Chapter2_Notes

# Inner and outer conductors r rs 2 11 ab m 3

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Unformatted text preview: ndence! Rs = π f µc (Ω) σc Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. (4) Electromagnetics I: Transmission lines 15 • What happens when σc → ∞? • Rs → 0 and R → 0. • For inductance per unit length we have L= µ ln 2π b a (H/m) (5) • G is shunt conductance, i.e. current ﬂowing between two conductors. 2πσ (S/m) (6) G= ln(b/a) • What happens if we have perfect dielectric material? Then σ = 0 and therefore G = 0. • C is the capacitance between two conductors. In this case C= 2π (F/m) ln(b/a) Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. (7) Electromagnetics I: Transmission lines 16 • Not only coax, but all TEM lines have the following relations: LC =µ , σ G = C (8) • If the insulating material between the conductors is air the transmission line is called and air line (free space parameters and σ = 0, G = 0). Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines N i(z, t) + v(z , t) N+1 R’∆z 17 i(z+∆z, t) L’∆z G’∆z C’∆z - + v(z + ∆z, t) ∆z Figure 8: ∆z section of a transmission line. Figure 2-8 1.3. Transmission line equations Now that we have an equivalent circuit (model) for transmission lines, what do we do? Start with a segment shown in Fig. 8 and write some Kirchhoﬀ’s equations for voltages and currents. • The voltage drop between nodes N and N + 1 gives the ﬁrst Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 18 equation: v (z, t) − R ∆z i(z, t) − L ∆z − ∂i(z, t) − v (z + ∆z, t) = 0 ∂t v (z + ∆z, t) − v (z, t) ∂ i(z, t) = R i(z, t) + L ∆z ∂t (9) (10) • If we take the limit as ∆z → 0 this becomes a diﬀerential equation: ∂v (z, t) ∂ i(z, t) − = R i(z, t) + L (11) ∂z ∂t Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 19 • The sum of the currents at node N + 1 gives the 2nd equation i(z, t) − i(z + ∆z, t) − iG − iC = 0 (12) Where the current in the resistor is (use Ohm’s law V = IR where R = 1/G), iG = v (z + ∆z, t) = G ∆z v (z + ∆z, t) 1/G ∆z (13) For the capacitor, start with equation relating voltage, charge and capacitance, V = Q/C or Q = CV , iC = C ∆z dv (z + ∆z, t) dt (14) Rearrange and divide by ∆z , i(z, t) − i(z +∆z, t) − G ∆z v (z +∆z, t) − C ∆z ∂v (z + ∆z, t) =0 ∂t (15) Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 20 i(z + ∆z, t) − i(z, t) ∂ v (z + ∆z, t) = G v (z +∆z, t)+C (16) ∆z ∂t and let ∆z approach zero, so we get a partial diﬀerential equation, ∂i(z, t) ∂ v (z, t) = G v (z, t) + C (17) − ∂z ∂t (note, typo in book on page 55 should say Eq. 2.15 becomes the s...
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