# Tutorial 11 – Link Budgets

1. Duy Nguyen says:

Thank you for making the tutorial available to public. The PDF file for tutorial 11 – Link Budgets is actually of tutorial 12.

2. Gunther March says:

I’m working on the Link Budget analysis for my master thesis.
I didn’t understand the differences between the two methods for Link budget analysis. There is the Eb/No method but also the SNR method to evaluate the link margin and all results.
The results of the two methods are different. I didn’t understand what is the difference between these methods and what result is good for me.

Sorry for the trouble. Thank you in advance.

• Charan L. says:

We use the EbNo method for digital signals such as QPSK, 8PSK, 16APSK, 32APSK etc. The metric of performance for these signals is the EbN0 vs. the Bit Error Rate (BER). So we first set a BER that is desired, say 10-6 and then determine what EbN0 would be needed to deliver the required BER. Depending on the code used, this may vary for a given modulation by 1-2 dB. The goal of a link budget where data is digital is to provide this minimum EbN0.

Analog signals do not have BER. They are judged instead by SNR, signal to noise ratio. For an analog signal such a FM signal set, or SSB signals, an average SNR and a peak SNR is usually specified based on number of signals sharing the bandwidth. The peak SNR can often be 10 to 14 dB higher than the average. The goal of the link budget for analog signals is to provide the peak SNR.

So the two methods you are talking about pertain to traffic types and will of course give completely different answers since there is no simple way to compare a FM/AM signal with a MPSK signal.

Charan Langton

3. Paul Zwicker says:

In section 4, where the FSL is written in dB form – the unit specific nature of the constant should be stated. This form is requires r to be in km and f to be in MHz. Especially since the example then uses the constant in the r[km] and f[GHz]. It just makes the material confusing for a second because of the inconsistency, but thank you for this material it has been a great refresher.

4. Keith Clark says:

I agree with Paul. On page 15 the equation for free space loss is only correct if “r” is measured in km and “f” is defined in MHz. If base units are used, “r” in metres and “f” in Hertz, then the constant is -147.56. This becomes a problem in Ex. 6. Firstly the height of the satellite is not defined in the question but assumed to be 22,000km as was used in Ex. 3. If a constant of -92.4 is used this means that “r” is defined in km and “f” in GHz. The penultimate line in Ex. 6 shows 20 Log (12×10^9), this is wrong as the units are in GHz so that line should simply read 20 Log (12). Doing the numbers correctly gives FSL = 200.6 dB not 196 dB.

Finally, Preceived = 60dBW + 52 – FSL dBW

If FSL = 196 dB (as shown in the example) then
Preceived = 60dBW + 52 – 196 dBW
Hence, Preceived = -84 dBW not -80 dBW as shown but this is using the wrong figures, FSL should be 200.6 dB not 196 dB

Therefore Preceived = 60dBW + 52 – 200.6 dBW

Having said all that, the tutorials are very helpful and not at all daunting making them very helpful. Hopefully you take my comments in the constructive way they were intended! 🙂

• Charan L. says:

Thank you. Very grateful for these. Got fix them.

Charan

5. Oliver Street says:

1) When analyzing noise the analog bandwidth and bit rate aren’t 1:1 equivalent, there are 2 bps/Hz, not 1bps/Hz. If anyone clings to the appeal of 1:1 show a 1Hz square wave is a 2bps alternating +/- binary sequence, and then show a 1Hz sine wave (claim the same RMS power without visually re-scaling) and integrate the energy over the half cycle bit interval to get an unambiguously equivalent +/- result.

2) The section on the BER and bandwidth of n-ary PSK is weak. Its simpler begin by showing QPSK as the sum of two BPSK signals created from a quadrature (90°, sine/cosine) offset copies of the carrier, where each has 1/2 the carrier power. (Its a simple way to make a QPSK modulator.) The relative bit energy for each of the two BPSK streams is more obviously -3.02dB. Since any n-ary PSK signal can be formed the same way, the bit enerygy relative to BPSK is 2/n or 10log10(2/n) dB. (Try to avoid describing the QPSK components as orthogonal for this purpose, it’s irrelevant when the analysis is extended to n-PPSK)

3) The section on coding gain falls into the same trap that many other do. The coding gain isn’t corrected for the coded channel bandwidth. This remains valid when, like the original applications for channel these codes, lowering the uncoded bit rate doesn’t produce a proportional increase in Eb/No because of the noise spectrum isn’t flat because of 1/f (and 1/f^2) noise.

Here’s a tutorial on coding gain that doesn’t fall into the bandwith trap http://ecee.colorado.edu/~mathys/ecen5682/slides/convperf99.pdf

4) The paragraph beginning with 10^-2 BER has some problems. It falls into the bandwidth coding gain trap, but it should recognize that its describing a link design constained by choices already made. Satellite channel bandwidth and downlink power budgets are typically designed for 10^-11 BER typical, 10^-6 minimum at full bandwidth including coding gain with a 50% rate reduction due to coding, and the raw channel BER is 10^-2 at the unencoded bitrate, and 10^-3 at the coded bit rate. A ground station designer can add antenna gain and transmit power, but the market has expectations based on coded channels.

5) Transponder links built on 10’s of MHz wide bent pipe TWT amplifier channels have been a staple of commercial geosynchronous SATCOM links. The detailed gain curve only applies to a particular transponder type. They aren’t representative of communication systems in general. Mentioning that transponder links often have a nonlinear gain curve and may both squelch weak signals and have significant AM suppression is enough to suggest that an analysis has to include the characteristics of components of the link.

6) It isn’t inappropriate for an introductory analysis to avoid the more complex topics like carrier phase noise, bit clock jitter, the effect of shaped modulation on bandwidth occupancy, and so on, but it might be appropriate to mention that these things exist and are required for a complete link budget.

6. Nathan says:

I think you have made a mistake with energy per bit.

Energy per bit = (Average Power) / bit rate = ((A_peak^2)/2) / bit rate = (A_rms^2) / bit rate

You refer to A as peak amplitude but are missing the division by 2

7. John Liston says:

I noticed a sign error on page 18 with regards to the conversion of C/N0 to C/N.

At the top of page 18 the equation reads “C/N0 = C/N x Bn” (not in dBs)

At the bottom of page 18 the equation reads “C/N = C/N0 + Bn” (in dBs)

This should read “C/N = C/N0 – Bn” (in dBs)

• Charan L. says:

Thank you. Yes, I will try to add it to the corrections.

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