An analog signal with a bandwidth of 36MHz is sampled at a frequency (fs) of 36,000,000 samples per second during the ADC (analo
1 answer:
Answer:
The signal would have experienced aliasing.
Step-by-step explanation:
Given that:
the bandwidth of the signal = 36MHz
= 36 × 10⁶ Hz
The sampling frequency = 36 × 10⁶ Hz
Suppose the sampling frequency is equivalent to the bandwidth of the signal, then aliasing will occur.
Therefore, according to the Nyquist criteria;
Nyquist criteria posit that if the sampling frequency is more above twice the maximum frequency to be sampled, a repeating waveform can be accurately reconstructed.
∴
By Nyquist criteria, for perfect reconstruction of an original signal, i.e. the received signal without aliasing effect;
Then,
∴
The signal would have experienced aliasing.
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P(r/w) is the probability of picking a red rose at first picking and a white rose at second picking.
P(w/r) is the probability of picking a white rose at first picking and a red rose at second picking.
P(r/w) =
×
= 
P(w/r) =
×
=
Notice that the second fraction is out of 18 because the second picking of rose will be out of 18 since the first rose is not replaced.
P(r/w) equals to P(w/r)
P(w/r) is the probability of picking a white rose at first picking and a red rose at second picking.
P(r/w) =
P(w/r) =
Notice that the second fraction is out of 18 because the second picking of rose will be out of 18 since the first rose is not replaced.
P(r/w) equals to P(w/r)
Answer: 0.2
Step-by-step explanation:
Given: The commuting time on a particular train is uniformly distributed over the interval (42,52).
∴ The probability density function of X will be :-
Thus, the required probability :-
Hence, the probability that the commuting time will be less than 44 minutes= 0.2