How to prove the bandwidth of intermodulation is related to its order ?

[criterion456]

Hi,

If 2 tones produce intermodulation(IMD)

fa, fb their bandwidth are : BWa and BWb

IMD2: fa +- fb
IMD3 : 2fa+-fb or 2fb+-fa

Then, the bandwidth of IMD2 is BWa + BWb
the bandwidth of IMD3 is (2BWa + BWb) or (2BWb + BWa)

Thus, higher order of IMD, wider its bandwidth

I know the conclusion is from convolution of Fourier transform.

Nevertheless, I don’t know the detailed proof.

Would you please tell me how can I know the detailed proof ?

Thanks a lot~!

 

[jiripolivka]

Hi,

If 2 tones produce intermodulation(IMD)

fa, fb their bandwidth are : BWa and BWb

IMD2: fa +- fb
IMD3 : 2fa+-fb or 2fb+-fa

Then, the bandwidth of IMD2 is BWa + BWb
the bandwidth of IMD3 is (2BWa + BWb) or (2BWb + BWa)

Thus, higher order of IMD, wider its bandwidth

I know the conclusion is from convolution of Fourier transform.

Nevertheless, I don’t know the detailed proof.

Would you please tell me how can I know the detailed proof ?

Thanks a lot~!

I think any degree of intermodulation generates an infinite series of harmonic combinations. There is no definite bandwidth but the power of individual spectrum lines increase.

As the spectrum bandwidth is usually defined to half-power or -3 dB, you can specify a reference level for it and then neglect all spectrum lines laying below that level.

 

[FvM]

I don't think that you need convolution to see why the bandwidth of intermodulation products is multiplied along with the center frequency of input bands. Just put band corner frequencies into the calculation.

 

[criterion456]

Okay~! Thanks you.