Electrical Length

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Electrical Length of the line in meters

We can express the physical length of the line in meters. However, in high-frequency electronics (microwave engineering), we usually convert this length to the fraction of a wavelength of a signal that is traveling on the line.

$\begin{eqnarray} l=N \lambda \end{eqnarray}$

Typically, but not always, $N$ is a fraction, for example, $N=\frac {1}{2}=0.5$ , $N=\frac {1}{4}=0.25$ or $N=\frac {1}{8}=0.125$ ; although it can be any The lenght of the line is then written as

$\begin{eqnarray} l=\frac{\lambda}{2} \\ l\frac{\lambda}{4} \\ l=\frac{\lambda}{8} \end{eqnarray}$

If the physical length of the line is $l=\frac {\lambda }{4}$ , we say: This line is quarter-wavelength long at 1GHz, meaning one-quarter of the wavelength fits on the line. We could also say that the line is 7.5cm long, as wavelength is $\lambda =30 \unit {cm}$ at $1$ GHz.

When we say quarter-wavelength long, we refer to the lines physical length at a specific frequency.

Electrical length of the line in degrees

The phase shift between input and output signal on a transmission line is $\Theta =\beta *l$ . $beta$ is called the phase constant. It represents the spatial frequency of the signal. $\Theta =\beta *L$ is the phase in degrees or radians (related is a time delay in seconds). $\Theta$ can be, for example $\Theta = 45^0$ , $\Theta = 90^0$ , $\Theta = 180^0$ . $\Theta$ is a function of frequency, because $\beta$ is a function of frequency. If $\Theta = 90^0$ , we say: The line is 90 degrees long at 1GHz, meaning the output signal at 1GHz will be shifted for $90^0$ with respect to the input signal. When we say $90^0$ , we refer to the lines electrical length, representing the number of degrees that the line introduces between the input and the output signal.

What is the electrical length of a 30 cm line in terms of the fraction of wavelength at 1 GHz? What is the electrical length of the line at 1GHz?

Wavelength at 1 GHz, assuming the wave is propagating in air is $\lambda =\frac {c}{f}=30$ cm. Since the line is also l=30 cm long, the length of the line in terms of wavelength is $\frac {l}{\lambda }=1$ , or $l=\lambda$ .

The electrical length of the line is $\theta =\beta l = \frac {2 \pi }{\lambda } \lambda = 2 \pi = 360^0$ .

What is the electrical length of a 15 cm line in terms of the fraction of wavelength at 1 GHz? What is the electrical length of the line at 1GHz?

Wavelenght at 1 GHz, assuming the wave is propagating in air is $\lambda =\frac {c}{f}=30$ cm. Since the line is 15 cm long, the length of the line in terms of wavelenth is $\frac {l}{\lambda }=\frac {1}{2}$ , or $l=\frac {\lambda }{2}$ .

The electrical length of the line is $\theta =\beta l = \frac {2 \pi }{\lambda } \frac {\lambda }{2} = \pi = 180^0$ .

What is the electrical length of a 7.5 cm line in terms of the fraction of wavelength at 1 GHz? What is the electrical length of the line at 1GHz?

Wavelength at 1 GHz, assuming the wave is propagating in air is $\lambda =\frac {c}{f}=30$ cm. Since the line is 7.5 cm long, the line’s length in terms of wavelenth is $l=\frac {\lambda }{4}$ .

The electrical length of the line is $\theta =\beta l = \frac {2 \pi }{\lambda } \frac {\lambda }{4} = \pi /2 = 90^0$ .