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Rule of thumb: Bandwidth of high speed signal

02 Jun 2014  | Eric Bogatin

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As the signal propagates down the channel, the rise time will increase due to frequency-dependent losses. This will decrease the bandwidth of the signal. The bandwidth at the Rx will be different than at the Tx.

In most channels, the attenuation scales roughly with frequency. If there is -3 dB attenuation at the Nyquist, there will be three times this, or -9 dB attenuation, for the 3rd harmonic, and -15 dB attenuation for the 5th harmonic. This means that all the higher harmonics will be dramatically attenuated compared to the first harmonic at the Nyquist. Figure 2 shows the eye diagram for a signal that has a very high bandwidth at the Tx, but passes through a lossy channel with -3 dB attenuation at the Nyquist. It looks almost like a sine wave at the Rx.

Figure 2: Eye diagram of a PRBS signal at the RX, passing through a channel with -3 dB attenuation at the Nyquist. The Nyquist is the highest sine wave frequency that survives (simulated with Agilent's ADS).

This is the origin of the second half of the rule of thumb. In a lossy channel, the first harmonic of the underlying clock is the highest frequency component that survives the losses, and the bandwidth of the data signal is the Nyquist:
At the Rx, the BW = ½ bitrate.
For example, at the Rx, the BW of a 10Gbit/s signal through a lossy channel will be about 5GHz.

Now you try it:

1. I want the bandwidth of my scope to be at least twice the bandwidth of my signal. To see the waveform at the Tx for a PCIe gen II signal, what is the minimum bandwidth scope I should use?

2. At the Rx, what is the bandwidth of a USB 3.1 signal passing through a long cable?

About the author
Eric Bogatin is Signal Integrity Evangelist at Teledyne LeCroy.

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