From "Radio Receiver Design" by Robert C. Dixon, 1998 (Marcel Drekker, Inc., New York; ISBN 0-8247-0161-5) p. 370, section 14.3, states

"Blocking dynamic range (BDR) is the difference, in dB, between the noise floor and a off-channel signal that causes 1 dB of gain compression in the receiver."

Dynamic blocking range is the difference between the weakest signal that can be perceived and the strongest signal that can be present without adversely effecting a weak signal. This “adverse effect” is about 1 dB of attenuation since this is the smallest change that can be heard by the human ear.

Hint: The difference between the noise floor and the incoming signal that is enough to recognize the intelligence in the signal.

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Dynamic range means the amplitude range of input signals that can be properly processed by the receiver. Insufficient dynamic range for strong and weak signals means the strong signals are no longer treated linearly, creating modulation products that spill over into the desired signals.

Silly Hint: Spurious is BAD.

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Mixing of signals to produce intermodulation products will occur in any non-linear electrical circuit. The devices in RF output stages are often run in non-linear modes for better efficiency. This means that they can also act as mixers, signals from other nearby transmitters will travel down the antenna feeds and mix with the signal being transmitted. One way to prevent this is to install circulators. These devices allow the transmitter output to reach the antenna but block signals traveling down the feedline from the antenna.

Hint: mix is in the correct answer.

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Installing a "Circulator" may reduce or remove intermodulation interference. When one port of a three-port circulator is terminated in a matched load, it acts as an isolator since a signal can travel in only one direction between the remaining ports.

The devices in RF output stages are often run in non-linear modes for better efficiency. This means that they can also act as mixers, signals from other nearby transmitters will travel down the antenna feeds and mix with the signal being transmitted. One way to prevent this is to insert a circulator configured as an isolator to block signals traveling down the feedline from the antenna from entering the RF output stages.

Hint: the answer is the only one that contains the word repeater

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The intermodulation-product signal is \(f_{mod}=146.70\:\text{MHz}\). One of the modulation signals is \(f_1=146.52\:\text{MHz}\). What is the other modulation signal?

The modulation equations are \(f_{mod}=2f_1-f_2\) and \(f_{mod}=2f_2-f_1\). Solve the modulation equations for \(f_2\).

Solving the equation \(f_{mod}=2f_1-f_2\) for \(f_2\) yields: \(f_2=2f_1-f_{mod}\) and solving the equation \(f_{mod} = 2f_2-f_1\) for \(f_2\) yields: \(f_2 = \frac{f_1 + f_{mod}}{2}\).

There are many possibilities for the intermodulation (IM) products resulting from mixing in the output stages of adjacent transmitters and each one is represented by a different formula. The lower odd order products are often (not always) the strongest. Here are the formulae for third order products generated by two adjacent transmitters at frequencies \(A\) and \(B\):

1. \(2A+B\)
2. \(2A-B\)
3. \(2B+A\)
4. \(2B-A\)

We can immediately eliminate products due to 1) and 3) since they will give products far away from the received IM frequency specified in this question. Let us plug the numbers into 2) and 4) and set the IM result to \(146.70\:\text{MHz}\) as detected by the receiver. Assume that the known transmitter is on frequency \(A\) and that the “other” transmitter is on \(B\).

Using formula 2) we have \(2 \times 146.52 - B = 146.70 \:\text{MHz}\)

Rearrange to get \(B = -146.70 + (2 \times 146.52) = 146.34\:\text{MHz}\).

Using formula 4) we have \(2 \times B - 146.52 = 146.70 \:\text{MHz}\). Rearrange to get \(B = \frac{146.70+146.52}{2} = 146.61 \:\text{MHz}\).

This gives the frequencies as in the correct answer ( \(146.34\:\text{MHz}\) and \(146.61 \:\text{MHz}\))

Hint: The sum of the two digits past the decimal equals 7 in the clue and correct answer.

• \(146.70\) -> \(7+0=7\)
• \(146.52\) -> \(5+2=7\)
• \(146.34\) -> \(3+4=7\)
• \(146.61\) -> \(6+1=7\)

Another hint: The midpoint between the two given frequencies is 146.61. It appears on only one answer.

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The term for the reduction in receiver sensitivity caused by a strong signal near the received frequency is desensitization.

This is caused by RF energy from a strong signal near the receive frequency entering pass-band of the receiver. One way to reduce this is to utilize additional filtering to eliminate signals outside of the receiver's desired bandwidth.

In the case of a repeater, the receiver must be adequately isolated from its transmitter, if normal receiver performance is to be expected. Isolation (dB) is required to protect the receiver from transmitter spurious and noise radiation and receiver desensitization.

Tip: Reduction in sensitivity --> desensitization

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Desensitization occurs when a receiver is unable to receive a weak signal because a stronger, unwanted signal at a nearby frequency is present. By DECREASING the RF bandwidth the unwanted signal is filtered out. Reduce= decrease WP4AES

SILLY HINT: Question has "desensitization" which may be seen as to reduce. Answer has "Decrease" which is also to reduce.

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According to Wikipedia: "Intermodulation is caused by non-linear behaviour of the signal processing being used." http://en.wikipedia.org/wiki/Intermodulation

Memorization: The question and the answer has the words Circuit and Circuits mentioned.

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There are some cases where a preselector might help.

To increase rejection of unwanted signals would be one case of the use of a preselector in communications receiver.

Hint: preselect rhymes with reject.

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The third order intercept point (IP3) is one measure of the intermodulation (IM) performance of a receiver. It is a measure of the tolerance of the receiver to strong signals outside the passband.

Hint: only the correct answer contains the word "theoretically"

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A third-order product of two signals may be within the band of the receiver.

Silly Hint: The most "likely" answer is the only option with the word "likely" in it :-]

Another memory tip: Question contains the word “receiver,” answer is the only one that contains the word “received.”

See also: http://en.wikipedia.org/wiki/Third-order_intercept_point

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The Link Margin of a system is an assessment of the gains and losses in a transmission path and how it compares to the receivers minimum discernable signal (receiver sensitivity) to allow for successful communication.

In general: $LM = \text{Rx Power}_{total} - \text{Rx Sensitivity} $

\(\text{Rx Power}_{total}\) is calculated by the sum of all gains and losses in the transmission path;

\begin{align} \text{Rx Power}_{total} &=\\ \text{Tx Power}&+ \text{Gain}_\text{Antenna}\\ &- \text{Loss}_\text{Cable}\\ &- \text{Loss}_\text{Path}\\ &- \text{SNR} \end{align}

So: \[40\text{dB} + 10\text{dB} - 3\text{dB} - 136\text{dB} - 6 \text{dB} = -95 \text{dB}\]

Receiver Sensitivity (Minimum discernable signal) is given: \begin{align} &= -103 dBm \end{align}

Making overall Link Margin: \begin{align} LM &= -95 dB - ( -103 dB)\\ &= +8 dB \end{align}

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The received signal level will be the sum of all gains and losses through the path at the receiver.

\begin{align} P_{rcvd} &= 40 dBm + 6 dBi + 3 dBi - 100 dB\\ &= -51 dBm \end{align}

In general for these types of questions with terms in dB, identify the gains and the losses provided in the question. Add each of the gains together, and subtract each of the losses to find the total power.

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Conversion of a dBm value ( decibels relative to a mW ) into Watts is calculated with formula:

\[P_{W} = 1 mW \times 10^{\frac{P_{dBm}}{10}}\]

To convert -100 dBm to Watts:

\begin{align} P_{W} &= 1 mW \times 10^{\frac{-100}{10}}\\ &= 0.001 \times 0.0000000001\\ &= 0.0000000000001 W = 0.1 pW \end{align}

If you'd rather avoid the decibel calculation, you can consider that:

\[-100 dB = 10^{\frac{-100}{10}} = 10^{-10}\] and \[1 mW = 10^{-3} W\]

therefore: \(10^{-3}W \times 10^{-10} = 10^{-13} W = 0.1 pW\)

Note: \(pico (p) == 10^{-12}\)

Memory Aid: One of these things is not like the others - all the incorrect answers are in microwatt units, while the correct choice is the only one expressed in picowatts.

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