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upsampling and adding delay to a sequence

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PChoppala 2012-1-18

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采纳的回答： Walter Roberson

在 MATLAB Online 中打开

I have a simple query.

Suppose i have a signal s = randn(1,10), if I want to add delay 100ms to this signal, what should i do?

I heard about upsampling the signal and then adding zeros, but can you help me with an example code

s = randn(1,8);
delay = 100ms;
s1 = upsample(s,?);
s_delay = [how many zeros, s1]?

Is that approach right?

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Walter Roberson 2012-1-19

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It is not possible to add a delay to a signal unless you know what the sampling frequency is.

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Okay, here is my actual prob

load handel.mat;

hfile = 'handel.wav';

[y,Fs] = wavread(hfile); % this will give me a broadband signal with 73113 samples and Fs = 8192;

I want to add a delay of 1.3 ms to this 'y'. How do I do that?

I assume I need to pad fix((Fs*1.3e-3)) zeros in the front, is that right?

In this case

fix(Fs*1.3e-3)% =10, so i need to pad 10 zeros in the beginning

x = vertcat(zeros(10,1), y(1:length(y)-10));

Is 'x' the 1.3ms delayed signal?

But I gather there is another method...wherein, we upsample the signal and low pass filter it...can you please help me with that?

Cheers

Walter Roberson 2012-1-19

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Express the intervals as fractions.

1.3 ms is 13/10000 .

Fs is 1/8192

solve as a fraction:

13/10000 = x/8192

which gives x = 6656/625

Cross-check: (1/10000)*(13*625) = 6656*(1/8192) gives 13/16 = 13/16 which is true.

Therefore, the signal must be upsampled by a factor of 6656, after which 1.3 ms will correspond to 13*625 = 8125 zeros.

Adding fix((Fs*1.3e-3)) zeros will *not* get you a 1.3 ms delay: it will get you a 1.22 ms delay, which is not a good enough approximation for sound processing purposes.

PChoppala 2012-1-19

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Agh, I understand it now. Thank you very much.

Please allow me to ask one more query.

1. The signal in the above quoted case 'y' has 73113 samples.

If I use

x = upsample(y,6656), I get the error

'Maximum variable size allowed by the program is exceeded'

What to do about that?

2. Fs is 8192, not 1/8192. Is it a mistake you typed in the 3rd statement?

3. Of the topic, how can we find the sampling frequency of a signal of the form s = randn(1,1000)?

Please help me as I am struggling with this concept for a couple of days.

Thanks a lot

Walter Roberson 2012-1-19

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2. Yes, sorry, the sampling interval is 1/Fs, 1/8192

1. I am mildly surprised that upsample said that the maximum variable size allowed by the program was exceeded, as the result would only be 3.6+ gigabytes, which would fit in to 32 bit memory if only everything else was out of the way. I would have thus expected "out of memory" instead of maximum variable size. In any case, Yup, it's too big for you to compute using upsample() unless you switch to the 64 bit version of MATLAB.

What to do about it? The usual way is to use a fractional delay filter instead of using upsample.

3. There is no sampling frequency for a signal defined by its samples. If someone from another of our departments were to send me a data file with 1000 samples, I would not know whether it was a measurement off of one of our attosecond (1E-18 second) lasers, or a measurement that had been done once a week for about 21 years.

Walter Roberson 2012-1-19

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Yup. But in your "actual prob" from your first comment, you do not have a random signal and you know the Fs from the wavread() command, so the question of sampling frequency for randn(1,1000) is not relevant to that situation.

PChoppala 2012-1-19

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Hi again,

Bear with me this one time.

1. Yeah, Fs for randn is not relevant. Sorry.

2. I read the document on design.fracdelay, but could not understand how to upsample a sequence or add delay to 'y', in the above stated context. I have not worked on filters or used the command.

Can you just give me an example of how to use the command to upsample or add delay?

I can start my coding once I can do that.

Please.

and thanks for the method, it will be a big relief for me, if I can get this going today!

Cheers

PChoppala 2012-1-19