Special cases again The input admittance of a transmission line for various special cases of termination at the load end is shown in Table 2.3 Case Admittance Open circuit load YI = jY0 tan βl Open Circuit λ/4 Line YI →∞ i.e. s/c Short circui...
Stokes theorem When a vector field intersects a non-closed surface S bounded by a contour C as shown in Figure 1.7, and the sense of the contour is chosen so that it is related to the sense of the surface area by the right hand rule, it has been shown by Stokes that Notice that the line integral...
Surface currents A surface current can be regarded as the limiting case when a finite amount of current I flows as shown in Figure 6.2 in a thin slab of area dimensions a and b and thickness t While t is small but not zero we may describe this situation in terms of a volume current density J or ...
TEM wave solutions We now note again that the equations 9.3 which we seek to solve are the same as those of the lossless case studied in Chaper 8, except that ω² has been replaced by ω²−jσ. Thus a corresponding set of solutions, with the same replacement, can be derived. We will quote below the ...
The Field and Material Vectors The names, usual symbols, and Standard International units for the four vectors used to describe electromagnetic fields in a general medium are E electric field intensity Vm−1 H magnetic field intensity Am−1 D electric flux density Cm−2 B magnetic flux density Wb...
The General Case in the Time Domain It is a simple matter, and should be taken as an exercise, to show that Maxwell's equations in integral form applied to these contours lead to the results Et2 − Et1 = 0 (6.1) Dn2 − Dn1 = ρs (6.2) ˆn × (Ht2 −Ht1) = Js (6.3 In the above equations the subscript...
The surface fi eld Since the reflection is not quite complete, there will be a small tangential component of electric field at the surface. We cannot calculate it yet, but we expect that it will be small. The magnetic field will, however, still be almost double the magnetic field corresponding t...
Total current fl ow Now we calculate the total current which flows across the x = 0 plane, per unit length of the y axis. This is K = Z ∞ 0 Jxdz = Z ∞ 0 (1 + j) δ H0e−(1+j)z/δdz = H0 (9.19) In the limit when δ → 0 as σ → ∞ this current becomes a surface current density spread over a...
Transfer along a line To find the input impedance of a line of length l and characteristic impedance Z0 terminated in a load impedance ZL, we normalise ZL with respect to Z0 to obtain zL, enter the Smith Chart at zL, move clockwise by an amount 2βl radians, and read off the nor malised input impe...
Transverse electromagnetic wave solutions We have above eight equations; four of them require and are satisfied by setting both longitudinal components to zero, i.e. Ez = 0 and Hz = 0. Because of these conditions the resulting waves are called Transverse Electromagnetic (abbreviated as TEM) waves...
