A newbie first question: fpga synchronous input
Hi all
this is my first time to send a question at this forum
it is also the first time to use Labview, I am totally newbie
My problem is:
I have Xilinx Spartan 3A FPGA with a design runs at 10 MHz " derived from the 50 MHz of the On-board clock"
I want to test my design by appling inputs to the FPGA and watching the output
so how to apply predefined synchronous inputs that are synchronized to the my design clock
thanks in advance
Good Afternoon msaleh87,
What supporting hardware are you using with your Xilinx Spartan 3A FPGA? The only NI hardware that the Xilinx Spartan 3A FPGA is found in are the sb-RIOs and the cRIO-9072 and cRIO-9074.
What Xilinx FPGA Chips Are Used by National Instruments RIO
Devices?
http://digital.ni.com/public.nsf/allkb/ED6FC9CF7B983CFD86256DCE0072E313?OpenDocument
Please elaborate on what you are trying to accomplish with this project. Also, please give some more details (such as source, speed, type, etc.) of your predefined synchronous inputs.
Regards,
Charlie Piazza
Staff Product Support Engineer, RF
National Instruments
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Hi, I have a Source Synchronous LVDS DDR input into a Kintex7, the launching clock is edge-aligned to the data and capture clock should capture on opposite edge (a launch on the rising edge should be captured by the falling edge). I have designed it to work at 100Mhz by compensating the clock insertion delay with a PLL (to save the MMCM for other purposes) using BUFH (the timing is not so tight to use BUFIO/BUFR). The PLL also centers the opposing edge on the data window by shifting -90°. Now the launching clock waveform is {5.0 0.0}, and the waveform generated by the PLL is {2.5 7.5}, this is reported correctly by Vivado. But when vivado analyses the setup path (I have set the proper set_input_delays) the following happens:
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Hello everyone,
Suppose I have a method which is a synchronized method (for example, method Foo in the following code block), and in this synchronized method another method which is not synchronized is invoked (for example, method Goo in the following code block). I am wondering when executing in Goo from Foo, whether we still have the synchronized feature (i.e. obtain the lock of "this" object).
synchronized public Foo()
Goo();
private Goo()
}Thanks in advance,
GeorgeKaj,
Hi, pls see my post above - I replied to the same question in the other forum and I was pretty sure about my answer until I found it different from ur reply here.
Here is my reply. Can u kindly guide me to what's wrong in it ?
if a lock on a sync-ed method extends to all methods being called from it, then the jvm (bcos it has no means of knowing until runtime the method calls from a sync method) has to lock all the methods in a class that has atleast one sync method.
What this would mean potentially is that the effect of sync-ing one method would be equal to sync-ing all the methods bcos any of the other methods could be potentially called from the sync-ed one.
The example below is co-erced, but serves to illustrate the point.
public class Test
synchronized public void Foo()
Goo();
synchronized public void Moo()
Goo();
notify();
private void Goo()
System.out.println(Thread.currentThread().getName()+" accessing goo");
try
if(Thread.currentThread().getName().equals("Th one"))
//force thread1 to wait , lets see if thread2 can access it while thread1's waiting on it
wait();
catch(Exception e){}
System.out.println(Thread.currentThread().getName()+" returning from goo");
//first thread
public class MyT implements Runnable
Test t;
public MyT(Test t)
this.t = t;
/* (non-Javadoc)
* @see java.lang.Runnable#run()
public void run()
t.Foo();
//second thread
public class MyT2 implements Runnable
Test t;
public MyT2(Test t)
this.t = t;
/* (non-Javadoc)
* @see java.lang.Runnable#run()
public void run()
t.Moo();
//Main
public class Main
public static void main(String[] args)
Test t = new Test();
Thread th = new Thread(new MyT(t),"Th one");
Thread th1 = new Thread(new MyT2(t), "Th two");
th.start();
th1.start();
And the o/p I got was
Th one accessing goo/*thread one in goo, is in wait() coz of the funny if clause there*/
Th two accessing goo /*there's no lock on goo, else with thread one in wait(), this wouldve been impossible*/
Th two returning from goo/*Thread 2 is over with Goo(), will go back and notify Thread 1*/
Th one returning from goo/*Thread 1 is notified, wakes up and returns :-)*/
tx,
ram.
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