I am reading about the resonant peaks in the RLC circuits. Stumbled upon this link.
Below is the frequency response of a series RLC circuit, voltage measured across the capacitor.
I just want to understand what is the blue trace and the red one. How is there a peak in the blue trace? And how to understand the intuitive behaviour of the RLC circuit to understand how the peak appears?
This is the circuit image is for your plot (also from the link in your question) shown for completion: -
Going back to your bode-type-plot...
I just want to understand what is the blue trace and the red one.
The blue trace is the true magnitude response of the output when the value of the resistor is quite low in value. I can't be numerically precise about this because it depends on the values of L and C. Nevertheless this is called the under-damped case and is typified by the resonant peak that you observe in blue.
The red trace is an approximate "straight-line" version of the bode plot that we sometimes use. These are often referred to as asymptotes and are just an approximation to the real response and don't include the resonant peaks.
I will also add that the graph in your image is for the square of the transfer function and I don't find that this is helpful at all. If you want to play with a real bode-plot then alter the values in the calculator in my basic website: -
The above gives the true response and all the incidental values associated with this type of low-pass filter.
And how to understand the intuitive behaviour of the RLC circuit to understand how the peak appears?
This can take a while to sink in but, a simple way to imagine it is when the value of resistance is very low. So, at resonance the L and C act as a series tuned circuit across the input source and, if R is very small in value there will be a large flow of current (much more than at other parts of the spectrum) and, this generates the large peak in output voltage.
Well, notice that the transfer function of that circuit is given by:
$$\mathscr{H}\left(\text{s}\right):=\frac{\displaystyle\text{V}_\text{o}\left(\text{s}\right)}{\displaystyle\text{V}_\text{i}\left(\text{s}\right)}=\frac{\displaystyle\frac{\displaystyle1}{\displaystyle\text{sC}}}{\displaystyle\frac{\displaystyle1}{\displaystyle\text{sC}}+\text{sL}+\text{R}}=\frac{\displaystyle1}{1+\text{CRs}+\text{CL}\text{s}^2}\tag1$$
So, for the magnitude we get:
$$\left|\mathscr{H}\left(\text{j}\omega\right)\right|=\frac{\displaystyle1}{\displaystyle\sqrt{\left(1-\text{CL}\omega^2\right)^2+\left(\text{CR}\omega\right)^2}}\tag2$$
The peak, \$\hat{\omega}\$, can be found at:
$$\frac{\displaystyle\partial\left|\mathscr{H}\left(\text{j}\hat{\omega}\right)\right|}{\displaystyle\hat{\omega}}=0\space\Longrightarrow\space\hat{\omega}=\frac{\displaystyle1}{\displaystyle\text{L}}\sqrt{\frac{\displaystyle\text{L}}{\displaystyle\text{C}}-\frac{\displaystyle\text{R}^2}{\displaystyle2}}\tag3$$
Where we have to assume that \$\displaystyle\text{CR}^2<2\text{L}\$, because otherwise the term in the square root will become negative and there will be no peak.
