Electronics and Communication Engineering - Networks Analysis and Synthesis - Discussion

Z(0) = 8.
2 resistance sum = 4 So, 8-4 = 4(R value).
1 capacitor sum = 1/2 so, 1/(8-(1/1/2)) (C value).

Here for capacitance, we use 1/c.

I don't know whether it's correct or not.

Best explanation will be that at s=0 i.e at w=0 capacitor will be have as an open circuit.

So Z(s) = 8 at s = 0 i.e 4+R+8.

R = 4 ohm.

Of course you can solve it completely also.

Can anyone explane the answer?

TRANSFER FUNCTION COMES OUT AS -: (2+2*R+s*R) + 3*[RC*s^2+s*(2*C+2*R*C+1)+2]
------------------------------------------------------------------------
RC*s^2+s*(2*C+2*R*C+1)+2.
From Denominator-:RC*s^2+s*(2*C+2*C*R+1)+2.
===== RC[ s^2 + s*(2/R+2+1/RC) + 2/RC].

And on comparing it from given equation(i.e. s^2+4*s+3 )
We get 3=2/RC ---> (1)
and from given option, only option (A) satisfy this condition.
that's why the answer is an option (A).

What if option b is 2/9 F and 4 ohm. Is it also an answer?

Please solve problem.

Can anyone explain this problem in detail?

Thank you @Satyam.

Thank you @Satyam.

Thanks @Satyam.