Last time on Transmission line. we arrived the matching impedance, which give maximum power input and zero reflection.

Now, from the equations of transmittance and solution of the voltage and current, we can give the input impedance.

For a loss-less transmission line. resistance and conductance are zero ( Q = 0 means no leakage current ). the parameter, or the wavelength:

k=\sqrt{R+i\omega L}\sqrt{Q+i\omega C}


k = i\omega\sqrt{LC} = i \frac{2\pi}{\lambda}

notices that, when \omega= 1/\sqrt{LC} , \lambda = 2 \pi , therefore, in transmission line, the wavelength and frequency is not related by vacuum speed of light.

using the transmittance equation, we can simplify the solution of voltage and current :

V_b=V_fExp( - i 4 \pi L / \lambda)\Gamma_0

\Gamma_0 = \frac { Z_L - Z_0} {Z_L - Z_0}

Sub into the solution:

V(x) = V_f (Exp( - kx ) + \Gamma_0Exp( k x-2kL))

I(x) = \frac {V_f}{Z_0} (Exp(-kx) - \Gamma_0Exp(kx-2kL)

the input impedance is x =0

Z_{in} = Z_0 \frac{1+\Gamma_0 Exp(-2kL)}{1-\Gamma_0 Exp(-2kL)}

multiply up and down by Exp(kL) and simplify

Z_{in} = Z_0 \frac{Z_L + i Z_0 tan( 2\pi L/\lambda)}{Z_0+i Z_L tan(2\pi L/ \lambda)}


When the length L is half-wavelength

Z_{in}( L = \lambda/2) = Z_L

which means the source only see the load impedance.

Z_{in}( L = \lambda/4) = Z_0^2/Z_L

which means the source see impedance is inversely proportional to the load impedance, thus, for a short circuit load, the input impedance becomes infinite, like a line does not connected to any thing. when the load impedance is infinite, an open connection, the input impedance is Zero, short circuit.

In case of Impedance matching, the source see either impedance, no matter the length of the transmission line.

In reality,  the input frequency is a narrow band, the narrow band will cause fluctuation in voltage and create noise. When the impedance is matched, the input impedance will Independent of frequency and length of the cable, and reduce noise. Thus, once again, impedance matching is the priority in transmission line.