Category: News
How do I proceed if QuickBooks payroll updates are not working?
I’m experiencing issues with QuickBooks payroll updates not working. Despite attempting to update, the system fails to download the latest updates. What could be causing this problem, and how can I resolve it?
I’m experiencing issues with QuickBooks payroll updates not working. Despite attempting to update, the system fails to download the latest updates. What could be causing this problem, and how can I resolve it? Read More
Resolving QuickBooks Error 6144 and 301 After Windows Update
How can I resolve QuickBooks Error 6144 and 301 that appeared after a recent Windows update? I’ve tried restarting my computer and updating QB, but the error persists. What steps should I follow to fix this issue?
How can I resolve QuickBooks Error 6144 and 301 that appeared after a recent Windows update? I’ve tried restarting my computer and updating QB, but the error persists. What steps should I follow to fix this issue? Read More
“New” to Viva Engage, with over 100 existing M365 Groups (private) – how to approach communities?
Hi,
I have used Yammer in the past (pre 2016) and have refrained and not encouraged usage in our organization due to the very disconnected user experience and also not having the ressources to manage yet another channel of communications.
I want revisit the current Viva Engage situation. I’ve let the Yammer Support reprovision our network for EU data residency, so it’s completely empty. By reading the documentation I would assume we’re about as aligend to M365 as is possible.
It seems like I can create Communities that are connected to M365 Groups by-default, but I haven’t found the other way around.
What am I todo with our existing 100 private M365 Groups that mostly represent organizational departments, project teams, larger organizational topics/interests?
Are we expected to re-create all these groups? Are there any useful “migration” paths/ideas for this scenario?
I’m really not a fan of M365 Groups clutter and also do not allow the creation of groups outside of administrator for this very reason.
any help or further links to other resources are high appreciated.
Hi, I have used Yammer in the past (pre 2016) and have refrained and not encouraged usage in our organization due to the very disconnected user experience and also not having the ressources to manage yet another channel of communications. I want revisit the current Viva Engage situation. I’ve let the Yammer Support reprovision our network for EU data residency, so it’s completely empty. By reading the documentation I would assume we’re about as aligend to M365 as is possible. It seems like I can create Communities that are connected to M365 Groups by-default, but I haven’t found the other way around. What am I todo with our existing 100 private M365 Groups that mostly represent organizational departments, project teams, larger organizational topics/interests? Are we expected to re-create all these groups? Are there any useful “migration” paths/ideas for this scenario?I’m really not a fan of M365 Groups clutter and also do not allow the creation of groups outside of administrator for this very reason. any help or further links to other resources are high appreciated. Read More
Trade Evista 1000 Reviews 2024: Trade Evista 1000 App Scam OR Not Trading AI Platform?
After diving into the functionalities and benefits of the Evista 1000 trading platform, it’s clear that this tool offers a robust solution for beginners eager to step into the trading world. With its user-friendly interface, comprehensive resources, and advanced tools adapted for novice traders, the Evista 1000 provides a solid foundation for anyone starting their trading journey.
Remember, the key to successful trading lies in continuous learning and practice. The Evista 1000 is designed to support your growth, making it easier to track markets, analyze trends, and execute trades efficiently. So, take your time to explore each feature, utilize the educational materials available, and start your trading experience with confidence. Happy trading!
After diving into the functionalities and benefits of the Evista 1000 trading platform, it’s clear that this tool offers a robust solution for beginners eager to step into the trading world. With its user-friendly interface, comprehensive resources, and advanced tools adapted for novice traders, the Evista 1000 provides a solid foundation for anyone starting their trading journey. Remember, the key to successful trading lies in continuous learning and practice. The Evista 1000 is designed to support your growth, making it easier to track markets, analyze trends, and execute trades efficiently. So, take your time to explore each feature, utilize the educational materials available, and start your trading experience with confidence. Happy trading! Read More
Optimizing Retrieval for RAG Apps: Vector Search and Hybrid Techniques
In our previous blog, we talked about LLMs and incorporating domain knowledge techniques such as Retrieval Augmentation Generation (RAG) to solve the issue of outdated knowledge.
In this blog we are going to dive into optimizing our search strategy with Hybrid search techniques. Common practices for implementing the retrieval step in retrieval-augmented generation (RAG) applications are;
Keyword search
Vector Search
Hybrid search (Keyword + Vector)
Hybrid + Semantic ranker
Optimal search strategy
Keyword search – Uses traditional full-text search methods – content is broken into terms through language-specific text analysis, inverted indexes are created for fast retrieval, and the BM25 probabilistic model is used for scoring.
Vector search – is best for finding semantically related matches, which is a fully supported pattern in Azure AI Search . Documents are converted from text to vector representations using an embedding model. Retrieval is performed by generating a query embedding and finding the documents whose vectors are closest to the query’s. We used Azure Open AI text-embedding-ada-002 (Ada-002) embeddings and cosine similarity.
Vector embeddings – An embedding encodes an input as a list of floating-point numbers. Different models output different embeddings, with varying lengths.
”dog” → [0.017198, -0.007493, -0.057982, 0.054051, -0.028336, 0.019245,…]
Vector similarity – We compute embeddings so that we can calculate similarity between inputs. The most common distance measurement is cosine similarity.
In the above image we can use Cosine similarity, a common way to measure vector similarity is by calculating the cosine similarity between two vectors. A cosine similarity value close to 1 indicates high similarity, while a value close to 0 indicates dissimilarity, this is in the context of natural language processing (NLP) and machine learning.
Let’s say we have word embeddings for “dog,” “woof,” and “cat.”
If we calculate the cosine similarity between the vectors for “dog” and “woof,” we might get a high value (close to 1) because they are related.
However, the cosine similarity between “dog” and “cat” would likely be lower (closer to 0) because they represent different animals with distinct characteristics. Remember that this concept extends beyond words; it applies to any vectors in an embedding space. Whether it’s words, images, or other data points, measuring similarity helps us understand relationships and make informed decisions in various applications
Hybrid search (Keyword + Vector) – combines vector search and keyword search, optimally using Reciprocal-Rank-Fusion for merging results and a Machine Learning model to re-rank results after
Hybrid + Semantic ranker – generative AI scenarios typically use the top 3 to 5 results as their grounding context to prioritize the most important results. AI Search applications work best with a calibrated relevance score that can be used to filter out low quality results. The semantic ranker runs the query and documents text simultaneously though transformer models that utilize the cross-attention mechanism to produce a ranker score.
A score of 0 represents a very irrelevant chunk, and a score of 4 represents an excellent one. In the chart below, Hybrid + Semantic ranking finds the best content for the LLM at each result set size. See code example on this repo by Pamela fox.
RAG with hybrid search
Using the above image, when user sends the prompt “do my company…” the embedding model creates vector representations of that text. These embeddings capture the semantic meaning allowing similarity comparisons between different pieces of text. These embedding models include word2vec, BERT, and GPT (Generative Pre-trained Transformer). Using hybrid search we are able to use keyword + vector search and sematic search to retrieve more accurate response from our source ie pdf.
RAG With Vector Databases
We can extend this robust retrieval for RAG with a vector Database. A vector database is specifically designed for first, efficient storage and retrieval of high-dimensional vectors. There two types of Vector database there two types of Vector databases
Integrated vector database
pure vector database
A pure vector database is designed to efficiently store and manage vector embeddings, along with a small amount of metadata; it is separate from the data source from which the embeddings are derived
A vector database that is integrated in a highly performant NoSQL or relational database provides additional capabilities. The integrated vector database in a NoSQL or relational database can store, index, and query embeddings alongside the corresponding original data. This approach eliminates the extra cost of replicating data in a separate pure vector database. Moreover, keeping the vector embeddings and original data together better facilitates multi-modal data operations, and enables greater data consistency, scale, and performance.
Integrated vector database on Azure
Azure Cosmos DB for MongoDB vCore
Azure Cosmos DB for PostgreSQL.
Azure Cosmos Db for NoSQL API – is under development and will be announced on May 2024.
Read more
Open-source vector databases
How to enable and use pgvector on Azure Cosmos DB for PostgreSQL
Build applications for free with Azure Cosmos DB for MongoDB (vCore) Free Tie
Querying in Azure AI Search
Transparency note: Azure AI Search
Code samples
Python notebook tutorial – Vector database integration through LangChain
Python RAG pattern – Azure product chatbot
Microsoft Tech Community – Latest Blogs –Read More
search interval in datetime
hi, I have a datetime array of several many days. I need to serch interval
example:
TimeA=22:00
TimeB=03:00
search datetime>=timeA and datetime<=timeB (22:00 ..22:01..22:02. etc……23:59 ..00:00…00:01.etc… 2:58..2:59 3:00)
So I want without checking the day to have the indices of that range in the whole array
(the times in datetime are already ordered in a correct chronological way so the next time is always after the previous one)hi, I have a datetime array of several many days. I need to serch interval
example:
TimeA=22:00
TimeB=03:00
search datetime>=timeA and datetime<=timeB (22:00 ..22:01..22:02. etc……23:59 ..00:00…00:01.etc… 2:58..2:59 3:00)
So I want without checking the day to have the indices of that range in the whole array
(the times in datetime are already ordered in a correct chronological way so the next time is always after the previous one) hi, I have a datetime array of several many days. I need to serch interval
example:
TimeA=22:00
TimeB=03:00
search datetime>=timeA and datetime<=timeB (22:00 ..22:01..22:02. etc……23:59 ..00:00…00:01.etc… 2:58..2:59 3:00)
So I want without checking the day to have the indices of that range in the whole array
(the times in datetime are already ordered in a correct chronological way so the next time is always after the previous one) search interval in datetime MATLAB Answers — New Questions
Converting a cell array to can message timetable
I am trying to convert a cell array into can message timetable format ,
% Given cell array
canMessageCell = {‘Timestamp 1546909042’, ‘Interface ”can0”’, ‘ID ”0B9”’, ‘Data [0,4,0,0,0,0,0,0]’};
% Extract the information from the cell array
timestampStr = canMessageCell{1};
idStr = canMessageCell{3};
dataStr = canMessageCell{4};
% Convert extracted strings to proper data types
timestamp = datetime(str2double(extractAfter(timestampStr, ‘Timestamp ‘)), ‘ConvertFrom’, ‘posixtime’);
id = hex2dec(extractBetween(idStr, ‘ID ”’, ””));
% Extract the data portion and convert it to a numeric array
dataChar = extractBetween(dataStr, ‘Data [‘, ‘]’);
data = str2num(dataChar{1}); % Convert cell array to character vector before using str2num
% Assume the CAN message is not an extended frame, not a remote frame, and has no error
isExtended = false;
isRemote = false;
hasError = false;
% Create a structure array with the expected field names
canMessageStruct = struct(‘TimeStamp’, timestamp, ‘ID’, id, ‘Data’, {data}, ‘Extended’, isExtended, ‘Remote’, isRemote, ‘Error’, hasError);
% Convert the structure array to a cell array
canMessageCell = {canMessageStruct};
% Load the DBC file (replace with the actual path to your DBC file)
dbcFile = ‘drivekit_kia_soul_ev.dbc’;
db = canDatabase(dbcFile);
% Convert the cell array to a canMessageTimetable
msgTimetable = canMessageTimetable(canMessageCell, db);
% Display the result
disp(msgTimetable);
I am following this syntax for converting my cell array into CAN message timetable db = canDatabase(‘myDatabase.dbc’)
msgTimetable = canMessageTimetable(canMsgs,db) but I am getting this error, Error using ()
Timetable row subscript must be a numeric, logical, or datetime array, or a timerange subscripter.
Error in canmessagetimetableconversion (line 33)
msgTimetable = canMessageTimetable(canMessageCell, db);
I dont know why because my timestamp is in datetime format as i converted it from posix format, my data is in uint8 format and I have created a struct for my timestamp, ID , Data , I have assumed Extended, Remote, Error to be false , if anyone can please help me in debugging this issue as i have been stuck on this , it would be greatly appreciated.I am trying to convert a cell array into can message timetable format ,
% Given cell array
canMessageCell = {‘Timestamp 1546909042’, ‘Interface ”can0”’, ‘ID ”0B9”’, ‘Data [0,4,0,0,0,0,0,0]’};
% Extract the information from the cell array
timestampStr = canMessageCell{1};
idStr = canMessageCell{3};
dataStr = canMessageCell{4};
% Convert extracted strings to proper data types
timestamp = datetime(str2double(extractAfter(timestampStr, ‘Timestamp ‘)), ‘ConvertFrom’, ‘posixtime’);
id = hex2dec(extractBetween(idStr, ‘ID ”’, ””));
% Extract the data portion and convert it to a numeric array
dataChar = extractBetween(dataStr, ‘Data [‘, ‘]’);
data = str2num(dataChar{1}); % Convert cell array to character vector before using str2num
% Assume the CAN message is not an extended frame, not a remote frame, and has no error
isExtended = false;
isRemote = false;
hasError = false;
% Create a structure array with the expected field names
canMessageStruct = struct(‘TimeStamp’, timestamp, ‘ID’, id, ‘Data’, {data}, ‘Extended’, isExtended, ‘Remote’, isRemote, ‘Error’, hasError);
% Convert the structure array to a cell array
canMessageCell = {canMessageStruct};
% Load the DBC file (replace with the actual path to your DBC file)
dbcFile = ‘drivekit_kia_soul_ev.dbc’;
db = canDatabase(dbcFile);
% Convert the cell array to a canMessageTimetable
msgTimetable = canMessageTimetable(canMessageCell, db);
% Display the result
disp(msgTimetable);
I am following this syntax for converting my cell array into CAN message timetable db = canDatabase(‘myDatabase.dbc’)
msgTimetable = canMessageTimetable(canMsgs,db) but I am getting this error, Error using ()
Timetable row subscript must be a numeric, logical, or datetime array, or a timerange subscripter.
Error in canmessagetimetableconversion (line 33)
msgTimetable = canMessageTimetable(canMessageCell, db);
I dont know why because my timestamp is in datetime format as i converted it from posix format, my data is in uint8 format and I have created a struct for my timestamp, ID , Data , I have assumed Extended, Remote, Error to be false , if anyone can please help me in debugging this issue as i have been stuck on this , it would be greatly appreciated. I am trying to convert a cell array into can message timetable format ,
% Given cell array
canMessageCell = {‘Timestamp 1546909042’, ‘Interface ”can0”’, ‘ID ”0B9”’, ‘Data [0,4,0,0,0,0,0,0]’};
% Extract the information from the cell array
timestampStr = canMessageCell{1};
idStr = canMessageCell{3};
dataStr = canMessageCell{4};
% Convert extracted strings to proper data types
timestamp = datetime(str2double(extractAfter(timestampStr, ‘Timestamp ‘)), ‘ConvertFrom’, ‘posixtime’);
id = hex2dec(extractBetween(idStr, ‘ID ”’, ””));
% Extract the data portion and convert it to a numeric array
dataChar = extractBetween(dataStr, ‘Data [‘, ‘]’);
data = str2num(dataChar{1}); % Convert cell array to character vector before using str2num
% Assume the CAN message is not an extended frame, not a remote frame, and has no error
isExtended = false;
isRemote = false;
hasError = false;
% Create a structure array with the expected field names
canMessageStruct = struct(‘TimeStamp’, timestamp, ‘ID’, id, ‘Data’, {data}, ‘Extended’, isExtended, ‘Remote’, isRemote, ‘Error’, hasError);
% Convert the structure array to a cell array
canMessageCell = {canMessageStruct};
% Load the DBC file (replace with the actual path to your DBC file)
dbcFile = ‘drivekit_kia_soul_ev.dbc’;
db = canDatabase(dbcFile);
% Convert the cell array to a canMessageTimetable
msgTimetable = canMessageTimetable(canMessageCell, db);
% Display the result
disp(msgTimetable);
I am following this syntax for converting my cell array into CAN message timetable db = canDatabase(‘myDatabase.dbc’)
msgTimetable = canMessageTimetable(canMsgs,db) but I am getting this error, Error using ()
Timetable row subscript must be a numeric, logical, or datetime array, or a timerange subscripter.
Error in canmessagetimetableconversion (line 33)
msgTimetable = canMessageTimetable(canMessageCell, db);
I dont know why because my timestamp is in datetime format as i converted it from posix format, my data is in uint8 format and I have created a struct for my timestamp, ID , Data , I have assumed Extended, Remote, Error to be false , if anyone can please help me in debugging this issue as i have been stuck on this , it would be greatly appreciated. cell array, can message timetable conversion MATLAB Answers — New Questions
Use of tensorprod function to compute multilinear forms
Hi everyone,
I have to compute the multilinear form of a tensor and a vector. Suppose for example we have a 4-dimensional tensor A and a vector v. The 4-linear form is the multiplication of A by v 4 times. The only way that I found to do this is to use "for" loops. I think there must be a way to use the "tensorprod" function to perform the operation more efficiently but I can’t find it. Thank you in advance to anyone who can help.
Cheers
n = 3;
A = rand(n,n,n,n);
v = rand(n,1);
for j = 1:n
for k = 1:n
G(:,k,j) = squeeze(v’*A(:,:,k,j));
end
end
for j = 1:n
J(:,j) = squeeze(v’*G(:,:,j));
end
res = v’*J*v;Hi everyone,
I have to compute the multilinear form of a tensor and a vector. Suppose for example we have a 4-dimensional tensor A and a vector v. The 4-linear form is the multiplication of A by v 4 times. The only way that I found to do this is to use "for" loops. I think there must be a way to use the "tensorprod" function to perform the operation more efficiently but I can’t find it. Thank you in advance to anyone who can help.
Cheers
n = 3;
A = rand(n,n,n,n);
v = rand(n,1);
for j = 1:n
for k = 1:n
G(:,k,j) = squeeze(v’*A(:,:,k,j));
end
end
for j = 1:n
J(:,j) = squeeze(v’*G(:,:,j));
end
res = v’*J*v; Hi everyone,
I have to compute the multilinear form of a tensor and a vector. Suppose for example we have a 4-dimensional tensor A and a vector v. The 4-linear form is the multiplication of A by v 4 times. The only way that I found to do this is to use "for" loops. I think there must be a way to use the "tensorprod" function to perform the operation more efficiently but I can’t find it. Thank you in advance to anyone who can help.
Cheers
n = 3;
A = rand(n,n,n,n);
v = rand(n,1);
for j = 1:n
for k = 1:n
G(:,k,j) = squeeze(v’*A(:,:,k,j));
end
end
for j = 1:n
J(:,j) = squeeze(v’*G(:,:,j));
end
res = v’*J*v; tensors, multilinear forms MATLAB Answers — New Questions
MATLAB code for ILEACH protocol in WSN reqired
I am working on WSN routing protocols simulation. I need matlab code for ILEACH protocol asap. If anyone has please do reply soon.I am working on WSN routing protocols simulation. I need matlab code for ILEACH protocol asap. If anyone has please do reply soon. I am working on WSN routing protocols simulation. I need matlab code for ILEACH protocol asap. If anyone has please do reply soon. for loop MATLAB Answers — New Questions
New Script keeps opening in new window instead of a tab
So, stupid question. I accidentally did something on my script file and now when I hit new script, it opens a new script in a new window instead of a new tab. Can someone tell me what I did and how to reverse it.
ThanksSo, stupid question. I accidentally did something on my script file and now when I hit new script, it opens a new script in a new window instead of a new tab. Can someone tell me what I did and how to reverse it.
Thanks So, stupid question. I accidentally did something on my script file and now when I hit new script, it opens a new script in a new window instead of a new tab. Can someone tell me what I did and how to reverse it.
Thanks script, live script MATLAB Answers — New Questions
Dealing With QuickBooks Company File Error 6190 and 816 How to Fix it
I’m encountering QuickBooks Company File Error 6190 and 816 when trying to access my company file. How can I resolve this issue? Any help would be appreciated!
I’m encountering QuickBooks Company File Error 6190 and 816 when trying to access my company file. How can I resolve this issue? Any help would be appreciated! Read More
Why Can’t open company file in QuickBooks Desktop After latest update?
I’m having trouble opening my company file in QuickBooks Desktop. Every time I try, I get an error message saying the file cannot be opened. I’ve tried restarting my computer and QuickBooks, but the problem persists. What could be causing this issue, and how can I resolve it?
I’m having trouble opening my company file in QuickBooks Desktop. Every time I try, I get an error message saying the file cannot be opened. I’ve tried restarting my computer and QuickBooks, but the problem persists. What could be causing this issue, and how can I resolve it? Read More
Why Is QuickBooks Direct Deposit Not Working on Window 10/11
I’m experiencing an issue where QuickBooks Direct Deposit is not working on Windows 10/11. It was functioning fine previously, but now it fails to process payments. What could be causing this problem, and how can I resolve it?
I’m experiencing an issue where QuickBooks Direct Deposit is not working on Windows 10/11. It was functioning fine previously, but now it fails to process payments. What could be causing this problem, and how can I resolve it? Read More
Why is My QuickBooks is Unable to open this company file Error 6190 Windows 10/11?
I’m encountering an error with QuickBooks: “Unable to open this company file Error 6190.” This issue is preventing me from accessing my company file, and it’s causing significant disruption to my business operations. What are the steps I can take to troubleshoot and resolve this error?
I’m encountering an error with QuickBooks: “Unable to open this company file Error 6190.” This issue is preventing me from accessing my company file, and it’s causing significant disruption to my business operations. What are the steps I can take to troubleshoot and resolve this error? Read More
Why I am Getting QuickBooks error 1603 when trying to Install or Update QB
I’m encountering QuickBooks error 1603 when trying to install or update QB. This error is causing significant disruption to my business operations. Could you provide a solution to fix this issue?
I’m encountering QuickBooks error 1603 when trying to install or update QB. This error is causing significant disruption to my business operations. Could you provide a solution to fix this issue? Read More
Resolve Quickbooks Desktop Error 6190 and 816 Windows 11
I’ve encountered QuickBooks Desktop Error 6190 and 816 on Windows 11. How can I resolve this issue? It’s urgent as it’s affecting my business operations. Any troubleshooting steps or solutions would be greatly appreciated.
I’ve encountered QuickBooks Desktop Error 6190 and 816 on Windows 11. How can I resolve this issue? It’s urgent as it’s affecting my business operations. Any troubleshooting steps or solutions would be greatly appreciated. Read More
Azure Monitor cost optimization using Azure Advisor
We in Azure Monitor are committed to assisting you in optimizing your budget allocation, making informed decisions about monitoring options, and discovering features and configurations that enable you to get more out of your infrastructure.
Azure Advisor is a free offering that can help you avoid problems and save money by providing you with proactive best practice guidance.
Consider configuring the low-cost Basic logs plan on selected tables:
This recommendation identifies an ingestion of more than 1 GB per month for tables eligible for the low-cost Basic Log data plan.
The Basic Log plan gives the you the ability to use search capabilities for debugging and troubleshooting at a much lower cost – Read more about Basic logs.
Consider Changing Pricing Tier:
Each Log Analytics workspace in Azure Monitor can have a different price tier.
This recommendation identifies Log Analytics workspaces that should be in a different pricing tier and tells which tier your workspace should change to – Read more about pricing tiers.
Consider removing unused restored tables:
If you use the restore function, you may forget to delete the restored tables after the operation is complete, resulting in high costs and frustration.
This recommendation detects restored data that has been active in your workspace for more than a week.
It is then recommends to delete the table to avoid unnecessary costs. Read more about restore tables.
How do I see my recommendations?
Log in to Azure Portal
New recommendations will be added in the next few months to help you reduce your overall cost in Azure Monitor and detect anomalies.
New recommendations will be added in the next few months to help you reduce your overall cost in Azure Monitor and detect anomalies.
Search for Advisor, in the search pane:
New recommendations will be added in the next few months to help you reduce your overall cost in Azure Monitor and detect anomalies.
Microsoft Tech Community – Latest Blogs –Read More
How do I plot a symbolic equation that I have created?
Hello everyone. I am about a month into a uni project and I am stuck on something that I cannot figure out, despite weeks of little bits of test code and Googling. I won’t share the whole thing, as it’s a project and currently about 350 lines. In general though, I have built something that solves the differential equations for an RLC circuit, showing all the working using symbolic maths.
So I get it down to the point where I have my symbol (Yt) which is equal to: val=1.0 – 0.2*exp(-1.0*t)*sin(5.0*t) – 1.0*exp(-1.0*t)*cos(5.0*t)
I have then successfully used fplot to draw this equation. However, I am now trying to demonstrate time invariance. The only way that I can find to do this is by creating a time vector, and then plotting the equation against time+1, similar to the test code below.
InputStartTime = 1
t=0:.005:5;
hold on;
test3 = -1.0*exp(-1.0*t).*cos(5.0*t)- 0.2*exp(-1.0*t).*sin(5.0*t) + 1
firstPoint = test3(1) % start point of function
plot(t+InputStartTime,test3)
plot([0 (t(1)+InputStartTime)], [0 firstPoint]) % Plot line from origin to start point of delayed function
However, and this is the key, test3 above is hardcoded. I cannot find a way of getting the sym into a format that I can use. If I try something like string, MATLAB then complains that the matrices are the "Incorrect dimensions for matrix multiplication". I then had to change the ‘*’ to ‘.*’ in the above equation. I have no idea how to take my solved equation from Symbolic maths and use it like this. Can anyone help please? This is how the formula looks once I have solved it, with yt being a "1×1 sym".
performing a string() of this gives: "1.0 – 0.2*exp(-1.0*t)*sin(5.0*t) – 1.0*exp(-1.0*t)*cos(5.0*t)" which shows that there is a * between exp() and sin/cos.Hello everyone. I am about a month into a uni project and I am stuck on something that I cannot figure out, despite weeks of little bits of test code and Googling. I won’t share the whole thing, as it’s a project and currently about 350 lines. In general though, I have built something that solves the differential equations for an RLC circuit, showing all the working using symbolic maths.
So I get it down to the point where I have my symbol (Yt) which is equal to: val=1.0 – 0.2*exp(-1.0*t)*sin(5.0*t) – 1.0*exp(-1.0*t)*cos(5.0*t)
I have then successfully used fplot to draw this equation. However, I am now trying to demonstrate time invariance. The only way that I can find to do this is by creating a time vector, and then plotting the equation against time+1, similar to the test code below.
InputStartTime = 1
t=0:.005:5;
hold on;
test3 = -1.0*exp(-1.0*t).*cos(5.0*t)- 0.2*exp(-1.0*t).*sin(5.0*t) + 1
firstPoint = test3(1) % start point of function
plot(t+InputStartTime,test3)
plot([0 (t(1)+InputStartTime)], [0 firstPoint]) % Plot line from origin to start point of delayed function
However, and this is the key, test3 above is hardcoded. I cannot find a way of getting the sym into a format that I can use. If I try something like string, MATLAB then complains that the matrices are the "Incorrect dimensions for matrix multiplication". I then had to change the ‘*’ to ‘.*’ in the above equation. I have no idea how to take my solved equation from Symbolic maths and use it like this. Can anyone help please? This is how the formula looks once I have solved it, with yt being a "1×1 sym".
performing a string() of this gives: "1.0 – 0.2*exp(-1.0*t)*sin(5.0*t) – 1.0*exp(-1.0*t)*cos(5.0*t)" which shows that there is a * between exp() and sin/cos. Hello everyone. I am about a month into a uni project and I am stuck on something that I cannot figure out, despite weeks of little bits of test code and Googling. I won’t share the whole thing, as it’s a project and currently about 350 lines. In general though, I have built something that solves the differential equations for an RLC circuit, showing all the working using symbolic maths.
So I get it down to the point where I have my symbol (Yt) which is equal to: val=1.0 – 0.2*exp(-1.0*t)*sin(5.0*t) – 1.0*exp(-1.0*t)*cos(5.0*t)
I have then successfully used fplot to draw this equation. However, I am now trying to demonstrate time invariance. The only way that I can find to do this is by creating a time vector, and then plotting the equation against time+1, similar to the test code below.
InputStartTime = 1
t=0:.005:5;
hold on;
test3 = -1.0*exp(-1.0*t).*cos(5.0*t)- 0.2*exp(-1.0*t).*sin(5.0*t) + 1
firstPoint = test3(1) % start point of function
plot(t+InputStartTime,test3)
plot([0 (t(1)+InputStartTime)], [0 firstPoint]) % Plot line from origin to start point of delayed function
However, and this is the key, test3 above is hardcoded. I cannot find a way of getting the sym into a format that I can use. If I try something like string, MATLAB then complains that the matrices are the "Incorrect dimensions for matrix multiplication". I then had to change the ‘*’ to ‘.*’ in the above equation. I have no idea how to take my solved equation from Symbolic maths and use it like this. Can anyone help please? This is how the formula looks once I have solved it, with yt being a "1×1 sym".
performing a string() of this gives: "1.0 – 0.2*exp(-1.0*t)*sin(5.0*t) – 1.0*exp(-1.0*t)*cos(5.0*t)" which shows that there is a * between exp() and sin/cos. symbolic, plot MATLAB Answers — New Questions
Unable to open the requested feature. Check your internet connection and proxy settings in MATLAB Web preferences and then try starting the feature again. Detailed information
out = parsim(simInput,’ShowSimulationManager’,’on’);
To check the parallel operation process of the simulation link file in real time with the above code, the following error window and error message occur.
I should be very grateful to you if you might help me.out = parsim(simInput,’ShowSimulationManager’,’on’);
To check the parallel operation process of the simulation link file in real time with the above code, the following error window and error message occur.
I should be very grateful to you if you might help me. out = parsim(simInput,’ShowSimulationManager’,’on’);
To check the parallel operation process of the simulation link file in real time with the above code, the following error window and error message occur.
I should be very grateful to you if you might help me. parallel computing toolbox, show simulation manager MATLAB Answers — New Questions
Drawing backward and forward points using intrinsic low dimensional manifold (ILDM)
I am trying to draw the backward and forward points of the slow invariant manifold (SIM) from the equilibrium point using the following code of Intrinsic low dimensional manifold (ILDM), but didn’t get all backward points. could anyone identify the mistake.
by using below code i am getting following graph which is not plotting the backward points
i am trying to get following result, but unfortunately there is a mistake in my code.
code:
% A + Z <=====> AZ
% AZ + B <=====> AB + Z
% Let A=c1, B=c2, Z=c3, AZ=c4 and AB=c5
% c1 + c3 <=====> c4 % c4 + c2 <=====> c5 + c3
clc;
clear
syms c1 c2 c3 c4 c5 x1 x2
syms k1p k2p k1n k2n %positive and negative k’s
%Stoichiometric Vectors
r1 = [-1; 0; -1; 1; 0];
r2 = [0; -1; 1; -1; 1];
%Parameters
k1p=1; k2p=0.5; k3p=1 ;
EP=[.5 .1 .1 .4 .1]; %Equilibrium Point
C1=EP(1); C2=EP(2); C3=EP(3); C4=EP(4); C5=EP(5);
%Finding k1n, k2n (((((((((((((((((((((((((((((((( %Reaction Rates
w1 = k1p*C1*C3 – k1n*C4;
w2 = k2p*C4*C2 – k2n*C5*C3;
%solving for k1n and k2n
k1n=solve(w1==0, k1n);
k2n=solve(w2==0, k2n);
%)))))))))))))))))))))))))))))))))))))))))))))))))
%Molecular Matrix[1 0 0 1 1; 0 1 0 0 1; 0 0 1 1 0]
bA = C1+C3+C4; %for A
bB = C2+C5; %for B
bZ = C3+C4; %for Z
% Reducing the system for species c1=A, c4=AZ;
c3 = bZ – c4;
c5 = bA – c1 – c4;
c2 = bB – c5;
%Reaction Rates
w1 = k1p*c1*c3 – k1n*c4;
w2 = k2p*c4*c2 – k2n*c5*c3;
%Using Law of Conservation of Mass
G = r1*w1 + r2*w2;
%Reduced System
F = [G(1); G(4)];
F = subs(F,[c1,c4],[x1,x2]);
EP = [C1; C4];
m=20; n=17; %no of points forward and backward to the equilibrium point
h=0.03; %constant increment
%Finding points above EP
P=EP; r=P;
for forward = 1:m
J = ILDM1D_jacobian(F,r);
b = ILDM1D_slowest(J);
r = P + h*b;
Q = ILDM1D_projection(J);
P = ILDM1D_manifold(Q,F,r);
allForward(forward,:) = P; % all Forward matrix stores all forward points
end
%Finding points below EP
P=EP; r=P;
for backward =1:n
J= ILDM1D_jacobian (F,r);
b = ILDM1D_slowest (J);
r = P + h*b;
Q= ILDM1D_projection (J);
ILDM1D_manifold (Q,F,r);
allBackward(backward,:) = P;
%allBackward matrix stores all backward points
end
%Plotting
plot (EP (1),EP (2), ‘ks’);hold on;
plot (allForward(:,1),allForward(:,2),’+k-‘);
plot (allBackward(:,2),allBackward(:,2),’*k-‘);
grid on;
axis ([0 1 0 .7]);
xlabel (‘A’); ylabel (‘AZ’);
legend(‘Eguilibrium point’, ‘Forward direction’, ‘Backword direction’)I am trying to draw the backward and forward points of the slow invariant manifold (SIM) from the equilibrium point using the following code of Intrinsic low dimensional manifold (ILDM), but didn’t get all backward points. could anyone identify the mistake.
by using below code i am getting following graph which is not plotting the backward points
i am trying to get following result, but unfortunately there is a mistake in my code.
code:
% A + Z <=====> AZ
% AZ + B <=====> AB + Z
% Let A=c1, B=c2, Z=c3, AZ=c4 and AB=c5
% c1 + c3 <=====> c4 % c4 + c2 <=====> c5 + c3
clc;
clear
syms c1 c2 c3 c4 c5 x1 x2
syms k1p k2p k1n k2n %positive and negative k’s
%Stoichiometric Vectors
r1 = [-1; 0; -1; 1; 0];
r2 = [0; -1; 1; -1; 1];
%Parameters
k1p=1; k2p=0.5; k3p=1 ;
EP=[.5 .1 .1 .4 .1]; %Equilibrium Point
C1=EP(1); C2=EP(2); C3=EP(3); C4=EP(4); C5=EP(5);
%Finding k1n, k2n (((((((((((((((((((((((((((((((( %Reaction Rates
w1 = k1p*C1*C3 – k1n*C4;
w2 = k2p*C4*C2 – k2n*C5*C3;
%solving for k1n and k2n
k1n=solve(w1==0, k1n);
k2n=solve(w2==0, k2n);
%)))))))))))))))))))))))))))))))))))))))))))))))))
%Molecular Matrix[1 0 0 1 1; 0 1 0 0 1; 0 0 1 1 0]
bA = C1+C3+C4; %for A
bB = C2+C5; %for B
bZ = C3+C4; %for Z
% Reducing the system for species c1=A, c4=AZ;
c3 = bZ – c4;
c5 = bA – c1 – c4;
c2 = bB – c5;
%Reaction Rates
w1 = k1p*c1*c3 – k1n*c4;
w2 = k2p*c4*c2 – k2n*c5*c3;
%Using Law of Conservation of Mass
G = r1*w1 + r2*w2;
%Reduced System
F = [G(1); G(4)];
F = subs(F,[c1,c4],[x1,x2]);
EP = [C1; C4];
m=20; n=17; %no of points forward and backward to the equilibrium point
h=0.03; %constant increment
%Finding points above EP
P=EP; r=P;
for forward = 1:m
J = ILDM1D_jacobian(F,r);
b = ILDM1D_slowest(J);
r = P + h*b;
Q = ILDM1D_projection(J);
P = ILDM1D_manifold(Q,F,r);
allForward(forward,:) = P; % all Forward matrix stores all forward points
end
%Finding points below EP
P=EP; r=P;
for backward =1:n
J= ILDM1D_jacobian (F,r);
b = ILDM1D_slowest (J);
r = P + h*b;
Q= ILDM1D_projection (J);
ILDM1D_manifold (Q,F,r);
allBackward(backward,:) = P;
%allBackward matrix stores all backward points
end
%Plotting
plot (EP (1),EP (2), ‘ks’);hold on;
plot (allForward(:,1),allForward(:,2),’+k-‘);
plot (allBackward(:,2),allBackward(:,2),’*k-‘);
grid on;
axis ([0 1 0 .7]);
xlabel (‘A’); ylabel (‘AZ’);
legend(‘Eguilibrium point’, ‘Forward direction’, ‘Backword direction’) I am trying to draw the backward and forward points of the slow invariant manifold (SIM) from the equilibrium point using the following code of Intrinsic low dimensional manifold (ILDM), but didn’t get all backward points. could anyone identify the mistake.
by using below code i am getting following graph which is not plotting the backward points
i am trying to get following result, but unfortunately there is a mistake in my code.
code:
% A + Z <=====> AZ
% AZ + B <=====> AB + Z
% Let A=c1, B=c2, Z=c3, AZ=c4 and AB=c5
% c1 + c3 <=====> c4 % c4 + c2 <=====> c5 + c3
clc;
clear
syms c1 c2 c3 c4 c5 x1 x2
syms k1p k2p k1n k2n %positive and negative k’s
%Stoichiometric Vectors
r1 = [-1; 0; -1; 1; 0];
r2 = [0; -1; 1; -1; 1];
%Parameters
k1p=1; k2p=0.5; k3p=1 ;
EP=[.5 .1 .1 .4 .1]; %Equilibrium Point
C1=EP(1); C2=EP(2); C3=EP(3); C4=EP(4); C5=EP(5);
%Finding k1n, k2n (((((((((((((((((((((((((((((((( %Reaction Rates
w1 = k1p*C1*C3 – k1n*C4;
w2 = k2p*C4*C2 – k2n*C5*C3;
%solving for k1n and k2n
k1n=solve(w1==0, k1n);
k2n=solve(w2==0, k2n);
%)))))))))))))))))))))))))))))))))))))))))))))))))
%Molecular Matrix[1 0 0 1 1; 0 1 0 0 1; 0 0 1 1 0]
bA = C1+C3+C4; %for A
bB = C2+C5; %for B
bZ = C3+C4; %for Z
% Reducing the system for species c1=A, c4=AZ;
c3 = bZ – c4;
c5 = bA – c1 – c4;
c2 = bB – c5;
%Reaction Rates
w1 = k1p*c1*c3 – k1n*c4;
w2 = k2p*c4*c2 – k2n*c5*c3;
%Using Law of Conservation of Mass
G = r1*w1 + r2*w2;
%Reduced System
F = [G(1); G(4)];
F = subs(F,[c1,c4],[x1,x2]);
EP = [C1; C4];
m=20; n=17; %no of points forward and backward to the equilibrium point
h=0.03; %constant increment
%Finding points above EP
P=EP; r=P;
for forward = 1:m
J = ILDM1D_jacobian(F,r);
b = ILDM1D_slowest(J);
r = P + h*b;
Q = ILDM1D_projection(J);
P = ILDM1D_manifold(Q,F,r);
allForward(forward,:) = P; % all Forward matrix stores all forward points
end
%Finding points below EP
P=EP; r=P;
for backward =1:n
J= ILDM1D_jacobian (F,r);
b = ILDM1D_slowest (J);
r = P + h*b;
Q= ILDM1D_projection (J);
ILDM1D_manifold (Q,F,r);
allBackward(backward,:) = P;
%allBackward matrix stores all backward points
end
%Plotting
plot (EP (1),EP (2), ‘ks’);hold on;
plot (allForward(:,1),allForward(:,2),’+k-‘);
plot (allBackward(:,2),allBackward(:,2),’*k-‘);
grid on;
axis ([0 1 0 .7]);
xlabel (‘A’); ylabel (‘AZ’);
legend(‘Eguilibrium point’, ‘Forward direction’, ‘Backword direction’) manifold, plot, intrinsic low dimensional manifold, chemical kinetics MATLAB Answers — New Questions