# Bandwidth Requirement

1. Jul 30, 2007

### jeejou

Hi,

I would like to know why do the digital signals require more bandwidth than the original analogue signals to convey the same information ? Is it something to do with the Nyquist Theorem or Shannon's Law ? And what about the harmonics ?

Thank you.

2. Jul 30, 2007

### mgb_phys

Generally because they operate at a lower signal to noise.
If you have a hifi audio stream with 20Khz bandwidth when sampled at say 16bits at 40Khz this uses more bits and so more bandwidth.
But assuming the analog signal only needs 20Khz assumes a signal to noise on the analogue signal of 16bits ( sorry can't do db in my head).
Generally bandwidth is cheaper than quality.

3. Jul 30, 2007

### Integral

Staff Emeritus
Square waves, the heart and blood of digital circuits, require a high band width. Consider that the Fourier series representation of a square wave consists of a sum of ALL odd harmonics. If you want a perfect square wave you need an infinite bandwidth.

4. Jul 31, 2007

### jeejou

Hello Friends,

Thanks for the reply. I am still confused with the whole Nyqusit theorem. Anyway, I understood about the sampling and also about the harmonics.

Thank you very much. Take care. Bye bye.

5. Jul 31, 2007

### rcgldr

The db range of the signal combined with the highest frequency handled would determine the bandwidth. The rate of volume changes allowed also determine bandwidth usage. If a signal just happens to be a square wave, then to perfectly reproduce it, you'd need infinite bandwidth even with analog transmission. Morse code transmitters have a controlled ramp up and down rate (several milliseconds) to reduce bandwidth usage. So do radio control transmitters.

For cd-roms, the data for each channel is sampled into 16 bit values, -32767 to +32767. Range is quoted at 96db's. Based on what I've read, +/- 1 would represent a 6db ac wave, +/- 2 a 12 db ac wave, +/- 4 a 18db ac wave, up to +/- 32768 = 96db ac wave, but actual max is just slightly less at +/- 32767. db = 6 x (1 + log2(value)). For best quality, the maximum volume on a cd-rom should be set to end up at 96db.

Getting back to analog bandwidth, high end power amplifiers include a maximum slew rate specification, the maximum rate of voltage change (generally expressed as volts per microsecond). The higher this rate, the closer to being able to reproduce a square wave, and effectively, chew up a lot of bandwidth. The only analog recorders that could handle this much bandwidth are high end tape decks. FM radio transmissions and records don't have this kind of bandwidth, so it's something that is straight out of a recording studio. Then again, there's the issue of finding speakers or microphones with this kind of reproduction capability (except for electronic devices that record direct to the tape recorder).

Last edited: Jul 31, 2007
6. Jul 31, 2007

### AlephZero

I think the OP (and perhaps some of the replies) are confusing "bandwidth" with "sampling rate" and perhaps also with "bit-rate" (or "information content").

The Sampling Theorem (Nyquist, Shannon) says that any bandwidth-limited signal, with bandwith from zero to f_max, can be reconstructed exactly from a samples taken at the rate of 2*f_max per second.

The sampling theorem assumes no amplitude quantization, and it is not a practical result because each sample value influences the entire signal from time minus infinity to plus infinity. Each sample value corresponds to a continuous function of the form sin(at)/t (where the constant a depends on the sample rate). This "sinc" function looks like a decaying oscillation, and is zero at all sample points except one.

So an bandwidth limited signal with bandwidth 20 kHhz requires a minimum 40,000 samples per second. The sample rate is 40,000 samples/sec, the bandwidth is 20 kHz.

You could just as well represent the same signal with a sample rate of 40 million samples per second if you wanted to - and for some purposes that's a useful thing to do. But the bandwidth of the signal is still 20 kHz.

Note that almost all "real world" signals (especially amplitude-modulated signals) are not bandwidth-limited, so the sampling theorem applies only approximately. That is one reason why professional digital audio uses sample rates of 96000 or 192000 samples/sec to represent audio signals, even though the human hearing range is usually assumed to be limited to 20 KHz.

All the above assumes that the amplitude of the signal and the samples is represented exactly. The errors introduced by quantizing the amplitude of the samples (to 16 bits for CD quality sound, for example) are also important, but those errors are not directly related to what the sampling theorem states.

Using 16 (or 24 or 32) bits of data to represent one sample doesn't increase the bandwidth by 16, 24, or 32 times, either. The number of bits per second to represent the signal (= bits-per-sample times samples-per-second) is called the bit-rate, not the bandwidth.

7. Jul 31, 2007

### rcgldr

FM broadcast in USA - 15khz max frequency (for both mono and stereo sub-bands), about 70 db dynamic range.

AM broadcast in USA -10khz seperation between "channels", but in USA stations are at least 20khz apart, and some stations are "clear", meaning they have a frequency not used anywhere else in the USA (like KOMA), allowing for a huge range of reception at night. Originally, AM stations had up to 20khz max frequency ("stealing" bandwidth from adjacent unused channels), but were later limited to 10khz by FCC. Dynamic range is about 50 db. There was a short period where stereo AM broadcasts with 15khz frequency range per channel was implemented. Most modern receivers don't implement good AM reception.

Both AM and FM broadcasting compress the dynamic range (quiet sounds become louder, while louder sounds remain the same). Receivers can reexpand the sound to get the dynamic range back. (This is also done for some types of tape recording). In a car, background noise while driving limits the pratical dynamic range to about 30db, and it's a good idea for cd-players in cars to include a compress feature.

taken at a rate > 2*f_max per second. Exactly 2 doesn't cut it. Just barely above 2 would take a very complicated filter. Cd-roms sample a 20khz max (cut off filter) signal at 44khz to improve accuracy and reduce the complexity of the digital to anallog filter, which must recontruct smooth sine waves from the sampled data. In addition, higher frequencies are an issue because it's getting close to 1/2 of the sampling rate, so higher frequencies are "emphasized" during recording, and "de-emphasized" when played back to reduce noise.

For digital sampling, higher sampling rates increases accuracy and reduces filter complexity. Higher number of bits per sample increases dynamic range and/or increases accuracy (better precision) as well.

Last edited: Jul 31, 2007
8. Jul 31, 2007

### chroot

Staff Emeritus
Square waves may be the heart and blood of digital circuits, but they are certainly not useful at all for digital transmission. All reasonably efficient digital modulation techniques use some form of sinusoids to represent digital information.

- Warren

9. Aug 1, 2007

### jeejou

10. Aug 7, 2007

### rcgldr

Normally, as the amount of information sent per second increases, so does the required bandwidth. However improvement in technology can increase this ratio. I'm not sure what the limits are.

Here is a link from Wiki about FM broadcasting, which takes 100khz of bandwidth to send mono and stereo audio, plus some digital information:

Standard analog broadcasts in the USA take up 6mhz:

HDTV manages to compress a digitized 1920 x 1080i x 60 fp/s image stream down to 19.39 megabits/ second stream of useable data, which is then squeezed into the same 6mhz bandwidth that the old analog channels use.

http://en.wikipedia.org/wiki/8VSB