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- 1. 27/01/2014 Objectives ECE 637 : MIC and Monolithic Microwave Integrated Circuits (MMIC) By Dr. Ghanshyam Kumar Singh ECE Department, SET Sharda University Course Outcomes After completing this course students will be able design Microstrip lines. analyse different types of dielectric materials and their effects. apply optimization technique for frequency behaviour of MMIC. solve the problems related to MMIC Designing. analyse various types of MMIC devices. Lecture 1 An Approximate Electrostatic Solution for Microstrip Line The Transverse Resonance Techniques Wave Velocities and Dispersion To introduce the design methodology of Microstrip Lines. To acquire knowledge of various typs of dielectric materials and their properties. To describe the frequency properties of MMIC. To explain the fundamentals of various types of MMIC devices. Lecture 1 TEM, TE and TM Waves Coaxial Cable Grounded Dielectric Slab Waveguides Striplines and Microstrip Line Design Formulas of Microstrip Line TEM, TE and EM Waves transmission lines and waveguides are primarily used to distribute microwave wave power from one point to another each of these structures is characterized by a propagation constant and a characteristic impedance; if the line is lossy, attenuation is also needed 1
- 2. 27/01/2014 TEM, TE and EM Waves structures that have more than one conductor may support TEM waves let us consider the a transmission line or a waveguide with its cross section being uniform along the zdirection b ρ a TEM, TE and EM Waves in a source free region, Maxwell’s equations can be written as ∇ × E = − jωµH, ∇ × H = jωεE Therefore, ∂E z + jβE y = − jωµH x , ( 3) ∂y ∂E − jβE x − z = − jωµH y , ( 4 ) ∂x ∂E y ∂Ex − = − jωµH z , ( 5 ) ∂x ∂y ∂H z + jβH y = jωεEx , ( 6 ) ∂y − jβH x − ∂H y ∂x − the electric and magnetic fields can be written as $ $ E( x, y, z) = [e t e t ( x, y ) + ze z ( x, y )]e − jβz − − (1) $ $ H( x, y, z) = [e t h t ( x, y ) + zh z ( x, y )]e − jβz − − ( 2) Where e t and ht are the transverse components and e t and h t are the longitudinal components TEM, TE and EM Waves each of the four transverse components can be written in terms of and , e.g., consider Eqs. (3) and (7): ∂E z + jβE y = − jωµH x ∂y ∂H z = jωεE y , ( 7) ∂x − jβH x − ∂H x = jωεE z , ( 8) ∂y TEM, TE and EM Waves each of the four transverse components can be written in terms of and , e.g., consider Eqs. (3) and (7): ∂E z + jβE y = − jωµH x ∂y − jβH x − TEM, TE and EM Waves ∂H z = jωεE y ∂x ∂H z = jωεE y ∂x TEM, TE and EM Waves ∂H z ∂E z ) / ( jβ ) = jωε ( − jωµH x − ∂x ∂y ∂E ∂H z ( k 2 − β 2 ) H x = jωε z − jβ ∂y ∂x − jβH x − Hx = j 2 kc ( ωε ∂E z ∂H z ) − − − − − − − ( 9) −β ∂y ∂x 2 k c = (k 2 − β 2 ) 2
- 3. 27/01/2014 TEM, TE and EM Waves Similarly, we have Hy = Ex = Ey = −j 2 kc −j 2 kc j 2 kc ( ωε (β ∂E z ∂H z ) − − − − ( 10 ) +β ∂x ∂y ∂E z ∂H z ) − − − − ( 11) + ωµ ∂x ∂y ( −β ∂E z ∂H z ) − − − ( 12) + ωµ ∂y ∂x k c is called the cutoff wavenumber TEM, TE and EM Waves now let us consider the Helmholtz’s equation 2 2 2 ∂ ∂ ∂ 2 ( ∇ 2 + k 2 ) E x = 0, 2 + 2 + 2 + k Ex = 0 ∂x ∂y ∂z ∂2 2 note that ∂z 2 → −β and therefore, for TEM wave, we have ∂2 ∂2 Ex = 0 + 2 ∂x ∂y 2 TEM, TE and EM Waves Knowing that ∇ 2 et = 0 and t ∇ • D = ε∇ t • e t = 0 ∇ 2Φ ( x, y) = 0 t , and we have 2 V12 = Φ 1 − Φ 2 = − ∫1 E • dl − − − − (14 ) while the current flowing on a conductor is given by I = ∫ H • dl − − − − (15) TEM, TE and EM Waves Transverse electromagnetic (TEM) wave implies that both E z and H z are zero (TM, transverse magnetic, H z =0, E z ≠ 0 ; TE, transverse electric, 0) E z =0, H z ≠ the transverse components are also zero unless k c is also zero, i.e., k 2 = ω 2µε = β 2 = ω µε = β TEM, TE and EM Waves this is also true for E y , therefore, the transverse components of the electric field (so as the magnetic field) satisfy the two-dimensional Laplace’s equation ∇ 2e t = 0 − − − − (13) t TEM, TE and EM Waves this is also true for E y, therefore, the transverse components of the electric field (so as the magnetic field) satisfy the two-dimensional Laplace’s equation ∇ 2e t = 0 − − − − (13) t C 3
- 4. 27/01/2014 TEM, TE and EM Waves Knowing that ∇ 2 e t = 0 and ∇ • D = ε∇ t • e t = 0 t , we have ∇ 2Φ ( x, y) = 0 TEM, TE and EM Waves we can define the wave impedance for the TEM mode: t Z TEM = the voltage between two conductors is given by 2 V12 = Φ 1 − Φ 2 = − ∫1 E • dl − − − − (14 ) while the current flowing on a conductor is I = ∫ H • dl − − − − (15 ) given by E x ωµ µ = = = η − − − ( 16 ) Hy β ε i.e., the ratio of the electric field to the magnetic field, note that the components must be chosen such that E x H is pointing to the direction of propagation C TEM, TE and EM Waves for TEM field, the E and H are related by h ( x, y ) = 1 $ z × e( x , y ) − − − − (17) Z TEM why is TEM mode desirable? a closed conductor cannot support TEM wave as the static potential is either a constant or zero leading to et = 0 if a waveguide has more than 1 dielectric, TEM mode cannot exists as cannot be zero in all regions why is TEM mode desirable? cutoff frequency is zero no dispersion, signals of different frequencies travel at the same speed, no distortion of signals solution to Laplace’s equation is relatively easy why is TEM mode desirable? sometime we deliberately want to have a cutoff frequency so that a microwave filter can be designed k ci = ( ε ri k 2 − β 2 ) 1/ 2 4
- 5. 27/01/2014 TEM Mode in Coaxial Line a coaxial line is shown here: b ρ V=Vo a V=0 the inner conductor is at a potential of Vo volts and the outer conductor is at zero volts TEM Mode in Coaxial Line use the method of separation of variables, we let Φ ( ρ, φ ) = R ( ρ )P( φ ) − − − − ( 21) substitute Eq. (21) to (18), we have 2 ρ ∂ ρ ∂R + 1 ∂ P = 0 − − − −(22) ∂φ 2 R ∂ρ ∂ρ P note that the first term on the left only depends on ρ while the second term only depends on φ TEM Mode in Coaxial Line TEM Mode in Coaxial Line the electric field can be derived from the scalar potential Φ; in cylindrical coordinates, the Laplace’s equation reads: 1 ∂ ∂Φ 1 ∂ 2Φ = 0 − − − − (18 ) ρ + ρ ∂ρ ∂ρ ρ 2 ∂φ 2 the boundary conditions are: Φ ( a , φ ) = Vo − − − ( 19 ), Φ (b, φ ) = 0 − − − − − ( 20 ) TEM Mode in Coaxial Line if we change either ρ or φ, the RHS should remain zero; therefore, each term should be equal to a constant ρ ∂ ρ ∂R = −k 2 − − − (23) ρ R ∂ρ ∂ρ 2 1 ∂ P = −k 2 − − − −(24), k 2 + k 2 = 0 − − − −(25) ρ φ φ ∂φ 2 P TEM Mode in Coaxial Line now we can solve Eqs. (23) and (24) in which only 1 variable is involved, the final solution to Eq. (18) will be the product of the solutions to Eqs. (23) and (24) boundary conditions (19) and (20) dictates that the potential is independent of φ, therefore k φ must be equal to zero and so as k ρ the general solution to Eq. (24) is Eq. (23) is reduced to solving ∂ ∂R ρ =0 ∂ρ ∂ρ P( φ ) = A cos( k φ φ ) + B sin( k φ φ ) 5
- 6. 27/01/2014 TEM Mode in Coaxial Line TEM Mode in Coaxial Line the solution for R(ρ) now reads the electric field now reads R ( ρ ) = C ln ρ + D Vo ∂Φ $ 1 ∂Φ $ $ e t ( ρ, φ ) = −∇ t Φ ( ρ, φ ) = − ρ +φ =ρ ∂ρ ρ ∂φ ρ ln b / a Φ ( ρ, φ ) = A ln ρ + B Φ ( a , φ ) = Vo = A ln a + B adding the propagation constant back, we have Φ ( b, φ ) = 0 = A ln b + B ⇒ B = − A ln b A = Vo / ln( b / a ) V e − jβz $ E( ρ, φ ) = e t ( ρ, φ ) e − jβz = ρ o − − − ( 27) ρ ln b / a Vo (ln ρ − ln b ) − − − ( 26 ) Φ ( ρ, φ ) = ln(b / a ) TEM Mode in Coaxial Line TEM Mode in Coaxial Line the magnetic field for the TEM mode $ H( ρ , φ ) = ρ the total current on the inner conductor 2π is 2π − jβ z Voe − jβz − − − ( 28) ρη ln b / a the potential conductors are between I a = ∫ H φ ( ρ, φ )adρ = 0 the two b Vab = ∫ Eρ (ρ, φ)dρ = Voe − jβ z − − − ( 29) TEM Mode in Coaxial Line the total current on the outer conductor is 2π − 2π − jβz 0 η ln(b / a ) Vo e = − Ia the characteristic impedance can be calculated as Zo = Vo η ln(b / a ) = − − − ( 31) Ia 2π Vo e − − − ( 30 ) the surface current density on the outer conductor is $ J s = − ρ × H( b , φ ) = a Ib = ∫ J szbdφ = η ln( b / a ) $ −z Vo e − jβz ηb ln( b / a ) TEM Mode in Coaxial Line higher-order modes exist in coaxial line but is usually suppressed the dimension of the coaxial line is controlled so that these higher-order modes are cutoff the dominate higher-order mode is TE11 mode, the cutoff wavenumber can only be obtained by solving a transcendental equation, the approximation k c = 2 / ( a + b ) is often used in practice 6
- 7. 27/01/2014 Surface Waves on a Grounded Dielectric Slab a grounded dielectric slab will generate surface waves when excited Surface Waves on a Grounded Dielectric Slab while it does not support a TEM mode, it excites at least 1 TM mode x d εr this surface wave can propagate a long distance along the air-dielectric interface it decays exponentially in the air region when move away from the air-dielectric interface Surface Waves on a Grounded Dielectric Slab for TM modes, from Helmholtz’s equation we have ∂ 2 ∂ 2 ∂ 2 2 2 + 2 + 2 + k Ez = 0 ∂y ∂z ∂x which reduces to ∂2 2 2 2 + ε r k o − β E z = 0, 0 ≤ x ≤ d ∂x ∂2 2 2 2 + k o − β E z = 0, d ≤ x ≤ ∞ ∂x Surface Waves on a Grounded Dielectric Slab the general solutions to Eqs. (32) and (33) are e z ( x, y ) = A sin k c x + B cos k c x,0 ≤ x ≤ d e z ( x, y ) = Ce hx + De − hx , d ≤ x ≤ ∞ the boundary conditions are tangential E are zero at x = 0 and x → ∞ tangential E and H are continuous at x = d z assume no variation in the y-direction which implies that ∂ / ∂y → 0 write equation for the field in each of the two regions match tangential fields across the interface Surface Waves on a Grounded Dielectric Slab Define ( ) ( 2 2 2 k c = ε rk o − β 2 , h2 = − ko − β 2 ) ∂2 2 2 + k c E z = 0,0 ≤ x ≤ d − − − ( 32) ∂x ∂2 2 2 − h E z = 0, d ≤ x ≤ ∞ − − − ( 33) ∂x Surface Waves on a Grounded Dielectric Slab tangential E at x=0 implies B =0 tangential E = 0 when x→ ∞ implies C=0 continuity of tangential E implies A sin k c d = De − hd − − − ( 34) tangential H can be obtained from Eq. (10) with H z = 0 7
- 8. 27/01/2014 Surface Waves on a Grounded Dielectric Slab Surface Waves on a Grounded Dielectric Slab tangential E at x=0 implies B =0 tangential E = 0 when x → ∞ C=0 continuity of tangential H implies implies εr −h De − hd − − − ( 35 ) A cos k c d = kc − h2 taking the ratio of Eq. (34) to Eq. (35) we have k c tan k c d = ε r h − − − ( 36 ) continuity of tangential E implies A sin k c d = De − hd − − − ( 34) Surface Waves on a Grounded Dielectric Slab ( 2 2 2 ) 2 ( 2 note that k c = ε r k o − β , h = − k o − β lead to 2 2 k c + h 2 = ( ε r − 1) k o − − − ( 37) Surface Waves on a Grounded Dielectric Slab 2 ) Eqs. (36) and (37) must be satisfied simultaneously, they can be solved for by numerical method or by graphical method Surface Waves on a Grounded Dielectric Slab Eq. (39) is an equation of a circle with a radius of ( ε r − 1)k o d , each interception point between these two curves yields a solution hd Eq.(38) r Eq. (39) π/2 π k cd to use the graphical method, it is more convenient to rewrite Eqs. (36) and (37) into the following forms: k c d tan k c d = ε r hd − − − ( 38 ) ( k c d) 2 + ( hd) 2 = ( ε r − 1)( k o d) 2 = r 2 − − − ( 39 ) Surface Waves on a Grounded Dielectric Slab note that there is always one intersection point, i.e., at least one TM mode the number of modes depends on the radius r which in turn depends on the d and ε r , k o h has been chosen a positive real number, we can also assume that k c is positive the next TM will not be excited unless r = ( ε r − 1)k o d = π 8
- 9. 27/01/2014 Surface Waves on a Grounded Dielectric Slab In general, TM n mode is excited if r = ( ε r − 1)k o d ≥ nπ the cutoff frequency is defined as ( ε r − 1)( 2πf c / c)d = nπ → fc = nc , n = 0,1, 2, .. .--- (40) 2d ε r − 1 Surface Waves on a Grounded Dielectric Slab once k c and h are found, the TM field components can be written as for 0≤x≤d E z = A sin k c xe − jβz − − − − ( 41) − jβ Ex = A cos k c xe − jβz − − − − ( 42) kc Hy = Surface Waves on a Grounded Dielectric Slab For d ≤ x ≤ ∞ E z = A sin k c de − h( x − d) e − jβz − − − − ( 44) − jβ Ex = A sin k c de − h( x − d) e − jβz − − − ( 45) h − jωε o Hy = A sin k c de − h( x− d) e − jβz − − − − ( 46 ) h similar equations can be derived for TE fields Striplines and Microstrip Lines the strip line was developed from the square coaxial coaxial rectangular line square coaxial flat stripline − jωε o ε r A cos k c xe − jβz − − − − ( 43) kc Striplines and Microstrip Lines various planar transmission line structures are shown here: stripline microstrip line slot line coplanar line Striplines and Microstrip Lines since the stripline has only 1 dielectric, it supports TEM wave, however, it is difficult to integrate with other discrete elements and excitations microstrip line is one of the most popular types of planar transmission line, it can be fabricated by photolithographic techniques and is easily integrated with other circuit elements 9
- 10. 27/01/2014 Striplines and Microstrip Lines Striplines and Microstrip Lines the following diagrams depicts the evolution of microstrip transmission line a microstrip line suspended in air can support TEM wave a microstrip line printed on a grounded slab does not support TEM wave the exact fields constitute a hybrid TM-TE wave when the dielectric slab become very thin (electrically), most of the electric fields are trapped under the microstrip line and the fields are essentially the same as those of the static case, the fields are quasi-static + + + two-wire line single-wire above ground (with image) microstrip in air (with image) microstrip with grounded slab Striplines and Microstrip Lines Design Formulas of Microstrip Lines one can define an effective dielectric constant so that the phase velocity and the propagation constant can be defined as β = k o ε e − − − ( 48 ) the effective dielectric constant is bounded by 1 < ε e < ε r , it also depends on the slab thickness d and conductor width, W design formulas have been derived for microstrip lines these formulas yield approximate values which are accurate enough for most applications they are obtained from analytical expressions for similar structures that are solvable exactly and are modified accordingly Design Formulas of Microstrip Lines Design Formulas of Microstrip Lines vp = c − − − − ( 47) εe or they are obtained by curve fitting numerical data the effective dielectric constant of a microstrip line is given by 1 ε + 1 εr − 1 εr = r + − − − − ( 49 ) 2 2 1 + 12d / W the characteristic impedance is given by for W/d≤ 1 Zo = 60 8d W ln + − − − − ( 50 ) ε r W 4d For W/d ≥ 1 Zo = 120 π − − ( 51) ε r [ W / d + 1.393 + 0.667 ln( W / d + 1.444 )] 10
- 11. 27/01/2014 Design Formulas of Microstrip Lines for a given characteristic impedance Z o and dielectric constant ε r , the W/d ratio can be found as W/d= 8e A e 2A − 2 − − − − ( 52) for W/d<2 Design Formulas of Microstrip Lines for a homogeneous medium with a complex dielectric constant, the propagation constant is written as 2 γ = α d + jβ = k c − κ 2 2 γ = k c − ω 2µ o ε o ε r (1 − j tan δ ) note that the loss tangent is usually very small Design Formulas of Microstrip Lines 2 ε −1 [ B − 1 − ln( 2B − 1) + r × 2ε r π for W/d > 2 0.61 {ln( B − 1) + 0.39 − }] − − − − ( 53) εr W/d= Where And 0.11 εr + 1 εr − 1 ( 0.23 + ) + 2 εr + 1 εr A= Zo 60 B= 377π 2Z o ε r Design Formulas of Microstrip Lines Note that (1 + x ) 1/ 2 = 1 + x / 2 where x is small therefore, we have 2 γ = kc − k 2 + jk 2 tan δ 2 2 kc − k2 − − − ( 54 ) 2 γ = k c − k 2 + jk 2 tan δ Design Formulas of Microstrip Lines 2 Note that jβ = k c − k 2 for small loss, the phase constant is unchanged when compared to the lossless case the attenuation constant due to dielectric loss is therefore given by k 2 tan δ Np/m (TE or TM) (55) αd = 2β Design Formulas of Microstrip Lines For TEM wave k = β , therefore αd = k tan δ Np/m (TEM) (56) 2 for a microstrip line that has inhomogeneous medium, we multiply Eq. (56) with a filling factor ε r ( ε e − 1) ε e ( ε r − 1) 11
- 12. 27/01/2014 Design Formulas of Microstrip Lines k ε tan δ ε r ( ε e − 1) = k o ε r ( ε e − 1) tan δ αd = o e 2 ε e ( ε r − 1) ε e ( ε r − 1) 2 (57) the attenuation due to conductor loss is given by Rs αc = Zo W (58) Np/m where R s = ωµ o / ( 2σ ) Design Formulas of Microstrip Lines note that for most microstrip substrate, the dielectric loss is much more significant than the conductor loss at very high frequency, conductor loss becomes significant R s is called the surface resistance of the conductor An Approximate Electrostatic Solution for Microstrip Lines d W y εr -a/2 x a/2 two side walls are sufficiently far away that the quasi-static field around the microstrip would not be disturbed (a >> d) An Approximate Electrostatic Solution for Microstrip Lines using the separation of variables and appropriate boundary conditions, we write ∞ nπx nπy sinh − − − ( 59 ),0 ≤ y ≤ d ∑ A n cos a a n = 1,odd ∞ nπx − nπy / a Φ ( x, y ) = e − − − ( 60 ), d ≤ y ≤ ∞ ∑ B n cos a n = 1,odd An Approximate Electrostatic Solution for Microstrip Lines we need to solve the Laplace’s equation with boundary conditions ∇ 2 Φ ( x, y ) = 0,| x| ≤ a / 2,0 ≤ y < ∞ t Φ ( x, y ) = 0, x = ± a / 2 Φ ( x, y ) = 0, y = 0, ∞ two expressions are needed, one for each region An Approximate Electrostatic Solution for Microstrip Lines the potential must be continuous at y=d so that A n sinh nπd = B n e − nπd / a a Φ ( x, y ) = note that this expression must be true for any value of n 12
- 13. 27/01/2014 An Approximate Electrostatic Solution for Microstrip Lines mπx An Approximate Electrostatic Solution for Microstrip Lines nπx π π a/2 cos dx = 0 due to fact that ∫− a / 2 cos a a if m is not equal to n Φ ( x, y ) = ∞ ∑ A n cos nπx sinh nπy the normal component of the electric field is discontinuous due to the presence of surface charge on the microstrip, E y = − ∂Φ / ∂y ∞ nπ nπx nπ y cos cosh ,0 ≤ y ≤ d ∑ An a a n = 1,odd a ∞ nπ nπ x nπd − nπ ( y − d)/ a cos sinh e , Φ ( x, y ) = ∑ A n a a a n = 1,odd − − − ( 61),0 ≤ y ≤ d Ey = − a a n = 1,odd ∞ nπx nπd − nπ ( y− d)/ a sinh e Φ ( x, y ) = − ( 62), ∑ A n cos a a n = 1,odd d≤ y≤ ∞ d≤ y≤ ∞ An Approximate Electrostatic Solution for Microstrip Lines multiply Eq. (63) by cos mπx/a and integrate from -a/2 to a/2, we have ρ s = ε o E y ( x, y = d + ) − ε o ε r E y ( x, y − ) ρs = ε o ∑ A n sinh nπd / a n = 1,odd the total charge on the strip is m πx 2 sin( mπW / 2a ) dx = a mπ / a ∞ nπ nπx nπd nπy cos (sinh ) dx = + ε r cosh ∑ An a a a n = 1,odd a nπ nπd nπd a / 2 mπx nπx (sinh ) ∫ cos cos dx = + ε r cosh ε o ∑ An a a a −a/2 a a ε o An nπ nπd n πd a (sinh ) ,m = n + ε r cosh a a a 2 An = 4a sin( mπW / 2a ) ( nπ ) 2 ε o [sinh( nπd / a ) + ε r cosh( nπd / a ) An Approximate Electrostatic Solution for Microstrip Lines the voltage of the microstrip wrt the ground plane is ∞ cos a/2 assuming that the charge distribution is given by ρ s = 1 on the conductor and zero elsewhere d V = − ∫0 E y ( x = 0, y )dy = ∫ −W/2 ∫− a / 2 ε o ∞ nπ nπx nπd nπd cos (sinh ) − − ( 63) + ε r cosh ∑ An a a a n = 1,odd a An Approximate Electrostatic Solution for Microstrip Lines W/2 a/2 ∫ − a / 2 ρsdx = the surface charge at y=d is given by the static capacitance per unit length is C= Q = V ∞ ∑ W 4a sin( mπW / 2a ) sinh( nπd / a ) 2 n = 1,odd ( nπ ) ε o [sinh( nπd / a ) + ε r cosh( nπd / a )] (64) this is the expression for ε r ≠ 1 W/2 ∫− W / 2 dx = W 13
- 14. 27/01/2014 An Approximate Electrostatic Solution for Microstrip Lines The Transverse Resonance Techniques the effective dielectric is defined as C εe = C Co , where o is obtained from Eq. (64) with ε r = 1 the transverse resonance technique employs a transmission line model of the transverse cross section of the guide the characteristic impedance is given by right at cutoff, the propagation constant is equal to zero, therefore, wave cannot propagate in the z direction Zo = εe 1 = v p C cC The Transverse Resonance Techniques it forms standing waves in the transverse plane of the guide the sum of the input impedance at any point looking to either side of the transmission line model in the transverse plane must be equal to zero at resonance The Transverse Resonance Techniques the characteristic impedance in each of the air and dielectric regions is given by k η k η k η Z a = xa o and Z d = xd d = xd o kd ε rko ko since the transmission line above the dielectric is of infinite extent, the input impedance looking upward at x=d is simply given by Z a The Transverse Resonance Techniques consider a grounded slab and its equivalent transmission line model to infinity d Za, kxa x εr z Zd,kxd The Transverse Resonance Techniques the impedance looking downward is the impedance of a short circuit at x=0 transfers Z + jZ o tan βl to x=d Z in = Z o L Z o + jZ L tan βl Subtituting Z L = 0, Z o = Z d , β = k xd , l = d , we have Z in = jZ d tan βd Therefore, k xa ηo k η + j xd o tan k xdd = 0 ko εrko 14
- 15. 27/01/2014 The Transverse Resonance Techniques Note that k xa = − jh , therefore, we have ε r h = k xd tan k xdd − − − ( 65 ) From phase matching, k yo = k yd which leads to 2 2 2 2 2 ε r k o − k xd = k o − k xa = k o + h 2 − − − ( 66 ) Eqs. (65) and (66) are identical to that of Eq. (38) and (39) Wave Velocities and Dispersion if the signal contains a band of frequencies, each frequency will travel at a different phase velocity in a nonTEM line, the signal will be distorted this effect is called the dispersion effect Wave Velocities and Dispersion a plane wave propagates in a medium at the speed of light 1 / µε Phase velocity, v p = ω / β , is the speed at which a constant phase point travels for a TEM wave, the phase velocity equals to the speed of light if the phase velocity and the attenuation of a transmission line are independent of frequency, a signal propagates down the line will not be distorted Wave Velocities and Dispersion if the dispersion is not too severe, a group velocity describing the speed of the signal can be defined let us consider a transmission with a transfer function of Z( ω ) = Ae − jβz = | Z( ω )| e − jψ Wave Velocities and Dispersion if we denote the Fourier transform of a timedomain signal f(t) by F(ω), the output signal at the other end of the line is given by fo ( t) = 1 ∞ j( ωt − ψ ) dω ∫ F( ω )| Z ( ω )| e 2π − ∞ π if A is a constant and ψ = aω, the output will be fo ( t) = Wave Velocities and Dispersion this expression states that the output signal is A times the input signal with a delay of a now consider an amplitude modulated carrier wave of frequency ω o s( t ) = f ( t ) cos ω o t = Re{f ( t ) e jω o t } ∞ 1 A ∫ F ( ω )e jω ( t − a ) dω = Af ( t − a ) 2π − ∞ π 15
- 16. 27/01/2014 Wave Velocities and Dispersion Wave Velocities and Dispersion jω t the Fourier transform of f ( t ) e o is given by S(ω ) = F( ω − ω o ) The output signal so ( t ) , is given by note that the Fourier transform of s(t) is equal to 1 {F( ω + ω o ) + F( ω − ω o )} 2 Wave Velocities and Dispersion if the maximum frequency component of the signal is much less than the carrier frequencies, β can be linearized using a Taylor series expansion β( ω ) = β( ω o ) + dβ | ( ω − ω o ) + ... dω ω = ω o note that the higher terms are ignored as the higher order derivatives goes to zero faster than the growth of the higher power of (ω − ω o ) so ( t ) = ∞ 1 Re ∫ AF(ω − ω o )e j( ωt − βz) dω 2π − ∞ π for a dispersive transmission line, the propagation constant β depends on frequency, here A is assume to be constant (weakly depend on ω) Wave Velocities and Dispersion with the approximation of &&& β ( ω ) ≈ β ( ω o ) + β'( ω o )( ω − ω o ) = β o + β'o ω ∞ 1 &&&z) so ( t ) = A Re{ ∫ F(&&&) e j( ωt − β o z− β'o ω dω } ω 2π π −∞ so ( t ) = ∞ 1 &&&( &&&} A Re{e j( ω o t − β o z) ∫ F(&&&)e jω t − β'o z) dω ω 2π π −∞ s o ( t ) = A Re{ f ( t − β'o z) e j( ω o t − β o z) } s o ( t ) = Af ( t − β'o z) cos( ω o t − β o z) − − − ( 67) Wave Velocities and Dispersion Eq. (67) states that the output signal is the time-shift of the input signal envelope Thank You!! the group velocity is therefore defined as 1 dω vg = β' o = |ω = ω o dβ 16

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