PG1995
Active Member
Hi
Q1: It is said that rising and falling edges, such as those of square wave, imply high frequency content. What does in really mean? For example, periodic square, sawtooth and triangular waves, all three have infinite number of harmonics of the fundamental frequency extending to infinity, so all three have high frequency content. Right? But a square wave has rising and falling edges which are more vertical than those of other two, so this should mean that a square has higher frequency content than the other two. But all three have harmonics extending to infinity. Please help me with this.
This is the reason I can think of. Although the harmonics extend to infinity but higher harmonics can be ignored without actually affecting the waveform because they have such a small amplitude. Let's take the case of a square and sawtooth waves. Perhaps, in case of a sawtooth wave all the harmonics above 1000 Hz can be ignored but ignoring the harmonics above 1000 Hz for a square wave will greatly affect the shape of the waveform.
Q2: Please have a look here.
Q3: In analog communication a medium should be able to support at least that much bandwidth as is required by a signal. For instance, suppose that we want to send the an audio baseband signal (speech and music) by a wire instead of transmitting it using AM then this would require that the wire at least support 5 kHz bandwidth. So, for analog communication the rule is simply that the medium should be able to support the bandwidth required by a baseband signal.
In case of digital communication the bandwidth is given in terms of data rate (bits per second). For example, if data rate is 500 bps, it means that the medium can 'transport' five hundred bits from one place to another in one second. The Shannon formula gives us relationship between bandwidth (range of frequencies) supported by a medium and data rate supported by it. If you need more data rate (wants to 'transports' more number of bits in one second) then you need to have more bandwidth assuming signal to noise ratio remains constant. What's the underlying reason for this relationship?
If I had been asked to explain this, I would go like this. (Please first note that my whatever I'm going to say is based on my reasoning given in last paragraph of Q1. If my reasoning is not correct there then whatever I'm going to say isn't also correct.)
For the sake of starting this discussion assume that a periodic square wave is used to send the bits (periodicity means that bits are sent like this 010101010101010101). When amplitude of the square wave is "1" it means bit "1" and when amplitude is "-1" it stands for bit "0". Now please have a look here. Do you agree with me?
In almost all practical situations the sequence of bits in non-periodic and it is represented by non-periodic square wave. It is my experience that non-periodic signals almost always require a large bandwidth. In other words, they always have significant higher harmonics. The more non-periodicity or randomness there is, the more bandwidth is required. We kind of say that the amount of non-periodicity or randomness is proportional to the required bandwidth. Do you think what I say is generally true?
You might find links #3 and #4 helpful for this question.
Q4: What's the relationship between bandwidth and noise? Does the noise decrease when bandwidth is decreased? In SSB-SC amplitude modulation, it is said that SSB-SC uses less bandwidth compared to DSB-SC therefore it accounts for less noise. I don't understand the reason for this relationship.
Thank you very much for the help and your time.
Regards
PG
Helpful Links:
1: https://www.music.mcgill.ca/~gary/307/week5/additive.html
2: https://www.matrixlab-examples.com/fourier-series.html
3: https://www.electro-tech-online.com/attachments/cs_56k_single-jpg.75258/
4: https://imageshack.com/a/img713/383/x4ix.jpg
5: https://imageshack.com/a/img18/2826/ab0c.jpg
Q1: It is said that rising and falling edges, such as those of square wave, imply high frequency content. What does in really mean? For example, periodic square, sawtooth and triangular waves, all three have infinite number of harmonics of the fundamental frequency extending to infinity, so all three have high frequency content. Right? But a square wave has rising and falling edges which are more vertical than those of other two, so this should mean that a square has higher frequency content than the other two. But all three have harmonics extending to infinity. Please help me with this.
This is the reason I can think of. Although the harmonics extend to infinity but higher harmonics can be ignored without actually affecting the waveform because they have such a small amplitude. Let's take the case of a square and sawtooth waves. Perhaps, in case of a sawtooth wave all the harmonics above 1000 Hz can be ignored but ignoring the harmonics above 1000 Hz for a square wave will greatly affect the shape of the waveform.
Q2: Please have a look here.
Q3: In analog communication a medium should be able to support at least that much bandwidth as is required by a signal. For instance, suppose that we want to send the an audio baseband signal (speech and music) by a wire instead of transmitting it using AM then this would require that the wire at least support 5 kHz bandwidth. So, for analog communication the rule is simply that the medium should be able to support the bandwidth required by a baseband signal.
In case of digital communication the bandwidth is given in terms of data rate (bits per second). For example, if data rate is 500 bps, it means that the medium can 'transport' five hundred bits from one place to another in one second. The Shannon formula gives us relationship between bandwidth (range of frequencies) supported by a medium and data rate supported by it. If you need more data rate (wants to 'transports' more number of bits in one second) then you need to have more bandwidth assuming signal to noise ratio remains constant. What's the underlying reason for this relationship?
If I had been asked to explain this, I would go like this. (Please first note that my whatever I'm going to say is based on my reasoning given in last paragraph of Q1. If my reasoning is not correct there then whatever I'm going to say isn't also correct.)
For the sake of starting this discussion assume that a periodic square wave is used to send the bits (periodicity means that bits are sent like this 010101010101010101). When amplitude of the square wave is "1" it means bit "1" and when amplitude is "-1" it stands for bit "0". Now please have a look here. Do you agree with me?
In almost all practical situations the sequence of bits in non-periodic and it is represented by non-periodic square wave. It is my experience that non-periodic signals almost always require a large bandwidth. In other words, they always have significant higher harmonics. The more non-periodicity or randomness there is, the more bandwidth is required. We kind of say that the amount of non-periodicity or randomness is proportional to the required bandwidth. Do you think what I say is generally true?
You might find links #3 and #4 helpful for this question.
Q4: What's the relationship between bandwidth and noise? Does the noise decrease when bandwidth is decreased? In SSB-SC amplitude modulation, it is said that SSB-SC uses less bandwidth compared to DSB-SC therefore it accounts for less noise. I don't understand the reason for this relationship.
Thank you very much for the help and your time.
Regards
PG
Helpful Links:
1: https://www.music.mcgill.ca/~gary/307/week5/additive.html
2: https://www.matrixlab-examples.com/fourier-series.html
3: https://www.electro-tech-online.com/attachments/cs_56k_single-jpg.75258/
4: https://imageshack.com/a/img713/383/x4ix.jpg
5: https://imageshack.com/a/img18/2826/ab0c.jpg
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