Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Register Log in

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

Status
Not open for further replies.
C

criterion456

Member level 2
Member level 2
Joined
Sep 13, 2015
Messages
50
Helped
6
Reputation
12
Reaction score
6
Trophy points
1,298
Activity points
2,022
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~!
 

criterion456 said:
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~!
Click to expand...

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.
 
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.
 
Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top