Hello! I would like to implement my own biquad, and to do that I need a delay of one and two samples. How do I get a one delay sample? As I have understood from the manual the way to do that is to set [block~ 1] and then use [delwrite~] and [delread~] in some way that I haven't figured out yet. Isn't there an easier way?

In jMax the way to do it is simply [delwrite~ del_name S 10] where the S stands for sample and I therefore get a delayline of 10 samples. Easy. Thats what I want to do in pd.

Any help?

/Ronny

in pd [delread~] is specified in milliseconds, so you just convert that to samples. the [samplerate~] external is essential here.

ok, but that is what I want to avoid. Specifying a delay of 1/44100=22.6757...microseconds feels quite inexact. How can I really be shure that I get one sample delay, it would depend on the roundoff in the time specified. I could get x[n] or x[n-1] depending on the accuracy of the delay. Hmm, do you get my point?? I guess that this is where [block~ 1] comes into the picture also.

In order of preference:

[biquad~] exists (Pd internal), you can feed it cooefficients of your choice, there are abstractions somewhere to calculate coefficients for different types of filter

[fexpr~] is an external that comes with Pd, lets you build all kinds of IIR filters

[z~] is an external (part of Zexy I think) that does sample-wise delay, so [z~] would be a one-sample delay.

If you're using [block~ 1], you could use [s~] and [r~] for feedback which will introduce a delay of one block (in this case a single sample). That doesn't help for the straight delay (without feedback) part, though.

Hi!

To make one-sample delay feedback I've tried [s~] and [r~] with [block~ 1].
It works, but PD (0.42.5-ext) says:

error: receive~ 1007.d: vector size mismatch
error: sigsend 1007.d: unexpected vector size

WTF?

Have you also looked at [rpole~]? (see the help patch for more info)

It's output is: y[n] = x[n] + a[n] * y[n-1]

So if you set the coefficient to 1 and added the output to an inverted version of your input you'll get a one sample delay, i.e.

[osc~ 440]
|\
| \
| \
| \
| \
| \
| \
| \
| \
| \
[rpole~ 1] [*~ -1]
| /
| /
| /
| /
| /
[+~ ]

Disclaimer, this is off the top of my head, apologies if it doesn't work.

Cheers,
Joe

Ahh crap sorry that won't work because it'll give you the previous output sample instead of the input sample. Hmmmm

Use [delwrite~] and [delread~] instead of [send~] and [receive~] to avoid those errors.

[tabsend~] and [tabreceive~] will also work for arbitrary block sizes

delwrite~/delread~ with 1000/samplerate delaytime works, but using cpole/zero is probably faster. there are biquad-filters using cpole/zero in frank barknechts rj-lib, or you can get a poles position from the coefficients with r=-a1*0.5, i=sqrt(a2-(a1*0.5)^2). (or b instead of a, depending on how your coefficients are named).

or you could use tabsend~ and bang~, and do all computations
using msg-objects... (i have not tried this, it is probably not very efficient)

bang~ has a 64 sample limit though, doesn't it? even if you set a lower block size, it still only bangs every 64 samples minimum.

that is right, but you can do whatever you like (64->until,...) on the 64sample
(or bigger) buffer you get at bang~.

oneSampleDelay.pd

revisiting this thread. Joe nearly had it...but instead of rpole~, using rzero~ you can get one sample delay

better to use cpole~ and czero~ to make your biquad, see [pd biquad~] in https://github.com/porres/LiveElectronicsTutorial/blob/master/Parte 6 - Filtros e Reverb/30.Filtros-II/3.Plano-Z/8.[biquad~].pd