how to plot y(t)

Tell me what is wrong with my code ? i didnt get the expected graph

Root Raised Cosine Filter..

clc;clear all;  close all;
fs=10;
Ts   = 1/fs;
c = 10;          
t = -1:0.01:1;
b=1;
T=1/fs;
Nt=-Ts/(2*b);
Pt=Ts/(2*b);
y=@(t) ((1/Ts)*(1+(b*((4/pi)-1)))) .*(t==0) +   ((b/(Ts*sqrt(2))) * (  ((1+2/pi)*(sin(pi/(4*b)))) + ((1-(2/pi))*cos(pi/(4*b))) )) ...
    .*((t==Nt)  | (t==Pt)) +   ((1/Ts).*((sin(pi*(t/Ts)*(1-b)) + (4*b*(t/Ts).*(cos(pi*(t/Ts)*(1+b))))) / (pi*(t/Ts).*(1-(4*b*(t/Ts)).^2))))  ...
    .*((t~=Nt)  & (t~=Pt) & (t~=0));
plot(t,y(t));

Answer 1

[KSSV]

Check the expressions thoroughly....not sure did I enter correct.

fs=10;
Ts   = 1/fs;
c = 10;          
t = -1:0.01:1;
b=1;
T=1/fs;
Nt=-Ts/(2*b);
Pt=Ts/(2*b);
h = zeros(size(t)) ;
for i = 1:length(t)
    if t(i) ==0 
        h(i) = (1/Ts)*(1+(b*((4/pi)-1))) ;
    elseif abs(t(i)) == Ts/(4*b)
        h(i) = b/(Ts*sqrt(2))*((1+2/pi)*sin(pi/(4*b))+(1-2/pi)*cos(pi/(4*b))) ;        
    else
        h(i) = 1/Ts*(sin(pi*t(i)/Ts*(1-b))+4*b*t(i)/Ts*cos(pi*t(i)/Ts*(1+b)))/(pi*t(i)/Ts*(1-(4*b*t(i)/Ts)^2)) ;
    end
end
plot(t,h);

[Anwin Kushal]

Thx i will check

[Anwin Kushal]

above method works

but in this code output should be zero due to the condition but it say NaN why??

Nt=-Ts/(4*b);
Pt=Ts/(4*b);
t=0
h = zeros(size(t)) ;
h=@(t) (((1/Ts).*((sin(pi*(t/Ts)*(1-b)) + (4*b*(t/Ts).*(cos(pi*(t/Ts)*(1+b)))))...
    / (pi*(t/Ts).*(1-(4*b*(t/Ts)).^2)))) ).*(t~=0)
h(t)

output:
t = 0
h = 
    @(t)(((1/Ts).*((sin(pi*(t/Ts)*(1-b))+(4*b*(t/Ts).*(cos(pi*(t/Ts)*(1+b)))))/(pi*(t/Ts).*(1-(4*b*(t/Ts)).^2))))).*(t~=0)

ans = NaN

[VBBV]

h=@(t) (((1/Ts).*((sin(pi*(t/Ts)*(1-b)) + (4*b*(t/Ts).*(cos(pi*(t/Ts)*(1+b)))))...
   ./ (pi*(t/Ts).*(1-(4*b*(t/Ts)).^2)))) ).*(t~=0);
%  ^^                     ^^
% example
t = 0;
y = 1/t % a very large number which cannot be known

y = Inf

To get the desired output, you need to suppy the vector t using your anonymous function

fs=10;
Ts   = 1/fs;
c = 10;          
t = -1:0.01:1 % give this vector

t = 1×201

-1.0000 -0.9900 -0.9800 -0.9700 -0.9600 -0.9500 -0.9400 -0.9300 -0.9200 -0.9100 -0.9000 -0.8900 -0.8800 -0.8700 -0.8600 -0.8500 -0.8400 -0.8300 -0.8200 -0.8100 -0.8000 -0.7900 -0.7800 -0.7700 -0.7600 -0.7500 -0.7400 -0.7300 -0.7200 -0.7100

b=1;
T=1/fs;
Nt=-Ts/(4*b);
Pt=Ts/(4*b);
h = zeros(size(t)) ;
h=@(t) (((1/Ts).*((sin(pi*(t/Ts)*(1-b)) + (4*b*(t/Ts).*(cos(pi*(t/Ts)*(1+b)))))...
    ./ (pi*(t/Ts).*(1-(4*b*(t/Ts)).^2)))) ).*(t~=0)

h = function_handle with value:
    @(t)(((1/Ts).*((sin(pi*(t/Ts)*(1-b))+(4*b*(t/Ts).*(cos(pi*(t/Ts)*(1+b)))))./(pi*(t/Ts).*(1-(4*b*(t/Ts)).^2))))).*(t~=0)

y = h(t);
idx = find(t == 0);
y(idx) = (1/Ts)*(1+(b*((4/pi)-1)));
plot(t,y)

Answer 2

[Luan]

I am a new intow, would anyone show me how to plot

y(t)= 1/3(2e^-t+1-3e^-2t)

Thanks