Hv(jω) Magnitude Calculation in non-standard circuits.

[DabaDuBa]

I was wondering how to calculate the magnitude and phase response of any circuit that is neither Low-Pass,High Pass nor Band-Pass. I am asking this because i haven't found anything relevant in any textbooks so far , since the only use standard filters.

e.g an AC circuit consisting of an AC source with internal resistance A , capacitor C , inductor L , resistance B (the load) such that :

Zc||(Zb + ZL)

Following the commonly suggested methodology , the equivalent Thevenin impedance would be :

[ (jωL)-(ω*ω*L*C*Za) + Za ] / [ 1 + j*ω*C*Za ]

and the Thevenin Voltage :
Vthev = Vsource * ( Zc / (Zc + Za))

Vload = Vthev *Zload / ( Zload + Zthev)

Therefore :


Vload = Vsource *( Zc * Zload) / [ ( Zc+Za)*(Zthev +Zload)]

where Hv(jω) is the coefficient in front of Vsource which will be quite complicated.


How can we calculate its magnitude and phase response ?

Thank you

 

[The Electrician]

I was wondering how to calculate the magnitude and phase response of any circuit that is neither Low-Pass,High Pass nor Band-Pass. I am asking this because i haven't found anything relevant in any textbooks so far , since the only use standard filters.

e.g an AC circuit consisting of an AC source with internal resistance A , capacitor C , inductor L , resistance B (the load) such that :

Zc||(Zb + ZL)

Following the commonly suggested methodology , the equivalent Thevenin impedance would be :

[ (jωL)-(ω*ω*L*C*Za) + Za ] / [ 1 + j*ω*C*Za ]

and the Thevenin Voltage :
Vthev = Vsource * ( Zc / (Zc + Za))

Vload = Vthev *Zload / ( Zload + Zthev)

Therefore :


Vload = Vsource *( Zc * Zload) / [ ( Zc+Za)*(Zthev +Zload)]

where Hv(jω) is the coefficient in front of Vsource which will be quite complicated.


How can we calculate its magnitude and phase response ?

Thank you

This is just a simple network analysis problem. To do the calculation, just substitute the numerical values for all those impedances (Za, Zb, Zc, Zthev, Zload, etc.) into the expression and work out the arithmetic. The arithmetic will, of course, involve complex numbers.