Scaling the values to specific range

Q=[ 0.0669
0.0383
0.0029
-0.0344
-0.0554
-0.0459
-0.0316
0.0086
0.0392
0.0650
0.0698
0.0478
0.0201
-0.0134
-0.0468
-0.0583
-0.0468
-0.0229
0.0115]
I want to scale the value of Q in range [-(2^13-1),(2^13-1)]
please help

6 个评论

Both [(2^13-1),(2^13-1)] are same.
sorry harshit
[-(2^13-1),(2^13-1)]
What limits, if any, are there on the input range of values?
I notice that all of your values, positive and negative, have 0 immediately after the decimal point. Is that characteristic? Should it be assumed that abs() of the values will be strictly less than 0.1 and that the output representation should take that into accout?
acutally walter those are output after performing dwt2 on a audio signal
I can't be bothered to research dwt2() to find out what the possible theoretical output ranges are for each possible class of input data. How about if you just state the value ranges and any specific encoding instructions ?
walter these are coefficients from dwt2
S=[ 0.0038
0.0019
0.0105
0.0048
0.0038
-0.0105
-0.0134
-0.0191
-0.0258
-0.0229
-0.0086
-0.0076
-0.0086
-0.0010
0.0038
0.0048
0.0019
-0.0038
-0.0086
-0.0029
0.0038
0.0086
0.0096
0.0124
0.0057
0.0067
-0.0010
0.0029
0.0086
0.0057]
from this i have to sample in that range i.e in 14 bit range .
for 16 bit i have performed
D1=int16(S*32768);please tell how to perform for 14 bit

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 采纳的回答

Harshit
Harshit 2012-11-21
D1=int16(S*32768) maps your any integer S*32768 outside the range to 32768 or for S>1 the output will be 32768(2^15) otherwise it will be round off to the nearest integer(You are assuming maximum non saturating input is 1 beyond this there will be saturation). You can see simply use max(round(2^13*S),2^13). Hope it will be fine.

13 个评论

I tried and i get all values as 8192 only
Sorry pat it should be min(round(2^13*S),2^13))
Max always be 8192 since 2^13 is always greater than s*2^13 when s<1
the output for 16 bit ranged from -32768 to 32768,same way i want to do for 14 vit from -8192 to 8192,i.e sampling the signal into 14 bit
It is. see D1=int16(S*32768) 32768=2^15 and that is why I am using 2^13. Also int16 just limit the value to 2^15 and my min is doing the same.
Check output for your S it is not -32768 this value is achieved for S=-1,1.
plz chk this
clc clear all [y,Fs,nbits] = wavread('bart.wav');
[cA,cH,cV,cD] = dwt2(y,'haar');
D1=int16(cA*32768);
min(D1) is -32768 max(D1 ) is 32767
clc clear all [y,Fs,nbits] = wavread('bart.wav');
[cA,cH,cV,cD] = dwt2(y,'haar');
W=min(round(2^13*cA),2^13);
min(W)=-12844
max(W)=8192
is it correct harshit
Ok finally here it is max(min(round(2^13*S),2^13),-2^13). The thing was that we didn't place a limit on negatives which I now did.
THANKS LOT HARSHIT ,finally i got it but one thing i get values for
cA as -8192 to 8192;
but for cH,cV,cD i get other values such as ,-78 71,-313 321,are these also correct
ok thanks a lot harshit my final question,i have performed operation on cA,if i am usind idwt2,have i to perform same operation for ch,cV,cD for effictive reconstruction,else performing on cA is enough
I will say you to do it because fixed point operations are done similarly.

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更多回答(1 个)

Qscaled = 2^13-1;
(Your permitted range consists of exactly one value, so all numbers must scale to that one value.)

2 个评论

sorry walter it is
[-(2^13-1),(2^13-1)]% a signed 14 bit
Jan
Jan 2012-11-21
编辑:Jan 2012-11-21
@Pat: Instead of adding comments, fixing this error by editing the question would be less confusing for the readers. So please fix this.

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