Question

The correction factors $F_1=\frac{sinh\left(\gamma l\right)}{\gamma l}$ and $F_2=\frac{tanh\left(\frac{\gamma l}{2}\right)}{\frac{\gamma l}{2}}$ which are complex numbers, have the units of _______

Short answer

Therefore, the correction factors are “dimensions less constants (per unit).”

Step-by-step solution

Determine the expression for the correction factors.

The expression for $sinh\left(\gamma l\right)$is,

$sin\left(\gamma l\right)=\frac{e^{\gamma l}-e^{-\gamma l}}{2}$

…… (1)

Here, $\gamma$ is the propagation constant and I is the length of the transmission line.

The expression for $tanh\left(\frac{\gamma l}{2}\right)$ is,

$tanh\left(\gamma l\right)=\frac{e^{\frac{\gamma l}{2}}-e^{-\frac{\gamma l}{2}}}{e^{\frac{\gamma l}{2}}+e^{-\frac{\gamma l}{2}}}$

…… (2)

Determine the unit of the correction factors.

The correction factor $f_1$is given by,

$F_1=\frac{\sin \left(\gamma l\right)}{\gamma l}$

Substitute$\frac{e^{\gamma l}-e^{-\gamma l}}{2}$ for$\sin \left(\gamma l\right)$ into above expression,

$F_1=\frac{e^{\gamma l}-e^{-\gamma l}}{2\left(\gamma l\right)}$

The correction factor $F_2$is given by,

$F_1=\frac{\tan \left(\frac{\gamma l}{2}\right)}{\frac{\gamma l}{2}}$

Substitute $\frac{e^{\frac{\gamma l}{2}}-e^{-\frac{\gamma l}{2}}}{e^{\frac{\gamma l}{2}}+e^{-\frac{\gamma l}{2}}}$for $\tan \left(\frac{\gamma l}{2}\right)$into above expression,

$\begin{array}{l}{F}_2=\frac{\frac{e^{\frac{\gamma l}{2}}-e^{-\frac{\gamma l}{2}}}{e^{\frac{\gamma l}{2}}+e^{-\frac{\gamma l}{2}}}}{\frac{\gamma l}{2}}\\ F_2=\frac{e^{\frac{\gamma l}{2}}-e^{-\frac{\gamma l}{2}}}{\left(e^{\frac{\gamma l}{2}}+e^{-\frac{\gamma l}{2}}\right)\left(\frac{\gamma l}{2}\right)}\end{array}$

From the above discussion, the correction factors are dimensions less constants (per unit).