If your signal is a pure sine then the reported rms should be around 0.707*amp. Here’s a printout of
a sine  wave with amp = 1, every 0.1 secs

 i   1 time     0.00227:     0.25806
 i   1 time     0.10204:     0.71042
 i   1 time     0.20181:     0.70922
 i   1 time     0.30159:     0.70452
 i   1 time     0.40136:     0.70336
 i   1 time     0.50113:     0.70735
 i   1 time     0.60091:     0.71098
 i   1 time     0.70068:     0.70929
 i   1 time     0.80045:     0.70458
 i   1 time     0.90023:     0.70333
 i   1 time     1.00000:     0.70728

once it settles down, it’s around 0.707.

So the rms relation to peak amp will depend on the waveform shape. A spiky signal will have a lower rms 
relatively to its peak amp than a “fuller” signal such as the sinewave. A pulse wave will for instance get
these values

 i   1 time     0.00227:     0.03676
 i   1 time     0.10204:     0.09732
 i   1 time     0.20181:     0.09803
 i   1 time     0.30159:     0.09869
 i   1 time     0.40136:     0.09943
 i   1 time     0.50113:     0.10011
 i   1 time     0.60091:     0.10070
 i   1 time     0.70068:     0.10157
 i   1 time     0.80045:     0.10214
 i   1 time     0.90023:     0.10284
 i   1 time     1.00000:     0.09737

The key thing here is that you are measuring an average of the amplitude over a (short) timespan, which
is different from measuring the max positive (or absolute) amplitude of the signal.

Victor
On 19 Dec 2013, at 15:53, Jacques Leplat  wrote:

> Hello CSounders,
> 
> I’ve been banging my head with the rms opcode. And wonder if someone could point me in the right direction.
> 
> The csd uses an “output instrument” to do the one and only actual “outs” in the csd.
> 
> ;-- Output Instrument
> instr outInstr
> 
> kfactorL = sin((gkPan/100) * $M_PI_2) * (gkVolume / 100)
> kfactorR = cos((gkPan/100) * $M_PI_2) * (gkVolume / 100)
> gaoutL = gaoutL * kfactorL
> gaoutR = gaoutR * kfactorR
> 
> outs gaoutL,gaoutR
> 
> kvl rms gaoutL
> kvr rms gaoutR
> if gklevelL < kvl then
>     gklevelL = kvl
> endif
> if gklevelR < kvr then
>     gklevelR = kvr
> endif
> printk2 gklevelL
> printk2 gklevelR
> gaoutL = 0
> gaoutR = 0
> 
> endin
> 
> When run, The printk2’s output lots of very small values (below 1.0), and yet the console also reports output values, which are much higher, 1.19064 being the highest.  And yet they are all below 1.0.
> 
> I have looked at the shapes.csd example, it does something like “gklevel = gklevel + ks”. Were I to add the k rate levels in the output instrument, I do not know when to stop adding.
> 
> This leads me to believe that I am missing a fundamental point here, I was expecting at least one of the printk2’s to be of the same magnitude as that reported by the console (i.e.: up to 1.19064).
> 
> Could someone please give me an idea where I should look?
> 
> 
> new alloc for instr 1:
> new alloc for instr 1:
> new alloc for instr 1:
> new alloc for instr outInstr:
> i2 0.00000
> i2 0.00000
> i2 0.00129
> i2 0.00129
> i2 0.00348
> i2 0.00348
> i2 0.00865
> i2 0.00865
> i2 0.01284
> i2 0.01284
> i2 0.02594
> i2 0.02594
> i2 0.04014
> i2 0.04014
> i2 0.04670
> i2 0.04670
> i2 0.06009
> i2 0.06009
> i2 0.07767
> i2 0.07767
> i2 0.09349
> i2 0.09349
> i2 0.10157
> i2 0.10157
> i2 0.10702
> i2 0.10702
> i2 0.12628
> i2 0.12628
> i2 0.13595
> i2 0.13595
> i2 0.14677
> i2 0.14677
> i2 0.16026
> i2 0.16026
> i2 0.17295
> i2 0.17295
> i2 0.18748
> i2 0.18748
> i2 0.20885
> i2 0.20885
> i2 0.24224
> i2 0.24224
> i2 0.25734
> i2 0.25734
> i2 0.30191
> i2 0.30191
> i2 0.31421
> i2 0.31421
> i2 0.34097
> i2 0.34097
> i2 0.34517
> i2 0.34517
> i2 0.35635
> i2 0.35635
> WARNING: Buffer overrun in real-time audio input
> i2 0.36647
> i2 0.36647
> i2 0.38649
> i2 0.38649
> i2 0.38656
> i2 0.38656
> i2 0.39982
> i2 0.39982
> i2 0.40219
> i2 0.40219
> i2 0.41401
> i2 0.41401
> B 0.000 .. 0.183 T 0.183 TT 0.183 M: 1.19064 1.19064
> number of samples out of range: 6 6
> WARNING: Buffer overrun in real-time audio input
> B 0.183 .. 0.344 T 0.344 TT 0.344 M: 0.83548 0.83548
> B 0.344 .. 0.510 T 0.510 TT 0.510 M: 0.60467 0.60467
> B 0.510 .. 0.759 T 0.759 TT 0.759 M: 0.52492 0.52492
> WARNING: Buffer overrun in real-time audio input
> B 0.759 .. 1.006 T 1.006 TT 1.006 M: 0.52469 0.52469
> B 1.006 .. 1.171 T 1.171 TT 1.171 M: 1.11965 1.11965
> number of samples out of range: 1 1
> B 1.171 .. 1.331 T 1.331 TT 1.331 M: 0.71890 0.71890
> B 1.331 .. 2.000 T 2.000 TT 2.000 M: 0.55596 0.55596
> Score finished in csoundPerformKsmps().
> inactive allocs returned to freespace
> end of score.	 overall amps: 1.19064 1.19064
> overall samples out of range: 7 7
> 0 errors in performance
> Elapsed time at end of performance: real: 3.324s, CPU: 0.406s
> closing device
> 87 2048 sample blks of 64-bit floats written to dac
> 
> 
> 

Thanks Victor,

After reading your reply it seemed worthwhile to try out max_k instead of rms. The result compared to rms is enlightening (to me anyway).

All the best,

Jacques

On 19 Dec 2013, at 16:45, Victor Lazzarini wrote:

> If your signal is a pure sine then the reported rms should be around 0.707*amp. Here’s a printout of
> a sine  wave with amp = 1, every 0.1 secs
> 
> i   1 time     0.00227:     0.25806
> i   1 time     0.10204:     0.71042
> i   1 time     0.20181:     0.70922
> i   1 time     0.30159:     0.70452
> i   1 time     0.40136:     0.70336
> i   1 time     0.50113:     0.70735
> i   1 time     0.60091:     0.71098
> i   1 time     0.70068:     0.70929
> i   1 time     0.80045:     0.70458
> i   1 time     0.90023:     0.70333
> i   1 time     1.00000:     0.70728
> 
> once it settles down, it’s around 0.707.
> 
> So the rms relation to peak amp will depend on the waveform shape. A spiky signal will have a lower rms 
> relatively to its peak amp than a “fuller” signal such as the sinewave. A pulse wave will for instance get
> these values
> 
> i   1 time     0.00227:     0.03676
> i   1 time     0.10204:     0.09732
> i   1 time     0.20181:     0.09803
> i   1 time     0.30159:     0.09869
> i   1 time     0.40136:     0.09943
> i   1 time     0.50113:     0.10011
> i   1 time     0.60091:     0.10070
> i   1 time     0.70068:     0.10157
> i   1 time     0.80045:     0.10214
> i   1 time     0.90023:     0.10284
> i   1 time     1.00000:     0.09737
> 
> The key thing here is that you are measuring an average of the amplitude over a (short) timespan, which
> is different from measuring the max positive (or absolute) amplitude of the signal.
> 
> Victor
>