# Basics of impedance matching

Discussion in 'Wireless & RF Design' started by simo_x, Apr 3, 2015.

1. ### simo_x Thread Starter Member

Dec 23, 2010
200
6
Hi all,

I started studying impedance matching and I am facing with a doubt with the basic circuit as shown below:

In the image, $\text i X_L$ should be $-\text i X_L$
I know that at the frequency of resonance for both components I should have:


\begin{align}
& R_G + \text jX_G + R_L - \text jX_L = R_G + R_L \\
& v_o = {v_i R_L \over R_G + R_L}
\end{align}

Given a frequency of 1 MHz and:

\begin{align}
& R_G = R_L = 50\,\Omega\\
& \omega_0 = 1\,\text{MHz}\\
& Z_L = \text i\omega L = 100\text i \Rightarrow L = 100\,\mu\text{H}\\
& Z_C = -{1 \over \text i\omega C} = -100\text i \Rightarrow C = 10\,\text{nF}
\end{align}

But I tried to put all this stuff into LTspice to see if my result should be OK, but I see that is not.
This is the voltage plot with respect to the frequency..

Of course the bode plot shows a cut off frequency at approximately 100 kHz.
I know there is something wrong, I think I misunderstood some concept..

Could you help me to understand ?
Thank you !

2. ### simo_x Thread Starter Member

Dec 23, 2010
200
6
I am a stupid. I did not consider the $2\pi$ coefficient .

3. ### alfacliff Well-Known Member

Dec 13, 2013
2,449
428
not stupid, just learning.

simo_x likes this.
4. ### simo_x Thread Starter Member

Dec 23, 2010
200
6
Yes, of course : )

Here is the screenshot. I also was visualizing the voltage in the wrong point of the network..

: )