Mixer output shown as red wave:


Can a band-pass filter be used to remove the low frequency shown as a blue sine wave and leave the rest high frequency just like the frequency shown as a green sine wave? If so, which band-pass filter provides the most proper output (most relatable with the green wave?) RC or RLC filters? Active or passive filters?

I want the filter to be compatible with 2 different frequency values to distinguish frequency-modulated waves while showing high impedance to lower or higher frequencies.

An RC bandpass filter outputs a low voltage of the red sine wave if I'm not mistaken

  • Bandpass filter's input - red sine wave
  • Bandpass filter's output - green sine wave
  • Removed frequency - blue sine wave

The circuit I got the screenshot:

enter image description here

The circuit I am up to:

enter image description here

  • \$\begingroup\$ Why would you want to do that? \$\endgroup\$ Commented Jun 18, 2022 at 22:21
  • \$\begingroup\$ I think it is good if you tell: 1. how much do you know about what exactly is bandpass filter. 2. What is the difference between active and passive filter. I think if you understand this two question then your problem will solve, am I right? \$\endgroup\$ Commented Jun 18, 2022 at 22:24
  • \$\begingroup\$ The red signal is not the sum of the blue and green signals. So, removing the blue signal from the red signal will not leave the green signal. \$\endgroup\$ Commented Jun 18, 2022 at 22:32
  • \$\begingroup\$ bandpass filters can be used to block undesired frequencies if the pass-band of the filter is narrow enough. However, as I mentioned, the red signal is not the sum of the blue and green signals. So when you pass the red signal through a bandpass filter, the result will not be the green signal. Where are you getting these signal images from? \$\endgroup\$ Commented Jun 18, 2022 at 22:37
  • \$\begingroup\$ Deleted cause I wanted to ask a new question electronics.stackexchange.com/questions/624113/… \$\endgroup\$
    – LucasSokol
    Commented Jun 18, 2022 at 23:00

1 Answer 1


A diode ring mixer used in your previous question can generate amplitude modulation suitable for radio-frequency transmission. The example shown uses a sine wave V_carrier, frequency of 50 MHz. This could also be a square wave source as well. Amplitude is about 1.5V RMS. These mixers work well when driven with a source resistance of 50 ohms. The carrier signal generator should be capable of delivering roughly +10 dBm - that's 10 milliwatts into 50 ohms.

The modulating signal here V_baseband is 1 MHz sine wave, amplitude of 0.2V with a 0.2V DC offset. When this generator is at 0V, no RF output appears. When amplitude rises to +0.2V, maximum RF output appears. These modulators can usually accept baseband input frequency of many megahertz. It needn't be a sine wave - digital signals can be applied too, as long as they can drive 50 ohms. Be aware that the carrier frequency should be far higher than the frequency of this baseband signal, to ease the requirement of the output filter needed to knock down harmonics.
schematic AM modulator

The output signal AM_radio_frequency_output contains harmonics, mostly at 3x carrier frequency (at 150 MHz) and above. These should be attenuated. This is usually done with a band-pass filter, centered at the carrier frequency (in this example 50 MHz).AM_radio_frequency_output waveform AM modulator spectrum showing all harmonics
These modulators are standard components made by various manufacturers which are to remain un-named. Many allow the modulating source to go down to DC, as shown in the schematic above. No need to build one yourself.
But be aware that output is feeble, and likely needs amplification before driving an antenna. And don't forget the bandpass filter, else those harmonics will cause problems for other spectrum users.
In this example, the bandpass filter should have a bandwidth of at least twice the modulating frequency (2 MHz). A low-pass filter would work too, with 50 MHz at the upper end of the pass-band, and 150 MHz in the stop-band. At these frequencies, LC filters are appropriate.


Your Answer

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

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