In the past, sensitivity was one of the most important receiver performance specifications. Today, instead of sensitivity, we speak of a receiver’s minimum discernible signal, or MDS. The MDS of a receiver is the minimum discernible signal. (E4C07) This is the weakest signal that a receiver will detect.
One parameter that affects receiver sensitivity is the noise figure. The noise figure of a receiver is the ratio in dB of the noise generated by the receiver compared to the theoretical minimum noise. (E4C04) Lowering the noise figure of a receiver would improve weak signal sensitivity. (E4C08)
A related specification is the noise floor. When we say that the noise floor of a receiver has a value of -174 dBm/Hz, it is referring to the theoretical noise at the input of a perfect receiver at room temperature. (E4C05) If a CW receiver with the AGC off has an equivalent input noise power density of -174 dBm/Hz, the level of an unmodulated carrier input to this receiver would have to be -148 dBm to yield an audio output SNR of 0 dB in a 400 Hz noise bandwidth. (E4C06)
A receiver’s selectivity is the result of a lot of things, including the filters a receiver has. 300 Hz is a desirable amount of selectivity for an amateur RTTY HF receiver. (E4C10)2.4 kHz is a desirable amount of selectivity for an amateur SSB phone receiver.(E4C11)
In addition to a 300 Hz filter and a 2.4 kHz filter, high-end receivers also have filters called roofing filters. A narrow-band roofing filter affects receiver performance because it improves dynamic range by attenuating strong signals near the receive frequency. (E4C13)
Back in the day, when superheterodyne receivers had intermediate frequencies, or IFs, in the 400 – 500 kHz range, image rejection was a problem. If there was a strong signal present on a frequency about two times the IF away from the frequency your receiver was tuned to, you might hear that signal. Accordingly, 15.210 MHz is a frequency on which a station might be transmitting if is generating a spurious image signal in a receiver tuned to 14.300 MHz and which uses a 455 kHz IF frequency. (E4C14)
One solution to this problem is to select an IF higher in frequency. One good reason for selecting a high frequency for the design of the IF in a conventional HF or VHF communications receiver is that it is easier for front-end circuitry to eliminate image responses. (E4C09) A front-end filter or pre-selector of a receiver can also be effective in eliminating image signal interference. (E4C02)
Another way to get rid of image signals is to use a narrow IF filter. An undesirable effect of using too wide a filter bandwidth in the IF section of a receiver is that undesired signals may be heard. (E4C12)
Because most modern transceivers use digital techniques to generate a local oscillator signal to tune a receiver, synthesizer phase noise might be a problem. An effect of excessive phase noise in the local oscillator section of a receiver is that it can cause strong signals on nearby frequencies to interfere with reception of weak signals. (E4C01)
Finally, here are two miscellaneous questions on receiver performance characteristics. Atmospheric noise is the primary source of noise that can be heard from an HF receiver with an antenna connected. (E4C15) Capture effect is the term for the blocking of one FM phone signal by another, stronger FM phone signal. (E4C03)
Elwood Downey, WB0OEW says
Rather than think of Noise Figure in terms of temperatures, I find it intuitive to think of it as just the difference between the SNR coming out and the SNR going into an amp. So for example, if a signal input has 10 db SNR, after going through an amp with 2 db NF it will be degraded to 8 db SNR on the output.
ivor says
“If a CW receiver with the AGC off has an equivalent input noise power density of -174 dBm/Hz, the level of an unmodulated carrier input to this receiver would have to be -148 dBm to yield an audio output SNR of 0 dB in a 400 Hz noise bandwidth. (E4C06)”
Is there any math to figuring out the -148 dBm?
David Ryeburn VE7EZM and AF7BZ says
If the input noise is -174 dBm/Hz that means that there would be -174 dBm of noise power, or 174 dB below a milliwatt, in a 1 Hz bandwidth. There would be 400 times as much noise power in a 400 Hz bandwidth. 400 times as much power is 10*log(400) = 10*2.6 = 26 dB more power (this is base 10 logarithm, not the natural “base e” logarithm), and -174 + 26 = -148. You have to get equal amounts of signal and noise power to get 0 dB signal to noise ratio.
This works because noise is randomly distributed across the spectrum. The noise in one 1 Hz slice of spectrum is not the same as the noise in the next 1 Hz slice of spectrum. If it were, i.e. if noise here were in phase with noise there, then you would add voltages instead of power and you’d get 160,000 times as much noise, not 400 times as much noise, in a 400 Hz slice of spectrum as you would get in 1 Hz. Of course this would make wide-band communication hopeless.
Dan KB6NU says
Thanks for the great explanation, David!
ivor says
Yes, thank you very much.
I’ll eye-ball this in the future. My method of eye-balling it will be:
10(log(400)) is obviously between 20 and 30. However it being log you know about half-way point is in the first third. So the number needs to be between 25 and 30 putting it between -144 and -149. Only answer D is in that range. So I won’t need a calculator for this question!
David Ryeburn VE7EZM and AF7BZ says
I didn’t need a calculator either. I learned early in high school that the base 10 logarithm of 2 is 0.30103. Later I learned that twice as much power is 3 dB more (which says the same thing), and every ham (I hope) knows about twice as much power and 3 dB. So 4 times as much is the same as twice as much, two times, and that means 6 dB. Same way, 8 times as much power would be 9 dB. So you don’t need a table of logarithms or a calculator or a computer for this one. Knowing twice as much power is 3 dB and ten times as much power is 10 dB takes care of lots of estimates. Save the heavy artillery for when you need it.
jk says
The units are not the same. So how can the numbers be added.
174 is dBm and 26 is dB. So how do you add 174 + 26 = -148