# What is the minimum signal strength I need to receive?

Let's say a transmitting station is using a 8W radio with a J pole antenna. The cable connecting the radio to the antenna is RG-58A/U and is 25ft long.

At 144.075MHz, the attenuation of the cable is 1.5dB. At 8W, the power from the radio is 9.03dB. Assuming no losses other than cable losses, the antenna is getting 7.53dB, or 5.66W.

Let's now say the receiving station, 4 miles away, has the same setup (8W radio, J pole, and RG-58A/U). Assuming the gain of the two J pole antennas is 2.15dBi, and using the Friis transmission equation, the receiving antenna is receiving 1.733e-8W.

With the cable losses of the receiving station being 1.5dB (1.41W), and the antenna receiving 1.733e-8W, it appears the received signal will never make it to the receiving station's radio.

Me and my buddy have this setup with Baofeng BF-F8HP radios. If we are both using J pole antennas, we cannot communicate with each other. If one of us is using a J pole and the other is using the stock rubber ducky, the rubber ducky user can receive the J pole user's transmissions. I'm assuming the reason we cannot communicate with each other if we are both using J poles is because of the cable losses.

Is this correct? If so, is there a way around this?

• How do you arrive at the statement: "With the cable losses of the receiving station being 1.5 dB (1.41 W), and the antenna receiving 1.733e-8 W, it appears the received signal will never make it to the receiving station's radio. I mean, what calculation did you do to compute the signal strength at the actual receiver RF Input? It seems that you are treating the loss as an absolute number whereas it is actually showing the percent loss of the signal. For example, a 3 dB loss cuts your input signal in half, it does not erase it totally. – K7PEH Oct 27 '15 at 16:07
• Gotcha. So with a received power of 1.73e-8W, or -77.63dB and a loss of 1.5dB, the power at the receiver is -77.63dB - 1.5dB = -79.13dB, or 1.22e-8W? – KM4NTK Oct 27 '15 at 16:22
• Comments are not for extended discussion, but only to discuss improvements to the question; this conversation has been moved to chat. – Kevin Reid AG6YO Oct 28 '15 at 23:30
• Excellent move. But is OP able to participate in chat with less than 20 points? – captcha Oct 29 '15 at 0:09
• It isn't a direct duplicate, but you definitely want to check out What is a link budget, and how do I make one? – user Oct 29 '15 at 14:32

At 8W, the power from the radio is 9.03dB.

With the cable losses of the receiving station being 1.5dB (1.41W), and the antenna receiving 1.733e-8W, it appears the received signal will never make it to the receiving station's radio.

This seems to be the crux of your misunderstanding. You can't convert watts into decibels, or decibels into watts. A decibel expresses a ratio of two things.

In the case of losses at the receiver, it expresses the ratio of the received power (before losses) to the output power (after losses).

In the case of dBi, it expresses the ratio of the radiant intensity of a given antenna in its direction of maximum gain to the same of a hypothetical antenna which radiates equally in all directions.

In the case of dBm, it expresses the ratio of some power to 1mW.

So if your transmitter is 8W (that is, 8000mW), we can convert that into dBm:

$$\require{cancel} \frac{8\:\mathrm W}{1\:\mathrm{mW}} = \frac{8000\:\cancel{\mathrm{mW}}}{1\:\cancel{\mathrm{mW}}} = 10 \cdot \log_{10}(8000)\:\mathrm{dBm} = 39\:\mathrm{dBm}$$

Notice how this is a ratio, and it is a unitless number because the unit (mW) cancelled in the second step. We add the "m" to "dB" to remember that the denominator is 1 mW.

We can convert 39 dB back into a ratio:

$$10^\frac{39}{10} = 7943$$

This ratio is telling us that something is 7943 times bigger than something else.

The "m" at the end of "dBm" reminds us that this is the proportion of some quantity in relation to 1 mW, so to get back to the original quantity we must multiply by 1 mW:

$$7943 \cdot 1\:\mathrm{mW} = 7941\:\mathrm{mW} \approx 8\:\mathrm W$$

The discrepancy is just rounding error. Really 8 W is closer to 39.03089986991944 dBm.

If your receiver losses are 1.5 dB, then we can calculate the corresponding ratio (and I'll use negative 1.5 dB, because it's a loss):

$$10^{-1.5 \over 10} = 0.71$$

Remember that's a ratio: to get to some quantity we have to multiply it by something. In this case we are comparing the power delivered to the receiver to the power received by the antenna. So if the antenna was receiving 0.01 mW, then we'd do this:

$$0.71 \cdot 0.01\:\mathrm{mW} = 0.0071\:\mathrm{mW}$$

With this in mind, try your calculations again, and I think you will get a more sensical answer.

• Thanks. If you read the comments you can see we already figured this out. – KM4NTK Oct 28 '15 at 17:05
• Yes, and now there's an answer which can be accepted. – Phil Frost - W8II Oct 28 '15 at 17:06
• No, because that wasn't the question. The question was if there was a way around the losses from the cable. – KM4NTK Oct 28 '15 at 17:07
• @KM4NTK feel free to write your own answer. – Phil Frost - W8II Oct 28 '15 at 17:07
• I'm the one asking the question... – KM4NTK Oct 28 '15 at 17:08

From the symptoms you state of not being able to hear transmissions from identical setups only 4 miles from each other, and then being able to have one person receive with the rubber duck antenna, it tells me your J-Pole antennas are not behaving ideally on the frequency you choose.

Consider that an antenna like a J-Pole has about a 10MHz bandwidth in the 70cm Ham band. The rubber duck antenna has a broader bandwidth. Prove this to yourself by comparing reception of your NOAA weather broadcast station using first the J-pole and then the rubber duck. I'll bet you can receive it with the rubber duck. The NOAA broadcast is about 162.5MHz which is far above the 2M Ham band yet the rubber duck will work and very likely the J-pole tuned for the Ham band will not work.

Consider that the two antennas may have their band sweetspot tuned to some frequency 20MHz away from your transmitting frequency as an example. The antenna itself may contribute a lot of loss. You begin by looking at your feedline loss but maybe at the transmit frequency you have a very high SWR meaning that most of the power is reflected back from the antenna at your chosen frequency.

Invest in an SWR meter and measure the SWR of your J-pole at your desired frequency. Compare the two J-Poles. You should be able to see an SWR between 1.0 and 1.5 and if you see some crazy high SWR >5 then you can blame the J-pole instead of the feedline. You may be able to "hunt" for a frequency that your J-pole likes by using your SWR meter, meaning a frequency that the J-pole is tuned to work.

Also consider that any obstructions like trees and houses and especially hills between you and your friend will contribute most of the loss you see. Fighting this problem by desiring more output wattage can be a losing battle. If you have reasonable line of sight between the two radios then likely the rubber duck antennas will work for you with only 1 Watt. You can prove this to yourself by driving to hills that are within line of sight and see how well it works.

None of your formulas will mean anything because you are not considering the path loose. What is the SWR on the J-poles at the frequency you are using? How high are they? If they are in your basement instead of on the roof, that's a problem. J-poles that have good SWR, at 30 ft, with flat ground between them should easily work 10 miles apart with 8 watts.

I ended up switching from a 25ft RG-58A/U cable with F connectors on each end to an 18ft RG-58A/U cable with PL-259 connectors on each end. I can now receive, using my J-pole, my buddy's signal when he is transmitting with his J-pole. However, there is a lot of noise. This fixed the reception issue, but I'm still experimenting to figure out what the minimum signal strength is required at the RF input of my radio.

Here is the answer to "What is the minimum signal strength I need to receive?"

Baofeng Sensitivity

• I assume that your radio has squelch control. Have you fiddled with that to remove the noise. An FM signal is called Full-Quieting if when the squelch is set so that you hear only the FM signal without any other noise. – K7PEH Oct 29 '15 at 20:49