On a neverending thread on mains cables @Fourlegs and @tuga got briefly side tracked on to S/PDIF and rf noise . While surfing around on this I came upon a post by John Siau of Benchmark. He was discussing not rf interference but jitter induced by miss matches in impedance between the sender and the cable and or the cable an the receiver and the signal reflections this causes.
It seem to me to illustrate a number of points often discussed on here. Cables can make a difference. The quality of the sending and receiving device can make a difference. As the impedance mismatch can impact jitter which may therefore be audible depending on the DAC and at least in this instance there is a credible scientific explanation. Anyway that's my conclusion.
"One correction to the original post:
The reflected signal can be either polarity. If the termination impedance is higher than the cable impedance, the reflection will be non-inverted. If the termination impedance is lower than the cable impedance, the reflection will be inverted.
Damage is done to the signal at the receiver after the signal has made 3 transits through the cable. The magnitude and polarity of the first reflection is determined by the impedance matching of the termination at the receiver. This first reflection will be reflected again if the drive impedance is mismatched to the cable. The magnitude and polarity of this reflection is determined by the matching of the source impedance to that of the cable. Each successive reflection is smaller than the previous reflection.
A perfect termination at the transmitter or the receiver, will eliminate 3rd transit reflections.
The 1.5 meter cable is long enough to allow completion of the rise or fall time before the third reflection reaches the receiver. A 1.5 m cable will have a transit time of 5.95 ns to 9.35 ns depending upon the velocity factor of coax used. The transit time through 1.5 meters of air would be about 5 ns (velocity factor of about 1). The velocity factor is determined by the insulating material used in the coax. Multiply the transit times by 3 to calculate the arrival time of the third transit. Multiply the transit time by 2 to arrive at the timing of the third transit relative to the original signal reaching the receiver. The first transit arrives at the receiver 5.95 to 9.35 ns after the original signal is generated. The third transit arrives 17.9 ns to 28 ns after the original signal is generated. The difference in arrival time is 2 transits which is 11.9 ns to 18.7 ns respectively. If the rise time of the original signal is less than 11.9 ns the reflections will arrive after the transition has completed and the reflections will have no impact on the timing of the data transition (assuming the use of the cable with the higher velocity factor). RG59 is commonly used for digital audio, and its lower velocity factor gives us 18.7 ns between the original signal and the reflected signal (at a cable length of 1.5 m). Slower cables give us more margin at 1.5 meters, but things start to get complicated when we look at longer cables.
As the cable gets longer, two things happen:
1: The rise times of the transitions will get longer due to high-frequency roll off. This makes clock recovery more difficult.
2: If 2 transits through the cable is equal to the period of data transitions, the third transit will arrive in sync with the next data transition. In AES or SPDIF digital audio streams, data transitions happen 128 times per sample. 128 x 44,100 = 5.6448 MHz. This means that there is a 177 ns spacing between data edges at a sample rate of 44.1 kHz. If the two-transit time is equal to 177 ns, the terminations need to be reasonably good. Using RG59, this unfortunate alignment happens when the cable length is 9.5 m long. Using the fastest coax, this problem would occur at a length of 14.9 meters. Divide these lengths by two for 88.2 kHz audio, and divide by two again for 176.4 kHz audio.
At 192 kHz, using the slowest coax, the reflections will align at a cable length of 2.2 meters. If we consider the fact that transitions have a rise time, interference will begin at shorter cable lengths. This interference will begin if the cable length is much over 1.5 meters long.
Bottom line: You can get away with really bad terminations (or no terminations at all) if your cable is 1.5 meters long. There is very little margin in either direction (longer or shorter). For this reason, cable length is a bad way to solve the problem. Nevertheless, a 1.5 meter cable can be used as a band-aid on a poorly designed system.
The best way to solve the problem is to use cables that have the correct impedance. The equipment must also have reasonably good terminations on either the transmitter or the receiver (preferably both). The receiving device should also have a good clock recovery system that provides jitter immunity. If these conditions are satisfied, the cable length is not critical.
Regards Andrew
It seem to me to illustrate a number of points often discussed on here. Cables can make a difference. The quality of the sending and receiving device can make a difference. As the impedance mismatch can impact jitter which may therefore be audible depending on the DAC and at least in this instance there is a credible scientific explanation. Anyway that's my conclusion.
"One correction to the original post:
The reflected signal can be either polarity. If the termination impedance is higher than the cable impedance, the reflection will be non-inverted. If the termination impedance is lower than the cable impedance, the reflection will be inverted.
Damage is done to the signal at the receiver after the signal has made 3 transits through the cable. The magnitude and polarity of the first reflection is determined by the impedance matching of the termination at the receiver. This first reflection will be reflected again if the drive impedance is mismatched to the cable. The magnitude and polarity of this reflection is determined by the matching of the source impedance to that of the cable. Each successive reflection is smaller than the previous reflection.
A perfect termination at the transmitter or the receiver, will eliminate 3rd transit reflections.
The 1.5 meter cable is long enough to allow completion of the rise or fall time before the third reflection reaches the receiver. A 1.5 m cable will have a transit time of 5.95 ns to 9.35 ns depending upon the velocity factor of coax used. The transit time through 1.5 meters of air would be about 5 ns (velocity factor of about 1). The velocity factor is determined by the insulating material used in the coax. Multiply the transit times by 3 to calculate the arrival time of the third transit. Multiply the transit time by 2 to arrive at the timing of the third transit relative to the original signal reaching the receiver. The first transit arrives at the receiver 5.95 to 9.35 ns after the original signal is generated. The third transit arrives 17.9 ns to 28 ns after the original signal is generated. The difference in arrival time is 2 transits which is 11.9 ns to 18.7 ns respectively. If the rise time of the original signal is less than 11.9 ns the reflections will arrive after the transition has completed and the reflections will have no impact on the timing of the data transition (assuming the use of the cable with the higher velocity factor). RG59 is commonly used for digital audio, and its lower velocity factor gives us 18.7 ns between the original signal and the reflected signal (at a cable length of 1.5 m). Slower cables give us more margin at 1.5 meters, but things start to get complicated when we look at longer cables.
As the cable gets longer, two things happen:
1: The rise times of the transitions will get longer due to high-frequency roll off. This makes clock recovery more difficult.
2: If 2 transits through the cable is equal to the period of data transitions, the third transit will arrive in sync with the next data transition. In AES or SPDIF digital audio streams, data transitions happen 128 times per sample. 128 x 44,100 = 5.6448 MHz. This means that there is a 177 ns spacing between data edges at a sample rate of 44.1 kHz. If the two-transit time is equal to 177 ns, the terminations need to be reasonably good. Using RG59, this unfortunate alignment happens when the cable length is 9.5 m long. Using the fastest coax, this problem would occur at a length of 14.9 meters. Divide these lengths by two for 88.2 kHz audio, and divide by two again for 176.4 kHz audio.
At 192 kHz, using the slowest coax, the reflections will align at a cable length of 2.2 meters. If we consider the fact that transitions have a rise time, interference will begin at shorter cable lengths. This interference will begin if the cable length is much over 1.5 meters long.
Bottom line: You can get away with really bad terminations (or no terminations at all) if your cable is 1.5 meters long. There is very little margin in either direction (longer or shorter). For this reason, cable length is a bad way to solve the problem. Nevertheless, a 1.5 meter cable can be used as a band-aid on a poorly designed system.
The best way to solve the problem is to use cables that have the correct impedance. The equipment must also have reasonably good terminations on either the transmitter or the receiver (preferably both). The receiving device should also have a good clock recovery system that provides jitter immunity. If these conditions are satisfied, the cable length is not critical.
Regards Andrew