In message <[EMAIL PROTECTED]>, David Kirkby writes: >The attenuator will reduce a 1W source to 250mW (6dB attenuation). >Assuming the load on the output of the attenuator is open or short (so >100% reflection), the power will be reduced by another factor of 4 (to >62.5mW) before it reaches the source. So only 6.25% is reflected back to >the source. That's a VSWR of less than 1.7.
Be aware that there is a big difference in impact on square and sine signals. Even a minor change in circumstances will push the zero-crossing of a sine wave signal one way or the other. This is because (compared to a square-wave) the slope at the zero-crossing is very low in terms of V/s. Also, if measuring squarewave signals, be sure to choose your trigger point wisely. Many digital drivers control for the V(il)/V(ih) voltages and neither is likely to be V(max)/2. -- Poul-Henning Kamp | UNIX since Zilog Zeus 3.20 [EMAIL PROTECTED] | TCP/IP since RFC 956 FreeBSD committer | BSD since 4.3-tahoe Never attribute to malice what can adequately be explained by incompetence. _______________________________________________ time-nuts mailing list [email protected] https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
