Test 1 - Signals and System | Electronics and Communication (ECE)
A test for Signals and System of Electronics and Communication (ECE)
Questions
A system is defined by its impulse response h (n) = 2nu (n - 2). The system is
- stable and causal
- causal but not stable
- stable but not causal
- unstable and non-causal
The Fourier series expansion of a real periodic signal with fundamental frequency f0 is given by gp (t) = $\displaystyle \sum_{n=-\infty}^\omega C_n e^{j^2 x n f_0 t}$it is given that C3 = 3 + j5. Then C-3 is
- 5+j3
- -3-j5
- -5+j3
- 3-j5
An input x (t) exp (- 2t) u (t) + $\delta$ (t - 6) is applied to an LTI system with impulse response h(t) u (t). The output is
- 1 - exp (- 2t)] u (t + 6)
- [1 - exp (- 2t)] u (t) + u (t - 6)
- 0.5 [1 - exp (- 2t)] u(t) + u (t + 6)
- 0.5 [1 - exp (- 2t)] u (t) + u (t - 6)
If x[n]=(1/3)| n | - (1/2)n u[n], then the region of convergence (ROC) of its Z-transform in the z- Plane will be Z - plane will be
- $\dfrac{1}{3}$<$\left | z \right|$< 3
- $\dfrac{1}{3}$<$\left | z \right|$< $\dfrac{1}{2}$
- $\dfrac{1}{2}$<$\left | z \right|$< 3
- $\dfrac{1}{3}$<$\left | z \right|$
The Fourier transform of a signal h(t) is H (j$\omega$) = (2 cos$\omega$) (sin2$\omega$)/$\omega$. The value of h(0) is
- ¼
- ½
- 1
- 2
The Fourier transform of a conjugate symmetric function is always
- imaginary
- conjugate anti-symmetric
- real
- conjugate symmetric
A discrete time linear shift - invariant system has an impulse response h [n] with h [0] = 1, h [1] = - 1, h [2] = 2 and zero otherwise. The system is given an input sequence x [n] with x [0] = x [2] = 1 and zero otherwise. The number of non zero samples in the output sequence y [n] and the value of y [2] are respectively
- 5, 2
- 6, 2
- 6, 1
- 5, 3
The ROC of z -transform of the discrete time sequence x (n) = $\left( \dfrac{1}{3} \right)^n u(n) - \left( \dfrac{1}{2} \right)^n$u (- n - 1) is
- $\dfrac{1}{3} \lt \left| z \right| \lt \dfrac{1}{2}$
- |z| > $\dfrac{1}{2}$
- |z| < $\dfrac{1}{3}$
- 2 < |z| < 3
The unit-step response of a system starting from rest is given by
c (t) = 1 - e-2t for t $\ge$ 0
The transfer function of the system is
- $\dfrac{1}{1+2s}$
- $\dfrac{2}{2+s}$
- $\dfrac{1}{2+s}$
- $\dfrac{2s}{1+2s}$
The trigonometric Fourier series of an even function does not have the
- dc term
- cosine terms
- sine terms
- odd harmonic terms
The transfer function of a discrete time LTI system is given by
H(z) = $\dfrac{ 2-\dfrac{3}{4}z^{-1} }{ 1 - \dfrac{3}{4}z^{-t} + \dfrac{1}{8}z^{-2} }$
Consider the following statements:
S1: The system is stable and causal for ROC:|z|>½
S2: The system is stable but not causal for ROC:|z|<¼
S3: The system is neither stable nor causal for ROC: ¼<|z|<½
Which one of the following statements is valid?
- Both S1 and S2 are true.
- Both S2 and S3 are true.
- Both S1 and S3 are true.
- S1, S2 and S3 are all true.
Let y[n] denote the convolution of h[n] and g[n], where h[n]= (1/2)n u[n] and g[n] is a causal sequence. If y[0] = 1 and y[1] = 1/2, then g[1] equals
- 0
- 1/2
- 1
- 3/2
A sequence x(n) has non-zero values as shown in the figure.
The Fourier transform of y(2n) will be
- $e^{-j2\omega} [cos4\omega + 2cos 2 \omega +2]$
- $[cos2\omega + 2cos \omega +2]$
- $e^{-j\omega} [cos2\omega + 2cos \omega +2]$
- $e^{j2\omega} [cos2\omega + 2cos \omega +2]$
A sequence x(n) has non-zero values as shown in figure.
The sequence
$y(n) =
\begin{cases}
x\left( \dfrac{n}{2} -1 \right) & \text{for n even} \\
0 & \text{for n odd}
\end{cases}
$
will be
The Fourier series of a real periodic function has only
(P) cosine terms if it is even
(Q) sine terms if it is even
(R) cosine terms if it is odd
(S) sine terms if it is odd
Which of the above statements are correct?
- P and S
- P and R
- Q and S
- Q and R
The first six points of the 8-point DFT of a real valued sequence are 5, 1 - j3,0,3 - j4, 0 and 3 + j4. The last two points of the DFT are respectively
- 0, 1 - j3
- 0, 1 + j3
- 1 + j3, 5
- 1 - j3, 5
Which of the following can be impulse response of a causal system?
Consider the z-transform X(z) = 5z2 + 4z-1 + 3; 0<|z| < $\infty$. The inverse z-transform x[n] is
- 5$\delta$[n + 2] + 3$\delta$[n] + 4$\delta$[n - 1]
- 5$\delta$[n - 2] + 3$\delta$[n] + 4$\delta$[n + 1]
- 5 u[n + 2] + 3 u[n] + 4 u[n - 1]
- 5 u[n - 2] + 3 u[n] + 4 u[n + 1]
The impulse response h (t) of a linear time invariant continuous time system is described by h (t) = exp ($\alpha$t) u (t) + exp ($\beta$t) u (- t) where u (- t) denotes the unit step function, and $\alpha$ and $\beta$ are real constants. The system is stable if
- $\alpha$ is positive and $\beta$ is positive
- $\alpha$ is negative and $\beta$ is negative
- $\alpha$ is positive and $\beta$ is negative
- $\alpha$ is negative and $\beta$ is positive
Let x (t) $\leftrightarrow$ X (j$\omega$) be Fourier Transform pair. The Fourier Transform of the signal x (5t − 3) in terms of X (j$\omega$) is given as
- $\dfrac{1}{5}e^{\dfrac{-j3\omega}{5}} \times \left( \dfrac{j\omega}{5} \right)$
- $\dfrac{1}{5}e^{\dfrac{j3\omega}{5}} \times \left( \dfrac{j\omega}{5} \right)$
- $\dfrac{1}{5}e^{\dfrac{-j3\omega}{5}} \times \left( \dfrac{j\omega}{5} \right)$
- $\dfrac{1}{5}e^{\dfrac{j3\omega}{5}} \times \left( \dfrac{j\omega}{5} \right)$
