6280 Views 25 Replies Latest reply: Oct 22, 2010 11:06 AM by BillDumke
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## Passband Intermodulation

I have read recently that 3rd order intermodulation terms will be be 3 times greater in bandwidth than the original bandwidth of either of the original two signals.  I tried to verify that, but am having some difficulty.  I was trying to use the Heaviside Unit Step functions, to create rectangular passbands.  I would appreciate any help on this.  Attached is a Mathcad 11 file.

Bill

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Oct 6, 2010 1:58 PM (in response to BillDumke)
Re: Passband Intermodulation

Don't you just need to put subscript f on variables p1 and p2 where they occur on the right-hand side of Kf?

Alan

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Oct 6, 2010 2:21 PM (in response to BillDumke)
Re: Passband Intermodulation

See attached.  Since I don't know much about passbands it might not make physical sense!  However, it works in Mathcad (M15) without any error messages.

Alan

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Oct 7, 2010 1:46 PM (in response to BillDumke)
Re: Passband Intermodulation

Take an example IMD term from the sinusoidal version: a1*a2*a2*cos[ (-2w2+w1)*t]. The generalization for rectangular spectra is to have a1 and a2 be (baseband) time functions rather than constants. They may have rectangular spectra or any shape.

Define the spectral bandwidth of a1(t) and a2(t) to be bw1 and bw2. Multiplying functions in time corresponds to convolution in freq. domain. Thus the sample term above has a bandwidth defined by the convolution [ a1(t) convolve a2(t) convolve a2(t) ]. The spectral bandwidths add when spectra are convolved. Therefore, this example term has a spectral bandwidth of bw1+2*bw2 centered at f = -2*f2+f1 . Each term in the expansion is treated similarly.  Clearly, the maximum spectral width of the third order terms is 3*bw1 or 3*bw2 , whichever is larger.

The same principles apply to higher order IMD.

In the sheet, the attempt to create rectangular spectra is wrong. See comments in attached.

Lou

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Oct 12, 2010 7:12 AM (in response to BillDumke)
Re: Passband Intermodulation

I don't have specific examples worked out, but it's fairly straightforward to do it (or look it up) for the case of a rectangular spectral envelope. Convolving a real, rectangular spectrum of width w with itself gives a triangular spectrum of total width 2w. Convolving again with the rectangular envelope gives a bell shaped spectrum (with parabolic segments) of total width 3w.

The usual caveats apply: specific cases with complex spectra may not come out like the textbook cases, any other IM compponents that may overlap can change the resulting total spectrum,etc.

Lou

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Oct 15, 2010 10:49 AM (in response to BillDumke)
Re: Passband Intermodulation

The function 'convol" in one of your sheets doesn't work for me. It's not in mcd11 as I have it; likely in an extension pack that you have. The latest has way too many terms to look at. Regardless, I put together a simple example of a signal with two comb spectral sections. The spectrum of the 3rd order IM products has the expected bell shape. You may be able to build on this to do fancier or more complicated models. It may also help guide you to determine why your examples don't show the expected shape.

Some general suggestions on the sheets and your approach. The statement I made earlier that the spectral shape of an IMD term is the convolution of the corresponding envelope terms is valid. If you are going to use the baseband envelopes (not explicitly defined in the attachment example), then the spectral position of that term needs to be separately calculated based on the carrier term. Alternately, you can calc the spectrum of the full signal, as per the attached example.

I recommend that you go through some simple examples by hand, the old fashioned way using pen and paper, so that you have a good understanding of the underlying math. The mcd sheets should be used to extend this to the more complex, real examples you are really interested in. It will be the basic understanding that will let you know if what the mcd sheet is reasonable, and whether the implemetation has errors.

Regarding implemetation, I personally prefer using vectors instead of range variables wherever possible. Also, I try to keep the coarseness of quantization (how many points to use)  - the conversion of a continuous problem (as here) into a discrete approximation - separate from the physical parameters of the problem. This makes it easier to change the accuracy of the approximation without changing the underlying physical parameters. In the attached example, The time resolution is defined by the sampling freq fs. The spectral bands are not defined by fs, but rather by their own parameters.

This is about all I can do for now; hope it helps.

Lou

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Oct 15, 2010 11:55 AM (in response to LouPoulo)
Re: Passband Intermodulation

Lou Poulo wrote:

The function 'convol" in one of your sheets doesn't work for me. It's not in mcd11 as I have it; likely in an extension pack that you have.

Hi Lou. Can find an implementation for mc11 at http://communities.ptc.com/servlet/JiveServlet/download/34110-3574/convol%20function.mcd in http://communities.ptc.com/message/34105

Regards. Alvaro.

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Oct 15, 2010 1:23 PM (in response to BillDumke)
Re: Passband Intermodulation

I assumed that the convolve was part of an extension pack. I have homemade convolve fcts., but I didn't have time to use them in a debug mode.

The two combs comprising the time function in the example have the same # of freq's since they are created with a single summation. Create whatever you want, using as many terms as you wish or any functional form. This is just the place to define your time function. It can have separate summations with distinct bandwidths and #'s of terms, and can be mixed with envelope*carrier terms. You can add noise as time functions here as well. There's really no limit.

I used half Hz resolution (or intended to) since I created signals with freq components at 1 Hz spacing with possible 0.5 Hz offset (from N/2). You may want to pick the sampling freq differently depending on the particular parameters of your problem. In any case, the example was intended to give you a starting point.

Lou

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Oct 18, 2010 2:57 PM (in response to BillDumke)
Re: Passband Intermodulation

With Tmax=2, the freq resolution ogf the analysis is 0.5 Hz, which is necessary if the freq components are not at integer freq's. Try your example 4 and change N to 11 and see what happens (lots of "leakage," as it's called).

My example has an incorrect offset formula for the lower band edges. The offset should be by (N-1)/2 and not N/2 as in the sheet. the total freq span of N components 1 Hz apart is (N-1) Hz, so the lower edge is the desired center fc - span/2 = fc - (N-1)/2. changing the summation limit just makes the combs have N+1 terms, and the offset formula is then correct for N+1 terms.

With 1 Hz spacing, the IM products will also have 1 Hz spacing. With Tmax=2, the analysis freq's are multiples of 0.5Hz up to fs, the sampling freq. With actual freq components at 1 Hz intervals, then every other point will be zero. The nonzero ones are the components of interest; you just have more info than you need when there are only integer freq's. With noninteger freq components, but with 1 Hz spacing, you're interested in the odd multiples of 0.5 Hz and not the even multiples. Unfortunately, you get all of them since thew DFT analysis freq set consists all the multiples of the fundamental repetition freq (0.5 Hz with Tmax=2).

The way the example is set up (with corrected offsets), if you have a four freq comb centered at 80 Hz with df= 1Hz, then the actual freq's are 78.5, 79.5, 80.5, and 81.5. If the set of analysis freq's doesn't include all of these (as when Tmax=1), then you get leakage. For a 3-freq comb, the freq's are 79,80,and 81- all integers - so Tmax = 1 works.

Unless you use windowing (a completely separate topic), and properly interpret the reults, you need to pick Tmax so that it represents an integral number of periods of the time signal. If the freq components are on a 0.5 Hz grid, then you need Tmax=2. If you have freq's of 1.1, 2.1, 4.2, 5.7, then you need Tmax = 10, since this is the shortest time over which all components are periodic. This is the foolproof way to get all the info. If a component =0, then there's no IM prduct at that freq. Oversampling rarely gets you into trouble.

Lou

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Oct 21, 2010 2:24 PM (in response to BillDumke)
Re: Passband Intermodulation

The changes made in the last example simply create combs that fit on the freq sampling grid (with no leakage) on consecutive samples, so there are no intervening freq's within a band that are zero. This can be done with fs=1 also, if the combs are constrained to use integer freq's (thus no leakage), and use successive samples (so no intermediate zeroes). This may be OK for the range of models that you plan to look at. However, you may want to model cases where the required sampling freq fs and the repeat time Tmax are such that the IM components don't fall on succesive samples. This is the case for these simple examples, but I don't think that constraining the sampling grid in this manner is good practice. Better to figure out how to allow for the intermediate zero components in the general case. If you can calulate the IM component spacing, and you also know the sampling grid, then you can average/max/select,etc. every N components to get only the correct IM locations.

Lou

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