Dear Shafquat Hussain,

I can give you some steps to solve your problem.

- Express you filter in S-domain H(s)

- Use the bilinear transform to transform H(s) to H(z)

- After getting H(z) you get h[n] the impulse response of filter

by transforming H(z) to h[n]

This what you need finally. The discrete time impulse response.

The other way is to get the impulse response in time domain h(t)

and then you simply discretize the time by sampling h(t)

by t= nTs,

You get h(nTs) = h[n]

You can also transform H(s) to time domain directly by inverse La Place transform to get h(t).

And then you can discretize it by sampling as I said before.

For more information please follow the hand outs:Presentation Introduction to digital filters- Hand outs

Best wishes

Dear Shafquat Hussain,

I can give you some steps to solve your problem.

- Express you filter in S-domain H(s)

- Use the bilinear transform to transform H(s) to H(z)

- After getting H(z) you get h[n] the impulse response of filter

by transforming H(z) to h[n]

This what you need finally. The discrete time impulse response.

The other way is to get the impulse response in time domain h(t)

and then you simply discretize the time by sampling h(t)

by t= nTs,

You get h(nTs) = h[n]

You can also transform H(s) to time domain directly by inverse La Place transform to get h(t).

And then you can discretize it by sampling as I said before.

For more information please follow the hand outs:Presentation Introduction to digital filters- Hand outs

Best wishes

You should introduce variables corresponding to current through the inductor I and capacitor voltage U. Let's say it is series LCR filter (picture attached) feed by voltage source Vin(t) and the output is voltage across R. You can write corresponding diff equations as

Vin(t) = L* dI/dt + U + R*I

C* dU/dt = I

We define output as

Vout(t) = R *I

Time discrete variant will be

Vin[k] = L *( I[k+1] - i[k])/dt + R *I[k] +U[k]

C *( U[k+1] - U[k] ) = dt*I[k]

Vout[k] = R *I[k]

You can solve it as (assuming some initial conditions e.g. I[0]=0 , U[0]=0 ):

I[k+1] = I[k] + dt*( Vin[k] - R*I[k] - U[k] )/L

U[k+1] =U[k] + dt*I[k]/C

Vout[k] = R *I[k]

Also, It is possible to use Impulse Invariance Method. First, it is obtained H(s). Then, H(z) and after, it is determined difference equation.

Here's a MATLAB function example for a series RLC (for voltage across the capacitor). You can modify it for other circuits (e.g. parallel), or currents/voltages. See Nikolay's answer.

%------------------------------------------------------------------------------

% FUNCTION Y = RLC_series(X,damping,N) -- series RLC filter (capacitor voltage)

% Y is output column vector of same length as X

% X is incoming signal

% DAMPING is damping (underdamped when 1, and critically

% damped when =1); xi = .5*RC/sqrt(LC) = pi*tau*nu

% N is natural period in units of time step h; N = 2*pi*sqrt(LC)/h

%------------------------------------------------------------------------------

% N = 1/(f0*h), DAMPING = df/(2*f0) (f0 is natural frequency)

%------------------------------------------------------------------------------

function y = RLC_series(x,xi,N)

a = 4*pi*xi/N; b = (2*pi/N)^2; c = 1+a+b;

y(1) = ( b*x(1) + (2+a)*x(1) - x(1) ) / c;

y(2) = ( b*x(2) + (2+a)*y(1) - x(1) ) / c;

for i=3:length(x)

y(i) = ( b*x(i) + (2+a)*y(i-1) - y(i-2) ) / c;

end

%------------------------------------------------------------------------------

return

%------------------------------------------------------------------------------

% TEST

h = 1e-6; % 1 MHz sampling rate

t = h*(1:1e4); % time

x = zeros(size(t)); x(200) = 1; % input impulse

%x = zeros(size(t)); x(200:end) = 1; % input step

f0 = 1e3; % 1 kHz natural frequency

N = 1/(f0*h);

damp1 = 1; % critical damping

damp2 = 0.1; % underdamped

damp3 = 2; % overdamped

y1 = RLC_series(x,damp1,N);

y2 = RLC_series(x,damp2,N);

y3 = RLC_series(x,damp3,N);

plot(t,y1,'r',t,y2,'b',t,y3,'k'), zoom on, grid on