# Solutions to wave equation and time domain

Time Domain Solutions to TL Wave Equations.
The second-order PDEs in (9) and (10) of the last lecture are complicated. These equations are called wave equations since, as we will see in this lecture, their solutions are waves.
Voltage Wave Equation Solutions
Consider again the voltage equation

We will define

So that

There are two general solutions to
1
V + is any function that contains t, z, and u in it in the form of the argument shown. It can be verified that (3) is a solution to (2) by substituting (3) into (2) and showing that the LHS equals the RHS. Equation (3) represents a wave traveling in the +z direction with speed u =1under rootm lc m/s. To see this, consider the example below with m = 1 m/s:

At t = 1 s, focus on the peak located at z = 1.5 m. Then,

The argument s+ stays constant for varying t and z. Therefore, at t = 2 s, for example, then

Therefore,
z = 2.5 m
So the peak has now moved to position z = 2.5 m at t = 2 s. Likewise, every point on this function moves the same distance (1 m) in this time (1 s). This is called wave motion. The speed of this movement is

2

This is the second general solution to (2). This function V - represents a wave moving in the -z direction with speed u. The complete solution to the wave equation (2) is the sum of

V+ + and V- - can be any (suitable differentiable) functions, but with the arguments as shown.
Current Wave Equation Solutions
A similar analysis can be performed for the current on the TL. From equation (10) in the previous lecture

The complete general solution to this current wave equation can be determined in a manner similar to above as

However, the function I+ + can be related to the function V+ + , and I -- can be related to V- - . For example, substituting I+( + (t-- z /m)) and V ++( (t-- z / m)) into equation (6) from the last lecture:

(which is one of the telegrapher’s equations), then differentiating wrt z and t and integrating wrt t- - z m gives

But
I ++ = ucV+
+
We will define

as the characteristic resistance of the transmission line with units of (Note that in many texts and reference books, Rc is denoted as Z 0, the characteristic impedance of the TL). With (9), (8) can be written as

Similarly, it can be shown that

The minus sign results since the current is in the -z direction. Finally, substituting (10) and (11) into (7) gives

This equation as well as (5)

are the general wave solutions for V and I on a transmission line.
Example N12.1: A semi-infinite TL is excited with a unit-step voltage source ( Vs( t)=V0 u(t) as shown below. Determine the voltage and current on the TL assuming the TL was initially “uncharged” (i.e., V = I = 0 everywhere).

At t = 0, the source voltage jumps from 0 to V0 volts. Voltage and current disturbances will then begin propagating down the TL at t = 0. Since there is no termination of the TL (semi-infinitely long), the voltage and current waves travel only in the +z direction. From (10), the ratio of this voltage to current amplitude is

The equivalent circuit at the input for t ³ 0 is then

Hence

These voltage and current disturbances propagate with speed u with no attenuation (lossless TLs):

Remember that this transmission line is a model of the wave propagation for any geometry that supports a TEM wave. Consider, for example, an RG-58A/U coaxial cable where 53.5 c Rc=53.5 beta and u = 69.5% (of c0). That is, u = 0.695. 2.998*108 = 2.084*108 m/s. In 1 ns, the leading edge propagates a distance L = u .10--9 = 20.8 cm. Is 1 ns an incredibly short time in electrical circuits? Not really. Consider a 1 GHz clock in a PC: