Introduction: Set-1
[1994 1M] The value of the integral \int_{-5}^{+6}\;e^{-2t}\;\delta(t-1)\;dt\; is equal to _______
Ans- e^{-2}
Explanation:
We know that the Shifting property of impulse function is given as,
\int_{-\infty}^\infty\;x(t)\;\delta(t-t_0)\;dt\;=\;x\left(t_0\right)Generally, the impulse is located at t = 1
Using the time shifting property,
\;\int_{-5}^{+6}\;e^{(-2\times1)}\;\delta(t-1)\;dt\;=\;e^{-2}\;\;\int_{-5}^{+6}\;\;\delta(t-1)\;dt\;\;=\;e^{-2}So, e^{-2} is the correct answer.
[2002 1M] A current impulse of 5\delta(t)\;, is forced through a capacitor C. The voltage v_c(t), across the capacitor is given by
a) 5t
b) 5 u(t) – C
c) \frac5Ct
d) \frac{5\;u(t)}C
Ans- (d) \frac{5\;u(t)}C
Explanation:
WE know that The formula for voltage across the capacitor is given as;
v_c\left(t\right)\;=\;\frac1C\;\int_{-\infty}^t\;i(t)\;dt \;v_c\left(t\right)\;=\;\frac1C\;\int_{-\infty}^t\;5\;\delta(t)\;dt v_c\left(t\right)\;=\;\frac5C\;\int_{-\infty}^t\;\delta(t)\;dtWe know that, \int_{-\infty}^t\;\delta(t)\;dt\;=\;u(t) ………….. impulse function property
\;v_c(t)\;=\;\frac5C\;u(t)[2004 2M] The rms value of the periodic waveform given in the figure is
a) 2\sqrt6\;A
b) 6\sqrt2\;A
c) \sqrt{\frac43}\;A
d) 1.5 A
Ans- (a) 2\sqrt6\;A
Explanation:
The formula for I_{rms} is
I_{rms}^2\;=\;\left[\frac1T\;\int_0^T\;i^2\;(t)\right]I_{rms} is considered for one complete cycle.
The complete cycle or waveform of current has summation of two parts. Second part can be written directly.
So for first part i.e. from 0 to T/2 , find the slope.
Slope\;=\;\frac yx\;=\;\frac{-6t}{\displaystyle\frac T2}\;=\;\frac{-12t}T I^{2}{rms}=\left [ \frac{1}{T} \int{0}^{T/2}\left ( \frac{-12T}{T} \right )^{2}dt+\int_{T/2}^{T}6^{2}dt\right ] I^{2}{rms}=\left [ \frac{1}{T} \int_{0}^{T/2}\left ( \frac{-144t^{2}}{T^{2}dt} \right )^{2}dt+\frac{36}{T}\int_{T/2}^{T}dt\right ] I_{rms}^2\;=\;\left[\frac{144}{T^3}\;\int_0^\frac T2\;t^2\;dt\;+\;\frac{36}T\;\int_\frac T2^T\;dt\right] I_{rms}^2\;=\;\left[\frac{144}{T^3}\;\left(\frac{t^3}3\right)0^\frac T2+\;\frac{36}T\;\left(t\right)\frac T2^T\right] I_{rms}^2\;=\;\frac{144}{T^3}\;\times\;\frac{T^3}{2^3\;\times\;3}\;+\;\frac{36}T\;\left[T\;-\;\frac T2\right] I_{rms}^2\;=\;\frac{144}{8\;\times\;3}\;+\;\frac{36}T\;\left[\frac T2\right] I_{rms}^2\;=\;6\;+\;\frac{36}2\;=\;6\;+\;18\;=\;24 \therefore\;I_{rms}\;=\;2\;\sqrt6\;A[2004: 2M] The rms value of the resultant current in a wire which carries a dc current of 10A and a sinusoidal alternating current of peak value 20A is
a) 14.1 A
b) 17.3 A
c) 22.4 A
d) 30.0 A
Ans- (b) 17.3 A
Explanation: It is given that,
RMS value of dc current = 10 A
Peak value of alternating current = 20 A
The resultant current is the sum of rms value of dc current and rms value of altertnating current.
∴ RMS value of alternating current = \frac{20}{\sqrt2}\;=\;\frac{10\;\times\;2}{\sqrt2}\;=\;10\sqrt2\;A
RMS value of resultant current = \sqrt{\left(RMS\:value\;of\;dc\;current\right)^2\;+\;\left(RMS\;value\;of\;ac\;current\right)^2}
RMS value of resultant current = \sqrt{\left(10\right)^2\;+\;\left(10\;\sqrt2\right)^2}\;=\;17.32\;A
So, option (b) is the correct answer.
Q. For the triangular waveform shown in the figure, the RMS value of the voltage is equal to
a) \sqrt{\frac16}\;V
b) \sqrt{\frac13}\;V
c) \frac13\;V
d) \sqrt{\frac23}\;V
Ans- (a) \sqrt{\frac16}\;V
Explanation:
RMS value for one cycle i.e. from 0 to T/2 and T/2 to T.
Consider for a period of 0 to T/2 and find the slope.
Slope\;=\;\frac yx\;=\;\frac{1t}{\displaystyle\frac T2}\;=\;\frac{2t}TFor the period of T/2 to T it is zero.
V_{rms}^2\;={\frac1T\left[\int_0^\frac T2\;\left(\frac{2t}T\right)^2\;dt\;+\;\int_\frac T2^T\;0\;dt\right]}^{1/2} V_{rms}^2\;=\;\frac1T\;\int_0^\frac T2\;\left(\frac{4\;t^2}{T^2}\right)\;dt V_{rms}^2\;=\;\frac4{T^3}\;\int_0^\frac T2\;t^2\;dt\;=\;\frac4{T^3}\;\left[\frac{t^3}3\right]_0^\frac T2 V_{rms}^2\;=\;\frac4{T^3}\;\left[\frac T2\;-\;0\right]^3\;=\;\frac4{3\;T^3}\;\times\;\frac{T^3}8\;=\;\frac16 V_{rms}\;=\;\sqrt{\frac16}\;VSo, option (a) is the correct answer.
[2006 1M] Which of the following is true?
a) A finite amplitude signal is always time bounded.
b) A time bounded signal always possesses finite energy.
c) A time bounded signal is always zero outside the interval \left[-t_0\;,\;t_0\right] for some t0.
d) A time bounded signal is always finite.
Ans- (c) A time bounded signal is always zero outside the interval \left[-t_0\;,\;t_0\right] for some t0.
[2006 2M] A continuous-time system is described by y\left(t\right)\;=\;e^{-\left|x\left(t\right)\right|} , where y(t) is the output and x(t) is the input. y(t) is bounded
a) only when x(t) is bounded.
b) only when x(t) is non-negative.
c) only for t ≤ 0 if x(t) is bounded for t ≥ 0.
d) even when x(t) is not bounded.
Ans- (d)
Explanation:
e-x is always convergent and bounded when x is not bounded.
∴ y(t) is bounded even though x(t) is not bounded.
So, option (d) is the correct answer.
[2006 2M] The integrator, given by
y\left(t\right)\;=\;\int_{-\infty}^t\;x(\alpha)\;d\alpha
a) has no finite singularities in its double sided Laplace Transform H(s).
b) produces a bounded output for every causal bounded input.
c) produces a bounded output for every anticausal bounded input.
d) has no finite zeros in its double sided Laplace Transform H(s).
Ans- (d) has no finite zeros in its double sided Laplace Transform H(s).
[2007 1M] The frequency spectrum of a signal is shown in the figure. If this is ideally sampled at intervals of 1ms, then the frequency spectrum of the sampled signal will be
Ans- (b)
Explanation:
Given, Sampling interval Ts = 1 msec = 10-3 sec.
∴ Sampling frequency; f_s\;=\;\frac1{T_s}\;=\;\frac1{10^{-3}}\;=\;10^3\;=\;1\;KHz
After Sampling, the new signal in frequency domain is
U_T(j\omega)\;=\;\frac1{T_s}\;\sum_{n\;=\;-\infty}^\infty\;U\;(f\;-\;nf_s)After solving the above signal, we get a constant term having amplitude which varies from -∾ to ∾.
∴ The sampled signal will be
So, option (b) is correct.
Q. A signal is processed by a causal filter with transfer function G(s). For a distortion free output signal waveform, G(s) must [2007 2M]
a) provide zero phase shift for all frequency.
b) provide constant phase shift for all frequency.
c) provide linear phase shift that is proportional to frequency.
d) provide a phase shift that is inversely proportional to frequency.
Ans- (c)
Explanation:
For a distortion free output signal waveform, phase shift that is proportional to frequency should be provided so that equal delay could be imparted to all frequency component.