how to plot conditional statement like when Vin=1, Vout=exp(-t/tn) and for Vin =0, Vout=1-exp(-t/tp) where Vin is a periodic square pulse.

i have tried as follows:

f=input('please enter the value of time');
t = 0:.00001*f:(1-.00001)*f;
fm=2/f;
Vin = .5+0.5*square(2*pi*fm*t);   
subplot(211);
plot(t,Vin)     
Vdd=5;      
Rn=2000;    
Rp=3000;    
C=0.1e-9;   
Tn=Rn*C;  
Tp=Rp*C;
V1=Vdd.*(exp(-t/Tn));
V0=Vdd*(1-exp(-t/Tp));
Vout=V1;
Vout(Vin<1)=V0(Vin<1);
subplot(212)
plot(t,Vout)

the problem is that for very small value of f, like f=.00000001 the desired graph is not obtained. Even for values like f=0.001, the gradual decay or rise that is expected is not obtained. please help.

[Azzi Abdelmalek]

What is the desired result?

[ANWESHA]

When Vin=1, Vout should be 0 with initial exponential decaying curve and if Vin=0 then Vout should be 1 with initial exponential rising curve.

Answer 1

[David Sanchez]

statement like when Vin=1, Vout=exp(-t/tn) and for Vin =0, Vout=1-exp(-t/tp):

if (Vin == 1)
  Vout=exp(-t/tn);
elseif (Vin == 0)
  Vout=1-exp(-t/tp);
end

[Azzi Abdelmalek]

This is not what he asked for

[ANWESHA]

this is the desired result

[ANWESHA]

I have also tried with the if and elseif statement.It is not working.

Answer 2

[Image Analyst]

Can't you just use a convolution, conv(), or filter()? Invert the signal and apply a kernel that only looks to the left. Seems pretty straightforward.

[ANWESHA]

Sorry sir, i didn't get you. Can you please elaborate. I have no idea how to apply kernel.

[Image Analyst]

See Star's code which uses filter(). It is fine. No need to worry about conv() now that you have the filter() code.