Test 2 - Signals and System | Electronics and Communication (ECE)

Description: A test for Signals and System of Electronics and Communication (ECE)
Number of Questions: 20
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Tags: Signals and System
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A signal $x(n) = sin(\omega_on + \phi)$ is the input to a linear time invariant system having a frequency response $H(e^{j\omega})$. If the output of the system is $Axn - n_0$, then the most general form of $\angle H (e^{j\omega})$ will be

  1. $-n_0 \omega_0 + \beta$ for any arbitrary real $\beta$

  2. $-n_0 \omega_0 + 2\pi k$ for any arbitrary integer $k$

  3. $n_0 \omega_0 + 2\pi k$ for any arbitrary integer $k$

  4. $-n_0 \omega_0 + \phi$


Correct Option: B
Explanation:

The unilateral Laplace transform of f(t) is $\dfrac{1}{s^2 + s +1}$. The unilateral Laplace transform of t f (t) is

    • $\dfrac{1}{ (s^2 + s +1)^2}$
    • $\dfrac{2s+1}{ (s^2 + s +1)^2}$
  1. $\dfrac{s}{ (s^2 + s +1)^2}$

  2. $\dfrac{2s+1}{ (s^2 + s +1)^2}$


Correct Option: D

In the following network, the switch is closed at t = 0- and the sampling starts from t = 0. The sampling frequency is 10 Hz.

The samples x (n), n = (0, 1, 2....) are given by

  1. 5 (1 - $e^{-0.05n}$)

  2. 5$e^{-0.05n}$

  3. 5 (1 - $e^{-\delta n}$)

  4. 5$e^{-\delta n}$


Correct Option: B
Explanation:

In the following network, the switch is closed at t = 0- and the sampling starts from t = 0. The sampling frequency is 10 Hz.

The expression and the region of convergence of the z −transform of the sampled signal are

  1. $\dfrac{5z}{z-e^5}, \left| z \right| \lt e^{-5}$

  2. $\dfrac{5z}{z-e^{0.05}}, \left| z \right| \lt e^{-0.05}$

  3. $\dfrac{5z}{z-e^{-0.05}}, \left| z \right| \lt e^{-0.05}$

  4. $\dfrac{5z}{z-e^5}, \left| z \right| \lt e^{-5}$


Correct Option: C
Explanation:

The Laplace transform of i(t) is given by I(s) = $\dfrac{2}{s(1+s)}$ As t $\rightarrow$$\infty$, the value of i(t) tends to

  1. 0

  2. 1

  3. 2

  4. $\infty$


Correct Option: C
Explanation:

The function x(t) is shown in figure. Even and odd parts of a unit-step function u(t) are respectively

  1. $\dfrac{1}{2}. \dfrac{1}{2} \times (t)$

  2. $-\dfrac{1}{2}. \dfrac{1}{2} \times (t)$

  3. $\dfrac{1}{2}. -\dfrac{1}{2} \times (t)$

  4. $-\dfrac{1}{2}. -\dfrac{1}{2} \times (t)$


Correct Option: A
Explanation:

Two systems H1 (z) and H2 (z) are connected in cascade as shown below. The overall output y(n) is the same as the input x(n) with a one unit delay. The transfer function of the second system H2 (z) is

  1. $\dfrac{ (1-0.6z^{-1}) }{z^{-1} (1-0.4z^{-1}) }$

  2. $\dfrac{ z^{-1} (1-0.6z^{-1}) }{ (1-0.4z^{-1}) }$

  3. $\dfrac{ z^{-1} (1-0.4z^{-1}) }{ (1-0.6z^{-1}) }$

  4. $\dfrac{ (1-0.4z^{-1}) }{ z^{-1}(1-0.6z^{-1}) }$


Correct Option: B
Explanation:

The impulse response h (t) of linear time - invariant continuous time system is given by h (t) = exp (- 2t) u (t) , where u (t) denotes the unit step function.

The output of this system, to the sinusoidal input x (t) = 2 cos 2t for all time t, is

  1. 0

  2. $2^{-0.25} cos(2t - 0.125\pi)$

  3. $2^{-0.5} cos(2t - 0.125\pi)$

  4. $2^{-0.5} cos(2t - 0.25\pi)$


Correct Option: D
Explanation:

The z-transform X [z] of a sequence x[n] is given by x[z] =$\dfrac{0.5}{1-2z^{-1}}$. It is given that the region of convergence of X[z] includes the unit circle. The value of x[0] is

    • 0.5
  1. 0

  2. 0.25

  3. 0.5


Correct Option: B
Explanation:

A system with transfer function H(z) has impulse response h(.), defined as h(2) = 1, h(3) = - 1 and h(k) = 0 otherwise. Consider the following statements.

S1 : H(z) is a low-pass filter. S2 : H(z) is an FIR filter.

Which of the following is correct?

  1. Only S2 is true

  2. Both S1 and S2 are false

  3. Both S1 and S2 are true, and S2 is a reason for S1

  4. Both S1 and S2 are true, but S2 is not a reason for S1


Correct Option: A
Explanation:

A continuous time LTI system is described by $\dfrac{d^2 y(t)}{dt^2} + 4 \dfrac{dy(t)}{dt} 3 y(t) = 2 \dfrac{dx(t)}{dt} + 4 \times (t)$. Assuming zero initial condition, the response y(t) of the above system for the input x(t) = e-2tu(t) is given by

  1. (et-e3t)u(t)

  2. (e-t-3-3t)u(t)

  3. (e-t+e-3t)u(t)

  4. (et+e3t)u(t)


Correct Option: B
Explanation:

The input and output of a continuous time system are respectively denoted by x (t) and y (t). Which of the following descriptions corresponds to a casual system?

  1. y (t) = x (t - 2) + x (t + 4)

  2. y (t) = (t - 4) x (t + 1)

  3. y (t) = (t + 4) x (t - 1)

  4. y (t) = (t + 5) x (t + 5)


Correct Option: C
Explanation:

Consider the function f(t) having Laplace transform F (s) = $\dfrac{\omega_0}{s^2 + \omega^2_0}$ Re [s] > 0

The final value of f(t) would be

  1. 0

  2. 1

  3. − 1 $\le f(\infty) \le 1$

  4. $\infty$


Correct Option: C
Explanation:

A system with input x [n] and output y [n] is given as y [n] = $\left( sin\dfrac{5}{6} \pi n \right)$x (n). The system is

  1. linear, stable and invertible

  2. non-linear, stable and non-invertible

  3. linear, stable and non-invertible

  4. linear, unstable and invertible


Correct Option: C
Explanation:

The impulse response h (t) of linear time - invariant continuous time system is given by h (t) = exp (- 2t) u (t) , where u (t) denotes the unit step function.

The frequency response H $(\omega)$of this system in terms of angular frequency$(\omega)$, is given by H $(\omega)$ =

  1. $\dfrac{1}{1+j2\omega}$

  2. $\dfrac{sin\omega}{\omega}$

  3. $\dfrac{1}{2+j\omega}$

  4. $\dfrac{j\omega}{2+j\omega}$


Correct Option: C
Explanation:

The output y(t) of a linear time invariant system is related to its input x(t) by the following equation: $y(t) = 0.5 \times (t-t_d + T) + x(t-t_d) + 0.5 \times (t-t_d-T)$. The filter transfer function $H(\omega)$ of such a system is given by

  1. $(1+cos\omega T) e^{-j\omega t_d}$

  2. $(1+ 0.5cos\omega T) e^{-j\omega t_d}$

  3. $(1+cos\omega T) e^{j\omega t_d}$

  4. $(1+ 0.5cos\omega T) e^{-j\omega t_d}$


Correct Option: A
Explanation:

Let x( t) and y( t) with Fourier transforms F (f) and Y( f) respectively be related as shown in Fig. Then Y( f) is

  1. -$\dfrac{1}{2} X(f/2) e^{-jxf}$

  2. -$\dfrac{1}{2} X(f/2) e^{-j2xf}$

  3. -$ X(f/2) e^{j2xf}$

  4. -$ X(f/2) e^{-j2xf}$


Correct Option: B
Explanation:

For an N-point FFT algorithm with N = 2m which one of the following statements is TRUE?

  1. It is not possible to construct a signal flow graph with both input and output in normal order.

  2. The number of butterflies in the mth stage is N/m.

  3. In-place computation requires storage of only 2N node data.

  4. Computation of a butterfly requires only one complex multiplication.


Correct Option: D
Explanation:

The 3-dB bandwidth of the low-pass signal e-t u(t), where u(t) is the unit step function, is given by

  1. $\dfrac{1}{2\pi} HZ$

  2. $\dfrac{1}{2\pi} \sqrt{ \sqrt{2} - 1HZ }$

  3. $\infty$

  4. 1 HZ


Correct Option: A
Explanation:

A causal LTI system is described by the difference equation 2y[n] = $\alpha$y[n-2] -2x [n] + $\beta$x[n-1]
The system is stable only if

  1. |$\alpha$| = 2, |$\beta$| < 2

  2. |$\alpha$| > 2, |$\beta$| > 2

  3. |$\alpha$| < 2, any value of $\beta$

  4. |$\beta$| < 2, any value of $\alpha$


Correct Option: C
Explanation:

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