[wdmaudiodev] Re: Default Sampling Rate

  • From: Matthew van Eerde <Matthew.van.Eerde@xxxxxxxxxxxxx>
  • To: "wdmaudiodev@xxxxxxxxxxxxx" <wdmaudiodev@xxxxxxxxxxxxx>
  • Date: Tue, 8 May 2012 15:43:36 +0000

It is normal that devices would not support KSPROPERTY_PIN_PROPOSEDATAFORMAT.  
If you download the Windows Driver Kit from Microsoft Connect, you can use KS 
Studio to see what properties any driver supports on various pins.

From: wdmaudiodev-bounce@xxxxxxxxxxxxx 
[mailto:wdmaudiodev-bounce@xxxxxxxxxxxxx] On Behalf Of Anton Shpakovsky
Sent: Monday, May 7, 2012 9:53 PM
To: wdmaudiodev@xxxxxxxxxxxxx
Subject: [wdmaudiodev] Re: Default Sampling Rate


I guess this property is unsupported by most devices, correct?
The fact is, that we've tried our code on several sound cards, but no success.

Obtaining the devicepath is correct for sure because we are able to request 
some other properties from the device.
Please, take a look at the code below.


// filling the KSProp structure

// and the KSPin structure
KSPin.PinId = 0;
KSPin.Property = KSProp;
KSPin.Reserved = 0;

// negotiating to the driver
hDevice = CreateFile(tczDevicePath, GENERIC_READ,

// requesting KSCOMPONENTID structure
            BOOL res = DeviceIoControl(hDevice, IOCTL_KS_PROPERTY,
                                                            &dwBytesRet, 0 );
            DWORD err = GetLastError();
// the error code is 1168 which stands for "ERROR_NOT_FOUND 1168 (0x490) 
Element not found."

We also tried to request KSPROPERTY_PIN_CTYPES data and it returned two pins 
The first was identified as "Recording Control" while the second appeared to be 

Are we doing something wrong or it's normal that some devices may not support 
this property?
Please advice.

Btw, from where it all started. We are implementing acoustic echo cancellation 
for our product and we found out
that the best result we can get only on specific frequencies that appear to be 
the native ones. When we request e.g. 22050Hz from the soundcard,
we guess a hardware resampling may occur because the sound card returns strange 
amount of samples per second - sometimes less than 22050, sometimes more. E.g. 
the 48000Hz works perfect so we though it may be the native frequency for the 

Thanks in advance,

On Mon, Apr 30, 2012 at 11:27 PM, Matthew van Eerde 
<Matthew.van.Eerde@xxxxxxxxxxxxx<mailto:Matthew.van.Eerde@xxxxxxxxxxxxx>> wrote:

If the device supports GET requests of the KSPROPERTY_PIN_PROPOSEDATAFORMAT 
property, you can assume that this is the device's preferred format.  If the 
device supports this, Windows will (usually) have that be the default format 
(though the user can still change the default to a different format.)

If the device does not support GET requests of the 
KSPROPERTY_PIN_PROPOSEDATAFORMAT, then it comes down to figuring out what 
formats the device supports.  Windows has a list of formats that it will try, 
and the device can expose KS data ranges with specific sample rates.

In addition to the below, you may come across devices that natively support 8 
kHz (so you will not need a conversion at all.)

From: wdmaudiodev-bounce@xxxxxxxxxxxxx<mailto:wdmaudiodev-bounce@xxxxxxxxxxxxx> 
[wdmaudiodev-bounce@xxxxxxxxxxxxx<mailto:wdmaudiodev-bounce@xxxxxxxxxxxxx>] on 
behalf of Finn William-A19681 
Sent: Monday, April 30, 2012 7:10 AM
To: wdmaudiodev@xxxxxxxxxxxxx<mailto:wdmaudiodev@xxxxxxxxxxxxx>
Cc: Finn William-A19681
Subject: [wdmaudiodev] Default Sampling Rate
Hi All,

I am not developing an audio device driver, but rather I am interested in 
finding out the default sampling rate for a given device. I am developing an 
application that must operate on 8 kHz sampled data. I currently support 
resampling to/from 48k, 44.1k, 22.05k, and 16k endpoint sampling rates. 
However, during testing, I came across a device using a 32 kHz sampling rate, 
and I want to figure out if a user intentionally changed the sampling rate to 
32 kHz or if it was the default sampling rate at installation.

TL;DR: Is there an easy way to find the default sampling rate for a given 
input/output device in Windows?


Best Regards,
Anton Shpakovsky

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