61. A discrete-time signal x[n] = δ[n - 3] + 2δ[n - 5] has z-transform X(z). If Y(z) = X(-z) is the z-transform of another signal y[n], then
62. For a periodic signal
$$v\left( t \right) = 30\sin 100t + 10\cos 300t + 6\sin \left( {500t + {\pi \over 4}} \right),$$
the fundamental frequency in rad/s is
$$v\left( t \right) = 30\sin 100t + 10\cos 300t + 6\sin \left( {500t + {\pi \over 4}} \right),$$
the fundamental frequency in rad/s is
63. The final value theorem is used to find the
64. The impulse response h[n] of a linear time-invariant system is given by
h[n] = u[n + 3] + u[n - 2] - 2u[n - 7],
where u[n] is the unit step sequence. The above system is
h[n] = u[n + 3] + u[n - 2] - 2u[n - 7],
where u[n] is the unit step sequence. The above system is
65. The inverse Laplace transform of the function $${{s + 5} \over {\left( {s + 1} \right)\left( {s + 3} \right)}}$$ is
66. If the Laplace transform of a signal y(t) is $$Y\left( s \right) = {1 \over {s\left( {s - 1} \right)}},$$ then its final value is
67. Specify the filter type if its voltage transfer function H(s) is given by
$$H\left( s \right) = {{K\left( {{s^2} + 1\omega _0^2} \right)} \over {{s^2} + \left( {{{{\omega _0}} \over Q}} \right)s + \omega _0^2}}$$
$$H\left( s \right) = {{K\left( {{s^2} + 1\omega _0^2} \right)} \over {{s^2} + \left( {{{{\omega _0}} \over Q}} \right)s + \omega _0^2}}$$
68. Input x(t) and output y(t) of an LTI system are related by the differential equation y"(t) - y'(t) - 6y(t) = x(t). If the system is neither causal nor stable, the impulse response h(t) of the system is
69. A stable linear time invariant (LTI) system has a transfer function $$H\left( s \right) = {1 \over {{s^2} + s - 6}}.$$ To make this system causal it needs to be cascaded with another LTI system having a transfer function H1(s). A correct choice for H1(s) among the following options is
70. A signal m(t) with bandwidth 500 Hz is first multiplied by a signal g(t) where
$$g\left( t \right) = \sum\limits_{k = - \infty }^\infty {{{\left( { - 1} \right)}^k}\delta \left( {t - 0.5 \times {{10}^{ - 4}}k} \right)} $$
The resulting signal is then passed through an ideal low pass filter with bandwidth 1 kHz. The output of the low pass filter would be
$$g\left( t \right) = \sum\limits_{k = - \infty }^\infty {{{\left( { - 1} \right)}^k}\delta \left( {t - 0.5 \times {{10}^{ - 4}}k} \right)} $$
The resulting signal is then passed through an ideal low pass filter with bandwidth 1 kHz. The output of the low pass filter would be
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