**3.1 Top reflection coefficient R**

**Figure 4** shows the upward-moving current wave *iu* arrives at the channel top (at the height h). The current wave is reflected at the open end of the lightning channel. For the current refection, the reflection point moves with the return stroke velocity *v* = *dh/dt* [13].

The increasing lightning channel creates the new channel segment *dh* during the time interval *dt*. The new channel segment is loaded by the charge *dQ*, composed of the charge of the upward moving current wave (*iu*) and the downward moving current wave (*id*). The charge density of the upward moving wave (*qu*) and the downward-moving wave (*qd*) is given by (*c*: speed of light):

$$q\_u = \frac{\dot{\imath}\_u}{\mathcal{c}} \tag{3}$$

$$q\_d = -\frac{i\_d}{c} \tag{4}$$

In Eq. (4), the negative sign is due to the current propagation in the opposite direction compared to the coordinate *z*. The charge *dQ* is given by:

$$dQ = q\_u dh + q\_d dh \tag{5}$$

Hence follows:

$$i = \frac{dQ}{dt} = q\_u \frac{dh}{dt} + q\_d \frac{dh}{dt} = q\_u v + q\_d v = i\_u \frac{v}{c} - i\_d \frac{v}{c} \tag{6}$$

The current is given by:

$$
\dot{\mathbf{u}} = \dot{\mathbf{u}}\_u + \dot{\mathbf{u}}\_d \tag{7}
$$

With Eqs. (6) and (7), the top reflection coefficient results in:

$$R = \frac{i\_d}{i\_u} = -\frac{c - v}{c + v} = -A \tag{8}$$

For example, if we assume the return stroke velocity *v* = *c*/3 = 100 m/μs, the top reflection coefficient is *R* = �0.5, i.e. the half of the current is reflected.

### **3.2 Source current** *iQ* **and channel base current** *iBase*

The TCS model uses the source current (*iQ*) as the input parameter. Because current data are only available from measurements at the striking point, converting the cannel-base current (*iBase*) into the source current (*iQ*) is required. This conversion is deduced in the following.

The channel-base current (*iBase*) is composed of the downward-moving current wave (iBase/d) and the upward-moving current wave (*iBase/u*). With *iBase/u* = *ρ*�*iBase/d*, it follows:

$$\dot{\mathbf{i}}\_{\text{Base}} = \dot{\mathbf{i}}\_{\text{Base}/\text{u}} + \dot{\mathbf{i}}\_{\text{Base}/\text{d}} = (\mathbf{1} + \rho)\dot{\mathbf{i}}\_{\text{Rate}/\text{d}} = \frac{\mathbf{1} + \rho}{\rho}\dot{\mathbf{i}}\_{\text{Base}/\text{u}}\tag{9}$$

The reflected current component ρ�iBase/d(t) arrives at the top of the lightning channel after the delay time T. Because at the time (*t+T*), the distance of propagation *cT* is equal to the height of the lightning channel *h* = *v* (*t+T*), the delay time *T* can be written as:

$$T = \frac{v}{c - v}t\tag{10}$$

After the reflection at the upper end of the lightning channel, the (reflected) current wave *Rρ*�*iBase/d*(*t*) moves down. The total downward moving current wave *id*(*h, t + T*) is composed of this reflected current wave and the current from the current source *iQ*(*t*). This current wave arrives at the ground after the delay time *T*.

At ground level, the downward moving current wave is given by:

$$i\_{Base/d}(t+\mathcal{T}T) = i\_Q(t+T) + R\rho \cdot i\_{Base/d}(t) \tag{11}$$

With Eqs. (8) and (10), it follows:

$$(t+2T) = \frac{t}{A} \tag{12}$$

$$\mathbf{r}(t+T) = \frac{t}{Ak} \tag{13}$$

The coefficient *k* is:

$$k = 1 + \frac{v}{c} \tag{14}$$

Thus, Eq. (11) can be rewritten as:

$$i\_{\text{Baste}/d}\left(\frac{t}{A}\right) - R\rho \cdot i\_{\text{Baste}/d}(t) = i\_Q\left(\frac{t}{Ak}\right) \tag{15}$$

Substituting the time *t*/*A* by the time *t*, it further follows:

$$i\_{\text{Base}/d}(t) - R\rho \cdot i\_{\text{Base}/d}(At) = i\_Q \left(\frac{t}{k}\right) \tag{16}$$

Now, the following iterative procedure is applied to Eq. (16). Between each iteration step, the equation is multiplied by the factor (*Rρ*) and the time is multiplied by the factor *A* resulting in the times *A*<sup>1</sup> *t*, *A*<sup>2</sup> *t*, *A*<sup>3</sup> *t* … :

$$R^1 \rho^1 i\_{\text{Base}/d}(A^1 t) - R^2 \rho^2 \cdot i\_{\text{Base}/d}(A^2 t) = R^1 \rho^1 i\_Q \left(\frac{t}{k} A^1\right) \tag{17}$$

$$R^2 \rho^2 \mathbf{i}\_{\text{Base}/d} \left( A^2 t \right) - R^3 \rho^3 \cdot \mathbf{i}\_{\text{Base}/d} \left( A^3 t \right) = R^2 \rho^2 \mathbf{i}\_Q \left( \frac{t}{k} A^2 \right) \tag{18}$$

$$R^3 \rho^3 i\_{\text{Base}/d} \left( A^3 t \right) - R^4 \rho^4 \cdot i\_{\text{Base}/d} \left( A^4 t \right) = R^3 \rho^3 i\_Q \left( \frac{t}{k} A^3 \right) \tag{19}$$

The addition of the series of equations gives the following formula:

$$i\_{\text{Base}/d}(t) = \sum\_{\nu=0}^{\infty} (\mathcal{R}\rho)^{\nu} \cdot i\_Q \left(\frac{t}{k} A^{\nu}\right) \tag{20}$$

Substituting the time *t* by the time *kt*, Eq. (20) can be rewritten:

$$i\_{Base/d}(kt) = i\_Q(t) + R\rho \cdot i\_Q(At) + \sum\_{\nu=2}^{\infty} (R\rho)^{\nu} i\_Q(A^{\nu}t) \tag{21}$$

Now, Eq. (21) is multiplied by the factor (�*Rρ*), and the time *t* is substituted by the time *At*. It results:

$$-\mathbf{R}\rho \cdot \mathbf{i}\_{\text{Rate}/d}(\mathbf{A}\mathbf{k}t) = -\mathbf{R}\rho \cdot \mathbf{i}\_Q(\mathbf{A}t) - \mathbf{R}\rho \cdot \sum\_{\nu=1}^{\infty} (\mathbf{R}\rho)^{\nu} \mathbf{i}\_Q(\mathbf{A}^{\nu+1}t) \tag{22}$$

and from it

$$-R\rho \cdot i\_{\text{Rate}/d}(Akt) = -R\rho \cdot i\_Q(At) - \sum\_{\nu=2}^{\text{es}} (R\rho)^{\nu} i\_Q(A^{\nu}t) \tag{23}$$

From adding Eqs. (21) and (23), it follows:

$$i\_Q(t) = i\_{Base/d}(kt) - R\rho \cdot i\_{Base/d}(Akt) \tag{24}$$

With Eqs. (9) and (24), the source current can be rewritten as a function of the channel-base current:

$$i\_Q(t) = \frac{\mathbf{1}}{\mathbf{1} + \rho} \left[ i\_{\text{Rate}}(kt) - R\rho \cdot i\_{\text{Rate}}(Akt) \right] \tag{25}$$

With Eqs. (9) and (21), the channel-base current can be rewritten as a function of the source current:

$$i\_{\text{Rate}}(\mathbf{t}) = (\mathbf{1} + \rho) \sum\_{\nu=0}^{\infty} (R\rho)^{\nu} i\_{\text{Q}} \left(\frac{\mathbf{t}}{\mathbf{k}} A^{\nu}\right) \tag{26}$$

Eqs. (25) and (26) are the fundamental equations of the TCS model. They provide the source current *iQ*(*t*) conversion into the channel-base current *iBase*(*t*) and vice versa.

#### **3.3 Lightning current along return stroke channel**

When the downward propagating current wave, *id*(*z, t*), starts from the height *z*, it arrives at the ground after the time delay *z/c*. With Eq. (9), it follows:

$$\dot{\mathbf{i}}\_d(\mathbf{z}, \mathbf{t}) = \dot{\mathbf{i}}\_{\text{Base}/d}(\mathbf{t} + \mathbf{z}/\mathbf{c}) = \frac{\mathbf{1}}{\mathbf{1} + \rho} \dot{\mathbf{i}}\_{\text{Base}}(\mathbf{t} + \mathbf{z}/\mathbf{c}) \tag{27}$$

In the opposite case, when an upward-moving current starts at the time (t-z/c) from ground level (*z* = 0), it arrives at time t at the height z. With Eq. (9), it follows:

$$\dot{\mathbf{u}}\_u(\mathbf{z}, \mathbf{t}) = \dot{\mathbf{u}}\_{\text{Rate}/u}(\mathbf{t} - \mathbf{z}/\mathbf{c}) = \frac{\rho}{\mathbf{1} + \rho} \dot{\mathbf{u}}\_{\text{Rate}}(\mathbf{t} - \mathbf{z}/\mathbf{c}) \tag{28}$$

From adding Eqs. (27) and (28), the total current results:

$$i(z,t) = \frac{1}{1+\rho} \left[ i\_{\text{Rate}} \left( t + \frac{z}{c} \right) + \rho \cdot i\_{\text{Rate}} \left( t - \frac{z}{c} \right) \right] \tag{29}$$

At the upper end of the return stroke channel (*z* = *h*), the current is given by:

$$i(h, t) = \frac{\mathbf{1}}{\mathbf{1} + \rho} \left[ i\_{\text{Base}} \left( t + \frac{h}{c} \right) + \rho \cdot i\_{\text{Base}} \left( t - \frac{h}{c} \right) \right] \tag{30}$$

With the channel height *h* = *vt,* it follows:

$$i(h, t) = \frac{1}{\mathbf{1} + \rho} \left[ i\_{\text{Base}}(kt) + \rho \cdot i\_{\text{Base}}(Akt) \right] \tag{31}$$

#### **3.4 Special case of no ground reflections**

The reflections at the ground are often ignored, and the ground reflection coefficient is set to *ρ* = 0. In this case, the relation between the current along the

lightning channel and the channel-base current can be simplified. From Eq. (29), it results for *z* ≤ *h*:

$$\dot{i}(z,t) = \dot{i}\_{Bas} \left(t + \frac{z}{c}\right) \tag{32}$$

From Eq. (25), the relation between the source current (*iQ*) and the channel-base current (*iBase*) follows to:

$$i\_Q(t) = i\_{Base}(kt)\tag{33}$$
