text:
# DataWeightsAndCombination

## Principles of Data Weighting

#### Visibility Weights

For optimal imaging performance it is critical that in a relative sense each visibility in the data have the correct weight after calibration -- that is, data with better sensitivity have more weight than data with less sensitivity. Formally, the post-calibration visibility weights should be equal to 1/σij2 where σij is the rms noise of a given visibility.

$\sigma_{ij} (Jy) =\frac{2k}{\eta_{q}\eta_{c}A_{eff}}$ $\sqrt{\frac{T_{sys,i} T_{sys,j}}{2\Delta\nu_{ch} t_{ij}}}$ $\times 10^{26},$

where:

k is Boltzmann's constant.

Aeff is the effective antenna area which is equal to the aperture efficiency (ηa) x the geometric area of the antenna (π r2). The aperture efficiency depends on the rms antenna surface accuracy and is about 0.75 for ALMA dishes.

ηq and ηc are the quantization and correlator efficiencies, respectively. These have values of about 0.88 and 0.96, respectively. See the ALMA Technical Handbook for more information.

Tsys,i is the system temperature for antenna i, and Tsys,j is the system temperature for antenna j

Δνch is the channel frequency width.

tij is the integration time per visibility.

Note, this equation can be extended from a single visibility to the expected imaging rms noise for an entire dataset by replacing Tsys,iTsys,j by the average Tsys2, tij by the total time on source, and the $\sqrt{2}$ in the denominator by $\sqrt{N(N-1)}$ , where N is the number of antennas. Additionally, $\sqrt{n_p}$ is included in the denominator, where np is the number of polarizations (1 for single pol and 2 for dual or full pol data).

In order to correctly combine and image data that have different Tsys, Δνch, tij, or antenna size it is essential to use visibility weights proportional to 1/σij2. The remainder of this guide deals with ensuring that the individual visibility weights are correct in your ALMA data.

#### Relative Overall Weights between Configurations

Additionally, when combining data from different antenna configurations, one will get optimal overall sensitivity to all spatial scales by matching the surface brightness sensitivity at each uv-distance. This can only be achieved by having time-on-source per configuration in the right proportion. This topic is covered in ALMA Memo 598. This memo informs the relative amount of time that ALMA observes a project with the 7m-array versus the 12m-array, and compact versus extended 12m-array configurations. However, since telescope time is expensive, one typically does not actually observe in the optimal proportion, in that case one will not fully realize the expected "impact" of adding the less sensitive configuration data. In that case, one may chose to "up-weight" the less sensitive configuration explicitly by changing its data weights, above and beyond 1/σij2. However, it should be noted that nothing come for free, and such a change will come at the expense of overall image sensitivity though it may very well be the optimal choice for your science case -- for example, if you are particularly interested in large scale structures you might apply extra up-weighting to the dataset(s) with the shortest baselines. Finding the optimal up-weighting is a matter of experimentation, but can easily be explored using the visweightscale parameter in concat. As a general rule of thumb extra up-weighting by more than a factor of 4 or so is not recommended. Before experimenting with this however, its always best to start with data that simply has the correct 1/σij2 weights. Additionally, some degree of relative re-weighting can also be accomplished simply by using the Briggs Robust parameter, without explicitly changing the visibility weights, and generally it is also a good idea to try this first before more aggressive up-weighting is attempted.

## Weights in CASA

A memo describing weights in CASA, in particular the significant changes that were made with CASA 4.2.2, can be found at CASA-data-weights.pdf

To summarize the situation specifically for ALMA data processed by the ALMA Project:

CASE 1) CASA 4.2.1 and earlier: Weights were calculated on a per spw bases and scaled by Nchan/[(Tsys(i) * Tsys(j)] using calwt=True at the applycal stage for Tsys table. Assuming that (i) there aren't any unflagged antennas with significantly low gain due to for example pointing errors, or hardware problems, and (ii) all the antennas have similar efficiencies -- usually good assumptions for ALMA data, then to good approximation the data weights are close to internally consistent and can produce good imaging results, but should not be combined with other data that have different Δνch, tij, or antenna size without further modification.
CASE 2) CASA 4.2.2 and later: Upon import data weights are scaled by 2ΔνchΔtij and then also scaled by 1/[(Tsys(i) * Tsys(j)] using calwt=True at the applycal stage for Tsys table (thus making them per channel weights). Additionally:
(A) For data calibrated by the 4.2.2 CASA Pipeline the weights are further modified by [gain(i)2 * gain(j)2] when the amplitude gain table is applied using calwt=True. Since the amplitude gains are directly proportional to the effective Antenna area, scaling the weights by the amplitude gains will take into account antenna size differences, and also down-weight antennas with comparatively low gain. Thus, these weights are correct (but see Absolute Accuracy of the Data Weights).
(B) For data manually calibrated in CASA 4.2.2, unfortunately calwt=False was still used to apply the antenna gain table, thus, these data have weights that are not correct in a relative sense when compared to other data with different antenna size by the factor [gain(i)2 * gain(j)2].
CASE 3) CASA 4.3 and later: Data calibrated in either the pipeline or manually will have correct weights (but see Absolute Accuracy of the Data Weights). An example of this situation is demonstrated in https://casaguides.nrao.edu/index.php/M100_Band3_Combine_4.3

## How Do I Know the Situation For My ALMA Data?

First, remember that good imaging performance requires that the relative weights be correct for all data included in the imaging. For all of the situations below, the data are sufficiently internally consistent that they can be imaged fine on their own or with other data that share the same situation and have the same data properties. Problems arise when data are combined that (a) do not share the same weight situation and/or differences in antenna size, visibility integration time, or channel bandwidth are not accounted for.

All datasets to be combined should be independently assessed to determine their weight situation and thus whether corrective action is required

• Cycle 0 and Cycle 1 data were reduced in 4.2.1 or earlier versions and correspond to Case 1 above. ACA 7m-array data were first offered in Cycle 1. If you want to combine 12m-array and 7m-array data taken during Cycle 1, it is very likely you need to correct the visibility weights before imaging.
• Cycle 2 data has a more confused range of possibilities (this includes actual Cycle 2 projects and carry-over projects or parts of projects from Cycle 1). Recall from Case 2 above, that the key factor that determines your situation for Cycle 2 is whether the data were pipeline or manually calibrated.
• Key dates:
• Start Cycle 2: June 1, 2014
• CASA 4.2.2 release date: Sept. 4, 2014
• Pipeline release date: Oct. 20, 2014
• The earlier in Cycle 2 your data was taken, the more likely it was manually calibrated, also if you have any very narrow spws, high frequency, etc it is more likely data were manually calibrated.
• If the data were manually calibrated early in Cycle 2 it is likely in Case 1 (certainly the case for data earlier than Sept. 2014). After Oct. 1, 2014 it is likely to be in Case 2B.
• How can I distinguish between pipeline and manual data reduction for Cycle 2 data?
• In your data delivery, the README file should say but there were some oversights on this front. A sure way to tell: look in the directory called script within your data delivery for a particular member_ouss. If you see a file with PPR*.xml, the data was calibrated by the pipeline. Otherwise it was done manually.

## What Are the Options for Adjusting the Weights for Older Reductions?

If the data weights are not correct in ALL the datasets you want to combine there are three options to correct the situation. These different methods carry different levels of pain/complexity depending on your situation. The situation can also be extra confusing if your data fall into multiple categories above. For example, it is not uncommon that in Cycle 2 the 12m-array data could be pipeline calibrated, but the 7m-array data done manually. Pay close attention to recommendations in Red and the Caveats for each option.

#### Option 1: Re-calibrate your data in CASA 4.2.2 or later

⇒ Option 1 is easiest to implement for Case 2B (Data observed in Cycle 2 manually calibrated in 4.2.2 or 4.3.1 with calwt=False for amplitude/flux gain table).
• All data imported in 4.2.2 (or later) will automatically adjust the weights by 2ΔνchΔtij. The Tsys application should also already be correct in your scripts. To correct for antenna size and weight by the gains you must change calwt=True for the amplitude table applycal.
• Where do I make the calwt change?
• As described in your README file, you can obtain a fully calibrated measurement set by executing the sciptForPI.py. For manually calibrated data, within the script directory you will also find scripts with the format uid*scriptForCalibration.py, one for each execution for that Scheduling Block -- the sciptForPI.py actually executes these scripts, thus this is where any change must be made.
• Toward the bottom of the uid*scriptForCalibration.py scripts (typically Step 17 or 18) you will find a step called # Application of the bandpass and gain cal tables. Change the calwt=F in all of the applycal calls to calwt=T.
Caveat 1: You must have the raw ALMA ASDM to run the sciptForPI.py (see the README file for more information).
Caveat 2: Most (all except early Cycle 0) ALMA manual calibration scripts have within them the CASA version used to create the script. For example, if your manual calibration script says the following, using any other version than 4.2.2 will give an error (even though using for example 4.3.1 would be fine in this case):
if re.search('^4.2.2', casadef.casa_version) == None:
sys.exit('ERROR: PLEASE USE THE SAME VERSION OF CASA THAT YOU USED FOR GENERATING THE SCRIPT: 4.2.2')

You must change the version number in the script to match the version you want to use or the script will not run.
Caveat 3: Scripts from earlier than 4.2.1 are likely to have commands that make them incompatible to run directly in later versions of CASA. It may be difficult for a non-expert to update the uid*scriptForCalibration.py script(s) to current syntax.

#### Option 2: Make an approximate overall relative correction:

⇒ Option 2 is easy to implement for Cases 1 and 2B, but is not as accurate as Options 1 or 3. However, it is typically adequate for most purposes, and is MUCH better than doing nothing.
• All ALMA data reduced in CASA should already have 1/Tsys2 weighting which accounts for many of the factors that make data from some baselines more sensitive than others within an individual dataset. If we assume that antennas with abnormally low gains (typically due to poor pointing, hardware failures) have already been flagged and all antennas meet the surface accuracy specifications, then the amplitude gains will be fairly constant from antenna to antenna, with the value dominated by the antenna size. Other parameters that can often be different between datasets include the Δνch, and tij.
• Thus, a reasonably good approximation for the relative weight scaling for data_a compared to data_b is:
• Case 1: (antenna_size_b/antenna_size_a)4 x (Δνch_b/Δνch_a) x (tij_b/tij_a)
• Case 2B: (antenna_size_b/antenna_size_a)4
• such an overall scale factor can be applied during the concat step using the visweightscale parameter. Note that because its a ratio, it doesn't matter if you use the antenna radius or diameter as long as you are consistent. An example is given below.
Caveat 1: This option requires that the datasets to be combined have only one (unflagged) antenna size per dataset. If this is not the case, then Options 1 or 3 must be used.
Caveat 2: This option also requires that both datasets to be combined are in the same weight situation (i.e. mixed Cases cannot be handled using simple overall relative scaling).

#### Option 3: Run the task statwt on your calibrated science target data

⇒ Option 3 is pretty easy to implement for both Cases 1 and 2B, and will account for (likely small) Gain variations from antenna to antenna within the individual datasets, whereas Option 2 does not. However, see Caveat 1, for very complex line emission it may be difficult to get a good result.
• This task attempts to assess the sensitivity per visibility and adjust the weights accordingly. It is very commonly used for JVLA data (including their pipeline).
Caveat 1: One must limit the calculation to line-free channels using the fitspw parameter. For complex line projects this can be painful, however, typically the line-free channels are already known from the continuum subtraction, and can be reused here. However, it is best to run statwt before continuum subtraction, and it should always be run on the fully channelized data before any spectral averaging is applied.
Caveat 2: statwt does not currently account for channel-dependent flagging. It is critical to explicitly avoid the edge channels for any TDM spws in the data (128 channels per spw) data using fitspw, (example: for a dataset with only TDM spws, fitspw='*,7~120').

## Example of Correcting Weights Using Option 2 (Relative Scaling)

Figure 1: M100 12m-array weights from 4.2.1 reduction.
Figure 2: M100 7m-array weights from 4.2.1 reduction.

Below we show what could have been done to approximately correct the weights in the M100 SV 7m-array and 12m-array data if it was reduced in CASA 4.2.1 (or earlier), i.e. Case 1 (For a complete treatment of the data processed in CASA 4.3 see https://casaguides.nrao.edu/index.php?title=M100_Band3_Combine_4.3).

In CASA 4.2.1 and earlier, the data weights are 1 upon import, later in the standard calibration procedure, applycal scales the weights by Nchan/[(Tsys(i) * Tsys(j)] because calwt=True for the Tsys table applycal. As an example, we plot the weights of 7m and 12m data imported in CASA 4.2.1. No averaging can be turned on when plotting the weights.

# In CASA
os.system('rm -rf 7m_WT.png 12m_WT.png')
plotms(vis='m100_12m_CO.ms',yaxis='wt',xaxis='uvdist',spw='0~2:200',
coloraxis='spw',plotfile='12m_WT.png')
#
plotms(vis='m100_7m_CO.ms',yaxis='wt',xaxis='uvdist',spw='0~2:200',
coloraxis='spw',plotfile='7m_WT.png')

The Tsys for the 12m data for the spw containing CO(1-0) ranges from about 100 to 150 K, and the number of channels per spw is 3840, yielding an expected range for the 4.2.1 weights of 0.38 to 0.17 in good agreement with Fig. 1.

As you can see from Figures 1 and 2, the weights are quite similar at this stage because the data were taken under similar weather conditions and hence Tsys, and this is the only thing the weights have been scaled by so far besides the number of channels per spw which is also the same.

Recall that the rms noise in a single channel for a single visibility is:

$\sigma_{ij} (Jy) =\frac{2k}{\eta_{q}\eta_{c}A_{eff}}$ $\sqrt{\frac{T_{sys,i} T_{sys,j}}{2\Delta\nu_{ch} t_{ij}}}$ $\times 10^{26},$

Beyond the obvious difference in the antenna dish sizes (Aeff), looking at the listobs output for these two datasets the same channel width was used but that the integration time per visibility is 10.1 sec for the 7m-array and 6.05 sec for the 12m-array. Since dish area is in the denominator of the radiometer equation and integration time per visibility is in the denominator, and assuming WT ∝ 1/σij2, the 7m weight should be scaled by: (7./12.)4 x (10.1/6.05) = 0.193 to account for the difference in telescope size and integration time per visibility.

12m data:

ObservationID = 0         ArrayID = 0
Date        Timerange (UTC)          Scan  FldId FieldName             nRows     SpwIds   Average Interval(s)    ScanIntent
10-Aug-2011/19:38:05.8 - 19:50:22.8    11      1 M100                      1500  [0,1,2,3]  [6.05, 6.05, 6.05, 6.05] [CALIBRATE_WVR#ON_SOURCE,OBSERVE_TARGET#ON_SOURCE]

Spectral Windows:  (4 unique spectral windows and 1 unique polarization setups)
SpwID  Name   #Chans   Frame   Ch0(MHz)  ChanWid(kHz)  TotBW(kHz) CtrFreq(MHz) BBC Num  Corrs
0              3840   TOPO  113726.419       488.281   1875000.0 114663.6750        1  XX  YY
1              3840   TOPO  111851.419       488.281   1875000.0 112788.6750        2  XX  YY
2              3840   TOPO  103663.431      -488.281   1875000.0 102726.1750        3  XX  YY
3              3840   TOPO  101850.931      -488.281   1875000.0 100913.6750        4  XX  YY

7m data:

ObservationID = 0         ArrayID = 0
Date        Timerange (UTC)          Scan  FldId FieldName             nRows     SpwIds   Average Interval(s)    ScanIntent
17-Mar-2013/04:44:04.3 - 04:50:43.7    11      1 M100                       261  [0,1,2,3]  [10.1, 10.1, 10.1, 10.1] [OBSERVE_TARGET#ON_SOURCE]

Spectral Windows:  (6 unique spectral windows and 1 unique polarization setups)
SpwID  Name                           #Chans   Frame   Ch0(MHz)  ChanWid(kHz)  TotBW(kHz) CtrFreq(MHz) BBC Num  Corrs
0      ALMA_RB_03#BB_1#SW-01#FULL_RES   4080   TOPO  101945.850      -488.281   1992187.5 100950.0000        1  XX  YY
1      ALMA_RB_03#BB_2#SW-01#FULL_RES   4080   TOPO  103761.000      -488.281   1992187.5 102765.1500        2  XX  YY
2      ALMA_RB_03#BB_3#SW-01#FULL_RES   4080   TOPO  111811.300       488.281   1992187.5 112807.1500        3  XX  YY
3      ALMA_RB_03#BB_4#SW-01#FULL_RES   4080   TOPO  113686.300       488.281   1992187.5 114682.1500        4  XX  YY
4      ALMA_RB_03#BB_1#SW-01#FULL_RES   4080   TOPO  111798.250       488.281   1992187.5 112794.1000        1  XX  YY
5      ALMA_RB_03#BB_2#SW-01#FULL_RES   4080   TOPO  113673.250       488.281   1992187.5 114669.1000        2  XX  YY

Figure 3: M100 7m-array (red) and 12m-array (black) weights after scaling by relative sensitivity.

NOTE: If these data had been manually calibrated in CASA 4.2.2 with calwt=F for the gain table applycal (i.e. Case 2B), then the overall scale factor to apply would be (7./12.)4 = 0.116, because the difference in visibility integration time would already be accounted for.

An easy way to perform this overall scaling is using the visweightscale parameter in concat.

# In CASA
# Concat and scale weights
os.system('rm -rf M100_Intcombo_0.193.ms')
concat(vis=['m100_12m_CO.ms','m100_7m_CO.ms'],
concatvis='M100_Intcombo_0.193.ms',
visweightscale=[1,0.193])

Now plot the concatenated weights to verify they are as expected.

# In CASA
os.system('rm -rf Intcombo_0.193_WT.png')
plotms(vis='M100_Intcombo_0.193.ms',yaxis='wt',xaxis='uvdist',spw='0~2:200',
coloraxis='spw',plotfile='Intcombo_0.193_WT.png')

These combined data with correct relative weights are now ready for imaging. See https://casaguides.nrao.edu/index.php?title=M100_Band3_Combine_4.3 for more on the imaging in CASA 4.3.

Note that the per spw 12m-array 4.2.1 derived weights are quite similar to those shown in M100_Band3_Combine_4.3 from CASA 4.3 (Figure 3) using the correct per channel weights. This is purely accidental... for this example 1/σij2 just happens to be approximately equal to Nchan/[(Tsys(i) * Tsys(j)] for an antenna with diamter=12m, Δνch=488.281 kHz, tij=6.048s, and 3840 channels.

## Example of Correcting Weights Using Option 3 (statwt)

Figure 4: M100 12m-array WT vs uvdist after calwt=False for amplitude gains.
Figure 5: M100 7m-array WT vs uvdist after calwt=False for amplitude gains.
Figure 6: M100 12m-array amplitude vs channel.
Figure 7: M100 7m-array amplitude vs channel.

This is an example of running statwt on the M100 SV data. The starting point was data calibrated as if it had been manually calibrated in CASA 4.2.2, i.e. CASE 2b calwt=False for the gains. Note for this starting point, the 7m data actually have higher weight than the 12m data. This is because the average Tsys were similar but the visibility integration time of the 7m data is 10.1 versus 6.048s for the 12m data (see listobs outputs in previous section). Imaging these data would work very poorly because as already described, the 7m should actually have weights ≈ 0.193 times lower than the 12m data.

os.system('rm -rf 12m_calwtF_WT.png')
plotms(vis='M100_Band3_12m_CalibratedData.cf.ms.fresh', spw='0',yaxis='weight',xaxis='uvdist',
plotfile='12m_calwtF_WT.png')

os.system('rm -rf 7m_calwtF_WT.png')
plotms(vis='M100_Band3_7m_CalibratedData.cf.ms.fresh', spw='3,5',yaxis='weight',xaxis='uvdist',
plotfile='7m_calwtF_WT.png')

As mentioned above in the Option 3 statwt caveats, it is essential to avoid strong line channels when calculating the statistics. What is meant by strong? At minimum, if you can see it in a uvplot it should not be included. The easiest way to check is to make a plot of the uv-data in channel space.

# In CASA
os.system('rm -rf 12m_amp_ch_Spw0.png')
plotms(vis='M100_Band3_12m_CalibratedData.cf.ms',
yaxis='amp',xaxis='channel',spw='0',avgtime='1e8',
avgscan=True,plotfile='12m_amp_ch_Spw0.png')

os.system('rm -rf 7m_amp_ch_Spw3_5.png')
plotms(vis='M100_Band3_7m_CalibratedData.cf.ms',
yaxis='amp',xaxis='channel',spw='5',avgtime='1e8',
avgscan=True,plotfile='7m_amp_ch_Spw5.png')

In statwt we need to avoid the line emission, while sampling the full spectral range. This is especially important in this example because the CO(1-0) transition is close to an oxygen line in the atmosphere -- this is what causes the noise to increase at higher channel numbers. Since we are deriving a spectrally averaged weight it would be unrealistic to only include the lower, less noisy channels.

# In CASA
statwt(vis='M100_Band3_12m_CalibratedData.cf.ms', spw='0',
fitspw='0:0~1800;2500~3839', datacolumn='data')

# In CASA
statwt(vis='M100_Band3_7m_CalibratedData.cf.ms', spw='3,5',
fitspw='3:0~1900;2500~3839,5:0~1900;2500~3839', datacolumn='data')

Now we can concat these datasets and make a plot to verify that the weights have the values expect.

Figure 8: M100 7m-array (purple) and 12m-array (black) weights after running statwt.
concat(vis=['M100_Band3_12m_CalibratedData.cf.ms','M100_Band3_7m_CalibratedData.cf.ms'],
concatvis='M100_Band3_statwt.ms')

os.system('rm -rf M100_statwt.png')
plotms(vis='M100_Band3_statwt.ms', spw='0,7,9',yaxis='weight',xaxis='uvdist',
coloraxis='spw',plotfile='M100_statwt.png')

These weights are in good agreement with those derived using the correct weights throughout CASA, i.e. Figure 3 in M100_Band3_Combine_4.3. However, as described in How_Accurate_Are_the_Weights section we actually expected them to come out a factor of 2 to 2.6 times higher, in order to account for the fact that these data were on-line Hanning smoothed, but not online channel averaged (see the next section for more information). This demonstrates the factor of about 2 descrepancy between the weights derived through CASA calibration and what is expected from statwt.

## Absolute Accuracy of the Data Weights

As mentioned several times in this guide, good imaging performance is only subject to correct relative weights, which is achieved as of CASA >4.2.2 provided calwt=True for the amplitude gain table. Even for Case 1 (4.2.1 and lower), the relative weights within a given dataset are reasonably accurate in most cases due to the 1/Tsys2 scaling. However, there are some instances, for example, uv-model fitting where one would like to use the visibility weights as an absolute measure of 1/σ2. Here we discuss the absolute accuracy of the visibility weights for options 1 and 2.

Option 1: All PI data taken with ALMA are Hanning smoothed online, but it is up to the user at proposal creation time to decide if additional channel averaging is done in the correlator (typically to reduce the data rate). Data that are Hanning smoothed but not channel averaged by the ALMA online system have a theoretical effective channel width 2.667 times that reported by for example listobs from the point of view of calculating sensitivity (it is a factor of 2 for the FWHM, i.e. channel resolution). The CASA initialization of 2ΔνchΔtij does not currently take this into account -- CASA does not have a way of knowing presently what the online system did, only how the data appears on disk, though this will change with future versions of the online software. Thus, presently CASA weights for this case are in principle factor of 2.667 times smaller than they should be from a sensitivity point of view. For online channel averaging of two or more channels, the difference between the channel width and the effective channel width diminishes rapidly. However, note that our tests suggest that the actual effective channel width is somewhat smaller than 2.667 by 10-30%, we are working to refine this value.
Option 3: For CASA <4.4 statwt does not take channel specific flagging into account, only flags that apply to all channels. Thus, the user must avoid flagged channels with the spwfit parameter. Additionally, strong line emission must be avoided. Provided these conditions are met, our tests suggest that there is still an ambiguity of order 2 in the weights produced by statwt compared to the expected values, even when (1) is taken into account.

Our investigation into these matters continues so if the absolute weights are important to you, check back here or send a helpdesk ticket with your inquiry.


text:
Q

# Help me answer: - Chemical Bonding and Molecular Structure - JEE Main-2

The decreasing order of bond angles in BF3, NH3, PF3 and $I_{3}^{-}$is :

• Option 1)

$I_{3}^{-} > NH_{3}> PF_{3}> BF_{3}$

• Option 2)

$I_{3}^{-} >BF_{3}> NH_{3}> PF_{3}$

• Option 3)

$BF_{3} >I_{3}^{-}}>PF_{3}>NH_{3}$

• Option 4)

$BF_{3} >NH_{3}>PF_{3}>I_{3}^{-}}$

124 Views

As we learned

S - P Hybridisation -

When one S - and one P - orbital belonging to the same main shell of an atom are mixed together to form two new SP - orbitals.

- wherein

shapes

Diagonal or Linear

sp2 Hybridisation -

When one s - and two p - orbitals of the same shell of an atom mix to form three new equivalent orbitals.The hybridised orbital is called sp2 orbital.

- wherein

Shape is Trigonal planar

sp3 hybridisation -

when one s - and three p - orbital belonging to the same shell of an atom mix together to form four new equivalent orbitals is called sporbital.

- wherein

Shape Tetrahedral

$Bond\: angle=120^{\circ}$

$Bond\: angle=107^{\circ}$

when size of central atom increases bond angle decreases

B.A of NH3   >    PH3

$Bond\: angle=180^{\circ}$

PF3 < NH3 < BF3 < $I_{3}^{-}$

Option 1)

$I_{3}^{-} > NH_{3}> PF_{3}> BF_{3}$

Option 2)

$I_{3}^{-} >BF_{3}> NH_{3}> PF_{3}$

Option 3)

$BF_{3} >I_{3}^{-}}>PF_{3}>NH_{3}$

Option 4)

$BF_{3} >NH_{3}>PF_{3}>I_{3}^{-}}$

Exams
Articles
Questions

format: plain


text:
# How do you find the product of the complex number and its conjugate 1 + 3i?

Dec 24, 2015

You can use the identity $z \overline{z} = \text{Re"(z)^2+"Im} {\left(z\right)}^{2}$ to find:

$\left(1 + 3 i\right) \left(1 - 3 i\right) = {1}^{2} + {3}^{2} = 1 + 9 = 10$

#### Explanation:

Notice that:

$\left(a + b i\right) \left(a - b i\right) = {a}^{2} - {\left(b i\right)}^{2} = {a}^{2} - {i}^{2} {b}^{2} = {a}^{2} + {b}^{2}$

So we can deduce:

$z \overline{z} = \text{Re"(z)^2+"Im} {\left(z\right)}^{2}$

for any Complex number $z$


text:
# Histograms with frequency density on the vertical axis

In a histogram with frequency density on the vertical axis, could the actual frequencies for each bar of the histogram come out not as whole numbers?

-

That depends: if the Y-axis is relative frequency then yes, it will be a fraction. Otherwise it should be the number of points in a particular bin.

-


text:
# Christoffel Symbols problem in GR

## Homework Statement

This is a problem in General Relativity, where I am trying to find the Christoffel symbols that correspond to a given metric. Any help would be greatly appreciated!

OK. I have been given the metric

$$ds^2 = (1+gx)^2 dt^2 - dx^2 - dy^2 - dz^2$$

and have been told to show that the non-vanishing Christoffel symbols are given by

$$\Gamma^1 _0 _0 = g(1+gx)$$

and

$$\Gamma^0 _1 _0 = \Gamma^0 _0 _1 = \frac{g}{(1+gx)}$$

I can get the first out OK but I'm not getting how to get the second.

## The Attempt at a Solution

So I wrote the Lagrangian out for the metric:

$$L\left(\frac{dx^\alpha}{d\sigma},x^\alpha\right) = - (1+gx)^2\frac{dt}{d\sigma}^2 +\frac{dx}{d\sigma}^2+\frac{dy}{d\sigma}^2+\frac{dz}{d\sigma}^2$$

which, after a little substitution work getting taus instead of sigmas, basically gives us the following equation:

$$\frac{d^2x}{d\tau^2} = -\frac{1}{2} . 2(1+gx).g \frac{dt}{d\tau}^2$$

...and comparing this to the definition of a C-symbol

$$\frac{d^2 x^\alpha}{d\tau^2} = -\Gamma ^\alpha _\beta _\gamma \frac{dx^\beta} {d\tau} \frac{dx^\gamma}{d\tau}$$

...it is pretty easy to see that

$$\Gamma^1 _0 _0 = g(1+gx)$$

However, I can't understand how to get the $$\Gamma^0 _1 _0 = \frac{g}{(1+gx)}$$ one out.

$$\Gamma^0 _1 _0$$ would mean that we were looking for the C-symbols in equations of the form...

$$\frac{d^2 x^0}{d\tau^2} = -\Gamma^0 _1_0 \frac{dx^1} {d\tau} \frac{dx^0}{d\tau}$$

...yes?

I don't see any way of comparing that to the Lagrangian to find the C-symbol $$\Gamma^0 _1 _0 = \Gamma^0 _0 _1 = \frac{g}{(1+gx)}$$

Any help? Have I made a complete meal out of this?! Thanks for your time...

CompuChip
Homework Helper
There is a well-known formula that allows you to calculate them directly:
$$\Gamma^i {}_{k\ell}=\frac{1}{2}g^{im} \left(\frac{\partial g_{mk}}{\partial x^\ell} + \frac{\partial g_{m\ell}}{\partial x^k} - \frac{\partial g_{k\ell}}{\partial x^m} \right)$$
where $g_{ij}$ can simply be read off and $g^{ij}$ are the coefficients of the inverse metric.

For the proof, you can refer to most GR or differential geometry texts.
Note that when you write it as
$$\Gamma^{\rho}_{\mu\nu} = \frac12 g^{\rho\sigma}\left( \partial_\nu g_{\sigma\mu} + \partial_\mu g_{\nu\sigma} - \partial_\sigma g_{\mu\nu} \right)$$
it's relatively easy to remember: there is only one upper index which goes in the inverse metric with a dummy variable, and then you cyclically permute the indices where the "canonical" order $\sigma\mu\nu$ has an opposite sign.

Thanks very much for your help!

Much appreciated.

George Jones
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Gold Member
I don't see any way of comparing that to the Lagrangian to find the C-symbol $$\Gamma^0 _1 _0 = \Gamma^0 _0 _1 = \frac{g}{(1+gx)}$$

For simple metrics like this, I quite like the Lagrangian method. You have used

$$\frac{\partial L}{\partial x} = \frac{d}{d \tau} \frac{\partial L}{\partial \frac{d x}{d \tau}}.$$

Now use

$$\frac{\partial L}{\partial t} = \frac{d}{d \tau} \frac{\partial L}{\partial \frac{d t}{d \tau}},$$

and differentiate with respect to $\tau$ the equation that you get.

George Jones
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Gold Member
Oops, I didn't mean to say
and differentiate with respect to $\tau$ the equation that you get.

Thanks a lot George - I appreciate the reply! I've now pushed through with both methods and they both work... funnily enough!

Why is it that when I try to solve it using the Lagrangian, my answer is off by a factor of 2 for the c-symbol $$\Gamma^0_{10}$$
Lagrange's equation for $$x^0$$ time component is
$$-\frac{d}{d\sigma}(\frac{\partial L}{\partial(\frac{\partial x^0}{\partial\sigma})})+\frac{\partial L}{\partial x^0}=0$$
where $$\frac{\partial L}{\partial x^0}=0$$ and $$\frac{\partial L}{\partial(\frac{\partial x^0}{\partial\sigma})}=\frac{1}{2}\frac{1}{L}(-2(1+gx^1)^2\frac{dx^0}{d\sigma}) =- \frac{1}{L}(1+gx^1)^2\frac{dx^0}{d\sigma}$$ as the Lagrangian $$L=\frac{d\tau}{d\sigma}$$, we can sub this in and also divide by L to convert the sigma's to tau's.
=>$$\frac{d}{d\tau}(\frac{d\sigma}{d\tau}(1+gx^1)^2\frac{dx^0}{d\sigma})=0$$
=>$$\frac{d}{d\tau}((1+gx^1)^2\frac{dx^0}{d\tau})=0$$
=>$$\frac{d}{d\tau}((1+gx^1)^2)\frac{dx^0}{d\tau}+(1+gx^1)^2\frac{d}{d\tau}(\frac{dx^0}{d\tau})=0$$
=>$$\frac{d^2x^0}{d\tau^2}=-\frac{2g}{1+gx^1}\frac{dx^1}{d\tau}\frac{dx^0}{d\tau}$$
so this shows $$\Gamma^0_{10}=\frac{2g}{1+gx^1}$$
So why do I have an extra factor of 2? T_T

cristo
Staff Emeritus
Why is it that when I try to solve it using the Lagrangian, my answer is off by a factor of 2 for the c-symbol $$\Gamma^0_{10}$$
$$0=\frac{d^2x^0}{d\tau^2}+\Gamma^0_{01}\frac{dx^1}{d\tau}\frac{dx^0}{d\tau}+\Gamma^0_{10}\frac{dx^0}{d\tau}\frac{dx^1}{d\tau}=\frac{d^2x^0}{d\tau^2}+2\Gamma^0_{01}\frac{dx^1}{d\tau}\frac{dx^0}{d\tau}$$
where the last equality is true since $$\Gamma^a_{bc}=\Gamma^a_{cb}$$


text:
# Partitions in the S-Box of Streebog and Kuznyechik

Abstract : Streebog and Kuznyechik are the latest symmetric cryptographic primitives standardized by the Russian GOST. They share the same S-Box, $\pi$, whose design process was not described by its authors. In previous works, Biryukov, Perrin and Udovenko recovered two completely different decompositions of this S-Box. We revisit their results and identify a third decomposition of. It is an instance of a fairly small family of permutations operating on 2m bits which we call TKlog and which is closely related to finite field logarithms. Its simplicity and the small number of components it uses lead us to claim that it has to be the structure intentionally used by the designers of Streebog and Kuznyechik. The 2m-bit permutations of this type have a very strong algebraic structure: they map multiplicative cosets of the subfield GF(2^m)* to additive cosets of GF(2^m)*. Furthermore, the function relating each multiplicative coset to the corresponding additive coset is always essentially the same. To the best of our knowledge, we are the first to expose this very strong algebraic structure. We also investigate other properties of the TKlog and show in particular that it can always be decomposed in a fashion similar to the first decomposition of Biryukov et al., thus explaining the relation between the two previous decompositions. It also means that it is always possible to implement a TKlog efficiently in hardware and that it always exhibits a visual pattern in its LAT similar to the one present in. While we could not find attacks based on these new results, we discuss the impact of our work on the security of Streebog and Kuznyechik. To this end, we provide a new simpler representation of the linear layer of Streebog as a matrix multiplication in the exact same field as the one used to define. We deduce that this matrix interacts in a non-trivial way with the partitions preserved by $\pi$.
Keywords :
Document type :
Journal articles
Domain :

Cited literature [36 references]

https://hal.inria.fr/hal-02396814
Contributor : Léo Perrin <>
Submitted on : Friday, December 6, 2019 - 11:19:38 AM
Last modification on : Monday, January 13, 2020 - 1:17:49 AM
Long-term archiving on: : Saturday, March 7, 2020 - 1:40:51 PM

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### Citation

Léo Perrin. Partitions in the S-Box of Streebog and Kuznyechik. IACR Transactions on Symmetric Cryptology, Ruhr Universität Bochum, 2019, 2019 (1), pp.302-329. ⟨10.13154/tosc.v2019.i1.302-329⟩. ⟨hal-02396814⟩

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If the integers a and n are greater than 1 and the product

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If the integers a and n are greater than 1 and the product of the first 8 positive integers is a multiple of a^n, what is the value of a ?

(1) a^n = 64
(2) n=6
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19 Jan 2008, 15:10
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D

the product of the first 8 positive integers is

$$P=1*2*3*4*5*6*7*8=2*3*2^2*5*(3*2)*7*2^3=2^6*3^2*5*7$$

$$2^6=64$$

1. only $$2^6$$ works. suff.

2. only $$2^6$$ works. suff.
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19 Jan 2008, 15:18
walker wrote:
D

the product of the first 8 positive integers is

$$P=1*2*3*4*5*6*7*8=2*3*2^2*5*(3*2)*7*2^3=2^6*3^2*5*7$$

$$2^6=64$$

1. only $$2^6$$ works. suff.

2. only $$2^6$$ works. suff.

Big walker, answer is B according to OA. Plus 1 for answering anyways though (can you take another crack at it for me).

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19 Jan 2008, 15:24
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dominion wrote:
If the integers A and N are greater than 1, and the product of the first 8 positive integers is a multiple of A^N, what is the value of A?

s1: A^N=64
s2: n=6

Thanks!

S1: A can be 2,4,8 thus n is 6,3,2, 8! is a multiple of any of these combinations.

S2: we have n=6. 8! has 8*7*6*5*4*3*2 ---> 2^7*3^2*5*7 n MUST be 2 or 8! won't be a multiple or be divisible by A^6.

Try 4^6 --> 2^12, this doesnt work. 3^6, there aren't enough 3's to cover this.

Thus s2 is suff.

B

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19 Jan 2008, 15:55
Hey gmatblackbelt, Im not following the logic above for statement 2.

We know that A^6 = (2^7*3^2*5)k ... where do we go from here ?

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19 Jan 2008, 16:53
pmenon wrote:
Hey gmatblackbelt, Im not following the logic above for statement 2.

We know that A^6 = (2^7*3^2*5)k ... where do we go from here ?

A^6 does not equal 2^7*3^2*5*7

8! is divisble by A^6 the only way that this could be is for A to equal 2. It cannot equal 1 since the main stem said A and N are bothg greater than 1.

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19 Jan 2008, 22:31
walker wrote:
D

the product of the first 8 positive integers is

$$P=1*2*3*4*5*6*7*8=2*3*2^2*5*(3*2)*7*2^3=2^6*3^2*5*7$$

$$2^6=64$$

1. only $$2^6$$ works. suff.

2. only $$2^6$$ works. suff.

All fine in your approach,except that you miss other possibilities in ( i ), which can be;

$$8^2, 4^3 etc.$$ Thus it's "not only" $$2^6$$ works. suff.

But, if n=6 is fixed as in statement ( ii ), then $$2^6*3^2*5*7$$, it is clear that A = 2.

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20 Jan 2008, 00:12
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I agree
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04 Jun 2010, 00:22
How should we solve this?
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04 Jun 2010, 04:14
dimitri92 wrote:
How should we solve this?

We know $$8! = 2^6 * 3^2 *5 *7$$
From question its given $$a^n *k = 8!$$

From St-1: $$a^n*K = 64$$. $$a$$ can be 2,4 or 8given a value of $$n$$ is greater than 1.

From St-2: $$a^6*K = 8!$$ -> a can only be 2.

Hence B

Last edited by cipher on 04 Jun 2010, 08:40, edited 1 time in total.

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dimitri92 wrote:
How should we solve this?

If the integers a and n are greater than 1 and the product of the first 8 positive integers is a multiple of a^n, what is the value of a?

Prime factorization would be the best way for such kind of questions.

Given: $$a^n*k=8!=2^7*3^2*5*7$$. Q: $$a=?$$

(1) $$a^n=64=2^6=4^3=8^2$$, so $$a$$ can be 2, 4, or 8. Not sufficient.

(2) $$n=6$$ --> the only integer (more than 1), which is the factor of 8!, and has the power of 6 (at least) is 2, hence $$a=2$$. Sufficient.

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Geronimo wrote:
If the integers a and n are greater than 1 and the product of the first 8 positive integers is a multiple of a^n, what is the value of a ?

(1) a^n = 64
(2) n=6

Prime factorization would be the best way for such kind of questions.

Given: $$a^n*k=8!=2^7*3^2*5*7$$. Q: $$a=?$$

(1) $$a^n=64=2^6=4^3=8^2$$, so $$a$$ can be 2, 4, or 8. Not sufficient.

(2) $$n=6$$ --> the only integer (more than 1), which is the factor of 8!, and has the power of 6 (at least) is 2, hence $$a=2$$. Sufficient.

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Re: How to solve this? [#permalink]

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15 Mar 2011, 23:49
Bunuel wrote:
dimitri92 wrote:
How should we solve this?

(2) $$n=6$$ --> the only integer (more than 1), which is the factor of 8!, and has the power of 6 (at least) is 2, hence $$a=2$$. Sufficient.

Hi bunuel,
Thank you for your instruction. I have another query:
how can we eliminate the possibility that: 8! is not a multiple of 4^6 (which means a#4) quickly?
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16 Mar 2011, 09:53
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MICKEYXITIN wrote:
Bunuel wrote:
dimitri92 wrote:
How should we solve this?

(2) $$n=6$$ --> the only integer (more than 1), which is the factor of 8!, and has the power of 6 (at least) is 2, hence $$a=2$$. Sufficient.

Hi bunuel,
Thank you for your instruction. I have another query:
how can we eliminate the possibility that: 8! is not a multiple of 4^6 (which means a#4) quickly?

8! has 2^7 as its factor. Thus maximum exponent of 4 is 3; $$(2^2)^3*2=(4)^3*2$$

Look into this for more on factorials;
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Re: How to solve this? [#permalink]

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16 Mar 2011, 20:14
From this we have three possibilities for a^n -> 2^7, 3^2 and 4^3 so a can be 2, 3 or 4

From (1) a^n = 64 => a^n = 2^6 or 4^3, so not sufficient

(2) n = 6, so we can rule out 3 or 4, as 2 is the only number with exponent > 6

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Re: How to solve this? [#permalink]

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16 Mar 2011, 20:48
I think 3 is ruled out isn't it? all the factorials greater than 5 are Even multiple of 10

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16 Mar 2011, 22:36
In (1) 3 raised to any integer can't be 64. In (2), 3 is ruled out because power/exponent of 3 < 6.
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20 Mar 2011, 00:48
fluke wrote:
MICKEYXITIN wrote:
Bunuel wrote:

Hi bunuel,
Thank you for your instruction. I have another query:
how can we eliminate the possibility that: 8! is not a multiple of 4^6 (which means a#4) quickly?

8! has 2^7 as its factor. Thus maximum exponent of 4 is 3; $$(2^2)^3*2=(4)^3*2$$

Look into this for more on factorials;

HI FLUKE,
thank you for your help. +1
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25 May 2011, 03:15
When looking at Statement 1, we know a^n = 64 and a and n are positive integers greater than 1, so a could be 2, 4 or 8 (since 2^6 = 4^3 = 8^2 = 64). The information in the stem isn't actually important here.

When we look at Statement 2, we know that 8! is divisible by a^6. If we prime factorize 8!, we find:

8! = 8*7*6*5*4*3*2 = (2^3)(7)(2*3)(5)(2^2)(3)(2) = (2^7)(3^2)(5)(7)

We need this prime factorization to be divisible by a^6 where a > 1. Looking at the prime factorization, the only possibility is that a = 2 (since for any other prime p besides 2, we can't divide the factorization above by p^6, nor can we divide the factorization above by (2^2)^6 = 2^12).
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# Why can't the orthogonal PN codes help in the near-far effect?

The CDMA near far problem arises because handsets may be anywhere within the particular cell boundaries. Some handsets will be close to the base station, whereas others will be much further away. In a free space scenario signals decay according a an inverse square law...signals within a cell will have a huge variation in signal strengths. For the receiver to be able to decode all the signals in the channel, they should ideally all be at the same signal strength

I got the above information on a website while trying to understand the near-far problem. I understood the basic concept that the mobile devices near the base station will have a higher amplitude when the signal reaches the base station as compared to the ones far away. But, can't the orthogonal PN sequences used in CDMA be able to handle this. As I recall, the sequences when ex-ored with each other will produce a string of all zeros. Note that this question has already been answered here. There It has been mentioned that, the mobile phones dont have orthogonal codes. What if they did ? would that solve the near-far problem ? If not, can I know why ? Basically, In short, If 2 mobile phone signals reach the base station at different levels of power but have Orthogonal codes, wouldnt that solve the near-far problem ?

• The correlators have finite dynamic range. – analogsystemsrf Jan 10 '19 at 4:24

the orthogonal PN sequences used in CDMA be able to handle this?

Orthogonal PN sequences could.

In reality, you don't get these – frequency and timing offsets, and trade-offs during the design phase mean that you only get approximately orthogonal sequences.

So, even if the cross-correlation between what TX1 and TX2 transmits is say -50 dB, but TX2 is 53 dB stronger at the receiver than TX1, then TX1's energy after correlation with the matching PN sequence is still only half of that of TX2's interference.

Well, the clue is in the last sentence. The receiver has to reliably detect the number of received signals (e.g. 1's in the code sequence), since they add up on the receiver end.

If there are two handsets, one with a received signal strength of say, 10 (disregarding exact units) and another one with a strength of 1, then we have total incoming signal strenghts of 0, 1, 10 and 11 (again, over-simplifying). That's pretty difficult to separate for a receiver as 4 different levels. Now if the ratio is 100:1, which is not unlikely with radios, the levels are 0, 1, 100 and 101; impossible to separate.

When the units are more or less equal in power, we get values 0, 1 and 2 (no, not 3). Much easier to separate and detect.

• Sort of. Extend your voting model to the simpler case of sinusoidal carriers (ie, classic narrowband radio), and think about the fact that it does work across huge power differences, provided that the codes are truly orthogonal and that the system remains linear. For a code receiver, linearity is a function of the instantaneous dynamic range of the pre-digital stages, in effect, if 10000 + 1 = 10001 or if it might compress or quantize to 10000. And as Marcus pointed out, orthogonality may be limited too. – Chris Stratton Jan 9 '19 at 16:50

CDMA is very unique in comparison with other multiplexing schemes because it doesn't (even in the ideal case) completely remove interference from other users. In an ideal time based (i.e TDMA) or an ideal frequency based (i.e FDMA) you can effectively remove all interference from other users, whilst in a CDMA scheme all you are doing is effectively scaling down other users' interference by a factor of $$\N\$$, where $$\N\$$ is the length of the spreading codes.

So if you have a two users $$\a\$$ and $$\b\$$ both transmitting to a common receiver using a common power level of $$\P\$$ but the power levels being received are $$\P_a\$$ and $$\P_b\$$ respectively (because one is near and the other is far away) then in an ideal FDMA system and transmitting over unit channel gain then you would have the normal SNR formula of

$$SNR_a = \frac{P_a}{P_N}$$

for user $$\a\$$ and a formula of

$$SNR_b = \frac{P_b}{P_N}$$

for user $$\b\$$. $$\P_N\$$ is the noise power. So it's basically a pure multiplexing scheme and so provided the two SNR values above are above your receiver's sensitivity then you should be able to decode them just fine. In an ideal CDMA scheme however the interference from other users is not removed but instead just scaled back by a factor of $$\N\$$ so the SNR for user $$\a\$$ would be

$$SNR_a = \frac{P_a}{\frac{P_b}{N} + P_N}$$

and for user $$\b\$$ it's

$$SNR_b = \frac{P_b}{\frac{P_a}{N} + P_N}$$

Now, if your orthogonal spreading sequence length $$\N\$$ is long enough this isn't an issue as this SNR will effectively be the same as in the first case. If on the other hand $$\N\$$ is not very long and you have one device near and the other one far then you will have $$\P_a >> P_b\$$ if user $$\b\$$ is far away. This can then cause $$\SNR_b\$$ to fall below the receive sensitivity of the receiver and make it hard for you to receive packets from $$\b\$$. You can mitigate this near far problem to a great extent by using long perfectly orthogonal PN sequences. In a lot of cases though PN sequences that are not perfectly orthogonal are used, the interference from other users is then reduced by a factor slightly lower than the PN sequence length. This exacerbates the "near far" problem.


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PIC with 32kHz external clock not running

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Not open for further replies.

Franchie

New Member
Hello,

I am using the 32kHz output of the MAXIM DS3234 Real-time clock as an external clock input for my PIC16F877A (in place of any oscillator crystal). As I understand the datasheet, the connection should be very simple, and the 32kHz signal should be connected to OSC1/CLKIN of the PIC directly. Similarly, the configuration bits have to be set to one of HS, XT or LP (LP is probably best in this case because of the low frequency, but HS should be great for debugging).

The problem is that it does not work: when programmed, nothing is executed. When in an ICD2 debug session, the session cannot be opened (Error ICD0083). However, everything works when a 20MHz clock is connected externally in the place of the 32kHz. Thus the problem is probably linked to the 32kHz signal.

Upon analysis, the signal appears good enough: there is little noise, its amplitude reaches 0.97 Vcc, and has a duty-cycle of roughly 30-40%. It fluctuates slightly when the ICD2CLK line is active, probably due to the routing, but this should be of little concern.

So if you have any ideas of what I might be doing wrong, please throw suggestions at me! Your help is greatly appreciated!
Franchie.

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ericgibbs

Well-Known Member
hi,
Does the 32KHz output of the RTC require a pullup resistor.?

Franchie

New Member
Hi, thanks for the response!
I don't believe that a pullup resistor is needed in this case, because the RTC has a "push-pull" output, and also because the waveform reaches the desired amplitudes... I have tried with a pullup resistor anyway, but this had little effect :S.

ericgibbs

Well-Known Member
Hi, thanks for the response!
I don't believe that a pullup resistor is needed in this case, because the RTC has a "push-pull" output, and also because the waveform reaches the desired amplitudes... I have tried with a pullup resistor anyway, but this had little effect :S.

hi,
Looking thru the DS3234 I see that you are correct, I use the DS1307 which does require a pullup.

Are you using a scope to measure the 32KHz on the OSC1 pin.?

EDIT:
Is there any clock signal on OSC2 output.?
The other common problem, are the same delay routine times being used for the 32KHz as the 20MHz.??

Last edited:

Franchie

New Member
I am indeed using a scope on the OSC1 pin, getting a pretty clean signal at around 32kHZ as expected (its an old oscilloscope, so its not possible to tell exactly).

b) No code is yet able to run, so that is not yet an issue...
a) Interesting point! I would have imagined that the output of OSC2 would have been a mirror of OSC1/CLKIN. However, it stays basically at Gnd, with regular 'capacitor charging curves', as if the clock signal was trying to charge a capacitor. It never reaches more than about 1/5 Vcc though... Is this the expected behaviour, or am I missing something?

Last edited:

Mr RB

Well-Known Member
Maybe you have a software bug. If something is timing out etc at the lower clock speed and the software appears to be not executing. Could you connect a 32kHz typical xtal and see if it runs then? That would be a quick test.

The output at OSC2 may not echo the signal at OSC1 if you are in HS mode?

Franchie

New Member
My software code is at its most basic level for the moment, it sets a port to output mode, and sets its value to 0xFF. Should be a valid test, I believe. So I believe that the error lies either in the clock signal (which looks valid) or the config bits. This would explain why programming works fine, but debug and execution fail. My current code is below for peer review, it is just a basic test code that does nothing useful yet.

Also Mr RB, thanks for the suggestion, being a poor under-equipped student, I don't have a crystal handy, but will try with a signal generator when I get access to one next week, it looks like a good idea...

So here's the code:
Code:
	list p=16f877a
include "P16F877A.inc"

__config _HS_OSC & _WDT_OFF & _LVP_OFF & _PWRTE_OFF & _CP_OFF & _WRT_OFF & _CPD_OFF & _BODEN_OFF

; MACRO
MOVLF   macro  literal,dest
movlw  literal
movwf  dest
endm

; START
org  0x0000             ;Reset vector
nop
goto  main

main
banksel		PORTD
MOVLF		0,		TRISD
MOVLF		0xFF,	PORTD
goto $end Thanks again so much for all your help! Franchie. birdman0_o Active Member Your code is wrong ; MACRO Code: MOVLF macro literal,dest [COLOR="Red"] movf literal,w[/COLOR] movwf dest endm You were writing the LSB of that RAM address Franchie New Member Are you sure? MOVLF as used in the code moves a litteral into a register. Therefore the steps to carry this out are (since it is not possible to do it in one step): 1) Place the litteral into a working register w, via movlw. 2) Move w to a register, via movwf In my opinion, using movf would only be valid if 'litteral' was a register, which it isn't. It would make 'litteral' not an immediate number, like it is used, but an address of a register. Surely this is incorrect, no? Unless the rules are somehow changed inside a "macro" construct (I had believed this to be a simple compile-time substitution of 'litteral' and 'dest'). Thanks for all your help, this all really helps narrow down the problem! birdman0_o Active Member Oh nvm. You are right, what you didn't do is switch back to bank0 (For PortD) ! birdman0_o Active Member Code:  list p=16f877a include "P16F877A.inc" __config _HS_OSC & _WDT_OFF & _LVP_OFF & _PWRTE_OFF & _CP_OFF & _WRT_OFF & _CPD_OFF & _BODEN_OFF ; MACRO MOVLF macro literal,dest movlw literal movwf dest endm ; START org 0x0000 ;Reset vector nop goto main main banksel [COLOR="Red"]TRISD[/COLOR] MOVLF 0, TRISD [COLOR="Red"] banksel PORTD[/COLOR] MOVLF 0xFF, PORTD goto$
end

Mr RB

Well-Known Member
Wouldn't that have generated a warning at assemble time??

birdman0_o

Active Member
A warning yes, but it would still compile! He probably just ignored it and was happy with the green 100%.

Franchie

New Member
Hello!

You are indeed all correct, I forgot to switch to bank 1. Since switching to bank 1 does not get rid of the warning (under my mplab anyway, maybe you have some idea how to avoid this?), I got used to ignoring those warnings, and since it was a test program I did not pay attention to it. birdman0_o is correct, the green bar is far too tempting .

But more humiliatingly, execute now works, which means my oscillator was probably fine all along, sorry for having made such an elementary mistake! :S

Strangely, debug mode still does not work, which is the reason why I jumped to the unfounded conclusion that it had to be a problem with the oscillator. I am not too bothered by this, since I can now use a normal usart for debugging, but out of curiosity, why would the debugger not be able to connect to the debug executive? I have switched to "debug" mode during compilation, flash it as usual, and it won't connect, are there any extra wires that need to be connected? Maybe the PIC16F877A simply does not support debugging?

Thank you all in any case, and I'm very sorry about it having been such a stupid elementary mistake :S.

Pommie

Well-Known Member
The debugger works at the same speed as the chip and 32k is veeerrrry slow.

Mike.

Franchie

New Member
Okay, so its basically so slow that it timed-out on connection? Makes sense then that it worked with a 20MHz clock...
Alright then, thank you all so much for all the help you have given me!

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Synth. Such ligands are called ambidentate ligands. • Generally this either involves, naming all the ligands followed by all the metals, in both cases using prefixes to indicate the number of each. pentamminenitrito-$$\kappa$$N-cobalt(2+). In many cases it is sufficient to reserve the use of formal IUPAC names for use in publications and to employ a comvenient shorthand naming system for everyday use. - e.g. Example of the application of the metal specification rules to a cationic and anionic platinum complexes.3. However, in the best systems each coordination sphere is named using the same rules as for mononuclear complexes -. 1998, 32, 98-107. Coordination Compound poses 2-3 questions every year in JEE Main.An important aspect of inorganic chemistry comprising questions from theory as well as numerals. Those who need a more thorough and accurate acquaintance with the full IUPAC nomenclature rules are encouraged to consult the IUPAC brief guide to inorganic nomenclature followed by complete guidelines, commonly known as the IUPAC red book. A complex is a substance in which a metal atom or ion is associated with a group of neutral molecules or anions called ligands.Coordination compounds are neutral substances (i.e. If you decide to dive right in to this section you might find it helpful to know that when applying nomenclature and formula rules most textbooks assume. When there are several different ways of attaching the metal and ligands, specifying the structural or stereoisomer, 2. systematically listing the ligands in a way that, as necessary, conveys information about how they are linked to the metal and their stereochemistry, 3. providing the identity of the metal and its oxidation state, or if the oxidation state is unclear, at least the overall charge on the complex. •If the name of the ligand itself has such a prefix, alternatives like bis-, tris-, etc., are used. Nomenclature of Coordination Compounds •Anionic ligands end in “o”; •neutral ligands are not changed. Table $$\sf{\PageIndex{1}}$$. coordination compounds. Prefixes are used to indicate the number of each ligand present. (C) Chelating ligands act like a Lobster claw in attaching to the metal via multiple sites. Stephen Contakes, Westmont College, to whom comments, corrections, and criticisms should be addressed. 2.5 Bridging ligands. Legal. Example: The compound in Scheme $$\sf{\PageIndex{VIIIA}}$$, [$$\mu$$-amido-$$\mu$$-hydroxo-octaamminedichromium(4+)] ion, [$$\mu$$-amido-$$\mu$$-hydroxo-bis(tetraamminechromium(III))] ion, [$$\mu$$-amido-$$\mu$$-hydroxo-bis(tetraamminechromium)(4+)] ion, may be named [tri-$$\mu$$-carbonyl-bis(tricarbonyliron)(0)], [tri-$$\mu$$-carbonyl-bis(tricarbonyliron(0))], or [tri-$$\mu$$-carbonyl-hexacarbonyldiiron)(0)], may be named di-$$\mu$$-chlorido-tetrachloridodicopper(II), di-$$\mu$$-chlorido-bis(dichlorocopper(II)), or di-$$\mu$$-chlorido-bis(dichlorocopper)(0). ... IUPAC Nomenclature of Coordination Compounds: Rules Of Nomenclature: The name of the compound is written in two parts (i) name of cation, and (ii) name of the anion. O.S.Pt + 1 x (-2) (for por; see table 9.2.2) + 1 x (-1) (for Cl-) = +0 (the complexes' charge) so O.S.Mn = +3 or Mn3+. Scheme $$\sf{\PageIndex{V}}$$. Consider bis{[($$\mu$$-2-mercaptoethyl)(2-mercaptoethyl)-methylthioethylaminato (2-)]Nickel(II)}. Figure $$\sf{\PageIndex{1}}$$. The atom within a ligand that is bonded to the central atom or ion is called the donor atom. • The purpose of these rules is to facilitate clear and precise communication among chemists. A perusal of the ligands in Table $$\sf{\PageIndex{1}}$$ reveals that several can bind to a metal in multiple ways. 2.5 Bridging ligands Bridging ligands are those bound to more than one central atom. For example, a more extensive IUPAC name for [Cu2Cl4($$\mu$$-Cl)2] in which the chloro ligands are individually and more completely specified would read di-$$\mu$$2-chlorido-tetrachlorido-1$$\kappa$$2Cl,2$$\kappa$$2Cl-dicopper(II). For example, η 2 describes a ligand that coordinates through 2 contiguous atoms. 2. Assigning metal oxidation states in a complex. To support those who wish to employ them they are given in this section. These cases involve specifying. Consequently, older sources often give an [Fe-Fe] after the name to reflect the presence of that bond. Figure $$\sf{\PageIndex{1}}$$. When naming an ambidentate ligand the atom through which it attaches to the metal is commonly specified after the ligand name using the italicized element symbol or, more formally a $$\kappa$$ followed by the italicized element symbol. Carboxylates might commonly act as monodentate ligands but dithiocarbamates more commonly bind metals through both sulfur atoms (Scheme $$\sf{\PageIndex{III}}$$) and are classified as bidentate. Figure $$\sf{\PageIndex{2}}$$. The ligand may be atomic or polyatomic. For a symmetrical complex like the [Cu2Cl4($$\mu$$-Cl)2] considered above it is enough to specify the existence of the two bridging and four terminal chloro ligands; there is no need to number the chloro ligands or to specify exactly which ones are involved in bridging or to clarify that the bridging involves $$\kappa$$2 coordination of the chloro ligands. Scheme $$\sf{\PageIndex{VIII}}$$. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. Missed the LibreFest? This does not mean that both oxygens of a carboxylate can never both bind to metal centers in a complex. Otherwise this work by Stephen Contakes is licensed under a Creative Commons Attribution 4.0 International License. [Pt(NH 3) 4]Cl 2 When writing the name, the ligands within the coordination sphere are written before the In the complex ion, neutral ligands are written before anionic ligands, and the formula of the whole complex ion is placed in square brackets. The prefixes to provide linkage and stereochemistry for ligands are given in Table$$\sf{\PageIndex{4}}$$. recognizing that in cases when a structure is particularly complex and a picture may make be particularly useful you should supply one (See the note below). being gracious towards the many professional inorganic chemists who adhere loosely to some of the rules you are about to learn. For compound C of Scheme $$\sf{\PageIndex{VIII}}$$. Because only one oxygen per carboxylate typically binds only one is counted when assigning a ligand's denticity. So in other words the ligand classifications presented here are just represent common binding modes. This chemistry video tutorial provides a basic introduction into naming coordination compounds. In anionic complexes of some metals a Latin-derived name is used instead of the element's English name. As shown in Scheme $$\sf{\PageIndex{IIB}}$$ complexes in which both carboxylates bind to a metal are known, and in fact are common in the active sites of some enzymes. The structures of bridging complexes can usually be inferred from their $$\mu$$-tagged ligands in their names and formulae. Specifically, Neutral and cationic ligand names are the same as the names of their neutral compounds with two caveats, names that involve spaces should either be put in parentheses or the spaces should be eliminated (preferred), For anionic ligands, the vowel at the end of their anion names is changed to an, When an ambidentate ligand is present the atom through which it is bound to the metal is indicated by giving either its element symbol or a $$\kappa$$ and its element symbol in italics after the ligand name. Use of the $$\mu$$ notation to specify bridging ligands in metal complexes.3. Name the following compounds and ions in both the Stock and Ewing-Bassett systems. Chemical structures of tfpb, dpp, 9-aca, and 4-bpmh. However, as the complex is no longer though to possess an Fe-Fe bond it is omitted here. Note $$\sf{\PageIndex{1}}$$: Sometimes the most helpful name to give a compound is 42. Here are the rules for nomenclature of coordination compounds: 1. [CoCl(NH3)5]Cl2 = pentamminechlorocobalt(III) chloride, [PtBr2(bpy)] = bipyridinedibromoplatinium(II), K[Ag(SCN)2] = potassium di-S-thiocyanatoargentate(I), [CoCl(NH3)5]Cl2 = pentamminechlorocobalt(2+) chloride, [PtBr2(bpy)] = bipyridinedibromoplatinium(0), K[Ag(SCN)2] = potassium di-S-thiocyanatoargentate(1-). Note that this complex was once thoguht to possess an Fe-Fe bond, in accordance with the predictions of the 18 electron rule. Bridging ligands are cited immediately after terminal ligands of the same kind, if any, and in increasing order of bridging multiplicity. or diamminesilver(I) dicyanidoargentate(I), or diamminesilver(1+) dicyanidoargentate(1-). Scheme $$\sf{\PageIndex{III}}$$. In this section the most common approaches as they are currently used by practicing chemists will be described. Naming coordination compounds involves a set of rules which includes the oxidation number of the central ion and the chemical composition of the ligand, focusing on the anionic and cationic properties of the molecule. Bridging ligands Bridging ligands connect two or more atoms i.e. Because of these factors it is technically more correct to say that carboxylates usually act as monodentate ligands and dithiocarbamates bidentate ones than it is to say that carboxylates are monodentate ligands and dithiocarbamates bidentate ones. Redrawn from this page describing the nomenclature of coordination complexes. Just as the structure of ethane may be more clearly conveyed as H. In speech, $$\eta^1$$=monohapto; $$\eta^2$$=dihapto; $$\eta^3$$=trihapto, etc. Before specifying the metal, the ligands are written as prefixes of the metal. 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Is omitted here one is central is made haphazardly Contakes is licensed under a Creative Commons 4.0... Tris-, etc., are used ) -2-mercaptoethyl ) ( 2-mercaptoethyl ) -methylthioethylaminato ( 2- ) ] Nickel II. In this section, etc thoguht to possess an Fe-Fe bond, in the list and... Ligand that coordinates through 2 contiguous atoms ethane may be named differently a... Id: 614d94c96dfa3fd8 • your IP: 209.197.191.91 • Performance & security by cloudflare, complete! \ ( \sf { \PageIndex { 2 } \ ) at least one ion is present of nomenclature formulae! Metal at any given time IUPAC red book examples of their application III ) for [ Ru ( )... See the IUPAC red book for details as such cases fall outside the of! That coordinates through 2 contiguous atoms to have their students name multinuclear complexes typically binds only one of the for... Iupac and common names and abbreviations to describe these ligands can be monoatomic well... 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text:
# [OS X TeX] Float Placement

Berend Hasselman bhh at xs4all.nl
Sat May 17 05:31:44 EDT 2008

On 17-05-2008, at 11:21, Richard J Benish wrote:

> It looks like our commands are pretty much the same. I have other
> figures in the same document that work fine using this approach. The
> problem figure is taller than the others, so I guess LaTeX doesn't
> want'' to put the body text in the proportionally smaller vertical
> space. I wish it were easier to control. Anyhow, Peter D has
> suggested a document that may shed some light.
>

The p placement option may be responsible for your problems.

Also, it may help to use the ! placement option as in

\begin{figure}[!tb]

! relaxes some restrictions of the h option.

Berend

> Thanks for the suggestions.
>
> RB
>
>
>
>> Perhaps I should have pointed out that I use:
>>
>> \begin{figure}[htbp]
>> \centering
>> \includegraphics[width=120mm,height=90mm]{Staff52.jpg}
>> \caption{Staff 1952 \tiny{Goulburnia}}
>> \end{figure}\\
>>
>> George
>> On 17 May 2008, at 12:58 PM, Richard J Benish wrote:
>>
>>> I have a float (a pdf figure) that takes up about two thirds of
>>> the height of a two column page. When I place it in the desired
>>> column, however, text above and below the float disappears,
>>> leaving a lot of unwanted blank space.
>>>
>>> In the "LaTeX Companion" I see that commands such as \intextsep
>>> may correct the problem, but I don't find any specific examples,
>>> so it's not clear exactly what to enter and where. The commands
>>> that result in the behavior described above are as follows:
>>>
>>>
>>> \begin{figure}[htbp]
>>> \begin{center}
>>> \includegraphics[width=0.48\textwidth]{RelSimGPS-Box-1-
>>> May-16-08.pdf}
>>> \label{default}
>>> \end{center}
>>> \end{figure}
>>>
>>>
>>> Any suggestions as to exactly how to use the \intextsep command or
>>> others to fix this problem?
>>>
>>> Muchas Gracias.
>>>
>>> Richard Benish
>>>
>>> TeX FAQ: http://www.tex.ac.uk/faq
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