# 4 khz pulsed signal

Status
Not open for further replies.

#### Tryui

Joined Sep 20, 2021
44
hello gentlemen I'm new to the forum and I hope I guessed the right section. So I am an electrician technician, but between my various jobs I have always had a doubt. If I have a dipole antenna with a total length of 6-7 cm, can I receive a signal with a frequency of 50 mhz but pulsed at for example 4 khz? that is, even if the signal is not perceived by the antenna because it is too long, the pulsation modulated by an audio can somehow be perceived ??

#### ronsimpson

Joined Oct 7, 2019
1,629
antenna with a total length of 6-7 cm, can I receive a signal with a frequency of 50 mhz
A antenna does not have to be resonant at the receiving frequency. The signal will be smaller. A too short or too long antenna will work.
It does not really matter if the RF is modulated or not. You will need to (optionally amplify) then demodulate the RF to get the 400hz audio back.

#### Tryui

Joined Sep 20, 2021
44
A antenna does not have to be resonant at the receiving frequency. The signal will be smaller. A too short or too long antenna will work.
It does not really matter if the RF is modulated or not. You will need to (optionally amplify) then demodulate the RF to get the 400hz audio back.
thanks for the answer but maybe I did not understand well, I also know that if a 50 mhz signal is too long and cannot be perceived by that antenna, but if that same signal is pulsed for example at 4 khz, the pulsations can be do you perceive from that antenna? in the sense, a small radar antenna of 6-7 cm can demodulate the pulsations and not the frequency ??

#### DickCappels

Joined Aug 21, 2008
7,953
4 kHz would be too low to do much.

When you mix (actually multiply) sine waves of two frequencies the result is the sum and difference of frequencies.

Let's say that your dipole works well at 8.5 gHz. If you multiply a 50 MHz sine wave by a 8.45 Ghz sine wave or an 8.55 GHz sine wave (either of which is commonly called the "carrier") your 50 MHz sine wave would show up at 8.5 GHz.

Resulting signal = carrier frequency ± modulation frequency, plus some amount of carrier depending on how you actually accomplished the mixing of the two signals.

Edit:
To help you picture this, here is the spectrum of a 38 KHz carrier modulated by a 1 kHz input signal. This particular mixer suppresses the carrier. This is the way the Left-Right audio signal is added to the Left + Right signal in FM stereo.

I also agree with @ronsimpson , you don't need to have the antenna resonant at the frequency you want to receive if you design the receiver with that in mind.

Last edited:

#### Tryui

Joined Sep 20, 2021
44
4 kHz would be too low to do much.

When you mix (actually multiply) sine waves of two frequencies the result is the sum and difference of frequencies.

Let's say that your dipole works well at 8.5 gHz. If you multiply a 50 MHz sine wave by a 8.45 Ghz sine wave or an 8.55 GHz sine wave (either of which is commonly called the "carrier") your 50 MHz sine wave would show up at 8.5 GHz.

Resulting signal = carrier frequency ± modulation frequency, plus some amount of carrier depending on how you actually accomplished the mixing of the two signals.
yes, I too had thought about beats precisely because I use analog signals, but this is not my question, surely we are not understanding each other, then the sine wave is one and it is at 50 mhz, this same wave is pulsed as in radar, with 4 khz pulses, I would like to understand if the pulsation is perceived by the receiving antenna and can be demodulated, so I am interested in the pulsation and not the 50 mhz wave which I already know is not perceived even by truck drivers.

#### crutschow

Joined Mar 14, 2008
28,156
The 4kHz pulsation of the 50MHz signal (not mHz) will have no effect on receiving the signal with an antenna.
You still need to detect 50MHz.

#### Papabravo

Joined Feb 24, 2006
17,235
When the modulation frequency is very small with respect to the carrier, as it is in this case,

$$4\;\text{kHz}\;/\;50\;\text{MHz}\;= 80\times10^{-6}$$

That is close to 7 orders of magnitude. It is the carrier frequency that determines the receiving antenna's response. The modulation frequency or method has no significant effect on the antenna's performance.

#### Tryui

Joined Sep 20, 2021
44
The 4kHz pulsation of the 50MHz signal (not mHz) will have no effect on receiving the signal with an antenna.
You still need to detect 50MHz.
ok then, so the important thing is that the carrier is perceived, if the carrier is not perceived then not even the pulsations are perceived. To solve the frequency problem I thought about the beats, as already mentioned by DickCappels, I state that I already knew the phenomenon, but tell me, in practice how can I get the sum of the frequencies from my transmitter? In other words, how can I get the beats?

#### Alec_t

Joined Sep 17, 2013
12,309
With a 50MHz carrier and 4kHz modulation the beat frequencies will be 50.004MHz and 49.996MHz. If you can't detect 50MHz then you can't detect the beats either.

#### Papabravo

Joined Feb 24, 2006
17,235
ok then, so the important thing is that the carrier is perceived, if the carrier is not perceived then not even the pulsations are perceived. To solve the frequency problem I thought about the beats, as already mentioned by DickCappels, I state that I already knew the phenomenon, but tell me, in practice how can I get the sum of the frequencies from my transmitter? In other words, how can I get the beats?
The underlying mathematics revolves around a trigonometric identity that you can derive when you multiply two sine(cosine) waves together in a circuit called a mixer. The mixer is a nonlinear circuit in which the output has components of different magnitudes at:
1. The carrier frequency - 50 Mhz in this case
2. The modulation frequency - 4 Mhz in this case
3. The sum of the two frequencies - 50 Mhz + 4 khz = 50.004 MHz
4. The difference of the two frequencies - 50 MHz - 4 kzhz = 49.996
That is how it works in the transmitter.

The same process is used in the receiver to demodulate the incoming RF. It is common for the transmitter to actually suppress (filter out) the RF energy of the carrier and one of the two sidebands and transmit only the Upper SideBand (USB) or the Lower SideBand (LSB). This is because both sidebands contain the same information and they are in a sense redundant.

Consider the USB at 50.004 MHz showing up at the receiver to be mixed with the output of a local oscillator of 50 MHz. Again we can identify four components in the output which are:
1. The 50 Mhz local oscillator
2. The amplified 50.004 Mhz input signal
3. The sum of the local oscillator and the input at 100.004 MHz
4. The difference of 4 kHz which is the modulating infomation
A simple low pass filter will eliminate the high frequency components and output just the modulating information.

This illustrates why the condition mentioned in post #7 is important. A receiving antenna behaves in essentially the same way for two frequencies that are close to each other. Another way of saying this is that if the ratio of two frequencies is close to 1, then any antenna will treat them the same way.

You can see this on a sophisticated piece of equipment called a Vector Network Analyzer. A low budget version of this is an antenna analyzer or a nanoVNA.

Results for NANOVNA (gigaparts.com)

#### Tryui

Joined Sep 20, 2021
44
The underlying mathematics revolves around a trigonometric identity that you can derive when you multiply two sine(cosine) waves together in a circuit called a mixer. The mixer is a nonlinear circuit in which the output has components of different magnitudes at:
1. The carrier frequency - 50 Mhz in this case
2. The modulation frequency - 4 Mhz in this case
3. The sum of the two frequencies - 50 Mhz + 4 khz = 50.004 MHz
4. The difference of the two frequencies - 50 MHz - 4 kzhz = 49.996
That is how it works in the transmitter.

The same process is used in the receiver to demodulate the incoming RF. It is common for the transmitter to actually suppress (filter out) the RF energy of the carrier and one of the two sidebands and transmit only the Upper SideBand (USB) or the Lower SideBand (LSB). This is because both sidebands contain the same information and they are in a sense redundant.

Consider the USB at 50.004 MHz showing up at the receiver to be mixed with the output of a local oscillator of 50 MHz. Again we can identify four components in the output which are:
1. The 50 Mhz local oscillator
2. The amplified 50.004 Mhz input signal
3. The sum of the local oscillator and the input at 100.004 MHz
4. The difference of 4 kHz which is the modulating infomation
A simple low pass filter will eliminate the high frequency components and output just the modulating information.

This illustrates why the condition mentioned in post #7 is important. A receiving antenna behaves in essentially the same way for two frequencies that are close to each other. Another way of saying this is that if the ratio of two frequencies is close to 1, then any antenna will treat them the same way.

You can see this on a sophisticated piece of equipment called a Vector Network Analyzer. A low budget version of this is an antenna analyzer or a nanoVNA.

Results for NANOVNA (gigaparts.com)
ok ok thanks papabravo for your answer so now I will give an example with a more normal situation, then you have the case of an analog fm transmission, if I have a transmitter that gives me a maximum frequency of 500 MHz, how can I increase in practice the frequency with the beats? So since this is a non-linear process the beats are created at the receiving antenna, so I transmit an RF at 500 mhz and then how do I increase the frequency in the two bands?

#### ronsimpson

Joined Oct 7, 2019
1,629
how can I increase in practice the frequency with the beats?
I think I do not understand. Will try an answer.
Say you have 500mhz and it is modulated by 4khz so you have 500mhz, 500,000,400hz and 499,999,600hz all together. Any you want 700mhz. Then insert that into a mixer with 200mhz and you will get out of the mixer:
(500mhz &+/-400hz) & (700mhz &+/-400hz) & (300mhz & +/-400hz)
Now filter off what you don't want. Band pass filter of 700mhz with a width of +/= 5khz.

#### Papabravo

Joined Feb 24, 2006
17,235
ok ok thanks papabravo for your answer so now I will give an example with a more normal situation, then you have the case of an analog fm transmission, if I have a transmitter that gives me a maximum frequency of 500 MHz, how can I increase in practice the frequency with the beats? So since this is a non-linear process the beats are created at the receiving antenna, so I transmit an RF at 500 mhz and then how do I increase the frequency in the two bands?
They are created at both the transmitting end and the receiving end. The numbers are different, and the desired output is different, but the process is the same on both ends.

#### DickCappels

Joined Aug 21, 2008
7,953
Which brings us to the question of why the length of the antenna was mentioned. It may have steered some of us off course.

#### Tryui

Joined Sep 20, 2021
44
I think I do not understand. Will try an answer.
Say you have 500mhz and it is modulated by 4khz so you have 500mhz, 500,000,400hz and 499,999,600hz all together. Any you want 700mhz. Then insert that into a mixer with 200mhz and you will get out of the mixer:
(500mhz &+/-400hz) & (700mhz &+/-400hz) & (300mhz & +/-400hz)
Now filter off what you don't want. Band pass filter of 700mhz with a width of +/= 5khz.
back then I originally wanted an apparatus that had a bandwidth from 0 to 5 ghz and with high powers for an all in one, but this is not possible, so I opted to have only the radar configuration, as regards the power l 'I have calculated and it is 800 watts but the frequency must be at least 1 ghz up to a maximum of 2 ghz. Also instead of using tubes to produce microwaves, even if it actually suits me, I don't use it because they are obsolete and so I chose solid state transitors to produce microwaves. On the internet I discovered a company called Ampleon who produce them. The chip closest to my needs has 1200 watts of power, but only 500 mhz of frequency, but since I will use analog fm signals, I can exploit the phenomenon of beats. The 4 khz discourse is a separate matter because those are the pulsations that I get with a class d amplifier. So I would like to increase the frequency with the beats, and the ronsimpson response is the most effective, so ip I have to use another chip for the additional 200mhz, but if I with only one chip (which cost a lot) I did an fm modulation, with a carrier of 500 mhz and a modulating one of 400 mhz ?? would I get in USB 900mhz? it is only this speech that grips me

#### Papabravo

Joined Feb 24, 2006
17,235
Only if the actual device is capable of producing useable power at the desired frequency. As you are no doubt aware, producing high power levels at high frequency is a notoriously tricky and difficult enterprise. This is not exactly my area of expertise, but I know many amateur radio operators who work in the range from 5 Ghz to 405 Ghz. They use a variety of "frequency multipliers" based on a 10 Mhz. reference oscillator, in a Styrofoam cooler, which they can compare to WWV for accuracy. They do this with what is essentially a VCO and a "phase locked loop", not a "mixer".

They usually mount rifle scopes on their dish antennas so they can establish an optimal "line of sight arrangement.

#### ronsimpson

Joined Oct 7, 2019
1,629
I think we need to start over.
"I want to make a 1ghz radar system..............."

#### Papabravo

Joined Feb 24, 2006
17,235
I think we need to start over.
"I want to make a 1ghz radar system..............."
For what purpose....exactly?
Shades of "the guy with the autopilot schematic".

#### ronsimpson

Joined Oct 7, 2019
1,629
For what purpose
It took 15 posts to find out this is not about 50mhz but ghz. Most people here that play radar will not read a 4khz post. I think you want to make 1 to 5ghz signals.

#### Papabravo

Joined Feb 24, 2006
17,235
It took 15 posts to find out this is not about 50mhz but ghz. Most people here that play radar will not read a 4khz post. I think you want to make 1 to 5ghz signals.
It is not just a question of frequency, but power levels as well. My precise question is: "What is the purpose of both the frequency AND the power level"?

Status
Not open for further replies.
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7733182311058044, "perplexity": 926.6227981549741}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358560.75/warc/CC-MAIN-20211128134516-20211128164516-00353.warc.gz"}