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Engineering Mathematics Question. (Involves Complex Numbers).

Any help with the question below would be very much appreciated and a worked solution would be appreciated even more! I think I have the correct answer for part a) and am on the right lines in part b) but I would love to be sure! Rep. will be awarded for good help!
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In a particular circuit the complex impedance, Z, is given in terms of the capacitance, C, inductance, L, resistance, R, and frequency, ω, by:

Z[w] = R + iωL + 1/(iωC)

C, L, R, ω are all real.

a)
Find in terms of L and C the value of the frequency, ω0, for which the impedance is real.

b)
Show that:

Z[w]/Z[ω0] = 1 + iQ(ω/ωo - ω0/ω.)

where Q is a real number.

c)
Find, in terms of ω0 and Q, the values of ω for which the ratio above has modulus √2.

d)
The complex voltage, V, and current, I, are related by V = Z[w]I. Find, in terms of Q, ω, ω0, the phase difference between V and I, i.e. Arg(V) - Arg(I).
Reply 1
Eddie K
In a particular circuit the
complex impedance, Z,
is given in terms of the
capacitance, C,
inductance, L,
resistance, R,
frequency, ω, by:

Z[w] = R + iωL + 1/(iωC)

C, L, R, ω are all real.

Find in terms of L and C the value of the frequency, ω0, for which the impedance is real.


Z = R + iωL + 1/(iωC)
Z(iωC) = R(iωC) + iωL(iωC) + 1
iZωC = iRωC - ω²LC + 1
iZωC - iRωC + ω²LC = 1
iCω(Z - R) + ω²LC = 1
Cω(i(Z - R) + ωL) = 1
i(Z - R) = 1/Cω - ωL
i(Z - R) = ω(1/C - L)
i(Z - R)/(1/C - L) = ω

Now investigate for which values of Z and R, the top part which is imaginary becomes real.

I think this is right.
Reply 2
Dekota
Z = R + iωL + 1/(iωC)
Z(iωC) = R(iωC) + iωL(iωC) + 1
iZωC = iRωC - ω²LC + 1
iZωC - iRωC + ω²LC = 1
iCω(Z - R) + ω²LC = 1
Cω(i(Z - R) + ωL) = 1
i(Z - R) = 1/Cω - ωL
i(Z - R) = ω(1/C - L)
i(Z - R)/(1/C - L) = ω

Now investigate for which values of Z and R, the top part which is imaginary becomes real.

I think this is right.


Unfortunately this does not help. The first part of the problem I have solved by making the two imaginary parts of the original function equal to 0. By doing this it is possible to isolate this part of the function as real in terms of C and L, as they require. This is the value for ω0. It is the second part of question and onwards which is proving difficult.
Eddie K
Unfortunately this does not help. The first part of the problem I have solved by making the two imaginary parts of the original function equal to 0. By doing this it is possible to isolate this part of the function as real in terms of C and L, as they require. This is the value for ω0. It is the second part of question and onwards which is proving difficult.

Maybe by posting someone will see my error. I cant get part(ii) :frown: Probably due to me being stupid more than anything else.

i)Z[w] = R + iωL + 1/(iωC)
we need this to be real
ie iwL+1/(iwC)=0
i(wL-1/(wC))=0
i(w^2CL-1)/(wc)=0
then Z(w) is real if w^2CL-1=0
choose w^2=1/CL and call this value wo
so we have wo^2=1/CL ...............(*)

ii) using this value for w0 we have Z(wo)=R
then Z(w)/Z(wo)=
1+1/R[ iωL + 1/(iωC)]
=1+i/R[{w^2CL-1}/{wC}]
using (*)
=1+i/R[{w^2-w0^2}/{wwo^2}
=1+i/RC[w/wo^2-1/w]...........not quite the expression, i have an extra wo.
=1+i/RCwo[w/wo-wo/w]...........taking the wo out seems a fudge to me but its still a real number
let Q=1/(RCwo) then Q is real and
Z[w]/Z[ω0] = 1 + iQ(ω/ωo - ω0/ω.)

iii)this had modulus root 2 when

Q(ω/ωo - ω0/ω.)=1 or -1
if Q(ω/ωo - ω0/ω.)=1
Q(w^2-wo^2)=wwo
solve for w
sim if Q(ω/ωo - ω0/ω.)=-1
remembering that w is real.

iv)
If V=Z(w)I
then arg(V)=arg(Z(w)I)
=arg(Z(w)+arg(I)
arg(V)-arg(I)=arg(Z(w))
arg(Z(w))=arctan[{ωL - 1/(ωC)}/R]
=arctan [Q(ω/ωo - wo/w)] from part(b)
Hope my attempt is somewhere close or enables you to do the question, i blame the hangover.
Reply 4
I'm going to give your methods a go - see if I can understand it and get answers as you have done. In the mean time, if anyone else can offer any help for the problem I would be most grateful!