[HBQRP] Tech message copied from "Larry's List"

[Ray McNally]

This message is forwarded in full with no editing.

Full credit goes to the author Tom Wheeler, who has explained the solution
very well.

I know we are not in KC, but I find a lot of great information flows
through Larry's list and would recommend subscribing.

73,
Ray - N5SEZ


Date: Sat, 20 Oct 2018 15:55:19 -0500
From: Laurance Staples <larrystaples at mac.com>
To: "Larry's List" <LarrysList at K0JPR.net>
Subject: [LarrysList] Solution to Tech Teaser given on 10/20/2018
        Casual  Friday Net
Message-ID: <FA7BCDCE-28A3-4D77-87B9-998069BE3618 at mac.com>
Content-Type: text/plain; charset=us-ascii

Solution to Tech Teaser given on 10/20/2018 Casual Friday Net

Related FCC Amateur Radio Question Pool Questions: E9E07, G9A09, G9A10
Topics: Reflection coefficient, VSWR, decibels

Problem Statement:

Dave has been looking for an RF attenuator to use in his hamshack. An RF
attenuator reduces the strength of a signal introduced into it by a
specified amount measured in decibels (See
https://en.wikipedia.org/wiki/Attenuator_(electronics)). Great news, Dave
sees a table with a pile of 10 dB attenuators at a great price. Bad news,
the seller says they're "untested" (which usually at a hamfest doesn't bode
well at all)! Dave has a hunch on how to do a quickie test on them,
however. He sees his friend Royce nearby. Royce always carries his trusty
multimeter in his pocket, and Dave uses the meter to measure the DC input
resistance at the input connector of the attenuator, which measures about
61 ohms, and at the output connector, which again measures about 61 ohms.
Dave decides the attenuator is probably working and buys it. How might Dave
have reached this conclusion?

Solution:

There are many ways of solving this problem. Since most attenuators are
rated from DC to an upper RF frequency, the behavior of the device with an
ohmmeter applied should parallel its RF performance, so the use of Royce's
multimeter to measure the input resistance of the device is valid at DC.
Since the resistance at both ports (or connectors) of the attenuator
measures 61 ohms, we know that the attenuator isn't outright burned out - -
both ports of a good attenuator should measure the same DC resistance. But
the resistance isn't 50 ohms, because there is no 50 ohm load placed on the
other end (Dave didn't have one to try). But we can infer the attenuation
value from the effective reflection coefficient the unit would show as
follows:

The reflection coefficient, G (Gamma), is easy to calculate based on the ZL
of 61 ohms Dave measured:

(1) G = (ZL-Z0)/(ZL+Z0) = (61-50)/(61+50) = +0.099099

Equation 1 tells us that for every forward RF volt that is fed into a 61
ohm resistor, 99 mV will be reflected back, given that the system
characteristic impedance Z0 is 50 ohms.

Think about this carefully: If we hook a 1 volt RF source to this
attenuator, 1 volt of forward RF voltage will result, and 99 mV of
reflected RF voltage will come back. Does that correspond with 10 dB of
attenuation? This takes some careful contemplation - let's ride along with
the RF signal wave.

1. The 1 volt forward voltage wave experiences 10 dB of attenuation as it
passes through the attenuator in the forward direction. At the far end of
the attenuator, the forward voltage wave is now 10 dB weaker (at least if
there's a 50 ohm load on the other end). That means that at the far end,
there's about 1 V * ( 10^(-10/20) ) volts, or  0.316 volts. (The minus sign
here shows attenuation instead of gain, and ^ means "raised to the power
of.")

The forward trip looks like this:

[ 1 volt ] --------------------------------------------------------->
[0.316 volt]

2. At the far end, there's no place for the 0.316 volt output signal to go
(there's an open circuit), so it reflects back into the attenuator in
phase. It makes a return trip back to the input of the attenuator, where it
again experiences a 10 dB reduction. So the total reduction in the wave is
20 dB, which should result in a backward wave on the input connector that's
exactly 0.1 (10%) of the original voltage.

The return trip looks like this:

[0.1 volt] <-------------------------------------------------------- [0.316
volt]

In other words, the reflection coefficient at the input of an ideal 10 dB
attenuator should be +0.1 (10%) if the output is open circuited.

Dave calculated an effective reflection coefficient of +0.099099, which is
very close to the ideal +0.1 value. That's why he concluded that the
attenuator was probably okay.

Why wasn't Dave's answer exactly +0.1? Because a real RF attenuator is made
out of a circuit with three resistors. This real-world network's actual
attenuation value depends on the termination resistance placed on the
attenuator output port. For attenuators with 10 dB or more reduction, less
than 1% error results from using Dave's approach.

Discussion

There are other good ways of checking this. Dave might have hunted for a 50
ohm resistor (or a good quality dummy load) to connect to the B side of the
attenuator. The input resistance should have then become very close to 50
ohms on the A side. Additionally, he could have connected a 1 volt DC
source at the input, and the same 50 ohm resistor on the output. Under that
condition, he would observe exactly 0.316 volts (31.6%, or 10 dB below the
DC source voltage value) across the 50 ohm load resistor if the attenuator
was operating correctly.


73,

Tom Wheeler, N0GSG
tom.n0gsg at gmail.com

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~``

Thanks, Tom, for the teaser and for the solution.

Larry, W0AIB