what is mean by 44k bit sram in a pt2399 echo chip?

[ballfire]

hello ...

i read about the pt2399 echo chip characteristics in a datasheet..,what is mean by that 44k bit sram?

thanks.

[DrNomis]

hello ...

i read about the pt2399 echo chip characteristics in a datasheet..,what is mean by that 44k bit sram?

thanks.

That's just a memory location inside the chip, it's like the ram sticks you have installed in your computer, in this case the sram can hold a maximum of 44,000 bits of info in it before it is full.....

[Dirk_Hendrik]

45056 bytes to be exact Simon. A "digital kilo" is 1024.

[ballfire]

hello..

thanks for the reply..good explanation..i appreciate it.

but..How?.. can you explain it when you using it as a diy delay pedal like the rebote delay pedal? do you mean that if you put some input from your guitar like playing some notes it feeds the memory 44,000 times? and the output from the amplifier creates many delays? or do you mean by the 44,000 is the count of feedbacks??

thanks..

[DrNomis]

45056 bytes to be exact Simon. A "digital kilo" is 1024.

Ah..right, thanks for correcting me Dirk....

I keep forgetting that a digital 1000 = 1024, my brain was obviously having a hissy fit yesterday mate...

[DrNomis]

hello..

thanks for the reply..good explanation..i appreciate it.

but..How?.. can you explain it when you using it as a diy delay pedal like the rebote delay pedal? do you mean that if you put some input from your guitar like playing some notes it feeds the memory 44,000 times? and the output from the amplifier creates many delays? or do you mean by the 44,000 is the count of feedbacks??

thanks..

No, actually what happens as far as I can tell, is that first the signal from the guitar is converted to digital via what's called an Analogue-To-Digital Converter, this digital code is stored in the 44kb sram, and then fed into a Digital-To-Analogue Converter where it is converted to an analogue signal which is played back to create the delay...

Maybe another member could possibly clarify things....

[mysticwhiskey]

Read up on analog-to-digital conversion. Here's my (possibly flawed) simplistic take on it:

Digital effects processors don't work on an analog signal, such as what comes out of your electric guitar or microphone. The first step is to convert the incoming analog signal to a digital representation. When converting an analog signal to digital, the analog signal's voltage is 'sampled' and converted to the closest digital representation. You need two parameters to do this: sampling resolution and sampling rate.

* The sampling resolution is how fine the digital representation of the signal is, and this is measured in 'bits'. Common values are 16, 24 and 48 bits. A higher value here means a more accurate representation, but at the cost of more storage (memory) required.

* The sampling rate is how often the analogue signal is sampled, and is measured in Hertz (typically kilohertz). A higher value here also means a more accurate representation of the original analog signal, and likewise at the cost of more storage.

For example, assuming an 8-bit 16kHZ sampling resolution and rate, this requires 8 * 16000 = 128000 bits = 125 Kbits of storage space for a 1 second sample.

So, now we know that a digital representation of analog audio needs a certain amount of bits, and it's these bits that are stored in memory (RAM). Getting back to the digital delay pedal: it's continually sampling the incoming audio and will replay it back after a delay, which is a variable period set by the 'delay' control. The pedal needs to store the last 'x' seconds of audio before it replays, so it needs a certain amount of memory to store that digital audio. The amount of memory required not only depends on the sampling rate and resolution, but also on how long the delay is:

* Sample resolution
* Sample rate
* Delay time

Total Storage Required (bits) = Resolution (bits) * Rate (Hz) * Time (seconds). The amount of memory you have available determines the constraints of the above three parameters.

The first two parameters are usually constrained by the ADC (analog to digital converter) chip that you're using. For the PT2399 this is built into the chip, and without referring to the data sheets I'm not sure what resolution and rate it uses. But for argument's sake, say it has a sample resolution of 8 bits, and a rate of 8KHz. Given your 45,056 bits of memory, the only unknown in the above equation is the maximum delay time:

Time = 45056 / (8 * 8000) = 0.704 seconds

So what does it all mean? I guess it boils down to this: The more memory you have available, the longer delay times you can acheive and/or you can have increased quality of audio. The less memory you have, either the delay time suffers or the quality of the delayed audio does.

[ballfire]

hello..

very well done explained..thanks .. i understand now how this thing works..

thank you very much..

[Niquel]

Hello,

I am new here. I know it is an old threat, but maybe someone could help me.
I was looking for its sample rate and resolution, which are not specified on the only datasheet I could find, and have found the previous post from mysticwhiskey, very interesting. But does anybody knows these specs?
Thanks in advance.

Miquel

[Dirk_Hendrik]

Sampling rate is the FCK value given in the delay times table. By altering the sampling rate the delay time changes.

Resolution? Indeed not given.

[Niquel]

Thank you Dirk.
Cheers!