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:

    impedanceMatch.png

    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:

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

    Given a frequency of 1 MHz and:
    <br />
\begin{align}<br />
& R_G = R_L = 50\,\Omega\\<br />
& \omega_0 = 1\,\text{MHz}\\<br />
& Z_L = \text i\omega L = 100\text i \Rightarrow L = 100\,\mu\text{H}\\<br />
& Z_C = -{1 \over \text i\omega C} = -100\text i \Rightarrow C = 10\,\text{nF}<br />
\end{align}<br />

    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..

    voltage_freq.png

    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..

    WinXp4.png

    : )
     
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