Find the Nyquist rate and Nyquist interval of sin(2πt).

\(\frac{1}{2}\) Hz, \(\frac{1}{2}\) sec

Consider the two continuous-time signals defined below:\({x_1}\left( t \right) = \left\{ {\begin{array}{*{20}{c}}{\left| t \right|, - 1 \le t \le 1\;}\\{0,\;otherwise\;}\end{array},} \right.{x_2}\left( t \right) = \left\{ {\begin{array}{*{20}{c}}{1 - \left| t \right|, - 1 \le t \le 1\;}\\{0,\;otherwise\;}\end{array}} \right.\)These signals are sampled with a sampling period of T = 0.25 seconds to obtain discrete time signals x1[n] and x2[n], respectively. Which one of the following statements is true?

The energy of x2[n] is greater than the energy of x2[n].

The energy of x1[n] is greater than the energy of x2[n].

The energy of x2[n] is greater than the energy of x2[n].

x1[n] and x2 have equal energies.

Neither x1[n] nor x2[n] is a finite-energy signal.

In flat-top sampling, a hold circuit is sometimes required. This hold circuit can be designed as a sampler followed by

A series resistance along with parallel RC circuit in shunt

If the sampling is carried out at rate higher than twice the highest frequency of the original single (fmax), then it is possible to receive the original signal from the sampled signal by passing it through

A high-pass filter with the cut-off frequency greater than fmax

A high-pass filter with the cut-off frequency equal to fmax

A low-pass filter with the cut-off frequency equal to fmax

A high-pass filter with the cut-off frequency greater than fmax

A low-pass filter with the cut-off frequency lesser than fmax

A band limited signal is sampled at Nyguist rate, The signal can be recovered by passing the samples through

ideal low pass filter with appropriate band width

PAM signals are constructed by using a low pass filter of passband slightly greater than baseband to avoid aliasing. This avoids distortion :

True if low pass filter has sharp cut off

Flat top pulses introduce envelope delay

Flat top pulses introduce amplitude distortion and delay

Nyquist frequency is double the folding frequency

Nyquist frequency is the same as the sampling frequency

Nyquist frequency is the same as folding frequency

Nyquist frequency is double the folding frequency

Nyquist frequency is half of oversampling rate

Let \({x_1}\left( t \right) \leftrightarrow {X_1}\left( \omega \right)\) and \({x_2}\left( t \right) \leftrightarrow {X_2}\left( \omega \right)\) be two signals whose Fourier Transforms are as shown in the figure below. In the figure, \(h\left( t \right) = {e^{ - 2\left| t \right|}}\) denotes the impulse response.For the system shown above, the minimum sampling rate required to sample y(t), so that y(t) can be uniquely reconstructed from its samples, is

\(4\left( {{B_1} + {B_2}} \right)\)

\(2\left( {{B_1} + {B_2}} \right)\)

\(4\left( {{B_1} + {B_2}} \right)\)

Assertion (A): Sampling theorem forms the basis for the relation between continuous-time signals and discrete-time signals.Reason (R): If the sampling is done at Nyquist rate, the signal x(t) can be reconstructed from the sample values x[n] = x[nTs]; −∞ ≤ n ≤ ∞.

Both (A) and (R) are true, but (R) is not the correct explanation of (A)

Both (A) and (R) are true and (R) is the correct explanation of (A)

Both (A) and (R) are true, but (R) is not the correct explanation of (A)

Directions: The following items consist of two statements, one labeled as “Assertion(A)” and the other labeled as “Reason(R)”. You are to examine the two statements carefully and decide if the Assertion (A) and the Reason (R) are individually true and if so whether the reason is a correct explanation of the assertion. Select your answer to these items using the codes given below and mark your answer accordingly.Assertion (A): Lesser number of bits per code are required due to less number of quantization levels is DPCM.Reason (R): In this case, the difference between two successive samples is quantized which does not differ much in amplitude.

Both (A) and (R) are true, but (R) is not the correct explanation of (A).

Both (A) and (R) are true and (R) is the correct explanation of (A).

Both (A) and (R) are true, but (R) is not the correct explanation of (A).

Assertion (A): Nyquist rate of sampling is the theoretical minimum sampling rate at which the signal can be sampled and still be reconstructed from its samples.Reason (R): When the Nyquist rate sampling is used, only an ideal low pass filter can be used to extract signal x(t) from sampled signal xs(t).Code:

Both (A) and (R) are true, but (R) is not the correct explanation of (A)

Both (A) and (R) are true and (R) is the correct explanation of (A).

Both (A) and (R) are true, but (R) is not the correct explanation of (A)

Assertion (A): A Non-Return to Zero (NRZ) type digital recording system is more common and efficient.Reason (R): It is possible to record twice the number of digits for the same number of pulses.

Both (A) and (R) are true, but (R) is not the correct explanation of (A).

Both (A) and (R) are true and (R) is the correct explanation of (A).

Both (A) and (R) are true, but (R) is not the correct explanation of (A).

113 + Mcqs in Sampling in Concrete Technology Page 1 McqOptions

1.	The loss in weight should not exceed ________ % when tested with Na2SO4 and ________ % with MgSO4 solution.
A.	12, 18
B.	18, 12
C.	10, 15
D.	15, 10
Answer» B. 18, 12

Discussion

2.	The mineral oil if present in mixing for concrete ______________
A.	Improves strength
B.	Reduces strength
C.	Gives more slump
D.	Gives a smooth surface
Answer» B. Reduces strength

Discussion

3.	When sand is fully dry then it’s volume is _________
A.	Equal
B.	Less
C.	More
D.	Can’t say
Answer» B. Less

Discussion

4.	Loss on Ignition (L.O.I.) is the loss in weight of cement after being heated to ____________
A.	1000°C
B.	100°C
C.	500°C
D.	1500°C
Answer» B. 100°C

Discussion

5.	Find the Nyquist rate and Nyquist interval for the signal f(t) = -10 sin ⁡40πt cos ⁡300πt.
A.	40 Hz, 40 sec
B.	340 Hz, 340 sec
C.	300 Hz, 300 sec
D.	340 Hz, \(\frac{1}{340}\) sec
Answer» E.

Discussion

6.	Determine the Nyquist rate of the signal x(t) = 1 + cos⁡ 2000πt + sin⁡ 4000πt.
A.	2000 Hz
B.	4000 Hz
C.	1 Hz
D.	6000 Hz
Answer» C. 1 Hz

Discussion

7.	The sampling frequency of a signal is Fs = 2000 samples per second. Find its Nyquist interval.
A.	0.5 sec
B.	5 msec
C.	5 sec
D.	0.5 msec
Answer» C. 5 sec

Discussion

8.	Find the Nyquist rate and Nyquist interval for the signal f(t) = rect(200t).
A.	∞ Hz, 0 sec
B.	0 Hz, ∞ sec
C.	∞ Hz, ∞ Hz
D.	0 Hz, 0 sec
Answer» B. 0 Hz, ∞ sec

Discussion

9.	Find the Nyquist rate and Nyquist interval for the signal f(t) = 1 + sinc300πt.
A.	300 Hz, 3 msec
B.	300 Hz, 3.3 msec
C.	30 Hz, 3 msec
D.	3 Hz, 3 msec
Answer» C. 30 Hz, 3 msec

Discussion

10.	Find the Nyquist rate and Nyquist interval for the signal f(t)=\(\frac{sin⁡500πt}{πt}\).
A.	500 Hz, 2 sec
B.	500 Hz, 2 msec
C.	2 Hz, 500 sec
D.	2 Hz, 500 msec
Answer» C. 2 Hz, 500 sec

Discussion

11.	Which of the following is the process of ‘aliasing’?
A.	Peaks overlapping
B.	Phase overlapping
C.	Amplitude overlapping
D.	Spectral overlapping
Answer» E.

Discussion

12.	Find the Nyquist rate and Nyquist interval of sinc[200t].
A.	200 Hz, \(\frac{1}{200}\) sec
B.	200 Hz, 200 sec
C.	\(\frac{1}{200}\) Hz, 200 sec
D.	100 Hz, 100 sec
Answer» B. 200 Hz, 200 sec

Discussion

13.	Find the Nyquist rate and Nyquist interval of Asinc[t].
A.	2 Hz, 2 sec
B.	1 Hz, 1 sec
C.	\(\frac{1}{2}\) Hz, 1 sec
D.	1 Hz, \(\frac{1}{2}\) sec
Answer» C. \(\frac{1}{2}\) Hz, 1 sec

Discussion

14.	Find the Nyquist rate and Nyquist interval of sinc[t].
A.	1 Hz, 1 sec
B.	2 Hz, 2 sec
C.	\(\frac{1}{2}\) Hz, 2 sec
D.	2 Hz, \(\frac{1}{2}\)sec
Answer» B. 2 Hz, 2 sec

Discussion

15.	Find the Nyquist rate and Nyquist interval of sin(2πt).
A.	2 Hz, \(\frac{1}{2}\) sec
B.	\(\frac{1}{2}\) Hz, \(\frac{1}{2}\) sec
C.	\(\frac{1}{2}\) Hz, 2 sec
D.	2 Hz, 2 sec
Answer» B. \(\frac{1}{2}\) Hz, \(\frac{1}{2}\) sec

Discussion

16.	Consider a real-value based-band signal x(t), band limited to 10 kHz. The Nyquist rate for the signal y(t) = x(t) \(x(1+ \frac{t}{2})\) is
A.	60 kHz
B.	30 kHz
C.	15 kHz
D.	20 kHz
Answer» C. 15 kHz

Discussion

17.	Consider the analog signal x(t) = 3 cos 100 πt. If the signal is sampled at 200 Hz, the discrete time signal obtained will be
A.	3 cos (πn/4)
B.	3 cos (πn/2)
C.	3 cos (πn)
D.	3 cos (πn/3)
Answer» C. 3 cos (πn)

Discussion

18.	In Digital Filters, how many interpolated data points are inserted between samples performing 4X over sampling?
A.	2
B.	3
C.	4
D.	5
Answer» D. 5

Discussion

19.	Consider the two continuous-time signals defined below:\({x_1}\left( t \right) = \left\{ {\begin{array}{{20}{c}}{\left\| t \right\|, - 1 \le t \le 1\;}\\{0,\;otherwise\;}\end{array},} \right.{x_2}\left( t \right) = \left\{ {\begin{array}{{20}{c}}{1 - \left\| t \right\|, - 1 \le t \le 1\;}\\{0,\;otherwise\;}\end{array}} \right.\)These signals are sampled with a sampling period of T = 0.25 seconds to obtain discrete time signals x1[n] and x2[n], respectively. Which one of the following statements is true?
A.	The energy of x1[n] is greater than the energy of x2[n].
B.	The energy of x2[n] is greater than the energy of x2[n].
C.	x1[n] and x2 have equal energies.
D.	Neither x1[n] nor x2[n] is a finite-energy signal.
Answer» B. The energy of x2[n] is greater than the energy of x2[n].

Discussion

20.	Calculate the minimum sampling rate to avoid aliasing when a continuous-time signal is given by x(t) = 5 cos 400πt
A.	100 Hz
B.	250 Hz
C.	400 Hz
D.	20 Hz
Answer» D. 20 Hz

Discussion

21.	Calculate the Nyquist rate for sampling when a continuous time signal is given by:x(t) = 5 cos 100πt +10 cos 200πt - 15 cos 300πt
A.	100 Hz
B.	150 Hz
C.	300 Hz
D.	600 Hz
Answer» D. 600 Hz

Discussion

22.	Given a continuous time signal s(t) = 5cos(200 πt). The minimum sampling rate required to avoid aliasing will be
A.	50 Hz
B.	100 Hz
C.	200 Hz
D.	400 Hz
Answer» D. 400 Hz

Discussion

23.	For the signal f(t) = 2 sin 9πt + sin 12 πt + sin 14 πt the minimum sampling frequency (in Hz) satisfying the Nyquist criterion is:
A.	14 sample / sec
B.	8 sample / sec
C.	12 sample / sec
D.	None of the above
Answer» B. 8 sample / sec

Discussion

24.	In flat-top sampling, a hold circuit is sometimes required. This hold circuit can be designed as a sampler followed by
A.	A shunt capacitor
B.	An envelope detector
C.	Parallel RC circuit
D.	A series resistance along with parallel RC circuit in shunt
Answer» B. An envelope detector

Discussion

25.	Find out the Nyquist rate of the following signal: x(t) = 3 sin 10πt + 100 sin 400πt – 9 cos 110πt.
A.	More than 5 Hz
B.	More than 200 Hz
C.	More than 110 Hz
D.	Less than 110 Hz
Answer» C. More than 110 Hz

Discussion

26.	If the sampling is carried out at rate higher than twice the highest frequency of the original single (fmax), then it is possible to receive the original signal from the sampled signal by passing it through
A.	A high-pass filter with the cut-off frequency equal to fmax
B.	A low-pass filter with the cut-off frequency equal to fmax
C.	A high-pass filter with the cut-off frequency greater than fmax
D.	A low-pass filter with the cut-off frequency lesser than fmax
Answer» C. A high-pass filter with the cut-off frequency greater than fmax

Discussion

27.	A bandpass signal occupies the bandwidth 390 kHz to 410 kHz. What minimum sampling frequency would you use from the options given below, so as to avoid aliasing?
A.	40 kHz
B.	820 kHz
C.	41 kHz
D.	800 kHz
Answer» D. 800 kHz

Discussion

28.	A 4 GHz carrier is amplitude-modulated by a low-pass signal of maximum cut off frequency 1 MHz. If this signal is to be ideally sampled the minimum sampling frequency should be nearly
A.	4 MHz
B.	4 GHz
C.	8 MHz
D.	8 GHz
Answer» B. 4 GHz

Discussion

29.	A band limited signal is sampled at Nyguist rate, The signal can be recovered by passing the samples through
A.	RC filter
B.	envelope detector
C.	PLL
D.	ideal low pass filter with appropriate band width
Answer» E.

Discussion

30.	Let \(x\left( t \right) = \;cos10\pi t + \cos \left( {30\pi t} \right)\) be sampled at 20Hz and reconstructed using an ideal LPF with fc of 20Hz. The frequency/frequencies present in reconstructed signal is/are
A.	5 Hz, 15 Hz only
B.	15 Hz only
C.	5 Hz only
D.	5, 10, 15 Hz only
Answer» B. 15 Hz only

Discussion

31.	If analog sampling frequency of a band limited signal is doubled then corresponding digital sampling frequency will be
A.	π
B.	2π
C.	\(\frac{\pi }{2}\)
D.	None of the above
Answer» B. 2π

Discussion

32.	A system has a sampling rate of 50,000 samples per second. The maximum frequency of the signal it can acquire to reconstruct is
A.	25 kHz
B.	50 kHz
C.	100 kHz
D.	10 kHz
Answer» B. 50 kHz

Discussion

33.	Let a signal x(t) = 3 cos (10πt) + cos (30πt) is sampled at a rate of 20 samples / second and reconstructed using on ideal low pass filter having cut off frequency of 20 Hz. The frequencies present in the reconstructed signal is / are.
A.	5 Hz and 15 Hz
B.	5 Hz, 10 Hz and 15 Hz only
C.	10 Hz and 15 Hz only
D.	5 Hz only
Answer» B. 5 Hz, 10 Hz and 15 Hz only

Discussion

34.	A signal is passed through a LPF with cut-off frequency 10 kHz. The minimum sampling frequency is:
A.	5 kHz
B.	10 kHz
C.	20 kHz
D.	30 kHz
Answer» D. 30 kHz

Discussion

35.	PAM signals are constructed by using a low pass filter of passband slightly greater than baseband to avoid aliasing. This avoids distortion :
A.	True for flat top pulses
B.	True if low pass filter has sharp cut off
C.	Flat top pulses introduce envelope delay
D.	Flat top pulses introduce amplitude distortion and delay
Answer» E.

Discussion

36.	Consider the sampling of the band-pass signal, whose spectrum is as shown below. What minimum sampling frequency would you use from the options given below, so as to avoid aliasing?
A.	24 Hz
B.	26.5 Hz
C.	48 Hz
D.	212 Hz
Answer» C. 48 Hz

Discussion

37.	In a DSP system:
A.	Nyquist frequency is the same as the sampling frequency
B.	Nyquist frequency is the same as folding frequency
C.	Nyquist frequency is double the folding frequency
D.	Nyquist frequency is half of oversampling rate
Answer» C. Nyquist frequency is double the folding frequency

Discussion

38.	Let \({x_1}\left( t \right) \leftrightarrow {X_1}\left( \omega \right)\) and \({x_2}\left( t \right) \leftrightarrow {X_2}\left( \omega \right)\) be two signals whose Fourier Transforms are as shown in the figure below. In the figure, \(h\left( t \right) = {e^{ - 2\left\| t \right\|}}\) denotes the impulse response.For the system shown above, the minimum sampling rate required to sample y(t), so that y(t) can be uniquely reconstructed from its samples, is
A.	\(2{B_1}\)
B.	\(2\left( {{B_1} + {B_2}} \right)\)
C.	\(4\left( {{B_1} + {B_2}} \right)\)
D.	\(\infty\)
Answer» C. \(4\left( {{B_1} + {B_2}} \right)\)

Discussion

39.	If the lower sideband overlaps the basebands, the distortion is called
A.	Cross-Over distortion
B.	Aliasing
C.	Crosstalk
D.	None of these
Answer» C. Crosstalk

Discussion

40.	A signal m(t) = 3 cos (6000 πt) + 4 cos (7000 πt) + 5 cos (10000 πt) to be sampled. What is the minimum sampling rate for bandpass sampling for truth full representation:
A.	2000 Hz
B.	3000 Hz
C.	5000 Hz
D.	10000 Hz
Answer» E.

Discussion

41.	Determine the Nyquist rate of signal x(t) = cos(200πt) × sin (300 πt) according to sampling theorem.(where ‘t’ indicates continuous-time domain)
A.	150
B.	100
C.	500
D.	250
Answer» D. 250

Discussion

42.	Consider the signal x(t) = cos(6πt) + sin(8πt), where t is in seconds. The Nyquist sampling rate (in samples/second) for the signal y(t) = x(2t + 5) is
A.	8
B.	12
C.	16
D.	32
Answer» D. 32

Discussion

43.	Assertion (A): Sampling theorem forms the basis for the relation between continuous-time signals and discrete-time signals.Reason (R): If the sampling is done at Nyquist rate, the signal x(t) can be reconstructed from the sample values x[n] = x[nTs]; −∞ ≤ n ≤ ∞.
A.	Both (A) and (R) are true and (R) is the correct explanation of (A)
B.	Both (A) and (R) are true, but (R) is not the correct explanation of (A)
C.	(A) is true, but (R) is false
D.	(A) is false, but (R) is true
Answer» B. Both (A) and (R) are true, but (R) is not the correct explanation of (A)

Discussion

44.	Directions: The following items consist of two statements, one labeled as “Assertion(A)” and the other labeled as “Reason(R)”. You are to examine the two statements carefully and decide if the Assertion (A) and the Reason (R) are individually true and if so whether the reason is a correct explanation of the assertion. Select your answer to these items using the codes given below and mark your answer accordingly.Assertion (A): Lesser number of bits per code are required due to less number of quantization levels is DPCM.Reason (R): In this case, the difference between two successive samples is quantized which does not differ much in amplitude.
A.	Both (A) and (R) are true and (R) is the correct explanation of (A).
B.	Both (A) and (R) are true, but (R) is not the correct explanation of (A).
C.	(A) is true, but (R) is false.
D.	(A) is false, but (R) is true.
Answer» B. Both (A) and (R) are true, but (R) is not the correct explanation of (A).

Discussion

45.	Let x(t) be a signal with Nyquist rate ω0. Determine the Nyquist rate for y(t) = x(t)cos(ω0t)
A.	ω0
B.	2ω0
C.	3ω0
D.	2π
Answer» D. 2π

Discussion

46.	An FM signal at 10.7 MHz. IF needs to be digitized for demodulation in a digital domain. If the bandwidth of this signal is 200 kHz, the maximum usable sampling frequency is
A.	200 kHz
B.	600 kHz
C.	400 kHz
D.	800 kHz
Answer» D. 800 kHz

Discussion

47.	Assertion (A): Nyquist rate of sampling is the theoretical minimum sampling rate at which the signal can be sampled and still be reconstructed from its samples.Reason (R): When the Nyquist rate sampling is used, only an ideal low pass filter can be used to extract signal x(t) from sampled signal xs(t).Code:
A.	Both (A) and (R) are true and (R) is the correct explanation of (A).
B.	Both (A) and (R) are true, but (R) is not the correct explanation of (A)
C.	(A) is true, but (R) is false.
D.	(A) false, but (R) is true
Answer» B. Both (A) and (R) are true, but (R) is not the correct explanation of (A)

Discussion

48.	Arrange the Nyquist sampling interval of the signal in descending order(A) sin c(300t)(B) sin c(300t) + sin c2(300t)(C) sin c(200t)(D) sin c(200t) + sinc2(200t)(E) sin c(200t) + sin c(500t)Choose the correct answer from the options given below:(1) (A), (C), (B), (D), (E)(2) (C), (D), (E), (A), (B)(3) (C), (A), (D), (E), (B)(4) (A), (E), (D), (C), (B)
A.	1
B.	2
C.	3
D.	4
Answer» D. 4

Discussion

49.	For a discrete memory less source containing K symbols, the upper bound for the entropy is
A.	1/K
B.	log2K
C.	log10 K
D.	log2 (1-K)
Answer» C. log10 K

Discussion

50.	Assertion (A): A Non-Return to Zero (NRZ) type digital recording system is more common and efficient.Reason (R): It is possible to record twice the number of digits for the same number of pulses.
A.	Both (A) and (R) are true and (R) is the correct explanation of (A).
B.	Both (A) and (R) are true, but (R) is not the correct explanation of (A).
C.	(A) is true, but (R) is false.
D.	(A) is false, but (R) is true.
Answer» B. Both (A) and (R) are true, but (R) is not the correct explanation of (A).

Discussion

Explore topic-wise MCQs in Concrete Technology.

The loss in weight should not exceed ________ % when tested with Na2SO4 and ________ % with MgSO4 solution.

The mineral oil if present in mixing for concrete ______________

When sand is fully dry then it’s volume is _________

Loss on Ignition (L.O.I.) is the loss in weight of cement after being heated to ____________

Find the Nyquist rate and Nyquist interval for the signal f(t) = -10 sin ⁡40πt cos ⁡300πt.

Determine the Nyquist rate of the signal x(t) = 1 + cos⁡ 2000πt + sin⁡ 4000πt.

The sampling frequency of a signal is Fs = 2000 samples per second. Find its Nyquist interval.

Find the Nyquist rate and Nyquist interval for the signal f(t) = rect(200t).

Find the Nyquist rate and Nyquist interval for the signal f(t) = 1 + sinc300πt.

Find the Nyquist rate and Nyquist interval for the signal f(t)=\(\frac{sin⁡500πt}{πt}\).

Which of the following is the process of ‘aliasing’?

Find the Nyquist rate and Nyquist interval of sinc[200t].

Find the Nyquist rate and Nyquist interval of Asinc[t].

Find the Nyquist rate and Nyquist interval of sinc[t].

Find the Nyquist rate and Nyquist interval of sin(2πt).

Consider a real-value based-band signal x(t), band limited to 10 kHz. The Nyquist rate for the signal y(t) = x(t) \(x(1+ \frac{t}{2})\) is

Consider the analog signal x(t) = 3 cos 100 πt. If the signal is sampled at 200 Hz, the discrete time signal obtained will be

In Digital Filters, how many interpolated data points are inserted between samples performing 4X over sampling?

Calculate the minimum sampling rate to avoid aliasing when a continuous-time signal is given by x(t) = 5 cos 400πt

Calculate the Nyquist rate for sampling when a continuous time signal is given by:x(t) = 5 cos 100πt +10 cos 200πt - 15 cos 300πt

Given a continuous time signal s(t) = 5cos(200 πt). The minimum sampling rate required to avoid aliasing will be

For the signal f(t) = 2 sin 9πt + sin 12 πt + sin 14 πt the minimum sampling frequency (in Hz) satisfying the Nyquist criterion is:

In flat-top sampling, a hold circuit is sometimes required. This hold circuit can be designed as a sampler followed by

Find out the Nyquist rate of the following signal: x(t) = 3 sin 10πt + 100 sin 400πt – 9 cos 110πt.

If the sampling is carried out at rate higher than twice the highest frequency of the original single (fmax), then it is possible to receive the original signal from the sampled signal by passing it through

A bandpass signal occupies the bandwidth 390 kHz to 410 kHz. What minimum sampling frequency would you use from the options given below, so as to avoid aliasing?

A 4 GHz carrier is amplitude-modulated by a low-pass signal of maximum cut off frequency 1 MHz. If this signal is to be ideally sampled the minimum sampling frequency should be nearly

A band limited signal is sampled at Nyguist rate, The signal can be recovered by passing the samples through

Let \(x\left( t \right) = \;cos10\pi t + \cos \left( {30\pi t} \right)\) be sampled at 20Hz and reconstructed using an ideal LPF with fc of 20Hz. The frequency/frequencies present in reconstructed signal is/are

If analog sampling frequency of a band limited signal is doubled then corresponding digital sampling frequency will be

A system has a sampling rate of 50,000 samples per second. The maximum frequency of the signal it can acquire to reconstruct is

Let a signal x(t) = 3 cos (10πt) + cos (30πt) is sampled at a rate of 20 samples / second and reconstructed using on ideal low pass filter having cut off frequency of 20 Hz. The frequencies present in the reconstructed signal is / are.

A signal is passed through a LPF with cut-off frequency 10 kHz. The minimum sampling frequency is:

PAM signals are constructed by using a low pass filter of passband slightly greater than baseband to avoid aliasing. This avoids distortion :

Consider the sampling of the band-pass signal, whose spectrum is as shown below. What minimum sampling frequency would you use from the options given below, so as to avoid aliasing?

In a DSP system:

If the lower sideband overlaps the basebands, the distortion is called

A signal m(t) = 3 cos (6000 πt) + 4 cos (7000 πt) + 5 cos (10000 πt) to be sampled. What is the minimum sampling rate for bandpass sampling for truth full representation:

Determine the Nyquist rate of signal x(t) = cos(200πt) × sin (300 πt) according to sampling theorem.(where ‘t’ indicates continuous-time domain)

Consider the signal x(t) = cos(6πt) + sin(8πt), where t is in seconds. The Nyquist sampling rate (in samples/second) for the signal y(t) = x(2t + 5) is

Assertion (A): Sampling theorem forms the basis for the relation between continuous-time signals and discrete-time signals.Reason (R): If the sampling is done at Nyquist rate, the signal x(t) can be reconstructed from the sample values x[n] = x[nTs]; −∞ ≤ n ≤ ∞.

Let x(t) be a signal with Nyquist rate ω0. Determine the Nyquist rate for y(t) = x(t)cos(ω0t)

An FM signal at 10.7 MHz. IF needs to be digitized for demodulation in a digital domain. If the bandwidth of this signal is 200 kHz, the maximum usable sampling frequency is

For a discrete memory less source containing K symbols, the upper bound for the entropy is

Assertion (A): A Non-Return to Zero (NRZ) type digital recording system is more common and efficient.Reason (R): It is possible to record twice the number of digits for the same number of pulses.

The loss in weight should not exceed % when tested with Na2SO4 and % with MgSO4 solution.