Test 4 - Signals and System | Electronics and Communication (ECE)
A test for Signals and System of Electronics and Communication (ECE)
Questions
A Hilbert transformer is a
- non-linear system
- non-causal system
- time-varying system
- low-pass 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
- 2
- 3
- 4
{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
- |X (k)|2
- $\dfrac{1}{N} \displaystyle \sum_{r = 0} ^{N-1} x (r) \times (k+r)$
- $\dfrac{1}{N} \displaystyle \sum_{r = 0} ^{N-1} x (r) \times (k+r)$
- 0
Given that F (s) is the one-sided Laplace transform of f (t). What is the Laplace transform of $\int_0^t f(\tau) d\tau$?
- F (s) - f (0)
- $\dfrac{1}{s}$F (s)
- $\displaystyle \int_0^s F(\tau) d\tau$
- $\dfrac{1}{s}$[F (s) - f (0)]
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.
- Causal, LP
- BIBO, LTI
- BIBO, Causal, LTI
- LP, LTI
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
- $\dfrac{1}{4}$
- 2
- 4
- $\dfrac{4}{3}$
If the Laplace transform of a signal y (t) is y(s) = $\dfrac{1}{s(s-1)}$, its final value is
- - 1
- 0
- 1
- unbounded
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
- 0
- 1
- $\infty$
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
- (0.2)n u[n]
- (0.2)n u[-n-1]
- - (0.2)n u[n]
- - (0.2)n u[-n-1]
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) = |
| P3 : Linear and time - invariant | R3 : y (t) = |
| R4 : y (t) = x (t - 5) |
- (P1, R1), (P2, R3), (P3, R4)
- (P1, R2), (P2, R3), (P3, R4)
- (P1, R3), (P2, R1), (P3, R2)
- (P1, R1), (P2, R2), (P3, R3)
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
- 4$\sqrt 2$ ej$\pi$n/4
- 4$\sqrt 2$ e-j$\pi$n/4
- 4 ej$\pi$n/4
- - 4 ej$\pi$n/4
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
- $\pi$, 2$\pi$
- 0.5$\pi$, 1.5$\pi$
- 0, $\pi$
- 2$\pi$, 2.5$\pi$
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
- P, S but not Q, R
- P, Q, S but not R
- P, Q, R, S
- Q, R, S but not P
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?
- y (.) is zero for all sampling frequencies $\omega_s$.
- y (.) is nonzero for all sampling frequencies $\omega_s$.
- y (.) is nonzero for $\omega_s$> 2, but zero for $\omega_s$< 2.
- y (.) is zero for$\omega_s$ > 2, but nonzero for $\omega_s$< 2.
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
- 5 (1- e-5t) u (t)
- 5 $\left( 1 - e^{-\dfrac{t}{5}} \right) u(t)$
- $\dfrac{1}{5}$(1- e-5t) u (t)
- $\dfrac{1}{5}$$\left( 1 - e^{\dfrac{t}{5}} \right) u(t)$
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 |
- E – 3, F – 2, G – 4, H – 1
- E – 1, F – 3, G – 2, H – 4
- E – 1, F – 2, G – 3, H – 4
- E – 1, F – 3, G – 4, H – 2
The Dirac delta function $\delta$ (t) is defined as
- $\delta$ (t)$ \begin{cases} 1 & t = 0 \\\\ 0 & otherwise \\\\ \end{cases} $
- $\delta$ (t) $ \begin{cases} \infty & t = 0 \\\\ 0 & otherwise \\\\ \end{cases} $
- $\delta$ (t) $ \begin{cases} 1 & t = 0 \\\\ 0 & otherwise \\\\ \end{cases} \qquad and \ \quad \displaystyle \int_{-\infty}^\infty \delta (t) dt = 1 $
- $\delta$ (t) $ \begin{cases} \infty & t = 0 \\\\ 0 & otherwise \\\\ \end{cases} \qquad and \ \quad \displaystyle \int_{-\infty}^\infty \delta (t) dt = 1 $