**You are now following this question**

- You will see updates in your followed content feed.
- You may receive emails, depending on your communication preferences.

# Received signal not on sampling time

6 views (last 30 days)

Show older comments

Hello all,

I have a sequence d of length N samples each of duration Ts. The channel I have will scale the received samples by a factor (1+a), which means that the received samples will not be on multiples of Ts but on multiples of Ts*(1+a). How can I capture this effect in MATLAB?

Thanks in advance

##### 0 Comments

### Accepted Answer

Star Strider
on 24 Apr 2014

##### 54 Comments

S. David
on 24 Apr 2014

Thanks for replying.

I don't want to resample a signal, but instead to transmit a vector of samples over a channel that scale the sample duration only. In other words, given a sequence d of N samples each of duration Ts, how to generate a received sequence of N samples each of duration Ts*(1+a)?

Thanks

Star Strider
on 24 Apr 2014

I’m still not certain I understand, so see if this does what you want:

s = repmat([zeros(1,10) ones(1,10) zeros(1,10)],1,5);

t = linspace(1,100,size(s,2));

a = 0.25;

t2 = t*(1+a);

figure(1)

subplot(2,1,1)

plot(t,s)

axis([min(t2) max(t2) ylim*1.1])

title('Sent')

grid

subplot(2,1,2)

plot(t2,s)

axis([min(t2) max(t2) ylim*1.1])

title('Received')

grid

If it doesn’t, please explain how it doesn’t.

S. David
on 24 Apr 2014

OK, let us begin again. Suppose I have the following code segment

Ts=10^-4;

N=512;

d=rand(N,1);

t=0:Ts:N*Ts;

and I want to make the sequence d aligned with sequence t, how can I do that? Can I use something like:

dt=interp1(t,d,t,'linear');

? If so, I can say with scaling it will be then:

dt=interp1(t,d,t*(1+a),'linear');

But is this true?

Star Strider
on 24 Apr 2014

I don’t understand t*(1+a). To me, it looks like you simply want to expand the signal, essentially what I did.

Do you want to sample the signal (1+a) more frequently, less frequently, (both look like simple resampling to me), or shift the signal by (1+a)?

Star Strider
on 25 Apr 2014

I understand about frequency = propagation velocity/ wavelength, and that different water densities would affect both, but I have no idea which of these (frequency or wavelength or both) dilates or compresses.

What about this is not just resampling on a different time base to account for changing frequency or wavelength? It seems to me that the frequency would increase (in denser water) by (1+a) and the wavelength would decrease by 1/(1+a). Upsample with interpolation for lower frequencies, or downsample with decimation for higher frequencies, and you’re there. The resample function will do either of these.

S. David
on 25 Apr 2014

Actually, a=v/c where v is the relative velocity between transceivers and c is the sound speed underwater. So, if the transmitted signal is s(t), the received signal then will be proportional to s(t[1+a]). So, it is more time compression/dilation rather than simple frequency shift.

Hence, I need to generate this received signal s(t[1+a]) and then do resampling at t/[1+a] to counteract the compression/dilation before detection.

Star Strider
on 27 Apr 2014

S. David
on 28 Apr 2014

Edited: S. David
on 28 Apr 2014

Unfortunately, I don't have any plot of what in mind, simply because I just know the math but not the MATLAB implementation.

Are you familiar with digital communications, how we generate digital data at specific rate fs=1/Ts, convert them to analog, transmit them over a communication channel, and then sampling it at the same rate fs to get the transmitted digital data?

Star Strider
on 28 Apr 2014

Star Strider
on 28 Apr 2014

Star Strider
on 28 Apr 2014

Edited: Star Strider
on 28 Apr 2014

Star Strider
on 1 May 2014

Did it solve your problem?

S. David
on 12 May 2014

Star Strider
on 12 May 2014

Here’s an example using interp1. I’m still not certain I understand what you’re doing, but this may help. Please experiment with it. Does it do what you want?

t = linspace(0, 4*pi, 35);

s = sin(t)+cos(3*t);

figure(1)

plot(t, s)

grid

t2 = linspace(0,5*pi,50);

s2 = interp1(t, s, t2, 'pchip', 'extrap');

figure(2)

plot(t2, s2)

axis([xlim -2 2])

grid

S. David
on 12 May 2014

I think it will resemble my case more if the time range changes while the number of points be fixed. Like this:

t = linspace(0, 4*pi, 35);

s = sin(t)+cos(3*t);

figure(1)

plot(t, s)

grid

t2 = linspace(0,4*pi*(1+0.024),35);

s2 = interp1(t, s, t2, 'pchip', 'extrap');

figure(2)

plot(t2, s2)

axis([xlim -2 2])

grid

I think your example in your second reply is clearer since it plots rectangular pulses. But there, you didn't align s to t2 as here.

Star Strider
on 13 May 2014

OK, change the first linspace length to 36 and :

s = [0 0 repmat([zeros(1,4) ones(1,4)], 1, 4) 0 0];

This is a square-wave sequence for you to experiment with. Change the interp1 method to 'linear' if you want. The method is probably not critical with this signal.

I’m simply trying things randomly since I still don’t understand what you’re doing. Everything I’ve suggested so far has gone for naught.

S. David
on 13 May 2014

Edited: S. David
on 13 May 2014

Basically I want to transmit an K-by-1 vector d over a channel with linear time varying impulse response:

h(tau;t)=sum_p h_p delta(tau-[tau_p-a*t])

The passband transmit signal is given by:

s(t)=Re{exp(j2*pi*fc*t) sum_k d_k g(t-kTs)}

where fc is the carrier frequency and g(t) is a rectangular pulse of duration Ts (the symbol time).

The baseband received signal in this case is given by:

v(t)=sum_k d_k sum_p h_p exp(-j2*pi*fc*tau_p) exp(j2*pi*fc*t*a) g(t[1+a]-kTs-tau_p)+w(t)

where w(t) is additive white Gaussian noise. So, you can see that the symbol duration has changed from Ts to Ts/(1+a) from the argument of g(t) in the transmit and receive signals. Eventually, I want to generate the signal v(t) from the vector d.

That is what I am trying to do. I do it in MATLAB like this:

for pp=1:P

yT=interp1(t,d,t*(1+a)-tau(pp),'cubic');

y=y+yT*h(pp)*exp(-1i*2*pi*fc*tau(pp)).*exp(1i*2*pi*fc.*t*a);

end

Is this correct?

Star Strider
on 13 May 2014

It is if it runs and produces the result you want.

Star Strider
on 13 May 2014

Edited: Star Strider
on 13 May 2014

Your h looks like a convolution, and s(t) like a cosine transform of some sort?

By the way, resample simply states that p and q must both be positive integers. You may be able to use the core MATLAB rats function to create a ratio of two integers for 1+a or 1/(1+a) that resample will accept.

To get the numerator and denominator from the rats function to use in resample, this will work:

x = 1.15077

y = rats(x);

xnumden = [y(1:findstr(y,'/')-1) ' ' y(findstr(y,'/')+1:end)];

xnd = str2num(xnumden)

Star Strider
on 14 May 2014

Star Strider
on 15 May 2014

S. David
on 15 May 2014

If we ignore the time scaling, the noise-free baseband received signal would be:

v(t)=sum_k d_k sum_p h_p exp(-j*2*pi*fc*tau_p) g(t-kTs-tau_p)

sampling this at t=nTs results in:

v(nTs)=sum_k d_k f_{n-k}

where

f_m=sum_p h_p exp(-j*2*pi*fc*tau_p) g(mTs-tau_p)

for m=0, ..., L, and L is the maximum delay in symbol time.

So, the discrete time equivalent signal can be written as:

v_n=sum_{m=0}^L f_m d_{n-m}

which is the convolution.

However, if I incorporate the time scaling, then

v_n=v(nTs)=sum_k d_k f_{n,n-k}

where

f_m=sum_p h_p exp(-j*2*pi*fc*tau_p) exp(j*2*pi*fc*n*Ts*a) g(mTs-[tau_p-a*n*Ts])

So, it appears to be convolution, however, each n has its own channel length. In this case, how can I make convolution for each received symbol separately? Is there any built-in function to do this?

Star Strider
on 15 May 2014

There are convolution functions in core MATLAB and the Signal Processing Toolbox, that also has modulation and demodulation functions. I don’t know if they’re relevant or not.

There might also be in the Communications System Toolbox. I don’t have it, so I haven’t explored it. (I do mostly simulation, modeling, parameter estimation, control, and signal processing.)

Star Strider
on 15 May 2014

Star Strider
on 15 May 2014

I assume so. I was describing it in terms of vector indices because that’s how I would code it.

S. David
on 15 May 2014

Edited: S. David
on 15 May 2014

Actually, I see what you were saying now. So, I don't need the time vector in this case. I define the transmit vector as:

dCP=[d(N-L+1:end) d];%Cyclic prefixed signal

Then I find the resampled signal as following:

a=0.004;

x = (1+a);

s = rats(x);

xnumden = [s(1:findstr(s,'/')-1) ' ' s(findstr(s,'/')+1:end)];

xnd = str2num(xnumden);

p=xnd(1);

q=xnd(2);

v=resample(dCP,p,q)

Now for the time shift, I do the following:

y1=[v zeros(1,L)];

y2=[zeros(1,L) v];

y=y1+y2;

where I assumed I have two time shift one at tau=0 and one at tau=L (maximum delay in symbol time). After finding the time shift, I do resampling again to recover the original data as:

x = 1/(1+a);

s = rats(x);

xnumden = [s(1:findstr(s,'/')-1) ' ' s(findstr(s,'/')+1:end)];

xnd = str2num(xnumden);

p=xnd(1);

q=xnd(2);

yRes=resample(y,p,q)

Do you think I am interpreting it right?

Star Strider
on 15 May 2014

Star Strider
on 15 May 2014

We’re beyond my expertise in digital communications. I very much hope it works!

S. David
on 16 May 2014

Edited: S. David
on 16 May 2014

You are awesome!! Using this approach my code works very well. I tried two cases: 1-) when a=0 (there is no time scaling), and 2-) when a=0.004. In both cases the noise-free received signal exactly matches its numerical equivalent.

However, when a~=0, the received signal contains the factor exp(1i*2*pi*fc*a*t). Now since I am not using t, how can I incorporate it in the received signal term?

Note: In the receiver, the received signal is multiplied by exp(-1i*2*pi*fc*a*t), to counteract the effect of Doppler shift, so my code works fine without incorporating exp(1i*2*pi*fc*a*t) in the received signal. However, I want to know how, because I will need it for more complicated channel models!!

Star Strider
on 16 May 2014

Edited: Star Strider
on 16 May 2014

Thank you!

The ‘extra’ factor seems to me a phase term of the sort I would expect as phase delay in analog filter output (other than a Gauss filter). I’m casting it in terms of my experience, so filters come quickly to mind. Does the received signal have characteristics of a filtered signal?

(Which brings to mind the possibility that you could consider your channel as a filter?)

S. David
on 16 May 2014

The channel is a Linear Time Varying filter with an impulse response

h(tau;t)=sum_p h_p delta(tau-[tau_p-a*t])

The passband transmitted signal is given by:

s(t)=Re{exp(j*2*pi*fc*t) sum_k d_k g(t-kTs)}

Passing s(t) through h(tau;t) (the convolution between both of them) and downconvert the signal to baseband with respect to the carrier frequency fc, the noise-free baseband received signal is given by:

v(t)=sum_k d_k sum_p h_p exp(-j*2*pi*fc*tau_p) * exp(j*2*pi*fc*t*a) g(t(1+a)-kTs-tau_p)

That is how the factor exp(j*2*pi*fc*t*a) is appeared!!

Star Strider
on 16 May 2014

S. David
on 16 May 2014

Passing the signal through the channel causes the phase shift. I've done it this way: First I defined the time vector t as

t=CP*Ts:Ts:(N+CP-1)*Ts

and then after time scaling and shifting the received signal y is multiplied by exp(1i*2*pi*fc*a.*t).

Thank you very much. This discussion was of a great help, I really appreciate it, and I hope we can continue discussion as I go, since although I am good at math, I am not that good in interpreting math to a MATLAB code.

Star Strider
on 16 May 2014

Edited: Star Strider
on 16 May 2014

My pleasure!

It was an education for me, too. My last communications course was a bit over 15 years ago (my interest being biomedical telemetry), and I haven’t done much with it since. I had to search through my library to find my textbooks and review them. I vaguely remember homework problems on the situation you’re dealing with, but obviously not in any great detail. I did discover my ancient MATLAB code for those assignments.

I’ll always be happy to provide whatever expertise I may have.

Star Strider
on 16 May 2014

My pleasure!

I’m happy to help.

S. David
on 3 Jun 2014

Edited: S. David
on 3 Jun 2014

Hi again,

When I have one factor a, the channel multiplies the signal by exp(j*2*pi*fc*t*a) and the receiver counteracts this by multiplying by exp(-j*2*pi*fc*t*a). So, I didn't have to include this in my code, since mathematically the effect of the channel and receiver processing cancels each others.

Now I am trying to resample the signal several times, where the received signal now is written as:

v(t)=sum_k d_k sum_p h_p exp(-j*2*pi*fc*tau_p) * exp(j*2*pi*fc*t*a_p) g(t*(1+a_p)-kTs-tau_p)

How can I deal with this in MATLAB?

t is defined as

t=0:Ts:(N-1)*Ts;

which has the same length as d. However, after resampling d will have p/q times the length of d!! I tried to do the following:

for cc=1:Np

x1 =(1+a(cc));

s1 = rats(x1);

xnumden1 = [s1(1:findstr(s1,'/')-1) ' ' s1(findstr(s1,'/')+1:end)];

xnd1 = str2num(xnumden1);

p1(cc)=xnd1(1);

q1(cc)=xnd1(2);

end

Ld=max(p1./q1)*N;

y=zeros(1,Ld);

for cc=1:Np

z=resample(d,p1(cc),q1(cc));

r=p1(cc)/q1(cc);

TsNew=Ts/r;

tNew=0:TsNew:(N-1)*Ts

y(1:length(z))=y(1:length(z))+z.*exp(1i*2*pi*fc*a(cc).*tNew);

end

The problem is tNew and z's sizes always differ by one!!

1- Is this approach correct in principle?

2- Why tNew and z are not of the same size as they should theoretically?

Thanks

Star Strider
on 4 Jun 2014

This is not my area of expertise, so I can only look at your code to see if I can find anything that might be a problem.

It will probably be best for you to define tNew by its beginning and end values (that you calculate) and then use the linspace function to define it to have the same length as z. That may be the most efficient (and least agonizing) way to equalise their lengths. I doubt if it will lead to significant loss of accuracy or precision in your subsequent signal processing.

Star Strider
on 4 Jun 2014

Yes. It may be ‘cheating’ a bit, but sometimes that’s necessary.

Star Strider
on 4 Jun 2014

### More Answers (1)

rifat
on 28 Apr 2014

I think the following script should do the work:

signal_vector= [1 1 1 1 1]; % Define your signal here

Ts=1/50; % Define sampling period here

a=1; % define 'a' here

N=length(signal_vector);

x=0:Ts:(N-1)*Ts;

new_x=0:(1+a)*Ts:(N-1)*(1+a)*Ts;

new_signal=interp1(x,signal_vector,new_x,'cubic'); % change the algorithm flag to

% 'pchip' or 'spline' if

% you need

##### 11 Comments

Star Strider
on 28 Apr 2014

There are. The method you use depends on what you’re interpolating and how you want to do it.

rifat
on 28 Apr 2014

rifat
on 28 Apr 2014

rifat
on 28 Apr 2014

John D'Errico
on 12 May 2014

### See Also

### Categories

### Tags

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!**An Error Occurred**

Unable to complete the action because of changes made to the page. Reload the page to see its updated state.

Select a Web Site

Choose a web site to get translated content where available and see local events and offers. Based on your location, we recommend that you select: .

You can also select a web site from the following list:

## How to Get Best Site Performance

Select the China site (in Chinese or English) for best site performance. Other MathWorks country sites are not optimized for visits from your location.

### Americas

- América Latina (Español)
- Canada (English)
- United States (English)

### Europe

- Belgium (English)
- Denmark (English)
- Deutschland (Deutsch)
- España (Español)
- Finland (English)
- France (Français)
- Ireland (English)
- Italia (Italiano)
- Luxembourg (English)

- Netherlands (English)
- Norway (English)
- Österreich (Deutsch)
- Portugal (English)
- Sweden (English)
- Switzerland
- United Kingdom (English)

### Asia Pacific

- Australia (English)
- India (English)
- New Zealand (English)
- 中国
- 日本Japanese (日本語)
- 한국Korean (한국어)