Electronics and Communication Engineering - Networks Analysis and Synthesis - Discussion

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

@Satyam is correct.

Put s = 0, where the capacitor is open circuited, then find the equivalent impedance. After that put s = 0 in given z (s) that is get z (0). Equate this with the previous result and get R and in option, there is only one option with R = 4. So A is the right answer.

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.

Add admittances of cap and res = (S/2)+1.

Then sum res = R+(2/S+2).

Add admittances = (CS)+(S+2/(R*(S+2)+2)).

Total = 3+(R*(S+2)+2))/(CS(R*(S+2)+2)) + (S+2));

For explanation refer to topic concept of network synthesis through long division method.

After continued fraction expansion you get;

3+{1/(s/6)+{1/(4)+{1/(s/2)+2}}}
C=1/6, R=4.

@Gopal Agarwal.

I didn't understand that. Please make it clear.

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

Could you please provide explanation for this by formula?