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A continuous-time signal has frequency content at f = 10 MHz, 50 MHz, and 70 MHz. The signal is sampled at a sampling frequency of 56 MHz and is then passed through a low-pass filter with a cutoff frequency of 15 MHz. The frequency content of the output of the filter will be:
- a)10 MHz
- b)10 MHz and 6 MHz
- c)10 MHz, 6 MHz, and 14MHz
- d)46 MHz
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
A continuous-time signal has frequency content at f = 10 MHz, 50 MHz, ...
Answer:
Frequency content of the continuous-time signal: 10 MHz, 50 MHz, and 70 MHz.
Sampling frequency: 56 MHz.
Low-pass filter cutoff frequency: 15 MHz.
Output frequency content after passing through the low-pass filter: 10 MHz, 6 MHz, and 14 MHz.
Explanation:
The Nyquist-Shannon sampling theorem states that a continuous-time signal can be perfectly reconstructed from its samples if the sampling frequency is greater than twice the highest frequency component of the signal. In this case, the highest frequency component of the signal is 70 MHz. Therefore, the sampling frequency of 56 MHz is sufficient to capture all the frequency content of the signal.
However, since the sampling frequency is less than twice the frequency of the highest component of the signal, aliasing can occur. Aliasing is the phenomenon where high-frequency components of the signal are folded back into the frequency band of the lower-frequency components.
To prevent aliasing, a low-pass filter is used to remove the high-frequency components of the signal before sampling. In this case, a low-pass filter with a cutoff frequency of 15 MHz is used.
The low-pass filter removes all frequency components above 15 MHz, including the 50 MHz and 70 MHz components of the original signal. The only frequency component that remains is the 10 MHz component.
However, since the original signal had frequency components at 10 MHz, 50 MHz, and 70 MHz, the low-pass filter output will also contain frequency components due to aliasing. These alias frequencies are given by:
- 2 × 56 MHz – 70 MHz = 42 MHz (alias of 70 MHz)
- 2 × 56 MHz – 50 MHz = 62 MHz (alias of 50 MHz)
Therefore, the frequency content of the output of the filter will be 10 MHz (due to the original 10 MHz component) and 6 MHz and 14 MHz (due to aliasing of the original 50 MHz and 70 MHz components, respectively). The correct answer is option 'C'.
Frequency content of the continuous-time signal: 10 MHz, 50 MHz, and 70 MHz.
Sampling frequency: 56 MHz.
Low-pass filter cutoff frequency: 15 MHz.
Output frequency content after passing through the low-pass filter: 10 MHz, 6 MHz, and 14 MHz.
Explanation:
The Nyquist-Shannon sampling theorem states that a continuous-time signal can be perfectly reconstructed from its samples if the sampling frequency is greater than twice the highest frequency component of the signal. In this case, the highest frequency component of the signal is 70 MHz. Therefore, the sampling frequency of 56 MHz is sufficient to capture all the frequency content of the signal.
However, since the sampling frequency is less than twice the frequency of the highest component of the signal, aliasing can occur. Aliasing is the phenomenon where high-frequency components of the signal are folded back into the frequency band of the lower-frequency components.
To prevent aliasing, a low-pass filter is used to remove the high-frequency components of the signal before sampling. In this case, a low-pass filter with a cutoff frequency of 15 MHz is used.
The low-pass filter removes all frequency components above 15 MHz, including the 50 MHz and 70 MHz components of the original signal. The only frequency component that remains is the 10 MHz component.
However, since the original signal had frequency components at 10 MHz, 50 MHz, and 70 MHz, the low-pass filter output will also contain frequency components due to aliasing. These alias frequencies are given by:
- 2 × 56 MHz – 70 MHz = 42 MHz (alias of 70 MHz)
- 2 × 56 MHz – 50 MHz = 62 MHz (alias of 50 MHz)
Therefore, the frequency content of the output of the filter will be 10 MHz (due to the original 10 MHz component) and 6 MHz and 14 MHz (due to aliasing of the original 50 MHz and 70 MHz components, respectively). The correct answer is option 'C'.
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Community Answer
A continuous-time signal has frequency content at f = 10 MHz, 50 MHz, ...
Calculation:
fs = Sampling Frequency = 56 MHz
The frequencies present at the input are:
fm1 = 10 MHz, fm2 = 50 MHz and, fm3 = 70 MHz
Once sampled, the frequency content at the output will be: fm ± fs
For, fm1 = 10 MHz, the output will contain frequencies of,
= fm1, fm1 ±, fm1 ± 2fs ….
= 10 MHz, 66 MHz, 122 MHz
For, fm2 = 50 MHz, the output will contain frequencies of
= fm2, fm2 ± fs, fm2 ± 2fs ….
= 50 MHz, 6 MHz, 106 MHz, 162 MHz
For, fm3 = 70 MHz, the output will contain frequencies of
= fm3, fm3 ± fs, fm3 ± 2fs ….
= 70 MHz, 14 MHz, 182 MHz ….
When these frequencies are passed through a low pass filter with a cutoff frequency of 15 MHz, the only remaining frequencies at its output will be 6MHz, 10 MHz and 14 MHz.
fs = Sampling Frequency = 56 MHz
The frequencies present at the input are:
fm1 = 10 MHz, fm2 = 50 MHz and, fm3 = 70 MHz
Once sampled, the frequency content at the output will be: fm ± fs
For, fm1 = 10 MHz, the output will contain frequencies of,
= fm1, fm1 ±, fm1 ± 2fs ….
= 10 MHz, 66 MHz, 122 MHz
For, fm2 = 50 MHz, the output will contain frequencies of
= fm2, fm2 ± fs, fm2 ± 2fs ….
= 50 MHz, 6 MHz, 106 MHz, 162 MHz
For, fm3 = 70 MHz, the output will contain frequencies of
= fm3, fm3 ± fs, fm3 ± 2fs ….
= 70 MHz, 14 MHz, 182 MHz ….
When these frequencies are passed through a low pass filter with a cutoff frequency of 15 MHz, the only remaining frequencies at its output will be 6MHz, 10 MHz and 14 MHz.
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A continuous-time signal has frequency content at f = 10 MHz, 50 MHz, and 70 MHz. The signal is sampled at a sampling frequency of 56 MHz and is then passed through a low-pass filter with a cutoff frequency of 15 MHz. The frequency content of the output of the filter will be:a)10 MHzb)10 MHz and 6 MHzc)10 MHz, 6 MHz, and 14MHzd)46 MHzCorrect answer is option 'C'. Can you explain this answer?
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