How can I use a ifft function to a symbolic variable ?

for designing optimum-L filter, I have used symbolic method.

 t=-100:0.001:100
 dt=t(2)-t(1)
 f=linspace(-1./(2.*dt),1./(2.*dt),length(t))
 if floor(N/2)<(N/2) % N is odd number
    k = (N-1)./2;
    syms x ohm;
    fx=0;

    for m=0:k
        a=2*m+1;
        fx=fx+a*legendreP(m,x);
    end

    L=(fx/(2^0.5*(k+1)))^2;
    y=int(L,-1,(2*ohm^2)-1);
    Gain=1/((1+y)^0.5);

I tried to use subs(Gain,ohm,f) and use ifft(ifftshift(Gain)) for getting impulse response function, but it failed. how can i do?

[Walter Roberson]

Where does the "if" end for "if floor" ?

Answer 1

[Karan Gill]

I'm assuming your error said:

Undefined function 'ifft' for input arguments of type 'sym'.

As the error message says, the output of "subs" is still symbolic. You need to convert it to double by using "double":

Gain = subs(Gain,ohm,f); % Gain is still symbolic
GainDouble = double(Gain); % now Gain is double and can be used by ifft

[Walter Roberson]

In theory, yes. In practice, the f vector is large enough that double() will spend well over half an hour on the computation. I do not know how long it would take to finish; I gave up and canceled it.