how can i quantize the normal distribution using uniform PCM?

sinusoidal signal, there is quantization code for 8 levels and 16 levels using PCM scheme. However, I want to do this for N(0,1) distribution as well. Do I just need to make changes in the "a" variable in this code?

echo on
t=[0:0.01:10];
a=sin(t);
[sqnr8,aquan8,code8]=u_pcm(a,8);
[sqnr16,aquan16,code16]=u_pcm(a, 16);
pause % Press a key to see the SQNR for N = 8.
sqnr8
pause % Press a key to see the SQNR for N = 16.
sqnr16
pause % Press a key to see the plot of the signal and its quantized versions.
plot(t,a,' -',t,aquan8,' - . ',t,aquan16,' -',t,zeros(1,length(t)))

function [sqnr,a_quan,code]=u_pcm(a,n)
%U_PCM  	uniform PCM encoding of a sequence
%       	[SQNR,A_QUAN,CODE]=U_PCM(A,N)
%       	a=input sequence.
%       	n=number of quantization levels (even).
%       	sqnr=output SQNR (in dB).
%		a_quan=quantized output before encoding.
%		code=the encoded output.

amax=max(abs(a));
a_quan=a/amax;
b_quan=a_quan;
d=2/n;
q=d.*[0:n-1];
q=q-((n-1)/2)*d;
for i=1:n
  a_quan(find((q(i)-d/2 <= a_quan) & (a_quan <= q(i)+d/2)))=...
  q(i).*ones(1,length(find((q(i)-d/2 <= a_quan) & (a_quan <= q(i)+d/2))));
  b_quan(find( a_quan==q(i) ))=(i-1).*ones(1,length(find( a_quan==q(i) )));
end
a_quan=a_quan*amax;
nu=ceil(log2(n));
code=zeros(length(a),nu);
for i=1:length(a)
  for j=nu:-1:0
    if ( fix(b_quan(i)/(2^j)) == 1)
	code(i,(nu-j)) = 1;
	b_quan(i) = b_quan(i) - 2^j;
    end
  end
end
sqnr=20*log10(norm(a)/norm(a-a_quan));

Answer 1

[Raikantwar Hrashikesh]

idk

Answer 2

[Yashwanth]

%ECE24063

function [quant,qlev] = uni(x,N)

x_max = max(x);

x_min = min(x);

N = 8;

delta = (x_max - x_min)/N;

R = x_min:delta:x_max;

quant = zeros(1,length(x));

qlev = zeros(1,length(x));

for i = 1:length(x)

for j = 1:(length(R)-1)

if x(i) >= R(j) && x(i) <= R(j+1)

quant(i) = R(j)+(delta/2);

qlev(i) = j-1;

end

end

end

end

function 1

Answer 3

[Yashwanth]

%ECE24063

function [quant,code] = uniformpcm1(x,N)

x_max = max(x);

x_min = min(x);

N = 8;

delta = (x_max - x_min)/N;

R = x_min:delta:x_max;

quant = zeros(1,length(x));

code = zeros(1,length(x));

for i = 1:length(x)

for j = 1:(length(R)-1)

if x(i) >= R(j) && x(i) <= R(j+1)

quant(i) = R(j)+(delta/2);

qlev(i) = j-1;

end

end

code = dec2bin(qlev);

end

end

2nd func

Answer 4

[Yashwanth]

%ECE24063

fs = 1000;

t = 0:1/fs:2;

x = sin(2*pi*5*t);

subplot(2,1,1)

plot(t,x)

title("sine signal")

xlabel("time")

ylabel("amplitude")

N = 8;

[quant,qlev] = uni(x,N);

subplot(2,1,2)

stem(t,quant)

title("Quantization Level")

xlabel("Number of Random Variables")

ylabel("quant")

display(qlev)

Answer 5

[Yashwanth]

%ECE24063

fs = 1000;

t = 0:1/fs:2;

x = sin(2*pi*5*t);

subplot(2,1,1)

plot(t,x)

title("sine signal")

xlabel("time")

ylabel("amplitude")

N = 8;

[quant,qlev] = uni(x,N);

[quant,code] = uniformpcm1(x,N);

subplot(2,1,2)

stem(quant)

title("Quantization Level")

xlabel("Number of Random Variables")

ylabel("code")

display(code)

Answer 6

[Yashwanth]

fs = 1000;

t = 0:1/fs:2;

x = sin(2*pi*5*t);

N = 8;

[quant,qlev] = uni(x,N);

[quant,code] = uniformpcm1(x,N);

P = mean(x.^2);

D = mean((x-quant).^2);

SQNR = P/D;

SQNR1 = 20*log(SQNR);

display(SQNR)

display(SQNR1)

Answer 7

[Yashwanth]

fs = 1000;

t = 0:1/fs:2;

x = sin(2*pi*5*t);

subplot(2,1,1)

stem(t,x)

title("Original signal")

title("t")

xlabel("x")

N = 8;

x_max = max(x);

x_min = min(x);

delta = (x_max - x_min)/N;

[quant,code] = uniformpcm1(x,N);

qlev = bin2dec(code);

q_r = min(x) + ((qlev).*delta) + (delta/2);

subplot(2,1,2)

stem(t,q_r)

title("Reconstructed signal")

title("t")

xlabel("q_r")