Frequency Selective Channel Coefficients in MATLAB

Hello all

How can I generate the following channel coefficients in MATLAB, where these coefficients correspond to a frequency selective channel:

f_m=sum_p h_p exp(-j*2*pi*fc*tau_p) g(mTs-tau_p), m=0,1,...,L-1

where g(t) is rectangular pulse of support duration Ts, and h_p, tau_p and fc are given.

Thanks in advance

[Youssef Khmou]

you have to change the title so that the specialists in the field will answer, put as example " radiation pattern" ," collected signals"..

Answer 1

[Youssef Khmou]

here is a start :

 Fs=800;
 fc=300; % Hz
 Ts=1/Fs;
 L=400;
 t=0:Ts:(L-1)*Ts;
 d=4; %  four signals per example,

tau is dependent on physical parameters, distance and coordinates, but i consider here a linear progressive phase, and the signals are sinusoidal :

 tau=(0:d-1);

 fm=0;
 for p=1:d
 fm=fm+exp(-j*2*pi*fc*tau(p))*sin(2*pi*t);
 end

 figure; plot(real(fm));

your assignment now is to replace the sin signals with the rectangular ones .

[S. David]

That is funny, because my problem was how to deal with rectangular pulse. I will try though, and tell me if it is right.

The vector t as you defined is actually a train of rectangular pulses. So, I guess the following:

fm=0;
for p=1:d
fm=fm+exp(-j*2*pi*fc*tau(p))*(t-tau(p));
end

Is that right?

[Youssef Khmou]

hi,

no, t is linear vector of time index, as for rectangular pulse, there is build-in function rectpuls(t,w), test this entire program :

Fs=800;
fc=300; % Hz
Ts=1/Fs;
L=400;
t=0:Ts:(L-1)*Ts;
d=4; %  four signals per example,
tau=(0:d-1)/4; % width of rectangular puls <=1.
SIG=zeros(d,L);
for n=1:d
   SIG(n,:)=rectpuls(t,tau(n));
end

 fm=0;
 for p=1:d
 fm=fm+exp(-j*2*pi*fc*tau(p))*SIG(p,:);
 end

plot the real part of fm, is it descending stair?

[S. David]

Indeed it is.

I read "help rectpuls" and it says that rectpuls(T,W) finds the amplitude value of the rectangular pulse (of unity value) at the points specified by the vector T with width W (which is one by default). Based on this information, I wrote the following code with actual parameters I am using for my communication system, and use some ideas from your codes. Note: fm should be a point not a vector, so I wrote another for loop of it.

fc=30*10^3;                 %Carrier frequency
B=1000;                     %The bandwidth
Ts=1/(2*B);                 %Sample time (Nyquist rate)
tau=[0 12 20].*10^-3;       %Path delays
h=[1 1 1];                  %Channel gains
Np=length(h);               %Number of distinct paths

L=round(tau(end)/Ts)+1;     %Channel length

fm=0;

for pp=1:Np
    for ll=1:L
        fm=fm+h(pp)*exp(-1i*2*pi*fc*tau(pp)).*rectpuls(ll*Ts-tau(pp),Ts);
    end
end

Is that correct?

[Youssef Khmou]

There is a problem with sample time, is fc the central frequency ([f1 fc f2]) or the lower frequency, anyway for wideband signals with spectrum ([f1 f2]) the sample time is T=1/(2*max(f1,f2)), in your case you only took B while signals contains fc>>B.

   Ts=1/(2*(B+fc));

there must a be meaning why fm is 1x1 rather than vector, are you trying to integrate fm? these are the issues you have to explore more.

[S. David]

I am working in baseband not bandpass, i.e.,: f1=0 and f2= 1 KHz. In this case T=1/(2*max(f1,f2))=1/(2*B). Right?

Yes, right, I was wrong. In this case fm is vector. So, the code becomes:

fc=30*10^3;                 %Carrier frequency
B=1000;                     %The bandwidth
Ts=1/(2*B);                 %Sample time (Nyquist rate)
tau=[0 12 20].*10^-3;       %Path delays
h=[1 1 1];                  %Channel gains
Np=length(h);               %Number of distinct paths

L=round(tau(end)/Ts)+1;     %Channel length
t=0:Ts:(L-1)*Ts;

fm=0;

for pp=1:Np
    fm=fm+h(pp)*exp(-1i*2*pi*fc*tau(pp)).*rectpuls(t-tau(pp),Ts);
end

Is it right now?

[S. David]

I am sorry, I don't understand. The sampling rate must be at least twice the highest frequency component, and I am working in baseband. So, why it is 2*(B+fc) and not 2*B? Am I missing something here?

Thanks

[Youssef Khmou]

you are right :

[Youssef Khmou]

if the problem is solved, accept the answer.

[S. David]

Thanks