to the time-dependent part so the solution reads where and are the constants of integration for the forward moving (+) and backward moving (−) waves, as in the prior section. When we recombine the time-dependent part we obtain the full solution: Since the equation for is the same form, it has a solution of the same form: where and are again constants of integration. The above equations are the wave solution for and . In order to be compatible, they must still satisfy the original differential equations, one of which is Substituting the solutions for and into the above equation, we get or Isolating distinct powers of and combining identical powers, we see that in order for the above equation to hold for all possible values of we must have: For the co-efficients of For the co-efficients of Since hence, for valid solutions require It can be seen that the constant , defined in the above equations has the dimensions of impedance (ratio of voltage to current) and is a function of primary constants of the line and operating frequency. It is called the “characteristic impedance” of the transmission line, and conventionally denoted by . which holds generally, for any transmission line. For well-functioning transmission lines, with either and both very small, or with very high, or all of the above, we get hence the characteristic impedance is typically very close to being a real number. Manufacturers make commercial cables to approximate this condition very closely over a wide range of frequencies. Alternative approach We follow an approach posted by Tim Healy. The line is modeled by a series of differential segments with differential series and shunt elements (as shown in the figure above). The characteristic impedance is defined as the ratio of the input voltage to the input current of a semi-infinite length of line. We call this impedance That is, the impedance looking into the line on the left is But, of course, if we go down the line one differential length the impedance into the line is still Hence we can say that the impedance looking into the line on the far left is equal to in parallel with and all of which is in series with and Hence: The added terms cancel, leaving The first-power terms are the highest remaining order. Dividing out the common factor of and dividing through by the factor we get In