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MIT 6 013 - Lecture Notes

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L13-1 EM GUIDANCE AND FILTERING Generic System Architecture: Signal Processing Signal Processing Guide Antenna AntennaGuideAmplification Microwave or optical tuning, resonance coupling waveguides microwave integrated circuits matching transitions Detection Systems fail at weakest link, so understand all parts Communications, bi-static radar—separately located systems Radar, lidar, data recording—co-located systems Passive sensing—uses receiver side onlyL13-2 MICROWAVE CIRCUITS Printed Circuits : Equivalent TEM line circuit:Difference Equations: ˆ ˆEz H z 0•= •= i(t,z) v(t,z) -+ z I(z) I(z+Δz) CΔz LΔz LΔz CΔz + -+ -Δz V(z) V(z+Δz) “TEM” ⇒ Wave Equation⇒ V(z+Δz) – V(z) = -jω LΔz I(z) I(z+Δz) – I(z) = -jω CΔzV(z) Limit as Δz → 0: 2 2 2 dV(z) LC V(z) 0 dz +ω = dV jL I dz dI jC V dz =−ω =−ωTEM PHASOR EQUATIONS ( d2 Wave Equation: dz2 +ω2LC)V(z) =0 Voltage Solution: V(z) = V+e-jkz + V-ejkz Test solution: [(-jk)2V+e-jkz + (jk)2V-ejkz] + ω2LC[V+e-jkz + V-ejkz] = 0 Passes test iff: Current I(z): Since: +-V+ +-V-Therefore: jL + −ω =Y (V e− jkz −V e+ jkz )o + − Line admittance: Y = k = ω LC = o ωL ωL L Zo C 1= TEM Equations: V(z) = V+e-jkz + V-ejkz I(z) = Yo(V+e-jkz –V-ejkz) k2 = ω2LC V(z) j L I(z) z ∂ =−ω ∂ jkz jkz 1I(z) jk(V e V e )− += − L13-3COMPLEX LINE IMPEDANCE Z(z) Impedance: Z(z) = V(z) R(z) = jX(z) I(z) + Z(z)I(z) + V(z) Zo-zResistance Reactance (−jkz jkz ) Z V e + V e 1+Γ(z) Z(z) = o +−jkz − jkz =Zo 1−Γ(z) EquivalentV e −V e+ − circuit where ΓL = Γ (z = 0)= V− V+ + − − + Γ =Γ jkz jkz 2jkz L Ve(z) eVe  Complex Reflection Coefficient Γ(z): Examples: Γ = 0 ⇒ Z(z) = Zo Γ = 1 ⇒ Z = ∞ Γ = -1 ⇒ Z = 0 Normalized Impedance Zn(z): [ ] [ ] [ ] [ ]o Z(z) n n Z n Z(z) 11 (z) Z (z) (z) 1 (z) Z(z) 1 −+Γ = Γ = −Γ + Definition: L13-4L13-5Z(z) = f(ZL, Zo, k, z):[ ][]ZZZ−Lo[1 + Γ(z)]2jkzSubstituting: Γ= Z( = Γ =ΓL into z) Zo; (z) eZZ+[1 ΓLo - (z)]Zj−LZta=onkzYields: Z(z) ZoZj−oZLtankz(z) TRANSFORMATIONSExample: Open Circuit, ZL= ∞:In general: -j∞ < Z= 0 when z = -λ/4, -3λ/4,…=∞when z = 0, -λ/2,…(-A) < +j∞(Yields ANY capacitance or inductanceat a SINGLE ω)Z(−=AA)−jZoocotk≅−jZ kA for kA<<1L/C 1 =−j =ω LCAjCω Acapacitor Coz = -Az0z-λ03λ−2λ−2Im{Zλ−4Open Short }circuitcircuitLEXAMPLES: Z(z) TRANSFORMATIONS L13-6 Example—Inductive Load, ZL = jωLo for z = -λ/4: Since: Therefore: Note: πλ π=− =− =− =−∞ λ −= =ω =λ λ π λ o o o o o 2 L o 2kz k , therefore tan(kz) 4 2 Z LC 1Z( ) = = (C = CL L ) Z jωL jωC Z(z) j L if 2, ,...[tan(-2 / ) = 0] A A A A Recall: L o o o L Z jZ tankz Z(z) Z Z jZ tankz − = − Zo 0 zλ/4 L? o Example – Source Transformation: s oTh o o s Z jZ tank Z Z Z jZ tank + = − A A = V (e + e ) = 2V cos k jk -jkAA A+ + Therefore: V = 2V = V Z /[(Z + Z )cos k ]A ASS ATh + Zs VA + -+ -Zo 0 z A ZTh Vs VTh ZA + -? A A S A o S A ZV= V where Z = -jZcot k Z+ Z AL13-7ALTERNATE APPROACH TO FINDING Z(z)[][]Algorithmic, rotate Γ−jkz jkzVe + VeZ(z) ==V(z) I(z) Z+−o−jkz jkzVe+−− Ve1(+Γ z)Z(z) = Zo1(−Γ z)+jkzVe2jkVΓ=(z)-z= ΓLLe= Γ(z), where Γ = Γ(z=0) =-jkzVVe++Z1n−Z(z)Γ=(z) Zn(z) Z1n+Zo(z):Γ-Plane Solution Method:ZL⇔ΓL⇔Γ(z) ⇔ Z(z)(3) (2) (1)Zn= Z/Zotoward generator(larger A)increasing AZZn= -jΓn= 0= Zn= jZZn= 1n= ∞Re{Γ}|ΓIm{Γ| = 1}Γ= 1VI(z)(z)ZLzZZo,c(z)0+-z = - A(3)(2)(1)(λ/2 ⇒ full rotation)e-2jkAgoes clockwise as A→∞jφ(e =φcos +jsinφ)-GAMMA PLANE ⇒ SMITH CHART Gamma Plane: Re{ZnΓ = -1 Z =1 n1 Toward generator (larger A) (λ/2 ⇒ full rotation) V(z) I(z) ZL z Z(z) Zo,c 0 + -z = - AA Γ = j} = 0 Im{Zn}=+j Γ = - j −Γ Γ = 1 Re{Zn}=1 Im{Zn}=0 Im{Zn} = -j Γ= 0 Zn = ∞ Re{Zn} = 3 +Γ 2jkz 2jk L L(z) e e−Γ =Γ =Γ A Smith Chart = Gamma Plane + ZZL ⇔ ZLn ⇔ΓL ⇔Γ(z) ⇔ Znn(z): [Z(z) − 1][1+ Γ (z) ]n = Γ (z) Z(z)  Z(z) = [Z(z) + 1] Z1− Γ (z) n no [ ] (z) ⇔ Z(z) L13-8MIT OpenCourseWare http://ocw.mit.edu 6.013 Electromagnetics and Applications Spring 2009 For information about citing these materials or our Terms of Use, visit:


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MIT 6 013 - Lecture Notes

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