The general formula for resonance is \$w_0=\frac{1}{LC}\$ . A simulation was built with such inductor and capacitor so the resonance will be at 1GHZ.However as you can see in the result ,the plot shows a very small Q factor resonance and its resonating not at a frequency which differs the formula. Where did i go wrong?
$$ w_0=\frac{1}{LC}=1 GHz $$
Well, notice that the impedance of your circuit is given by:
$$\left|\space\underline{\text{Z}}_{\space\text{i}}\right|\left(\omega\right)=\frac{\displaystyle1}{\displaystyle\sqrt{\left(\frac{\displaystyle1}{\displaystyle\text{R}}\right)^2+\left(\omega\text{C}-\frac{\displaystyle1}{\displaystyle\omega\text{L}}\right)^2}}\tag1$$
Now, we can see that \$\$ is at a maximum when:
$$\omega\text{C}-\frac{\displaystyle1}{\displaystyle\omega\text{L}}=0\space\Longrightarrow\space\omega:=\hat{\omega}=\frac{1}{\sqrt{\text{CL}}}\tag2$$
Which gives:
$$\left|\space\underline{\text{Z}}_{\space\text{i}}\right|\left(\hat{\omega}\right)=\text{R}\tag3$$
So, now we can solve for the cut-off frequencies:
$$\left|\space\underline{\text{Z}}_{\space\text{i}}\right|\left(\omega\right)=\frac{\displaystyle\text{R}}{\displaystyle\sqrt{2}}\space\Longrightarrow\space\omega:=\omega_\pm=\frac{\displaystyle1\pm\sqrt{1+\frac{\displaystyle4\text{CR}^2}{\text{L}}}}{\displaystyle2\text{CR}}\tag4$$
So, for the bandwidth we get:
$$\mathcal{B}:=\left|\omega_+-\omega_-\right|=\frac{\displaystyle1}{\displaystyle\text{CR}}\sqrt{1+\frac{\displaystyle4\text{CR}^2}{\displaystyle\text{L}}}\tag5$$
And for the quality factor we get:
$$\mathcal{Q}:=\frac{\displaystyle\hat{\omega}}{\displaystyle\mathcal{B}}=\frac{\displaystyle\frac{1}{\sqrt{\text{CL}}}}{\displaystyle\frac{\displaystyle1}{\displaystyle\text{CR}}\sqrt{1+\frac{\displaystyle4\text{CR}^2}{\displaystyle\text{L}}}}=\text{R}\cdot\sqrt{\frac{\displaystyle\text{C}}{\displaystyle\text{L}+4\text{CR}^2}}\tag6$$
Using your values, we get:
$$\mathcal{Q}=50\cdot\sqrt{\frac{\displaystyle10^{-9}}{\displaystyle10^{-9}+4\cdot10^{-9}\cdot50^2}}=\frac{\displaystyle50}{\displaystyle\sqrt{10001}}\approx0.499975\tag7$$
Why does S-param resonance behavior differ from formula?
Because you have not used the correct formula. You stated this: -
\$w0=\dfrac{1}{LC}=1GHz\$
A more suitable formula is this (note the square root of L and C): -
$$\omega_0=\dfrac{1}{\sqrt{LC}}$$
If you enter the values of 1 nF & 1 nH you get a natural resonant frequency of 159.15 MHz. The above formula yields radians per second and, to get hertz, you must divide by \$2\pi\$.
For the frequency -
The formula you quote gives you the radian frequency. 1 Grad/s = 1G/2pi Hz, roughly the 150 to 160 MHz we can read off the graph.
For the Q -
The inductor has a resistance of 1 ohm. This is probably a default the simulator assigned. As its reactive impedance is only 1 j.ohm at the resonant frequency, the Q will be in the order of 1.
Hint, if you do increase the Q by decreasing that inductor resistance, or increasing the inductance, you will need to improve the frequency resolution around the resonant frequency to be able to see a sharp dip in the S11.
Should you set the inductor resistance to 0 and get an infinite Q, think about what S11 you would expect with the port seeing an effective open circuit at resonance.
