random signal delay

2012 2 27
2 个回答
更新时间：2013 10 31
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0 个投票

clc
clear all
close all
fs=10000; % sampling frequency in Hz
f1=500; % frequency of signal in Hz
f2=200;
f3=700;
t_duration=1; %duration of LONG signal in seconds
t = 0:1/fs:t_duration-1/fs; %sampling points
x = sin(2*pi*f1*t)+cos(2*pi*f2*t)+sin(2*pi*f3*t);
d=0.000079734452; % time delay in seconds
I can delay it mathematically as follows:
s1=sin(2*pi*f1*(t-d))+cos(2*pi*f2*(t-d))+sin(2*pi*f3*(t-d));
How can I do the same (mathematically) if:
x=randn(size(t));
I need to produce the same set of random values each time. I have gone through doc rng; but Iam unable to do it.
Please illustrate with an example.
thank you.

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回答（2 个）

Honglei Chen
Honglei Chen 2012-2-27

0 个投票

There are two questions in your post so I'm not sure if I understand exactly what you look for. I'll include my thoughts below:
1. To get the same random value each time, set a seed using rng, for example
rng(1);
randn(10,1);
rng(1);
randn(10,1);
You should get the same random values for two trials.
2. Now if you want to delay the random signal, you can always append zero in front. The number of zeros you insert depends on your sampling frequency and required delay.
x = randn(10,1);
x_delay = [0;x]; % delayed by 1 sample

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zozo
zozo 2012-2-27
Sir, my problem is delay being a non-integer multiple of sampling frequency.
As mentioned in my code,
d=0.000079734452; % time delay in seconds
how can I implement this mathematically into 'randn' like I did with 's1' above?
(Note: Iam doing this to cross check if my frequency domain delaying technique works perfectly for any random signal and any random delay 'd').
Honglei Chen
Honglei Chen 2012-2-27
I suspected a little for this but I wasn't sure. Given this is what you want to do, I have a question for you. What do you think the delayed signal should be? Now this is a signal you created out of randn, which is a white noise. By definition, the value at the next moment has nothing to do with the value at the current moment, so I would argue that the delayed version is a complete different randn vector. Therefore, IMHO, I don't think this is a good test signal.
zozo
zozo 2012-2-27
Ok..Can you suggest an alternative test, apart from the one I used above in my code?
Honglei Chen
Honglei Chen 2012-2-27
Personally I think your test case is fine, if you really want, you can try a triangle signal or even a rectangular signal. Just be prepared that there will be distortion in the result.

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vijay
vijay 2013-10-31

0 个投票

do u have modified code for this

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zozo
zozo
2012-2-27

回答：

vijay
vijay
2013-10-31

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