Wave Guides
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Wave Guides Lecture37: Electromagnetic Theory Professor D. K. Ghosh, Physics Department, I.I.T., Bombay Rectangular Waveguide In the last lecture, we have coonsidered the case of guided wave between a pair of infinite conducting planes. In this lecture, we consdier a rectangular wave guide which consists of a hollow pipe of infinite extent but of rectangular cross section of dimension ππ × ππ. The long direction will be taken to be the z direction. y x b a z Unlike in the previous case ππ⁄ππππ is not zero in this case. However, as the propagation direction is along the z direction, we have equations as ππ ππππ → −πΎπΎ and ππ ππππ → ππππ. We can write the Maxwell’s curl πππ»π»π§π§ + πΎπΎπ»π»π¦π¦ = πππππππΈπΈπ₯π₯ ππππ πππ»π»π§π§ −πΎπΎπ»π»π₯π₯ − = πππππππΈπΈπ¦π¦ ππππ πππ»π»π¦π¦ πππ»π»π₯π₯ − = πππππππΈπΈπ§π§ ππππ ππππ The second set of equations are obtained from the Faraday’s law, (these can be written down from above by πΈπΈ ↔ π»π», ππ → −ππ πππΈπΈπ§π§ + πΎπΎπΈπΈπ¦π¦ = −πππππππ»π»π₯π₯ ππππ πππΈπΈπ§π§ −πΎπΎπΈπΈπ₯π₯ − = −πππππππ»π»π¦π¦ ππππ πππΈπΈπ¦π¦ πππΈπΈπ₯π₯ − = −πππππππ»π»π§π§ ππππ ππππ 1 As in the case of parallel plate waveguides, we can express all the field quantities in terms of the derivatives of πΈπΈπ§π§ andπ»π»π§π§ . For instance, we have, πππππππΈπΈπ₯π₯ = πΎπΎπ»π»π¦π¦ + = which gives, so that where οΏ½ππππππ − πππ»π»π§π§ ππππ πππ»π»π§π§ πππΈπΈπ§π§ πΎπΎ οΏ½πΎπΎπΈπΈπ₯π₯ + οΏ½+ ππππ ππππ ππππππ πΎπΎ πππΈπΈπ§π§ πππ»π»π§π§ πΎπΎ 2 οΏ½ πΈπΈπ₯π₯ = + ππππππ ππππ ππππππ ππππ πΈπΈπ₯π₯ = − πΎπΎ πππΈπΈπ§π§ ππππππ πππ»π»π§π§ − 2 ππ 2 ππππ ππ ππππ (1) ππ 2 = πΎπΎ 2 + ππ2 ππππ The other components can be similarly written down, πΎπΎ πππΈπΈπ§π§ ππππππ πππ»π»π§π§ − 2 ππ 2 ππππ ππ ππππ ππππππ πππΈπΈπ§π§ πΎπΎ πππ»π»π§π§ − π»π»π₯π₯ = 2 ππ ππππ ππ 2 ππππ ππππππ πππΈπΈπ§π§ πΎπΎ πππ»π»π§π§ − 2 π»π»π¦π¦ = − 2 ππ ππππ ππ ππππ πΈπΈπ¦π¦ = − (2) (3) (4) As before, we will look into the TE mode in detail. Since πΈπΈπ§π§ = 0, we need to solve for π»π»π§π§ from the Helmholtz equation, ππ 2 ππ 2 οΏ½ 2 + 2 + ππ 2 οΏ½ π»π»π§π§ (π₯π₯, π¦π¦) = 0 (5) πππ₯π₯ πππ¦π¦ Remember that the complete solution is obtained by multiplying with ππ −πΎπΎπΎπΎ +ππππππ . We solve equation (5) using the technique of separation of variables which we came across earlier. Let π»π»π§π§ (π₯π₯, π¦π¦) = ππ(π₯π₯)ππ(π¦π¦) Substituting this in (5) and dividing by ππππ throughout, we get, 1 ππ 2 ππ 1 ππ 2 ππ 2 + ππ = − ≡ πππ¦π¦2 ππ πππ¦π¦ 2 ππ πππ₯π₯ 2 where πππ¦π¦ is a constant. We further define πππ₯π₯2 = ππ 2 − πππ¦π¦2 2 We now have two second order equations, ππ 2 ππ + πππ₯π₯2 ππ = 0 πππ₯π₯ 2 ππ 2 ππ + πππ¦π¦2 ππ = 0 πππ¦π¦ 2 The solutions of these equations are well known ππ(π₯π₯) = πΆπΆ1 cos πππ₯π₯ π₯π₯ + πΆπΆ2 sin πππ₯π₯ π₯π₯ ππ(π¦π¦) = πΆπΆ3 cos πππ¦π¦ π¦π¦ + πΆπΆ4 sin πππ¦π¦ π¦π¦ This gives, π»π»π§π§ (π₯π₯, π¦π¦) = πΆπΆ1 πΆπΆ3 cos πππ₯π₯ π₯π₯ cos πππ¦π¦ π¦π¦ + πΆπΆ1 πΆπΆ4 cos πππ₯π₯ π₯π₯ sin πππ¦π¦ π¦π¦ +πΆπΆ2 πΆπΆ3 sin πππ₯π₯ π₯π₯ cos πππ¦π¦ π¦π¦ + πΆπΆ2 πΆπΆ4 π π π π π π πππ₯π₯ π₯π₯ sin πππ¦π¦ π¦π¦ The boundary conditions that must be satisfied to determine the constants is the vanishing of the tangential component of the electric field on the plates. In this case, we have two pairs of plates. The tangential direction on the plates at π₯π₯ = 0 and π₯π₯ = ππ is the y direction, so that the y component of the electric field πΈπΈπ¦π¦ = 0 at π₯π₯ = 0, ππ . Likewise, on the plates at π¦π¦ = 0 and π¦π¦ = ππ, πΈπΈπ₯π₯ = 0 at π¦π¦ = 0, ππ We need first to evaluate πΈπΈπ₯π₯ and πΈπΈπ¦π¦ using equations (1) and (2) and then substitute the boundary conditions. Since πΈπΈπ§π§ = 0, we can write eqn. (1) and (2) as ππππππ πππ»π»π§π§ πΈπΈπ₯π₯ = − 2 ππ ππππ ππππππ πππ»π»π§π§ πΈπΈπ¦π¦ = − 2 ππ ππππ ππππππ πΈπΈπ₯π₯ = − 2 [−πΆπΆ1 πΆπΆ3 k y cos πππ₯π₯ π₯π₯ sin πππ¦π¦ π¦π¦ + πΆπΆ1 πΆπΆ4 πππ¦π¦ cos πππ₯π₯ π₯π₯ cos πππ¦π¦ π¦π¦ ππ −πΆπΆ2 πΆπΆ3 k y sin πππ₯π₯ π₯π₯ sin πππ¦π¦ π¦π¦ + πΆπΆ2 πΆπΆ4 πππ¦π¦ sin πππ₯π₯ π₯π₯ cos πππ¦π¦ π¦π¦ ππππππ [−πΆπΆ1 πΆπΆ3 k x sin πππ₯π₯ π₯π₯ cos πππ¦π¦ π¦π¦ − πΆπΆ1 πΆπΆ4 πππ₯π₯ sin πππ₯π₯ π₯π₯ sin πππ¦π¦ π¦π¦ ππ 2 +πΆπΆ2 πΆπΆ3 k x cos πππ₯π₯ π₯π₯ cos πππ¦π¦ π¦π¦ + πΆπΆ2 πΆπΆ4 πππ₯π₯ ππππππ πππ₯π₯ π₯π₯ sin πππ¦π¦ π¦π¦] πΈπΈπ¦π¦ = − Since πΈπΈπ¦π¦ = 0 at π₯π₯ = 0, we must have πΆπΆ2 = 0 and then we get, ππππππ πΈπΈπ¦π¦ = − 2 [−πΆπΆ1 πΆπΆ3 k x sin πππ₯π₯ π₯π₯ cos πππ¦π¦ π¦π¦ − πΆπΆ1 πΆπΆ4 πππ₯π₯ sin πππ₯π₯ π₯π₯ sin πππ¦π¦ π¦π¦] ππ Further, since πΈπΈπ₯π₯ = 0 at y=0, we have πΆπΆ4 = 0. Combining these, we get, on defining a constant πΆπΆ = πΆπΆ1 πΆπΆ3 3 ππππππ πΆπΆπππ¦π¦ cos πππ₯π₯ π₯π₯ sin πππ¦π¦ π¦π¦ ππ 2 ππππππ πΈπΈπ¦π¦ = 2 πΆπΆπππ₯π₯ sin πππ₯π₯ π₯π₯ cos πππ¦π¦ π¦π¦ ππ πΈπΈπ₯π₯ = and π»π»π§π§ = πΆπΆ cos πππ₯π₯ π₯π₯ cos πππ¦π¦ π¦π¦ We still have the boundary conditions, πΈπΈπ₯π₯ = 0 at π¦π¦ = ππ and πΈπΈπ¦π¦ = 0 at π₯π₯ = ππto be satisfied. The former gives πππ¦π¦ = and ππππ ππ ππππ while the latter gives πππ₯π₯ = , where m and n are integers. Thus we have, ππ ππππ ππππππ οΏ½ π₯π₯οΏ½ cos οΏ½ π»π»π§π§ = πΆπΆ cos οΏ½ ππ ππ ππ 2 = πππ₯π₯2 + πππ¦π¦2 = οΏ½ However,ππ 2 = ππ2 ππππ + πΎπΎ 2 , so that, πΎπΎ = οΏ½οΏ½ ππππ 2 ππππ 2 οΏ½ +οΏ½ οΏ½ ππ ππ ππππ 2 ππππ 2 οΏ½ + οΏ½ οΏ½ − ππ 2 ππππ ππ ππ For propagation to take place, πΎπΎ must be imaginary, so that the cutoff frequency below which propagation does not take place is given by ππππ = 1 ππππ 2 ππππ 2 οΏ½οΏ½ οΏ½ + οΏ½ οΏ½ ππ ππ √ππππ The minimum cutoff is for TE1,0 (or TE0,1 ) mode which are known as dominant mode. For these modes πΈπΈπ₯π₯ (or πΈπΈπ¦π¦ ) is zero. TM Modes We will not be deriving the equations for the TM modes for which π»π»π§π§ = 0 . In this case, the solution for πΈπΈπ§π§ , becomes, ππππ ππππ πΈπΈπ§π§ = πΈπΈπ§π§0 sin οΏ½ οΏ½ sin οΏ½ οΏ½ ππ ππ As the solution is in terms of product of sine functions, neither m nor n can be zero in this case. This is why the lowest TE mode is the dominant mode. For propagating solutions, we have, where, π½π½ = οΏ½ππ 2 ππππ − οΏ½ ππππ = ππππ 2 ππππ 2 οΏ½ − οΏ½ οΏ½ = οΏ½ππππ οΏ½ππ 2 − ππππ2 ππ ππ 1 ππππ 2 ππππ 2 οΏ½οΏ½ οΏ½ + οΏ½ οΏ½ ππ ππ √ππππ 4 We have, ππ2 = π½π½ 2 ππππ + ππππ2 . Differentiating both sides, we have, The group velocity of the wave is given by ππ 1 ππππ = π½π½ ππππ ππππ 2 ππππ π½π½ 1 ππ2 √πππποΏ½ππ 2 − ππππ οΏ½1 − ππ = = = ππππ ππππππ ππππππ ππ 2 √ππππ which is less than the speed of light. The phase velocity, however, is given by ππ ππ 1 1 = π£π£ππ = = 2 π½π½ √πππποΏ½ππ 2 − ππππ2 √ππππ οΏ½1 − ππ ππ2 π£π£ππ = It may be noted that π£π£ππ π£π£ππ = 1 ππππ ππ , which in vacuum equal the square velocity of light. For propagating TE mode, we have, from (1) and (4) using πΈπΈπ§π§ = 0 πΈπΈπ₯π₯ ππππππ ππππ ππ = = =οΏ½ πΎπΎ π½π½ π»π»π¦π¦ ππ 1 οΏ½1 − ππ ππ2 ππ 2 ≡ ππ ππππ where ππ ππππ is the characteristic impedance for the TE mode. It is seen that the characteristic impedance is resistive. Likewise, πΈπΈπ¦π¦ = −ππ ππππ π»π»π₯π₯ Power Transmission We have seen that in the propagating mode, the intrinsic impedance is resistive, indicating that there will be average flow of power. The Poynting vector for TE mode is given by 1 οΏ½β∗ ⟩ ⟨ππ⟩ = Re⟨πΈπΈοΏ½β × π»π» 2 1 = ReοΏ½πΈπΈπ₯π₯ π»π»π¦π¦∗ − πΈπΈπ¦π¦ π»π»π₯π₯∗ οΏ½ 2 1 οΏ½|πΈπΈπ₯π₯2 | + οΏ½πΈπΈπ¦π¦2 οΏ½οΏ½ = 2ππ ππππ Substituting the expressions for the field components, the average power flow through the x y plane is given by ππ ππ 1 ⟨ππ⟩ = ∫ ⟨ππ⟩ππππ ππππ = οΏ½ οΏ½ οΏ½|πΈπΈπ₯π₯2 | + οΏ½πΈπΈπ¦π¦2 οΏ½οΏ½ππππ ππππ 2ππ ππππ 0 0 2 2 2 1 πΆπΆ ππ ππ ππππ 2 ππ ππ ππππ ππππ οΏ½οΏ½ οΏ½ οΏ½ οΏ½ cos 2 οΏ½ π₯π₯οΏ½ sin2 οΏ½ π¦π¦οΏ½ ππππ ππππ = 4 2ππ ππππ ππ ππ ππ ππ 0 0 ππ ππ ππππ ππππ ππππ 2 + οΏ½ οΏ½ οΏ½ οΏ½ sin2 οΏ½ π₯π₯οΏ½ cos 2 οΏ½ π¦π¦οΏ½ ππππ ππππ οΏ½ ππ ππ ππ 0 0 5 = 1 πΆπΆ 2 ππ2 ππ2 ππππ 2 ππππ 2 ππππ οΏ½οΏ½ οΏ½ οΏ½ οΏ½ οΏ½ + 2ππ ππππ ππ 4 ππ 4 ππ Impossibility of TEM mode in Rectangular waveguides We have seen that in a parallel plate waveguide, a TEM mode for which both the electric and magnetic fields are perpendicular to the direction of propagation, exists. This, however is not true of rectangular wave guide, or for that matter for any hollow conductor wave guide without an inner conductor. We know that lines of H are closed loops. Since there is no z component of the magnetic field, such loops must lie in the x-y plane. However, a loop in the x-y plane, according to Ampere’s law, implies an axial current. If there is no inner conductor, there cannot be a real current. The only other possibility then is a displacement current. However, an axial displacement current requires an axial component of the electric field, which is zero for the TEM mode. Thus TEM mode cannot exist in a hollow conductor. (for the parallel plate waveguides, this restriction does not apply as the field lines close at infinity.) Resonating Cavity In a rectangular waveguide we had a hollow tube with four sides closed and a propagation direction which was infinitely long. We wil now close the third side and consider electromagnetic wave trapped inside a rectangular parallopiped of dimension ππ × ππ × ππ with walls being made of perfector conductor. (The third dimension is taken as d so as not to confuse with the speed of light c). Resonanting cavities are useful for storing electromagnetic energy just as LC circuit does but the former has an advantage in being less lossy and having a frequency range much higher, above 100 MHz. The Helmholtz equation for any of the components πΈπΈπΌπΌ of the electric field can be written as ∇2 πΈπΈπΌπΌ = −ππ2 πππππΈπΈπΌπΌ We write this in Cartesian and use the technique of separation of variables, πΈπΈπΌπΌ (π₯π₯, π¦π¦, π§π§) = πππΌπΌ (π₯π₯)πππΌπΌ (π¦π¦)πππΌπΌ (π§π§) Introducing this into the equation and dividing by πππΌπΌ (π₯π₯)πππΌπΌ (π¦π¦)πππΌπΌ (π§π§), we get, 1 ππ 2 πππΌπΌ 1 ππ 2 πππΌπΌ 1 ππ 2 πππΌπΌ + + = −ππ2 ππππ πππΌπΌ πππ¦π¦ 2 πππΌπΌ πππ§π§ 2 πππΌπΌ πππ₯π₯ 2 Since each of the three terms on the right is a function of an independent variable, while the right hand side is a constant, we must have each of the three terms equaling a constant such that the three constants add up to the constant on the right. Let, ππ 2 πππΌπΌ = −πππ₯π₯2 πππΌπΌ πππ₯π₯ 2 6 ππ 2 πππΌπΌ = −πππ¦π¦2 πππΌπΌ πππ¦π¦ 2 ππ 2 πππΌπΌ = −πππ§π§2 πππΌπΌ πππ§π§ 2 such that πππ₯π₯2 + πππ¦π¦2 + πππ§π§2 = ππ2 ππππ. As each of the equations has a solution in terms of linear combination of sine and cosine functions, we write the solution for the electric field as πΈπΈπΌπΌ = (π΄π΄πΌπΌ cos πππ₯π₯ π₯π₯ + π΅π΅πΌπΌ sin πππ₯π₯ π₯π₯)(πΆπΆπΌπΌ cos πππ¦π¦ π¦π¦ + π·π·πΌπΌ sin πππ¦π¦ π¦π¦)(πΉπΉ cos πππ§π§ π§π§ + πΊπΊπΌπΌ sin πππ§π§ π§π§) Tha tangential component of above must vanish at the metal boundary. This implies, 1. At π₯π₯ = 0 and at π₯π₯ = ππ , πΈπΈπ¦π¦ and πΈπΈπ§π§ = 0 for all values of π¦π¦, π§π§, 2. At π¦π¦ = 0 and π¦π¦ = ππ, πΈπΈπ₯π₯ and πΈπΈπ§π§ = 0 for all values of π₯π₯, π§π§, 3. At π§π§ = 0 and z=d, πΈπΈπ₯π₯ and πΈπΈπ¦π¦ = 0 for all values of π₯π₯, π¦π¦ Let us consider πΈπΈπ₯π₯ which must be zero at π¦π¦ = 0, π¦π¦ = ππ, π§π§ = 0, π§π§ = ππ. This is posible if πΈπΈπ₯π₯ = πΈπΈπ₯π₯0 (π΄π΄π₯π₯ cos πππ₯π₯ π₯π₯ + π΅π΅π₯π₯ sin πππ₯π₯ π₯π₯) sin πππ¦π¦ π¦π¦ sin πππ§π§ π§π§ with πππ¦π¦ = ππππ ππ and πππ§π§ = ππππ ππ . Here m and n are non-zero integers. In a similar way, we have, with πππ₯π₯ = ππππ ππ We now use πΈπΈπ¦π¦ = πΈπΈπ¦π¦0 sin πππ₯π₯ π₯π₯ (πΆπΆπ¦π¦ cos πππ¦π¦ π¦π¦ + π·π·π¦π¦ sin πππ¦π¦ π¦π¦) sin πππ§π§ π§π§ πΈπΈπ§π§ = πΈπΈπ§π§0 sin πππ₯π₯ π₯π₯ sin πππ¦π¦ π¦π¦ (πΉπΉπ§π§ cos πππ§π§ π§π§ + πΊπΊπ§π§ sin πππ§π§ π§π§) , with ππ being non-zerointeger. πππΈπΈπ₯π₯ πππΈπΈπ¦π¦ πππΈπΈπ§π§ + + =0 ππππ ππππ ππππ This relation must be satisfied for all values of ππ, ππ, ππ within the cavity. This requires, πΈπΈπ₯π₯0 (−π΄π΄π₯π₯ πππ₯π₯ sin πππ₯π₯ π₯π₯ + π΅π΅π₯π₯ πππ₯π₯ cos πππ₯π₯ π₯π₯) sin πππ¦π¦ π¦π¦ sin πππ§π§ π§π§ + πΈπΈπ¦π¦0 sin πππ₯π₯ π₯π₯(−πΆπΆπ¦π¦ πππ¦π¦ sin πππ¦π¦ π¦π¦ + π·π·π¦π¦ πππ¦π¦ cos πππ¦π¦ π¦π¦) sin πππ§π§ π§π§ + πΈπΈπ§π§0 sin πππ₯π₯ π₯π₯π₯π₯π₯π₯π₯π₯ πππ¦π¦ π¦π¦(−πΉπΉπ§π§ πππ§π§ sin πππ§π§ π§π§ + πΊπΊπ§π§ πππ§π§ cos πππ§π§ π§π§) = 0 ∇ ⋅ πΈπΈοΏ½β = Let us choose some special points and try to satisfy this equation. Let π₯π₯ = 0, π¦π¦, π§π§ arbitrary, This requires π΅π΅π₯π₯ = 0. Likewise, taking π¦π¦ = 0, x, z arbitrary, we require π·π·π¦π¦ = 0 and finally, π§π§ = 0, π₯π₯, π¦π¦ arbitrary gives πΊπΊπ§π§ = 0. With these our solutions for the components of the electric field becomes, πΈπΈπ₯π₯ = πΈπΈπ₯π₯0 cos πππ₯π₯ π₯π₯ sin πππ¦π¦ π¦π¦ sin πππ§π§ π§π§ πΈπΈπ¦π¦ = πΈπΈπ¦π¦0 sin πππ₯π₯ π₯π₯ cos πππ¦π¦ π¦π¦ sin πππ§π§ π§π§ πΈπΈπ§π§ = πΈπΈπ§π§0 sin πππ₯π₯ π₯π₯ sin πππ¦π¦ π¦π¦ cos πππ§π§ π§π§ where we have redefined our constants πΈπΈπ₯π₯0 , πΈπΈπ¦π¦0 and πΈπΈπ§π§0 . 7 We also have, ππππ ππππ ππππ , πππ¦π¦ = , πππ§π§ = ππ ππ ππ the integers ππ, ππ, ππ cannot be simultaneously zero for then the field will identically vanish. Note further that since ∇ ⋅ πΈπΈοΏ½β = 0 must be satisfied everywhere within the cavity, we must have, πππ₯π₯ = calculating the divergence explicitly from the above expressions, πππ₯π₯ πΈπΈπ₯π₯0 + πππ¦π¦ πΈπΈπ¦π¦0 + πππ§π§ πΈπΈπ§π§0 = 0 This requires that πποΏ½β is perpendicular to πΈπΈοΏ½β . One can see that the modes within a cavity can exist only with prescribed “resonant” frequency corresponding to the set of integers ππ, ππ, ππ 1 1 ππππ 2 ππππ 2 ππππ 2 2 οΏ½οΏ½ οΏ½ + οΏ½ οΏ½ + οΏ½ οΏ½ οΏ½ ππ = ππ ππ √ππππ ππ Though there is nothing like a propagation direction here, one can take the longest dimension (say d) to be the propagation direction. We can have modes like πππΈπΈππ,ππ ,ππ corresponding to the set ππ, ππ, ππ for which πΈπΈπ§π§ = 0. For this set we get, πΈπΈπ₯π₯ = πΈπΈπ₯π₯0 cos πππ₯π₯ π₯π₯ sin πππ¦π¦ π¦π¦ sin πππ§π§ π§π§ πΈπΈπ¦π¦ = πΈπΈπ¦π¦0 sin πππ₯π₯ π₯π₯ cos πππ¦π¦ π¦π¦ sin πππ§π§ π§π§ with πππ₯π₯ πΈπΈπ₯π₯0 + πππ¦π¦ πΈπΈπ¦π¦0 = 0. The magnetic field components can be obtained from the Faraday’s law, πππΈπΈπ¦π¦ 1 πππ§π§ οΏ½− οΏ½= πΈπΈ sin πππ₯π₯ π₯π₯ cos πππ¦π¦ π¦π¦ cos πππ§π§ π§π§ −ππππππ ππππ ππππππ π¦π¦0 πππ§π§ 1 πππΈπΈπ₯π₯ οΏ½ οΏ½=− πΈπΈ cos πππ₯π₯ π₯π₯ sin πππ¦π¦ π¦π¦ cos πππ§π§ π§π§ π»π»π¦π¦ = −ππππππ ππππ ππππππ π₯π₯0 πππΈπΈπ¦π¦ πππΈπΈπ₯π₯ 1 1 οΏ½ − οΏ½=− (πΈπΈ ππ − πΈπΈπ₯π₯0 πππ¦π¦ ) cos πππ₯π₯ π₯π₯ cos πππ¦π¦ π¦π¦ sin πππ§π§ π§π§ π»π»π§π§ = −ππππππ ππππ ππππππ π¦π¦0 π₯π₯ ππππ π»π»π₯π₯ = , Wave Guides Lecture37: Electromagnetic Theory Professor D. K. Ghosh, Physics Department, I.I.T., Bombay 8 Tutorial Assignment 1. A rectangular waveguide has dimensions 8 cm x 4 cm. Determine all the modes that can propagate when the operating frequency is (a) 1 GHz, (b) 3 GHz and (c) 8 GHz. 2. Two signals, one of frequency 10 GHz and the other of 12 GHz are simultaneously launched in an air filled rectangular waveguide of dimension 2 cm x 1 cm. Find the time interval between the arrival of the signals at a distance of 10 m from the common place of their launch. 3. What should be the third dimension of a cavity having a length of 1 cm x 1 cm which can operate in a TE103 mode at 24 GHZ? Solutions to Tutorial Assignments 1. The cutoff frequency for (m,n) mode is ππππ = of the lowest cutoffs are as under (in GHz) : m 0 0 0 1 1 1 1 2 2 2 3 3 3 4 4 5 n 1 2 3 0 1 2 3 0 1 2 0 1 2 0 1 0 ππ 2ππ Cutoff frequency (GHz) 3.75 7.5 11.25 1.875 4.192 7.731 11.405 3.75 5.30 8.39 5.625 6.76 9.375 7.5 8.38 9.375 9 2 2 2 2 οΏ½οΏ½ππππ οΏ½ + οΏ½ππππ οΏ½ = 1.5 × 1010 οΏ½ππ + ππ . Some ππ ππ 64 16 At 1 GHz, no propagation takes place. At 3 GHz only TE10 mode propagates (recall no TM mode is possible when either of the indices is zero.) At 8 GHz, we have TE01, TE02, TE10, TE11, TE12, TE20, TE21, TE30, TE31, TE40, TM11, TM12, TM21 and TM31, i.e. a total of 14 modes propagating. 2. The cutoff frequency is given by ππππ = ππ ππ 2 ππ 2 οΏ½οΏ½ οΏ½ + οΏ½ οΏ½ = 7.5 πΊπΊπΊπΊπΊπΊ 2 ππ ππ ππ 2 The propagation of the wave is given by the group velocity π£π£ππ = ππ οΏ½1 − οΏ½ ππ οΏ½ , which is 1.98x 108 ππ 8 and 2.07x10 m/s respectively for f=10 GHz and 12 GHz respectively. Thus the difference in speed is 9x 106 m/s, resulting in a time difference of approximately 10-5 s in travelling 10m. 3. The operating frequency is given by 1 1 ππππ 2 ππππ 2 ππππ 2 2 οΏ½οΏ½ οΏ½ + οΏ½ οΏ½ + οΏ½ οΏ½ οΏ½ ππ = ππ ππ √ππππ ππ Substituting ππ = 1, ππ = 0, ππ = 3, we get ππ = 2.4 cm. Wave guides Lecture37: Electromagnetic Theory Professor D. K. Ghosh, Physics Department, I.I.T., Bombay 10 Self Assessment Questions 1. A rectangular, air filled waveguide has dimension 2 cm x 1 cm. For what range of frequencies, there is a “single mode” operation in the guide? 2. The cutoff frequency for a TE10 mode in an air filled waveguide is 1.875 GHz. What would be the cutoff frequency of this mode if the guide were to be filled with a dielectric of permittivity 9ππ0 ? 3. In an air filled waveguide of dimension 2 cm x 2 cm, the x component of the electric field is given by 2ππππ οΏ½ sin(ππππ − 100π§π§) πΈπΈπ₯π₯ = −8 sin οΏ½ ππ Identify the propagating mode, determine the frequency of operation and find π»π»π§π§ and πΈπΈπ§π§ . Solutions to Self Assessment Questions 1. The cutoff frequency for (m,n) mode is ππππ = ππ 2ππ 2 2 2 οΏ½οΏ½ππππ οΏ½ + οΏ½ππππ οΏ½ = 1.5 × 1010 οΏ½ππ + ππ2 . ππ ππ 4 Substituting values, the cutoff for (1,0) mode is 7.5 GHz and for (2,0) is 15 GHz. The frequency for (0,1) is also 15 GHz. All other modes have higher cut off. Thus in order that only one mode propagates, the operating frequency should be in the range 7.5 < ππ < 15 GHz. In this range, only TE10 mode operates. Recall that there is no TM mode with either of the indices being zero. 2. The cutoff frequency is proportional to 1 √ππππ . Thus the frequency would be reduced by a factor of 3 making it 625 MHz. 3. Since one of the standing wave factor is missing, one of the mode indices is zero. Thus it is a TE mode. The electric field of TEmn mode is given by (with a=b) ππππππ ππππππ ππππππ οΏ½ sin οΏ½ οΏ½ sin(π½π½π½π½ − ππππ) πΈπΈπ₯π₯ = 2 πΆπΆπππ¦π¦ cos οΏ½ ππ ππ ππ Thus the mode is TE02 mode. From the relation πππ₯π₯2 + πππ¦π¦2 + π½π½ 2 = 100ππ, π½π½ = 100, ππ2 = (100ππ)2 + (100)2 ππ 2 11 ππ 2 ππ 2 . Given πππ₯π₯ = 0, πππ¦π¦ = 2ππ 0.02 = the operating frequency is 15.74 GHz. Since πππ₯π₯ = 0, πΈπΈπ¦π¦ = 0 . Further, by definition for TE mode πΈπΈπ§π§ = 0. Here we have ππ 2 = πππ₯π₯2 + πππ¦π¦2 = πππ¦π¦2 . The ratio so that οΏ½ πππ¦π¦ ππ 2 π»π»π§π§ οΏ½= = = 0.0025 πΈπΈπ₯π₯ πππππππ¦π¦ ππππ π»π»π§π§ = 0.020 cos οΏ½ 2ππ π¦π¦οΏ½ cos (ππππ − 100π§π§) A/m ππ 12
