0
\$\begingroup\$

I was going through radio transmission, and looking at a handout from Analog Devices on Heterodyne structure. Here is layout from this handout.

enter image description here

What I understand that in "MOD" (refer the layout above) level the actual signal is mixed with the carrier frequency, so that my baseband signal is moved to carrier frequency on which we will do the transmission. In that case why we need two mixers after the "MOD" level to further upconvert the frequency.

Example: to transmit at 200 MHz, my carrier frequency would already be 200MHz inside the "MOD" stage. Then why need to further upconvert this frequency via these mixers.

I have thought on it a lot, but still could not figure a satisfactory answer.

Your guidance will be greatly appreciated.

ADDITION:

During "MOD" stage, we do add the signal with some sort of sine function, say $$sin(2\pi f_ct)$$.

In that case is this $$f_c$$ the carrier frequency (200 MHz for my example), clearly it can't be, it has to be much much lower than that, and then we use the mixer to move it to 200MHz. Then during modulation stage, why do we call the $$f_c$$ as carrier frequency, aren't we suuposed to transmit the signal at this frequency??

\$\endgroup\$

1 Answer 1

Reset to default
2
\$\begingroup\$

What I understand that in "MOD" level the actual signal is mixed with the carrier frequency

That is not the case, MOD stands for Modulator which is a block that applies a certain modulation to a signal. For example, FM modulation for an audio signal or OFDM for a datastream for WiFi.

At the output of the MOD you would then have a baseband signal. You cannot transmit this signal as the frequencies are too low. This is where the Mixer comes in, it mixes the baseband signal up in frequency so that it can be transmitted at for example 200 MHz.

In your diagram there are 2 mixers and an IF stage. This can be used to first mix up to 10 MHz (right mixer), then filter that signal, for example only let frequencies between 9 and 11 MHz pass, attenuate the rest.

Why then not mix directly to 200 MHz and filter there, saving a mixer ?

Well, a 2 MHz wide filter at 10 MHz has a Q factor of 10/2 = 5 which is easy to make.

At 200 MHz that would be 200/2 = 100 and that is much harder to make, more sensitive to drift, proper tuning etc. It would need to be better than 1% accurate. All in all not so easy and for sure expensive.

Then in the left mixer mix this 10 MHz IF (and modulated) signal with 190 MHz (or 210 MHz) resulting in 200 MHz for the RF amplifier.

\$\endgroup\$
4
  • \$\begingroup\$ On a side note @FakeMoustache - during modulation stage we do multiply the signal with say $$sin(2\pi f_ct)$$, then I believe, this $$f_c$$ is not the 200Mhz one, but rather a low one? Can you please clarify. \$\endgroup\$
    – niki_t1
    Jan 26, 2017 at 12:46
  • \$\begingroup\$ Usual communications parlance is that a Modulator takes a Baseband signal and generates an IF (or the carrier directly) from it. I can't think of any modulation type for which you'd describe the output as a baseband. Unless of course the MOD block above stands for the Modulation signal, which could be baseband, which is then simply AM'd or DSBM'd onto the first LO in the first mixer which is the modulator. \$\endgroup\$
    – Neil_UK
    Jan 26, 2017 at 12:48
  • 1
    \$\begingroup\$ That $$sin(2\pi f_c)$$ is FM modulation. You could use $$f_c = 200 MHz$$ but then the RF signal would span between 0 and 200 MHz, which you do not want. You want $$f_c = 1 MHz$$ so that the modulated signal is between 0 and 1 MHz and mix that up to 200 MHz +/- 1 MHz (so 199 MHz - 201 MHz). \$\endgroup\$
    – Bimpelrekkie
    Jan 26, 2017 at 13:00
  • \$\begingroup\$ @Neil_UK I can't think of any modulation type for which you'd describe the output as a baseband I can, it is commonly used in almost all "digital" radios (WiFi, Bluetooth, LTE etc) as these are usually zero IF transceivers so the baseband (coming out of I&Q DACs) is filtered and fed into the mixers and upconverted to RF. In essence the modulator would then be a digital circuit. \$\endgroup\$
    – Bimpelrekkie
    Jan 26, 2017 at 13:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.