First, k=1 from circuit inspection: The parallel combination of L and C is an open circuit (infinite impedance) at wo=1/sqrt(LC). At this frequency, no current will flow through R and hence Vo=Vi.
Second, the Q factor is RCwo or, equivalently, the second expression you gave.
If you work further on the original expression you will certainly get it.
It is possible to prove mathematically that the Q factor of this band-pass RC circuit is limited to: $$ 0<Q<0.5 $$
First, we need to remember that the standard form of the transfer function for a second-order band-pass filter is:
$$H(s)=\frac{K(2ζω_n)s}{(s^2+2ζω_ns+ω_n^2 )}$$
Where:
- \$K\$ is the maximum gain of the filter
- \$ζ\$ is the damping ratio
- \$ω_n\$ is the undamped natural frequency
And the transfer function based on component values is:
$$H(s)=\frac{sC_1 R_1}{C_1 C_2 R_1 R_2 \cdot s^2+(C_1 R_1+C_2 R_1+C_2 R_2)s+1}$$
To prove that 0 < Q < 0.5, I will start by setting up an equation that, when solved, give the poles (roots of the denominator) of the above transfer function:
$$C_1 C_2 R_1 R_2 s^2+(C_1 R_1+C_2 R_1+C_2 R_2)s+1=0$$
Note that this expression is a quadratic equation in the form \$ as^2+bs+c=0\$. Its discriminant is given by \$Δ=b^2-4ac\$. So, we have:
$$Δ=(C_1 R_1+C_2 R_1+C_2 R_2)^2-4⋅C_1 C_2 R_1 R_2$$
Assuming that \$ C_1>0\$, \$ C_2>0\$, \$R_1>0\$ and \$R_2>0\$, it follows that \$Δ>0\$ for any choice of components.
Now, doing the same process but using the standard form this time, we have:
$$s^2+2ζω_n s+ω_n^2=0$$
$$Δ=(2ζω_n )^2-4ω_n^2=4ζ^2 ω_n^2-4ω_n^2=4ω_n^2⋅(ζ^2-1)$$
But we already know that \$Δ>0\$. So, using the standard form:
$$4ω_n^2⋅(ζ^2-1)>0$$
The term \$4ω_n^2\$ is surely positive. Then, to satisfy the above inequality, its necessary that
$$(ζ^2-1)>0$$
Before proceeding, it's important to remember one more thing:
$$Q≜\frac{1}{2ζ} \implies ζ=\frac{1}{2Q}$$
Then, by making the substitution, we have:
$$\left(\frac{1}{2Q}\right)^2-1>0 \implies \left|\frac{1}{2Q}\right|>1$$
Analyzing the expression of Q factor for the filter, we have, doing the same assumptions made above with \$Δ\$:
$$Q=\frac{\sqrt{C_1C_2R_1R_2}}{C_1R_1+C_2R_1+C_2R_2}>0$$
Thus, we can finally conclude that:
$$\frac{1}{2Q}>1 \implies 0<Q<0.5$$
Best Answer
You could use the Laplace transform, and the 'bode' command in Matlab:
\$G(s)=\frac{RCs}{s^2LC+sCR+R}\$
Then in Matlab, immediate mode, and using the component values given in the OP: :
num=[1 0];
den=[0.02 1 1];
g=tf(num, den);
bode(g)