## Introduction

This article explains the calculation steps of Guard Band, for the different LTE channels’ Bandwidth.

“A guard band is a narrow frequency range that separates two ranges of wider frequency. This ensures that simultaneously used communication channels do not experience interference, which would result in decreased quality for both transmissions.

The guard band concept applies to wired and wireless communications. It also simplifies the process of signal filtering for hardware, software or both”

The Guard Band in LTE has defined to be 10% of available bandwidth. As known, LTE technology offers several available channels bandwidth:

- 1.4 MHz
- 3 MHz
- 5 MHz

- 10 MHz
- 15 MHz
- 20 MHz

This means that different channels bandwidth will have different Guard Band, as shown in the following Table.

Furthermore, it must to be clear in your mind that LTE protocol is defined to be an OFDMA System and channels are multiplexed in Resource Elements (RE) and Resource Block (RB).

*Table 1: Bandwidth, RB number, Useful Band and Guard Band for LTE*

From Table. 1 it should be clear that only in the case of 1,4 MHz bandwidth the guard band does not correspond to 10%. In fact, it is equal to **320 KHz** or** 22.85%**

Therefore, following it is possible to find an explanation about this difference between channel 1,4 MHz and the others.

### Guard Band calculation for 1.4 MHz bandwidth channel

As we described before, for an LTE channel we will consider a Guard Band equal to 20% of the available bandwidth, distributed to 10% on the lower limit and 10% on the upper limit.

Therefore, the Useful Band available is:

Since an RB has a spectral occupancy of 180 kHz (an RB is equal to 12 RE, each with 15 kHz) we can calculate the effective number of Resource Block assigned to this channel.

We have to remind the measure units in LTE are RE, RB and channel bandwidth. So, to calculate the right number of RB, we must to round down to the nearest number:

Now, we have an oddment of 0,22, that means:

Therefore, the guard band for to 1.4 MHz channel is equal to:

which is equal to 22,85% of 1,4 MHz.

Finally the effective Useful Band is:

### Guard Band calculation for all other bandwidth channel

Now it should be clear that using the following formula in all other bandwidth cases, the number of RB is an integer. It means that the bandwidth is a multiple of RB plus the Guard Band (10%).

## 34 Responses

Hi,

thanks for sharing this .Very good explanation.

Br

Ashok

Hi Ashok,

thanks for your post!

We are happy to share our articles and we are going to publish more.

Please share

Stay Tuned!

BR

Hi Michele Roselli,

Very Nice article indeed thanks for sharing

Calculation for same 5MHZ band giving me result of Bg=1039khz which is not right how it can be corrected.

Hi Shaan,

for calculating the guard band to 5MHz channel (and the other channels except that one to 1.4 MHz) you have only to consider the 10% of the available bandwdith, distributing to 5% on the lower limit and 5% on the upper limit.

So, the correct steps are showed as follow.

The Useful Band available is:

Bu = 5 MHz – (250 kHz * 2) = 4500 kHz

where

Bg = 500 kHz <=> 250 kHz + 250 kHz

which is equal to 10% to 5MHz.

Therefore in this case the number of RB is:

# RB = Bu / 1RB = 4500 kHz / 180 kHz = 25

I hope I explained well this point and clarified your doubt.

Thanks for your comment!!

BR

Great.

Was having the same question. Explained well

1 more question, Why is case of higher bandwidth except 1.4MHz we have to consider 10% gaurd band? Is it some kind of standard or just we are assuming to make calculations?

Hi Muhammad,

the channel bandwidths in LTE are a trade-off between in- and out-of-channel distortions. Since LTE1800 is located between GSM1800 frequency bands and the LTE1800 bandwidth is 1.4 MHz or 3 MHz, a minimum guard band of 200 kHz is needed. With the 1.4 MHz channel bandwidth a 10% guard band would not be enough, therefore the LTE physical layer specifies to assume the configuration with 6 RBs and a larger guard band

(LTE collocation with GSM/WiMAX/UMTS – Guard band and Usable Bandwidth – Technical Tweak)

thanks for this nice article, this answers some of the questions I had about LTE 1.4 MHz guard bands.

Let’s imagine that you have 1.4 MHz of spectrum available. Would this mean that you could populate it with 1.4 MHz LTE without any additional guard band?

Hi IOTC360,

generally speaking the answer is yes!

Anyway, the boundaries conditions could change the answer… I mean that it is a need to know all the system setup.

e.g. The additional guard band could be a need when there are multiple technologies in the same band. LTE and GSM systems (@1800 MHz), could need an adjustment.

Thanks for your comment,

BR

Thanks a lot Michele, again great post!

Share our posts and do not hesitate to ask us more article!!!

I think you forgot to deduct the DC subcarrier, i.e. the subcarrier located at the dead center of the carrier which doesn’t carry any modulation and is used by UE to locate the center of the carrier frequency. See figure 5.6.1 of 3GPP 36.101.

So, subtract 7.5KHz from the above calculation (assuming subcarrier spacing of 15KHz).

Hi Liu,

you are right, thank you very much for your contribution!

BR

Hi Michele,

Thanks for your efforts!

Is it possible to explain as to WHY the 10% is chosen for others and 20% chosen for 1.4 MHz? I see the calculation as to arrive at the number of RBs but I fail to see why they chose the specific % as guard band. Why not more or why not less?

Hi Arun,

considering the 1.4MHz total band and dividing it for the occupation of a RB (180 kHz), it gets:

1400/180 = 7.77777… -> That is 7 RB with a surplus of 0.77777…

But 0.7 * 180 = 126 kHz which is less than 140 kHz (corresponding to 10% of the total bandwidth).

At this point, to increase the Guard Band and bring it to at least 10%, the only way you can take is to reduce the number of RBs by a unit.

The result is then 1.77777… of surplus (in the previous case was 0.77777…) corresponding to 1.77777… * 180 = 320 kHz Guard Band.

This value is approximately 22.85% of the total band calculated also in the demonstration in the article.

Furthermore, if you apply the same flow to other channels, you will get the correct bandwidth results shown in the table.

Let us know!

Thanks,

MR

Hi Michele,

Thanks for sharing. It’s a great post.

In your latest reply,

“1400/180=7.77777… -> That is 7 RB with surplus of 0.77777…

But 0.7*180=126 kHz which is less than 140 kHz (corresponding to 10% of the total bandwidth)”

I found that, 0.7*180=126 kHz is correct.

However, 0.77777…*180=140 kHz (equal to 10% of 1.4MHz)

That is, if we choose full RB as 7 RBs for 1.4MHz channel bandwidth.

Bu=7*180=1260 kHz

Bg=140 kHz => 70 kHz + 70 kHz

which is equal to 10% of 1.4MHz.

In this case, the 10% rule of guard bands would be able to apply for all channel bandwidth (1.4MHz, 3MHz, 5MHz, 10MHz, 15MHz)

But still, 3GPP defines full RB=6RBs for 1.4MHz.

Is it possible to explain why 6RB is chosen instead of 7RB?

Thanks a lot.

Eileen

The values of the LTE channel bandwidths are a compromise between in- and out-of-channel distortions and were extensively studied in 3GPP. When LTE1800 is located between GSM1800 frequency bands and the LTE1800 bandwidth is 1.4 MHz or 3 MHz, a minimum guard band of 200 kHz is needed. With the 1.4 MHz channel bandwidth a 10% guard band would not be enough, therefore the LTE physical layer specifies to assume the configuration with 6 RBs and a larger guard band (https://technicaltweak.wordpress.com/2017/07/09/lte-collocation-with-gsmwimaxumts-guard-band-and-usable-bandwidth/).

hi Michele,

this answer of your clear all my doubt..thanks alot!

Hi Michele Roselli,

This is a great post, and cleared my doubt about the minimum bandwidth of LTE a little bit.

This is a top-down calculation, assuming 1.4M is “legit”.

My question is, since 1.08 is the maximum occupied bandwidth, why not define the minimum bandwidth as 1.08*1.1=1.2M? So that the guard band will be 50K each? Would it be even more logical?

Cyrus

Hi Cyrus,

the 1.4 MHz and 3 MHz channel bandwidth options have been chosen to facilitate a multitude of CDMA2000 and GSM migration scenarios.

Furthermore, the values of the LTE channel bandwidths are a compromise between in- and out-of-channel distortions and were extensively studied in 3GPP.

In fact, the basic OFDM spectrum comprises only slowly decaying sidelobes and efficient usage of the spectrum requires the use of filtering to effectively confine the out-of-band emissions. Such filters require a certain amount of transition bandwidth in order to be practical and to consume only a small amount of the cyclic prefix duration [1].

When LTE1800 is located between GSM1800 frequency bands and the LTE1800 bandwidth is 1.4 MHz or 3 MHz, a minimum guard band of 200 kHz is needed. With the 1.4 MHz channel bandwidth a 10% guard band would not be enough, therefore the LTE physical layer specifies to assume the configuration with 6 RBs and a larger guard band [2].

REFERENCES[1] H. Holma, A. Toskala – “LTE for UMTS – OFDMA and SC-FDMA Based Radio Access”, p. 286 (https://books.google.it/books?id=uhr3KwSww2kC&pg=PA286&lpg=PA286)

[2] https://technicaltweak.wordpress.com/2017/07/09/lte-collocation-with-gsmwimaxumts-guard-band-and-usable-bandwidth/

Hi! This is very helpful! Does this formula apply to both FDD and TDD LTE?

Yes, the possible configurations are shown in 3GPP TS 36.104 (http://www.qtc.jp/3GPP/Specs/36104-800.pdf)

Here you can find an overview: https://image.slidesharecdn.com/anritsulteguide-150714132105-lva1-app6891/95/anritsu-lte-guide-12-638.jpg

Hi,

A great explanation indeed, Was searching it for a long time. Also explains how to calculate no of RBs.

Dear,

Thank you for your explanation

for LTE 20M, what could be the impact if I consider only 500 khz as a guard Bad for each side and re use 1M for DCS1800 ?

Since 500 kHz is enough for LTE 10M and the same filter is used (Same RF) for LTE 20M and LTE10M, reusing the remaining 1M will help a lot for DCS1800.

WHat you think

Hi Fahmi,

I think your consideration is valid because using DCS1800 carriers in this way is very similar to use NB-IoT carriers in “guard-band” mode.

An offset of 200 kHz should be enough, because the NB-IoT carrier in a Guard band scenario is one additional 180 kHz Physical Resource Block (PRB) consisting of 12 OFDM sub-carriers (tones) with 15 kHz frequency separation.

I have 2 questions:

1.Why is case of higher bandwidth except 1.4Mhz we have to consider at 10% but not any other percantage? Is it some kind of standard or we prove it?

2.For LTE 15MHZ has 1.5Mhz for guard band ,what could be the impact if I consider only 250Khz as a guard band for each side (the same filter with LTE 5Mhz ) and reusing the remaining 1Mhz for LTE to increase bandwith LTE?

Thanks!

PS: Question 2: Case GSM and LTE in the same 1800MHZ,bandwidth 20MHz, LTE 1800Mhz is located between GSM 1800MHZ (LTE 15Mhz and GSM 5Mhz(2.5Mhz each side))

Hi Vinh,

the percentage for the guard bands (except for 1.4 MHz bandwidth) was defined as a standard, because this choice simplifies the allocation of resource blocks (RB). In fact, with 10% of bandwidth used as guard band, the effective used bandwidth is a multiple of 180 kHz, that is the width of one RB.

Anyway, you can still use an arbitrary guard band as long as it is greater than 200 kHz, that is the minimum requirement to avoid interferences. Hence you can consider only 250 kHz as a guard band for each side without important issues (it is very similar to use NB-IoT carriers in “guard-band” mode).

I have 1 question

When LTE use |BW 20MHz? Why GB for each side is 1000KHz? Not 200KHz?

Hi Nam, 20MHx means 20E6 (20.000KHz) and 10% of it is 2E6 or 2MHz (2.000KHz). This is the amount of

Guard Band.Now, if you split in two, you have 1MHz (1.000KHz) each side. Refer to Graph 1.

I hope this will clarify your doubt!

Thanks and Share!

Calculate PRB in a channel bandwidth of 1.4 MHz & 3 MHz (in 15 KHz subcarrier

spacing

Hi Ravi,

We understand you are asking for the number of Physical Resource Block (PRB) for these Channel Bandwidth of 1.4 and 3 MHz (15 KHz subcarrier spacing); this has ben already reported in Table 1 of the article (second row below “Number of RB in Frequency domain”) and the calculation rules have been explained as well.

Channel bandwidth (MHz) 1.4 3.0 5.0 10.0 15.0 20.0

Number of RB in Frequency Domain (Mhz) 6.0 15.0 25.0 50.0 75.0 100.0

The general rule is dividing the useful BW for the PRB bandwidth (180 KHz), where the useful Bandwidth is the nominal Bandwidth without the Guard Band (10% of the nominal Bandwidth). E.g. for the Channel Bandwidth 3 MHz we get:

(3000-300) = 2700 MHz -> 2700/180 = 15 (Number of RB or PRB)

the lower 1.4 MHz Channel makes an exception as the Gard Band cannot be lower than 200 KHz, so we have to increase the Guard Band to a 22,85% (320 KHz double sided) as we have to round the division result to the lower integer value (6, in this case; the result will be the same if we consider exactly the lower Guard Band limit of 200 KHz, that is almost 14% of nominal Bandwidth)

I hope this answer your question, otherwise please better detail.

Thank you

Calculate Guard Band in channel bandwidth of 5 MHz & 10 MHz ( in 15 KHz subcarrier spacing)?

Hi Ravi,

You can find Guard Band for Channel Bandwidth 5 MHz and 10 MHz (15 KHz subcarrier spacing) in the table 1 reported in the article:

Guard Band on Each Side (KHz) 160 150 250 500 750 1000

This has been calculated as 10% of nominal Bandwidth, then divided in two sub-band at the lower and upper part of the spectrum, each respectively 250 or 500 KHz for 5 MHz and 10 MHz Channels.

I hope this answer your question, otherwise please better detail.

Thank you

The post itself and the comments/discussion was so fruitful.

Thanks to Author and commentators as well to make this article beautiful.