
DIt is not periodic

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Correct Answer: B. 2π

Explanation: The DTFT is periodic in angular frequency with period 2π because e^(-j(ω+2π)n)=e^(-jωn) for integer n.

Reference: Standard signals and systems: dtft

Question 565

Signals & SystemsStability

A causal rational continuous-time system has poles at -2 and +1. It is BIBO stable?

AYes, because one pole is negative
BNo, because a pole lies in the right half-plane
CYes, because it is causal
DAlways stable if rational

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Correct Answer: B. No, because a pole lies in the right half-plane

Explanation: For a causal rational CT system, all poles must lie strictly in the left half-plane for BIBO stability.

Reference: Standard signals and systems: stability

Question 566

Signals & SystemsDTFT

The DTFT of delta[n] is?

A0
B1
Ce^(-jω)
D2πdelta(ω) only

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Correct Answer: B. 1

Explanation: By definition, the only nonzero sequence sample is at n=0, so X(e^(jω))=1.

Reference: Standard signals and systems: dtft

Question 567

Signals & SystemsStability

A causal discrete-time system has poles at z=0.4 and z=0.8. It is BIBO stable?

AYes
BNo
COnly if both poles are negative
DOnly if a pole is at z=1

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Correct Answer: A. Yes

Explanation: For a causal rational DT system, stability requires all poles to lie strictly inside the unit circle; both 0.4 and 0.8 satisfy this.

Reference: Standard signals and systems: stability

Question 568

Signals & SystemsDTFT

If x[n] is delayed by n0 samples, its DTFT is multiplied by?

Ae^(-jωn0)
Be^(jωn0)
Cn0
Djω

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Correct Answer: A. e^(-jωn0)

Explanation: The DTFT time-shift property is x[n-n0] ↔ e^(-jωn0)X(e^(jω)).

Reference: Standard signals and systems: dtft

Question 569

Signals & SystemsStability

A causal discrete-time system has a pole at z=1.2. Its BIBO stability is?

AStable
BUnstable
CMarginally stable in every sense
DIndependent of pole location

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Correct Answer: B. Unstable

Explanation: For a causal rational DT system, a pole outside the unit circle prevents the unit circle from being in the ROC, so it is not BIBO stable.

Reference: Standard signals and systems: stability

Question 570

Signals & SystemsZ-Transform

If h[n]=(0.5)^n u[n], H(z) is?

A1/(1-0.5z^-1)
B1/(1+0.5z^-1)
Cz^-1/0.5
D1-0.5z^-1

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Correct Answer: A. 1/(1-0.5z^-1)

Explanation: The Z-transform of a^n u[n] is 1/(1-az^-1), with |z|>|a|.

Reference: Standard signals and systems: z-transform

Question 571

Signals & SystemsFrequency Response

For h[n]=(0.5)^n u[n], H(e^(jω)) is obtained by replacing z^-1 with?

Ae^(-jω)
Be^(jω)
Cjω
D1/jω

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Correct Answer: A. e^(-jω)

Explanation: On the unit circle z=e^(jω), hence z^-1=e^(-jω).

Reference: Standard signals and systems: frequency response

Question 572

Signals & SystemsFrequency Response

The DC gain of H(z)=1/(1-0.5z^-1) is H(e^(j0)) equal to?

A0.5
B1
C2
D4

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Correct Answer: C. 2

Explanation: At ω=0, e^(-jω)=1, so H=1/(1-0.5)=2.

Reference: Standard signals and systems: frequency response

Question 573

Signals & SystemsDTFT

For a real-valued sequence x[n], its DTFT has conjugate symmetry given by?

AX(e^(-jω))=X*(e^(jω))
BX(e^(-jω))=-X*(e^(jω)) always
CX(e^(jω))=0
DX(e^(jω)) is always purely real

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Correct Answer: A. X(e^(-jω))=X*(e^(jω))

Explanation: A real time-domain sequence has a conjugate-symmetric DTFT: X(e^(-jω))=X*(e^(jω)).

Reference: Standard signals and systems: dtft

Question 574

Signals & SystemsLTI Frequency Response

For a stable LTI system, a bounded sinusoidal input produces a steady-state sinusoidal output at?

AThe same frequency
BTwice the input frequency
CZero frequency only
DA random frequency

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Correct Answer: A. The same frequency

Explanation: An LTI system does not create new frequencies from a single sinusoidal input in steady state; it changes only amplitude and phase at that frequency.

Reference: Standard signals and systems: lti frequency response

Question 575

Signals & SystemsSystem Response

A causal first-order system has H(s)=5/(s+5). Its unit-step response is?

A(1-e^(-5t))u(t)
B5e^(-5t)u(t)
Ce^(-t)u(t)
D5u(t)

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Correct Answer: A. (1-e^(-5t))u(t)

Explanation: For a unit step X(s)=1/s, Y(s)=5/[s(s+5)]=1/s-1/(s+5), giving y(t)=(1-e^(-5t))u(t).

Reference: Standard signals and systems: system response

Question 576

Control SystemsControl Fundamentals

A control system that uses feedback from the output is generally called?

AOpen-loop system
BClosed-loop system
CUncontrolled system
DIsolated system

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Correct Answer: B. Closed-loop system

Explanation: A closed-loop system compares output-related feedback with the reference and uses the resulting error in control action.

Reference: Standard control systems: control fundamentals

Question 577

Control SystemsFeedback

Which is a common advantage of negative feedback in a control system?

AReduced sensitivity to parameter variations
BInfinite output for every input
CElimination of all components
DGuaranteed zero error for every system

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Correct Answer: A. Reduced sensitivity to parameter variations

Explanation: Negative feedback generally reduces sensitivity to plant parameter variations and can improve accuracy and robustness when properly designed.

Reference: Standard control systems: feedback

Question 578

Control SystemsFeedback

In a standard negative-feedback system, the error signal is commonly?

AReference minus feedback signal
BReference plus feedback signal always
COutput multiplied by input
DFeedback divided by reference

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Correct Answer: A. Reference minus feedback signal

Explanation: At the summing junction of a standard negative-feedback loop, e(t)=r(t)-b(t).

Reference: Standard control systems: feedback

Question 579

Control SystemsTransfer Functions

For forward transfer function G(s) and negative feedback H(s), the closed-loop transfer function C(s)/R(s) is?

AG(s)/(1+G(s)H(s))
BG(s)/(1-G(s)H(s))
CH(s)/(1+G(s))
DG(s)H(s) only

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Correct Answer: A. G(s)/(1+G(s)H(s))

Explanation: For standard negative feedback, the closed-loop transfer function is G/(1+GH).

Reference: Standard control systems: transfer functions

Question 580

Control SystemsFeedback

For unity negative feedback, H(s) is equal to?

A0
B1
CG(s)
Ds

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Correct Answer: B. 1

Explanation: Unity feedback means the feedback transfer function is H(s)=1.

Reference: Standard control systems: feedback

Question 581

Control SystemsTransfer Functions

The transfer function of a linear time-invariant system is defined, with zero initial conditions, as?

ALaplace transform of output divided by Laplace transform of input
BInput divided by output
COutput plus input
DDerivative of output only

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Correct Answer: A. Laplace transform of output divided by Laplace transform of input

Explanation: The transfer function is G(s)=C(s)/R(s) under zero initial conditions.

Reference: Standard control systems: transfer functions

Question 582

Control SystemsTransfer Functions

The poles of a transfer function are the values of s that make its?

ANumerator zero
BDenominator zero
CGain always unity
DInput zero

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Correct Answer: B. Denominator zero

Explanation: Poles are the roots of the denominator polynomial after pole-zero cancellations are accounted for.

Reference: Standard control systems: transfer functions

Question 583

Control SystemsTransfer Functions

For G(s)=10/(s+5), the pole is located at?

As=+5
Bs=-5
Cs=0
Ds=-10

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Correct Answer: B. s=-5

Explanation: The denominator is zero when s+5=0, giving the pole s=-5.

Reference: Standard control systems: transfer functions

Question 584

Control SystemsTransfer Functions

For G(s)=(s+2)/(s+4), the finite zero is at?

As=-2
Bs=-4
Cs=2
Ds=4

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Correct Answer: A. s=-2

Explanation: The numerator becomes zero at s=-2.

Reference: Standard control systems: transfer functions

Question 585

Control SystemsTime Response

A first-order system has transfer function 1/(Ts+1). The parameter T represents its?

ATime constant
BDamping ratio
CNatural frequency
DSteady-state gain only

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Correct Answer: A. Time constant

Explanation: T is the first-order time constant and determines how quickly the exponential transient decays.

Reference: Standard control systems: time response

Question 586

Control SystemsTime Response

For a first-order system with time constant T, the unit-step response reaches approximately 63.2% of its final value at?

At=T
Bt=2T
Ct=5T
Dt=T/10

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Correct Answer: A. t=T

Explanation: The normalized first-order step response is 1-e^(-t/T); at t=T it equals 1-e^-1≈0.632.

Reference: Standard control systems: time response

Question 587

Control SystemsTime Response

A first-order system is commonly considered essentially settled to within about 2% after approximately?

AT
B2T
C4T
D10T

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Correct Answer: C. 4T

Explanation: For a standard first-order response, e^-4≈0.0183, so about 4 time constants corresponds to the 2% settling criterion.

Reference: Standard control systems: time response

Question 588

Control SystemsControl Fundamentals

A control system in which the control action is independent of the output is called?

AOpen-loop system
BClosed-loop system
CAdaptive system only
DNonlinear feedback system

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Correct Answer: A. Open-loop system

Explanation: In an open-loop control system, the output is not fed back to influence the control action.

Reference: Standard control systems: control fundamentals

Question 589

Control SystemsSecond-Order Systems

A second-order system with ζ=1 is?

AUnderdamped
BCritically damped
COverdamped
DUnstable necessarily

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Correct Answer: B. Critically damped

Explanation: ζ=1 corresponds to critical damping.

Reference: Standard control systems: second-order systems

Question 590

Control SystemsSecond-Order Systems

A second-order system with ζ>1 is?

AUnderdamped
BUndamped
COverdamped
DCritically damped

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Correct Answer: C. Overdamped

Explanation: A damping ratio greater than one gives an overdamped response.

Reference: Standard control systems: second-order systems

Question 591

Control SystemsSecond-Order Systems

A second-order system with ζ=0 has, in the ideal standard model, a response that is?

AUndamped
BCritically damped
COverdamped
DFirst order

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Correct Answer: A. Undamped

Explanation: With zero damping, the standard second-order system has sustained natural oscillation in the ideal model.

Reference: Standard control systems: second-order systems

Question 592

Control SystemsSecond-Order Systems

For the characteristic equation s²+6s+25=0, the undamped natural frequency ωn is?

A3 rad/s
B5 rad/s
C6 rad/s
D25 rad/s

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Correct Answer: B. 5 rad/s

Explanation: Comparing with s²+2ζωn s+ωn², ωn²=25, so ωn=5 rad/s.

Reference: Standard control systems: second-order systems

Question 593

Control SystemsSecond-Order Systems

For s²+6s+25=0, the damping ratio ζ is?

A0.3
B0.6
C1.0
D1.2

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Correct Answer: B. 0.6

Explanation: 2ζωn=6 and ωn=5, hence ζ=6/(2×5)=0.6.

Reference: Standard control systems: second-order systems

Question 594

Control SystemsTime Response

For a stable underdamped second-order system, increasing damping ratio generally causes percentage overshoot to?

AIncrease
BDecrease
CRemain exactly unchanged
DBecome infinite

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Correct Answer: B. Decrease

Explanation: For 0<ζ<1, greater damping ratio generally reduces percentage overshoot.

Reference: Standard control systems: time response

Question 595

Control SystemsSteady-State Error

Steady-state error is the difference between reference input and output-related response as time approaches?

AZero only
BInfinity
COne second always
DThe natural period only

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Correct Answer: B. Infinity

Explanation: Steady-state error is the limiting error after transients have died out, typically e_ss=lim(t→∞)e(t) when the limit exists.

Reference: Standard control systems: steady-state error

Question 596

Control SystemsSecond-Order Systems

The standard second-order characteristic polynomial is commonly written as?

As²+2ζωn s+ωn²
Bs²+ζs+ωn
Cs+ωn²
Ds²+2ωn

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Correct Answer: A. s²+2ζωn s+ωn²

Explanation: The standard second-order denominator is s²+2ζωn s+ωn².

Reference: Standard control systems: second-order systems

Question 597

Control SystemsSecond-Order Systems

In a standard second-order system, ζ denotes the?

ADamping ratio
BNatural frequency
CStatic gain
DTime delay

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Correct Answer: A. Damping ratio

Explanation: The symbol ζ (zeta) denotes damping ratio.

Reference: Standard control systems: second-order systems

Question 598

Control SystemsSecond-Order Systems

In a standard second-order system, ωn denotes the?

AUndamped natural frequency
BDamping ratio
CDC error
DFeedback factor

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Correct Answer: A. Undamped natural frequency

Explanation: ωn is the undamped natural frequency, usually expressed in rad/s.

Reference: Standard control systems: second-order systems

Question 599

Control SystemsSecond-Order Systems

A second-order system with 0<ζ<1 is classified as?

AUnderdamped
BCritically damped
COverdamped
DUndamped only

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Correct Answer: A. Underdamped

Explanation: A damping ratio between zero and one gives an underdamped response.

Reference: Standard control systems: second-order systems

Question 600

Control SystemsSteady-State Error

For G(s)=10/(s+2) with unity feedback, the steady-state error to a unit-step input is?

A1/5
B1/6
C1/10
D0

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Correct Answer: B. 1/6

Explanation: Kp=5, so ess=1/(1+Kp)=1/6.

Reference: Standard control systems: steady-state error