Previous Year Questions- Introduction of C.T. and D.T. Signals

Q1: If the energy of a continuous-time signal x(t) is E and the energy of the signal 2x(2t - 1) is cE, then c is ______ (rounded off to 1 decimal place).      (2024)
(a) 1.2
(b) 2
(c) 2.6
(d) 3.2
Ans: (b)
Sol: 
Q2: Consider the discrete-time system T₁ and T₂ defined as follows:
Which one of the following statement is true?     (2024)
(a) $𝑇1$T₁ and T₂ are BIBO stable.
(b) T₁ and T₂ are not BIBO stable.
(c) T₁ is BIBO stable butT₂ is not BIBO stable.
(d) T₁ is not BIBO stable but T₂ is BIBO stable.  
Ans: (d)
Sol: Let,
Let,
T₁ is not BIBO stable but T₂ is BIBO stable.

Q3: Suppose signal y(t) is obtained by the time-reversal of signal x(t), i.e.,  y(t) = x(-t), -∞
ltt
lt∞. Which one of the following options is always true for the convolution of x(t) and y(t) ?       (2024)
(a) It is an even signal.
(b) It is an odd signal.
(c) It is a causal signal.
(d) It is an anti-causal signal.
Ans: (a)
Sol: even function.

Q4: The period of the discrete-time signal x[n] described by the equation below is N = ___ (Round off to the nearest integer).     (2023)
(a) 16
(b) 32
(c) 48
(d) 64
Ans: (c)
Sol: Given :

Q5: For the signals x(t) and y(t) shown in the figure, z(t) = x(t) ∗ y(t) is maximum at t = T₁. Then T₁ in seconds is (Round off to the nearest integer).       (2023)
(a) 2
(b) 3
(c) 4
(d) 5
Ans: (c)
Sol: Using convotution property
z(t) = x(t) ∗ y(t)
For max. overlap

Q6: Two discrete-time linear time-invariant systems with impulse responses h₁[n] = δ[n - 1] + δ[n + 1] and  h₂[n] = δ[n] + δ[n - 1] are connected in cascade, where δ[n] is the Kronecker delta. The impulse response of the cascaded system is       (2021)
(a) $𝛿[𝑛-2]+𝛿[𝑛+1]$δ[n - 2] + δ[n + 1]
(b) $𝛿[𝑛-1]𝛿\left[𝑛\right]+𝛿[𝑛+1]𝛿[𝑛-1]$δ[n - 1]δ[n] + δ[n + 1]δ[n - 1]
(c) δ[n - 2] + δ[n - 1] + δ[n] + δ[n + 1]
(d) δ[n] δ[n - 1] + δ[n - 2] δ[n + 1]
Ans: (c)
Sol: By applying z -transform
By applying inverse ZT,

Q7: Suppose for input x(t) a linear time-invariant system with impulse response h(t) produces output y(t), so that x(t) ∗ h(t) = y(t). Further, if  x(t)∣ ∗ ∣h(t)∣ = z(t), which of the following statements is true?         (2020)
(a) For all t ∈ (-∞, ∞), z(t) ≤ y(t)
(b) For some but not all t ∈ (-∞, ∞), z(t) ≤ y(t)  
(c) For all t ∈ (-∞, ∞), z(t) ≥ y(t)
(d) For some but not all t ∈ (-∞, ∞), z(t) ≥ y(t)
Ans: (c)
Sol: Case-1:
case 2:
then, y(t) = z(t)
Thus, z(t) ≥ y(t), for all 't' .

Q8: The symbols, a and T, represent positive quantities, and u(t) is the unit step function. Which one of the following impulse responses is NOT the output of a causal linear time-invariant system?      (2019)
(a) $𝑒+𝑎𝑡𝑢\left(𝑡\right)$e^+atu(t)
(b) $𝑒-𝑎(𝑡+𝑇)𝑢\left(𝑡\right)$e^-a(t+T)u(t)
(c) $1+𝑒-𝑎𝑡𝑢\left(𝑡\right)$1+e^-atu(t)
(d) e^-a(t-T)u(t)
Ans: (c)
Sol: a and T represents positive quantities.
u(t) is unit step function.
h(t) = 1+e^-atu(t), is non-causal
Here '1' is a constant and two sided so the system will be non-causal, because for causal system,

Q9: The signal energy of the continuous-time signal x(t) = [(t - 1)u(t - 1)] - [(t - 2)u(t - 2)] - [(t - 3)u(t - 3)] + [(t - 4)u(t - 4)] is     (2018)
(a) 11/3
(b) 7/3
(c) 1/3
(d) 5/3
Ans: (d)
Sol: x(t) = r(t - 1) - r(t - 2) - r(t - 3) + r(t - 4)

Q10: The mean square value of the given periodic waveform f(t) is_________      (SET-2 (2017))
(a) 12
(b) 24
(c) 6
(d) 3
Ans: (c)
Sol: Mean square value = power of f(t)
Mean square value

Q11: Consider the system with following input-output relation y[n] = (1 + (-1)ⁿ)x[n] where, x[n] is the input and y[n] is the output. The system is       (SET-1 (2017))
(a) invertible and time invariant
(b) invertible and time varying
(c) non-invertible and time invariant
(d) non-invertible and time varying
Ans: (d)
Sol: Given relationship,
y(n) = [1+(-1)ⁿ]x(n)
Time invariance test: 
Since, y(n-1) ≠ y′(n)
So, the system is time invariant.
Invertibility test:
Thus, we are getting many to one mapping between input and output. So, the system is non-invertible.

Q12: The value of  is the Dirac delta function, is     (SET-1  (2016))
(a) 1/2e
(b) 2/e
(c) 1/e²
(d) 1/2e²
Ans: (a)
Sol: To find the value of
Since,  above integral can be written as

Q13: The function shown in the figure can be represented as     (SET-1 (2014))
(a)
(b)
(c)
(d)
Ans: (a)
Sol: The given function can be realized as follow:
Combining the function (i0, (ii), (iii) and (iv), we get the given function
Therefore,

Q14: A zero mean random signal is uniformly distributed between limits -a and +a and its mean square value is equal to its variance. Then the r.m.s value of the signal is      (2011)
(a) a/√3
(b) a/√2
(c) a√2
(d) a√3
Ans: (a)

Q15: Given a sequence x[n], to generate the sequence y[n] = x[3 - 4n], which one of the following procedures would be correct ?      (2008)
(a) First delay x[n] by 3 samples to generate z₁[n], then pick every 4^th sample of z₁[n] to generate z₂[n], and than finally time reverse  z₂[n] to obtain y[n].
(b) First advance x[n] by 3 samples to generate z₁[n], then pick every 4^th sample of z₁[n] to generate z₂[n], and then finally time reverse  z₂[n] to obtain y[n]
(c) First pick every fourth sample of x[n] to generate v₁[n], time-reverse v₁[n] to obtain v₂[n], and finally advance v₂[n] by 3 samples to obtain y[n]
(d) First pick every fourth sample of x[n] to generate v₁[n], time-reverse v₁[n] to obtain v₂[n], and finally delay v₂[n] by 3 samples to obtain y[n]
Ans: (b)
Sol: y[n] = x[3 - 4n] = x[-4n + 3]
So to obtain y[n], we first advance x[n] by 3 unit.
i.e. z₁[n] = x[n + 3]
Now we will take every fourth sample of z₁[n]
i.e. z₂[n] = z₁[4n] = x[4n + 3]
Now reverse (time reverse) z₂[n] will give
 y[n] = z₂[-n] = x[-4n + 3]

Q16: A continuous-time system is described by y(t) = e^-∣x(t)∣, where y(t) is the output and x(t) is the input. y(t) is bounded      (2006)
(a) only when x(t) is bounded
(b) only when x(t) is non-negative
(c) only for t ≥ 0 if x(t) is bounded for t ≥ 0
(d) even when x(t) is not bounded
Ans: (d)

Q17: Which of the following is true?      (2006)
(a) A finite signal is always time bounded
(b) A time bounded signal always possesses finite energy
(c) A bounded signal is always zero outside the interval [-t₀, t₀ ] for some t₀
(d) A time bounded signal is always finite
Ans: (d)

Q18: What is the rms value of the voltage waveform shown in figure?     (2002)
(a) (200/π) V
(b) (100/π) V
(c) 200V
(d) 100V
Ans: (d)
Sol: RMS value

Q19: A current impulse 5δ(t), is forced through a capacitor C. The voltage V_c(t), across the capacitor is given by       (2002)
(a) 5t
(b) 5u(t) - C
(c) (5/C) (t)
(d) 5u(t)/C
Ans: (d)
Sol: