Test 4 - Signals and System | Electronics and Communication (ECE)

Description: A test for Signals and System of Electronics and Communication (ECE)
Number of Questions: 17
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Tags: Electronics and Communication Engineering - EC Signals and System GATE(ECE)
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A Hilbert transformer is a

  1. non-linear system

  2. non-causal system

  3. time-varying system

  4. low-pass system


Correct Option: A
Explanation:

A Hilbert transformer is a non-linear system.

Given f(t) = L–1 $\left[ \dfrac{3s+1}{s3 + 4s2 + (K-3)s} \right]$. If $\displaystyle lim_{x \rightarrow \theta}$f(t) = 1, then the value of K is

  1. 1

  2. 2

  3. 3

  4. 4


Correct Option: B
Explanation:

{x (n)} is a real - valued periodic sequence with a period N. x (n) and X (k) form N - point Discrete Fourier Transform (DFT) pairs. The DFT Y (k) of the sequence y (n) = $\dfrac{1}{n} \displaystyle \sum_{r = 0} ^{N-1} x (r) \times (n+r)$is

  1. |X (k)|2

  2. $\dfrac{1}{N} \displaystyle \sum_{r = 0} ^{N-1} x (r) \times (k+r)$

  3. $\dfrac{1}{N} \displaystyle \sum_{r = 0} ^{N-1} x (r) \times (k+r)$

  4. 0


Correct Option: A
Explanation:

Given that F (s) is the one-sided Laplace transform of f (t). What is the Laplace transform of $\int_0^t f(\tau) d\tau$?

  1. F (s) - f (0)

  2. $\dfrac{1}{s}$F (s)

  3. $\displaystyle \int_0^s F(\tau) d\tau$

  4. $\dfrac{1}{s}$[F (s) - f (0)]


Correct Option: B
Explanation:

Consider a system whose input x and output y are related by the equation y (t) = $\displaystyle \int_{-\infty} ^\infty x(t - \tau) g(2\tau) d\tau$, where h (t) is shown in the graph.

Which of the following four properties are possessed by the system? BIBO : Bounded input gives a bounded output. Causal : The system is causal. LP : The system is low pass. LTI : The system is linear and time-invariant.

  1. Causal, LP

  2. BIBO, LTI

  3. BIBO, Causal, LTI

  4. LP, LTI


Correct Option: B
Explanation:

Here $h(t) \ne 0 \ for \ t \lt 0.$ Thus system is non causal. Again any bounded input $x(t)$ gives bounded output $y(t)$. Thus it is BIBO stable.

Let $x(n) = \left( \dfrac{1}{2} \right) ^ n u(n), \ y(n) = x^2(n)$ and $Y (e^{j\omega})$ be the Fourier transform of y(n)  Then $Y(e^{j0})$ is

  1. $\dfrac{1}{4}$

  2. 2

  3. 4

  4. $\dfrac{4}{3}$


Correct Option: D
Explanation:

If the Laplace transform of a signal y (t) is y(s) = $\dfrac{1}{s(s-1)}$, its final value is

    • 1
  1. 0

  2. 1

  3. unbounded


Correct Option: D
Explanation:

The unit impulse response of a system is h (t) = e-t , t$\ge$ 0 For this system, the steady-state value of the output for unit step input is equal to

    • 1
  1. 0

  2. 1

  3. $\infty$


Correct Option: C
Explanation:

The z-transform of a system is H(z) = $\dfrac{z}{z-0.2}$. If the ROC is |z| < 0.2, then the impulse response of the system is

  1. (0.2)n u[n]

  2. (0.2)n u[-n-1]

    • (0.2)n u[n]
    • (0.2)n u[-n-1]

Correct Option: D
Explanation:

Let (x) t be the input and (y) t be the output of a continuous time system. Match the system properties P1, P2 and P3 with system relations R1, R2, R3, R ||| |---|---| | Properties| Relations| | P1 : Linear but NOT time - invariant| R1 : y (t) = t2 x (t)| | P2 : Time - invariant but NOT linear| R2 : y (t) = |t| x (t)| | P3 : Linear and time - invariant| R3 : y (t) = |x (t)|| | | R4 : y (t) = x (t - 5)|

  1. (P1, R1), (P2, R3), (P3, R4)

  2. (P1, R2), (P2, R3), (P3, R4)

  3. (P1, R3), (P2, R1), (P3, R2)

  4. (P1, R1), (P2, R2), (P3, R3)


Correct Option: B
Explanation:

The impulse response H [n] of a linear time invariant system is given as $h[n] = \begin{cases} -2\sqrt 2 & n=1, -1 \\ 4\sqrt 2 & n=2, -2 \\ 0 & otherwise \end{cases}

$ If the input to the above system is the sequence ej$\pi$n/4, then the output is

  1. 4$\sqrt 2$ ej$\pi$n/4

  2. 4$\sqrt 2$ e-j$\pi$n/4

  3. 4 ej$\pi$n/4

    • 4 ej$\pi$n/4

Correct Option: D
Explanation:

The signal x (t) is described by x (t) =$ \begin{cases} 1 & for \ -1 \le t \le + 1 \\ 0 & otherwise \end{cases}

$ Two of the angular frequencies at which its Fourier transform becomes zero are

  1. $\pi$, 2$\pi$

  2. 0.5$\pi$, 1.5$\pi$

  3. 0, $\pi$

  4. 2$\pi$, 2.5$\pi$


Correct Option: A
Explanation:

Let P be linearity, Q be time-invariance, R be causality and S be stability. A discrete time system has the input-output relationship, Y(n) = $ \begin{cases} x(n), & n \ge 1 \\ 0 & n=0 \\ x(n+1), & n \le -1 \end{cases}

$ where x(n) is the input and y(n) is the output. The above system has the properties

  1. P, S but not Q, R

  2. P, Q, S but not R

  3. P, Q, R, S

  4. Q, R, S but not P


Correct Option: A
Explanation:

An LTI system having transfer function $\dfrac{s^2 +1}{s^2 +2s + 1}$ and input x (t) = sin (t + 1) is in steady state. The output is sampled at a rate $\omega_s$rad/s to obtain the final output {x (k)}. Which of the following is true?

  1. y (.) is zero for all sampling frequencies $\omega_s$.

  2. y (.) is nonzero for all sampling frequencies $\omega_s$.

  3. y (.) is nonzero for $\omega_s$> 2, but zero for $\omega_s$< 2.

  4. y (.) is zero for$\omega_s$ > 2, but nonzero for $\omega_s$< 2.


Correct Option: A

The frequency response of a linear, time-invariant system is given by H(f) = $\dfrac{5}{1 + j10\pi f}$ The step response of the system is

  1. 5 (1- e-5t) u (t)

  2. 5 $\left( 1 - e^{-\dfrac{t}{5}} \right) u(t)$

  3. $\dfrac{1}{5}$(1- e-5t) u (t)

  4. $\dfrac{1}{5}$$\left( 1 - e^{\dfrac{t}{5}} \right) u(t)$


Correct Option: B
Explanation:

Match the following and choose the correct combination:

 
Group 1 Group 2
E. continuous and aperiodic signal 1. Fourier representation is continuous and periodic
F. continuous and periodic signal 2. Fourier representation is discrete and periodic
G. discrete and aperiodic signal 3. Fourier representation is continuous and periodic
H. discrete and periodic signal 4. Fourier representation is discrete and periodic
  1. E – 3, F – 2, G – 4, H – 1

  2. E – 1, F – 3, G – 2, H – 4

  3. E – 1, F – 2, G – 3, H – 4

  4. E – 1, F – 3, G – 4, H – 2


Correct Option: D
Explanation:

For continuous and aperiodic signal, Fourier representation is continuous and periodic.

For continuous and periodic signal, Fourier representation is discrete and periodic.

For discrete and aperiodic signal, Fourier representation is discrete and periodic.

For discrete and periodic signal, Fourier representation is discrete and periodic.

The Dirac delta function $\delta$ (t) is defined as

  1. $\delta$ (t)$ \begin{cases} 1 & t = 0 \\ 0 & otherwise \\ \end{cases}

    $

  2. $\delta$ (t) $ \begin{cases} \infty & t = 0 \\ 0 & otherwise \\ \end{cases}

    $

  3. $\delta$ (t) $ \begin{cases} 1 & t = 0 \\ 0 & otherwise \\ \end{cases}

    \qquad and \ \quad

    \displaystyle \int_{-\infty}^\infty \delta (t) dt = 1 $

  4. $\delta$ (t) $ \begin{cases} \infty & t = 0 \\ 0 & otherwise \\ \end{cases}

    \qquad and \ \quad

    \displaystyle \int_{-\infty}^\infty \delta (t) dt = 1 $


Correct Option: D
Explanation:

Dirac delta function $\delta (t)$ is defined at $t = 0$ and it has infinite value a $t=0$. The area of dirac delta function is unity.

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