Plotting Impedance of Parallel RLC Circuit

[RockyRoad]

Homework Statement

Plot |z| vs. f(Hz) of the circuit. R=100, C=600 pf = 6E-12 F, L=10E-6 H . All elements of the circuit are in parallel. Also identify the resonant frequency.

Homework Equations

ZR=R
ZL=jwL
ZC=1/(jwC)=-j/(wC)
w=2(pi)f
w0=1/sqrt(LC)
f0=1/(2(pi)sqrt(LC))
G=1/R

The Attempt at a Solution

Z1=ZR
Z2=ZL
Z3=ZC
Zeq= 1/((1/R)+1/(jwL)+jwc)

Switching to admittance and skipping a few steps, it can be found that
Y=G + j * 2 * pi * f * C * (1 - (f0 / f) ^ 2)
Z=R + 1 / (j * 2 * pi * f * C * (1-(f0 / f) ^ 2))
Z=100 + 1/(j * 2 * pi * f * 600E-12 *(1 - (2054681.48 / f) ^ 2)

To plot, i used a few lines in matlab:

f=linspace(0,4000000,100000);
x=1i.*2.*pi.*(600.^(-12));
z=100+(1./(x.*f.*(1-((2054681.48./f).^2))));
plot=semilogx(z);

The plot shows me a vertical line at f=100

My professor told us that the graph should be bell shaped, and i have no idea where I've gone wrong. Anyone see a mistake?

 

[vela]

For one thing, 1/(a+b) isn't equal to 1/a + 1/b, which is what you did when you went from the admittance back to the impedance.

 

[RockyRoad]

Ok, change it to this then:

Z=1/(100 + j * 2 * pi * f * 600E-12 *(1 - (2054681.48 / f) ^ 2))

The plot is still a vertical line.

 

[vela]

Well, that's not quite right either. I'll leave it to you, however, to fix your algebra mistakes.

I don't know Matlab, but I'm guessing you're letting f start from 0. Two problems with that. At f=0, you're dividing by zero in your expression. Also, when f=0, Z=0 since the inductor is a short, so the log of Z is undefined.

 

[The Electrician]

In this expression:

x=1i.*2.*pi.*(600.^(-12));
...^

is that lower case i that I've pointed to supposed to represent SQRT(-1)?

If it is, that's a problem. You can't plot expressions that have an imaginary component. You have to plot the magnitude, or modulus, of the impedance expression.

Some mathematical packages have a command such as Abs[] that can do it. Or you can do it yourself by plotting SQRT(Z * Conjugate(Z)).