ELEC3115 Electromagnetic Engineering Part B – High-frequency electromagnetics ELEC3115 – part B – Andrea Morello 1 Special relativity Information cannot travel faster than the speed of light ELEC3115 – part B – Andrea Morello 2 ≈ 3 × 108 m/s (in vacuum) Light (and therefore information) travels at most 30 cm in 1 nanosecond Important at high frequency! Clock speed 3 GHz ELEC3115 – part B – Andrea Morello 3 Modern CPUs are multi-core, because information cannot travel fast enough (compared to the clock period) from one end of the chip to the other! So the chip needs to be sliced up into more compact “cores”. 100 ps < 3 cm propagation 2 cm A simple example ELEC3115 - part B - Andrea Morello 4 + R = 100 I V = 1 V How much current flows in the cable? = = 10 mA A tricky example ELEC3115 - part B - Andrea Morello 5 How much is the current now? ≠ the pulse generator cannot “know” the resistance at the other end of a 3 m long cable, if it sends out a pulse that only lasts for 1 ns !! + V R = 100 I 3 m t 1 V 1 ns Electrical length ELEC3115 - part B - Andrea Morello 6 Consider a system whose purpose is to handle or transmit signals that vary significantly over a timescale , and has a physical size . In the time , light can propagate at most to a distance . The electrical length of the system is = In the example before, = 10. The aim of part 2 of this course is to understand the behaviour of systems with ≫ Lumped elements ELEC3115 - part B - Andrea Morello 7 When we analyse this circuit, we assume that each circuit element has “concentrated” (lumped) properties. For example, a resistor has only the property of having a resistance between its two terminals. Its behaviour in a circuit is such that, when passing a current I, it imposes a voltage drop V = RI across its terminals. This is only true if the physical size of the resistor is small compared to the distance travelled by light within the timescale of relevance to the problem! Also, we assume that nodes connected by a perfect wire have the same voltage, e.g. the collector of Q6 and the base of Q15. Implicitly, this means we assume that Q6 and Q15 are “within the same light cone” for the signal. Distributed elements ELEC3115 - part B - Andrea Morello 8 In a coaxial cable, you can measure: • The resistance of each conductor from one end to the other • The inductance of each conductor from one end to the other • The capacitance between the inner and outer conductors • The conductance between the inner and outer conductors These parameters are not lumped! They are distributed all along the length of the cable. If you double the length, each parameter doubles. NEW: How do we analyse a circuit with distributed elements? Lumped elements approximation ELEC3115 - part B - Andrea Morello 9 Consider a very short piece of cable, of length z. We can approximate it with lumped elements R z L z C z G z R = series resistance per unit length ( / m) L = series inductance per unit length (H / m) C = shunt capacitance per unit length (F / m) G = shunt conductance per unit length (S / m) Note: R and L are the combined resistance and inductance of the two conductors Note: in a perfect cable R = G = 0 However, L and C are never zero Typically: C 100 pF / m L 250 nH / m (, ) ( + ∆, ) (, ) ( + ∆, ) N Telegrapher’s equations ELEC3115 - part B - Andrea Morello 10 Kirchoff’s voltage law: + ∆, = , − ∆ , − ∆ (, ) − + ∆, − , ∆ = , + (, ) Kirchoff’s current law at node N: + ∆, = , − ∆ + ∆, − ∆ + ∆, − + ∆, − , ∆ = + ∆, + ( + ∆, ) Telegrapher’s equations ELEC3115 - part B - Andrea Morello 11 Taking the limit ∆ → 0: − , = , + , (1) − , = , + , (1) For sinusoidal signals at frequency (phasors), use Fourier transform: − = + (2) − = + (2) These are called Telegrapher’s equations Wave equations ELEC3115 - part B - Andrea Morello 12 Take / of (2a), then substitute (2b) in it: 2 2 = + + = 2() 3 Same for (2b): 2 2 = + + = 2() 3 Where we have defined the propagation constant: = + + = + (4) Eqns. (3) are wave equations. Propagating waves ELEC3115 - part B - Andrea Morello 13 Eqns. (3) are solved by waves propagating in opposite directions: = 0 + − + 0 − + 5 = 0 + − + 0 − + 5 0 + and 0 + are the amplitude of the voltage and current associated with a wave the propagates along the positive direction (“left to right”) 0 − and 0 − are the amplitude of the voltage and current associated with a wave the propagates along the negative direction (“right to left”) Notes: • 0 + and 0 − depend on the initial and boundary conditions. • 0 +/− and 0 +/− are not independent of each other! Propagating waves ELEC3115 - part B - Andrea Morello 14 Wave propagation is best understood in the time domain. Consider for simplicity a “perfect” (lossless) cable, with = = 0. Then, = = (recall Eq. (4)) Consider a wave propagating left to right 0 − = 0 : , = = 0 + − (6) At = 0, = 0 we obviously have 0,0 = 0 + However, we also have 0 + at every place and time where = , because − = 1 there and then. , = 0 + Phase velocity ELEC3115 - part B - Andrea Morello 15 We deduce that Eqn. (5) describes the voltage in space and time associated with a wave that propagates with velocity = = 1 (7) is called phase velocity. You can think of it as the speed at which the peak of the wave propagates in space. If = 100 pF/m and = 250 nH/m, then = 1 10−10×2.5×10−7 = 2 × 108 m/s i.e. 2/3 of the speed of light in vacuum. Wavelength ELEC3115 - part B - Andrea Morello 16 For a wave with angular frequency , i.e. period = 2 , we have , = 0 + and 0, = 0 + because = = 2 ⇒ − = −2= −0= 1 After a time T equal to the period (in time) of the wave, the wave has moved a distance = 2 = = (8) is the wavelength. It represent the period (in space) of the wave. Characteristic impedance ELEC3115 - part B - Andrea Morello 17 Calculate (2a) using (5a): = − + = −0 + − + 0 − + = + 0 + − − 0 − + (9) Comparing (9) with (5b) we deduce that: 0 + = 0 + + 0 − = −0 − + Characteristic impedance ELEC3115 - part B - Andrea Morello 18 Therefore: 0 + 0 + = + = + + + = + + We define the characteristic impedance of the cable as: 0 = + + (10) The characteristic impedance is the ratio of the amplitudes of voltage and current in the propagating wave. If the cable is lossless ( = = 0) and = 100 pF/m, = 250 nH/m, then 0 = 2.5 × 10−7 10−10 = 50 Ω The answer to the earlier question ELEC3115 - part B - Andrea Morello 19 The current is: = 0 If 0 = 50 Ω, then = 20 mA. This is what we called 0 + before. It is completely independent of what value of resistance is attached at the other end of that cable! + V R = 100 I 3 m t 1 V 1 ns Voltage and current on a line ELEC3115 - part B - Andrea Morello 20 Recall Eqns. (9) and (10): = + 0 +− − 0 −+ (9) 0 = + + (10) Plug these into (5b): = 0 + − + 0 − + : = 0 + 0 − − 0 − 0 + 11 Notice that 0 = 0 + 0 + = − 0 − 0 − The minus sign in the front of the expressions for the backwards-propagating waves is because we define the currents as positive going left to right. Lossless transmission line ELEC3115 - part B - Andrea Morello 21 Recall Eq. (4) = + + = + If there is no resistive dissipation along the cable, and no leakage current between the two conductors, i.e. = = 0, then: = 0 = Attention! This is the opposite of what you are used to in low-frequency circuit analysis! Here, the characteristic impedance is purely real when you have no resistive components! Terminated lossless line ELEC3115 - part B - Andrea Morello 22 ~ + l 0 = = Propagation delay: = / Voltage and current on the line: = 0 +− + 0 −+ 12 = 0 + 0 − − 0 − 0 + 12 = 0 = − Boundary conditions ELEC3115 - part B - Andrea Morello 23 At the load ( = 0), from Eqns. (1
2): = 0 = 0 + + 0 − (13) = 0 = 0 + + 0 − = 0 + − 0 − 0 (13) The load impedance imposes a boundary condition: = Using Eqns. (13): = 0 0 + + 0 − 0 + − 0 − (14) Solving for 0 −: 0 + − 0 − = 0 0 + + 0 − 0 − = − 0 + 0 0 + (15) Voltage reflection coefficient ELEC3115 - part B - Andrea Morello 24 From Eq. (15) we can calculate the ratio between the voltages of the reflected wave and the incident wave: Γ = 0 − 0 + = − 0 + 0 (16) is the voltage reflection coefficient. It is often expressed in terms of the normalized load impedance = 0 : Γ = − 1 + 1 (17) We can rewrite Eqns. (12) as: = 0 + − + Γ + 18 = 0 + 0 − − Γ + 18 Open, shorted, and matched line ELEC3115 - part B - Andrea Morello 25 From Eq. (16), we find some useful limiting cases: • = ∞ ⇒ Γ = 1 open line: the incident voltage is fully reflected. • = 0 ⇒ Γ = −1 short-circuited line: the incident voltage is fully reflected, but with opposite sign. • = 0 ⇒ Γ = 0 matched line: the load impedance matches the cable (characteristic) impedance. No reflections occur. Note: in a lossless line, 0 is a real number, but the load impedance might be complex, so in general Γ is complex too. Γ = Γ 0 ≤ Γ ≤ 1 is called phase angle Propagation in the time domain ELEC3115 - part B - Andrea Morello 26 Telegrapher’s equations (1) for lossless line, in the time domain: − , = , − , = , They yield a wave equation: 2(, ) 2 = 2(, ) 2 With general solution: , = + − + − + Where = 1 Reflections in the time domain ELEC3115 - part B - Andrea Morello 27 For a lossless line, Γ is frequency-independent. If the load impedance is purely real (resistive), then Γ is also real. Under these assumptions, the relation Γ = − + = − 0 + 0 holds also in the time domain. We consider a line with a generator at one end, and a resistive load at the other. The internal impedance of the generator is real, ~ + l Γ = − 0 + 0 Γ = − 0 + 0 Reflection diagram ELEC3115 - part B - Andrea Morello 28 z t l 2 Time-domain reflectometry ELEC3115 - part B - Andrea Morello 29 It is often difficult to locate the exact place where a cable is broken or bent, a connector is loose, etc… However, each of these faults causes a local discontinuity in the characteristic impedance reflections Time-domain reflectometry is a method where a step (or pulse) generator sends a pulse down the line, and a fast oscilloscope records the reflected signal. The delay between the pulse and the reflection tells you the distance between the fault and the generator! Power along a line ELEC3115 - part B - Andrea Morello 30 Consider a lossless line. Recall Eqns. (18) = 0 + − + Γ + 18 = 0 + 0 − − Γ + 18 The average power at any point is: = 1 2 ∗ (19) Recall that these are phasors, i.e. we are considering steady-state sinusoidal signals of the form , = cos( + ) and so on. In other words, is the peak value of the sinusoid. “Average power” means “averaged over one period of the sinusoid”. That’s why there is a factor ½ in Eq. (19). = 1 2 0 + 2 0 − + Γ + + − Γ∗ − = = 1 2 0 + 2 0 1 − Γ∗ −2+ Γ +2 − Γ 2 Power along a line ELEC3115 - part B - Andrea Morello 31 = 1 2 0 + 2 0 1 − Γ 2 (20) Recall that we are looking at steady-state sinusoidal signals (we used the phasors representation). Eq. (20) has a simple meaning: the time-averaged power flowing along the line is constant (i.e. independent of z) and consists of : • The incident power 1 2 0 + 2 0 sent towards the load, • Minus a reflected power 1 2 Γ 2 0 + 2 0 coming back from the load if it is mismatched, i.e. Γ ≠ 0. Maximum power at the load ELEC3115 - part B - Andrea Morello 32 Γ 2 From Eq. (20) follows that the maximum power is obtained when Γ = 0 Matching the load to the characteristic impedance of the line has two beneficial effects: 1. Eliminates transients in the time domain 2. Ensures the maximum power is delivered to the load. = 0 = − Return loss ELEC3115 - part B - Andrea Morello 33 = −20 Log Γ (21) The return loss expresses in dB the fraction of power that the line “loses to the load”, i.e. delivers without reflection. It is often quoted by suppliers of microwave components Γ = 0 ⇒ = ∞ (all power in the line is delivered to the load) Γ = 1 ⇒ = 0 (all power in the line stays on the line, i.e. returns back towards the generator because of reflection) Standing waves ELEC3115 - part B - Andrea Morello 34 Although the power is constant along the line, the voltage and the current are not necessarily so. Using Eq. (18a): () = 0 +− 1 + Γ +2 = 0 + 1 + Γ +2 At a point = − backwards from the load, and recalling that Γ = Γ (−) = 0 + 1 + Γ (−2) (22) (−) is maximum where (−2)= 1 ⇒ = 2 (−) is minimum where (−2)= −1 ⇒ = − 2 A pattern of minima and maxima repeats itself with period = Standing waves Standing waves ELEC3115 - part B - Andrea Morello 35 Example: = = 150 Ω; 0= 50 Ω ⇒ Γ = −0 +0 = 1 2 = 1 2 0 () 0 + 0 + 1 + Γ 0 + 1 − Γ = − 4 = − 2 Recall Eq. (8): = 2 ⇒ = 2 the standing wave pattern repeats every half-wavelength Voltage Standing Wave Ratio (VSWR) ELEC3115 - part B - Andrea Morello 36 The VSWR is the ratio between the maximum and the minimum voltage in the standing wave pattern: = = = 1 + Γ 1 − Γ (23) Ideally we want = 1 (Γ = 0, matched line). Since 0 ≤ Γ ≤ 1, then 1 ≤ ≤ ∞ Example: short-circuited line, Γ = −1 ⇒ = ∞ 0 + 0 + 1 + Γ () Slotted line ELEC3115 - part B - Andrea Morello 37 The slotted line is a simple tool to probe the voltage along a line, by sliding a probe through a slot. This allows the extraction of the VSWR, and from there the reflection coefficient, etc… Input impedance ELEC3115 - part B - Andrea Morello 38 We learned that: 1. The time-averaged power along the line is constant 2. The voltage along the line can have a standing-wave pattern if the load is mismatched The current must oscillate with a pattern opposite that of the voltage, to keep constant The impedance seen looking into the line must oscillate accordingly Input impedance of the line at position = −, seen looking towards the load: = (−) (−) = 0 + + Γ − 0 + 0 − Γ − = 0 1 + Γ −2 1 − Γ −2 (24) Special case: matched line ELEC3115 - part B - Andrea Morello 39 Γ = 0; = 1 = 0 + ∀ = 0 + 0 ∀ () = 0 ∀ An impedance-matched line is often called “flat”: the voltage, current and impedance are the same no matter where you look along the line. = 0 () () = 0 = − () Special case: shorted line ELEC3115 - part B - Andrea Morello 40 Γ = −1; = ∞ = = 0 + − − + = −20 + sin = 20 + sin = 0 + 0 − + + = 20 + 0 cos = 20 + 0 cos = −0 tan = 0 tan (25) The input impedance of a shorted line is purely imaginary. = 0 () () = 0 = − () Special case: shorted line ELEC3115 - part B - Andrea Morello 41 () 20 + 1 −1 − 4 − 2 − 3 4− ()0 20 + 1 −1 () 0 Special case: open line ELEC3115 - part B - Andrea Morello 42 Γ = 1; = ∞ = = 0 + − + + = 20 + cos = 20 + cos = 0 + 0 − − + = − 20 + 0 sin = 20 + 0 sin = 0 cot = −0 cot (26) The input impedance of an open line is purely imaginary. = ∞ () () = 0 = − () Special case: open line ELEC3115 - part B - Andrea Morello 43 () 20 + 1 −1 − 4 − 2 − 3 4 − ()0 20 + 1 −1 () 0 Line calibration ELEC3115 - part B - Andrea Morello 44 Shorted and open lines are often used as fixed-point calibrations to extract the characteristic impedance 0 and the phase factor of an unknown line. Take the product of Eqns. (25) and (26): = −0 tan × 0 cot = 0 2 0 = (27) Notice that Eq. (27) is independent of : it doesn’t matter where you measure the impedances, as long as you have a short and open at the end of the line. Taking the ratio: tan = − (28) Here, knowing you can extract Half-wavelength line ELEC3115 - part B - Andrea Morello 45 In Eq. (24), set = Τ 2: ±2 ൗ 2 = ±2= 1 = 0 1 +
Γ 1 − Γ = 0 1 + − 0 + 0 1 − − 0 + 0 = 0 + 0 + − 0 + 0 − + 0 = 0 2 20 = The input impedance of a half-wavelength line is the same as the impedance of the load. It’s as if the line weren’t there at all! This holds for any integer multiples of Τ 2 Quarter-wavelength line ELEC3115 - part B - Andrea Morello 46 In Eq. (24), set = Τ 4: ±2 ൗ 4 = ±= −1 = 0 1 − Γ 1 + Γ = 0 1 − − 0 + 0 1 + − 0 + 0 = 0 + 0 − + 0 + 0 + − 0 = 0 20 2 = 0 2 The input impedance of a quarter-wavelength line is inversely proportional to the impedance of the load. A Τ line transforms an open circuit into a short circuit and vice versa. This still holds when adding any integer multiples of Τ 2. Quarter-wave transformer ELEC3115 - part B - Andrea Morello 47 A Τ 4 line is sometimes used for impedance matching. Suppose I have a line with impedance 0 , a load ≠ 0 , and I can insert a Τ 4 line of impedance 0 between the two. 0 0 4 4 = 0 2 If I choose 0 = 0 , then = 0 and the line is matched. This technique is useful in stripline circuits, where 0 is easily set by the geometry of the strip. Microstrip ELEC3115 - part B - Andrea Morello 48 0 is easily changed by varying the width of the upper conductor. Smith chart ELEC3115 - part B - Andrea Morello 49 The Smith chart was invented in 1939 by P. H. Smith at Bell labs, as a graphical tool to help solving transmission line problems. It’s a graphical representation of the reflection coefficient is polar coordinates Γ = Γ = Γ + Γ Since 0 ≤ Γ ≤ 1 the Smith chart is contained within a circle of radius 1 Γ Γ Γ Normalized impedance ELEC3115 - part B - Andrea Morello 50 Recall the normalized impedance: = 0 : Γ = − 1 + 1 = Γ (17) Solving for : = 1 + Γ 1 − Γ Separating the real and imaginary parts: = + Γ = Γ + Γ + = 1 + Γ + Γ 1 − Γ − Γ Multiplying numerator and denominator by 1 − Γ + Γ: = 1 − Γ 2 − Γ 2 1 − Γ 2 + Γ 2 29 ; = 2Γ 1 − Γ 2 + Γ 2 (29) Parametric equations ELEC3115 - part B - Andrea Morello 51 Eqns. (29) can be rearranged as parametric equations, where and (the impedance) are expressed in terms of Γ and Γ (the horizontal and vertical axis on the Smith chart): Γ − 1 + 2 + Γ 2 = 1 1 + 2 (31) Γ − 1 2 + Γ − 1 2 = 1 2 (31) Notice that Eqns. (31) are of the same form as: − 0 2 + − 0 2 = 2 Which describes a circle of radius , centred at the point 0, 0 (31a) is the resistance circle (31b) is the reactance circle Resistance circles ELEC3115 - part B - Andrea Morello 52 Γ − 1 + 2 + Γ 2 = 1 1 + 2 ⇒ 0 = 1 + ; 0 = 0; = 1 1 + Γ Γ All resistance circles touch the point (1,0) to their right, and are centred along the positive real axis. The radius is 0 for = ∞ (open circuit) and 1 for = 0 Note: = 0 is not necessarily a short circuit! It could be a pure inductor or capacitor. However, there is only one way to make = ∞ Reactance circles ELEC3115 - part B - Andrea Morello 53 Γ − 1 2 + Γ − 1 2 = 1 2 ⇒ 0 = 1; 0 = 1 ; = 1 Γ Γ All reactance circles are centred along the vertical line Γ = 1 and touch the (1,0) point. The radius is equal to the vertical shift of the centre. There’s no sense in drawing the reactance circles outside the circle of unit radius, since there is no physical value of compatible with that. Γ = 1 Full Smith chart ELEC3115 - part B - Andrea Morello 54 Resistance and reactance circles always cross orthogonally to each other Reflection coefficient on the Smith chart ELEC3115 - part B - Andrea Morello 55 Given a (normalized) impedance = + : • The reflection coefficient Γ is found immediately by marking the intersection of the and (point P). • Γ is the length of the segment, relative to the radius of the chart: Γ = OP/OR. • is found by reading the angle scale around the circle labelled “angle of reflection coefficient” Reflection coefficient on the Smith chart ELEC3115 - part B - Andrea Morello 56 Example: = 2 + 1 P O R Γ = 0.4526.6 Input impedance ELEC3115 - part B - Andrea Morello 57 Recall Eq. (24) for the input impedance of a transmission line, at a distance l from the load: = 0 1 + Γ −2 1 − Γ −2 (24) Normalizing it by the characteristic impedance, and substituting Γ = Γ : = 0 = 1 + Γ (−2) 1 − Γ (−2) (32) Eq. (32) is identical to Eq. (29), after decreasing the phase of the reflection coefficient by an angle 2. On the Smith chart, the input impedance is found by rotating clockwise by an angle on the circle of radius , starting from the point P determined by . Rotations around the Smith chart ELEC3115 - part B - Andrea Morello 58 The Smith chart has four scales around its periphery. From the outside inwards: 1. Wavelengths towards the generator: this is a relative scale, used to perform clockwise rotations of angle 2 to find the input impedance at distance l “backwards”, from the load towards the generator. 2. Wavelengths towards the load: this is a relative scale, used to perform anti-clockwise rotations of angle 2 to find the input impedance at distance l “forwards”, from the generator towards the load. 3. Angle of reflection coefficient: used to find (see example before). 4. Angle of transmission coefficient. We won’t use it. Recall that = 2 . When = 2 , the rotation 2 covers an angle 2, i.e. you go a full circle around the chart. We saw this before – a half-wavelength line acts as if it weren’t there! Radially scaled parameters ELEC3115 - part B - Andrea Morello 59 Under the Smith chart you find several horizontal scales that can be used to find the parameters of the system: With a ruler, measure the length of the segment OP, where P is the point representing the normalized impedance on the Smith chart, then use that length to find out: • Standing wave ratio SWR • Reflection coefficient, E or I (the “P” refers to power, Γ 2) • Return Loss • Etc… Example ELEC3115 - part B - Andrea Morello 60 Consider the following circuit: Normalized load impedance: = 2 + 1.5 = 100 + 75 Ω 0 = 50 Ω 0.2 At the load ELEC3115 - part B - Andrea Morello 61 = 30 From the scale “angle of reflection coefficient” Parameters ELEC3115 - part B - Andrea Morello 62 SWR = 3.8 RTN loss = 4.7 dB Γ = 0.59 At the input ELEC3115 - part B - Andrea Morello 63 = 0.42 − 0.57 = −114 All other parameters stay the same because they only depend on Γ The black circle centred at the origin, Γ = const. is sometimes called “S circle”, because S (= VSWR) is constant along it. Impedance admittance conversion ELEC3115 - part B - Andrea Morello 64 Recall the effect of introducing a /4 line (page 46): = 0 2 In normalized form: = 1 = Where is the normalized admittance of the load. On the Smith chart: A rotation of 180o on the Smith chart converts a normalized impedance into the corresponding normalized admittance. Sometimes called “flipping the chart”. After the 180o rotation you can interpret the values on Smith chart, as is, as representing admittances instead of impedances. Combined Impedance - admittance chart ELEC3115 - part B - Andrea Morello 65 You can find combined Smith charts that plot both impedance and admittance, with different colours. Red circles: impedance Blue circles: admittance Here you can read directly z and y without “flipping the chart”. However, this chart becomes a bit messy, so we won’t use it. Impedance matching networks ELEC3115 - part B - Andrea Morello 66 An impedance matching network is a circuit that presents an input impedance = 0 even when connected to a load ≠ 0. Therefore, the network “transforms” the unmatched load into something that appears matched. ~ + matching network 0 We will normally assume = 0 Matching on the Smith chart ELEC3115 - part B - Andrea Morello 67 On the Smith chart, a matched load correspond to a point at the origin, Γ = 0. So we need to find a “path” to go from our unmatched impedance, point P, to the origin, point O. In terms of normalized impedance, we want to achieve = 1 + 0. Recall that, by simply adding a piece
of transmission line, we can rotate around the chart on the circle Γ = const. Basic idea: Step 1: Add a length of line that brings you to intersect the = 1 circle. Now you have matched the real part of the impedance, but the reactance might still be nonzero. Step 2: Add a reactance that cancels the residual reactance after matching the resistive component. This can be done by adding something in series or in parallel to the line. Single-stub matching ELEC3115 - part B - Andrea Morello 68 A piece of transmission line of length l (called “stub”), terminated by a short or an open circuit, is added in parallel or in series to the line (called “feedline”), at a distance d away from the load Short-circuited stub in parallel ELEC3115 - part B - Andrea Morello 69 It is often easier to insert stubs in parallel, both in coaxial cables and in microstrip. Also, a short circuit is usually more “perfect” than an open circuit, where fringing fields and other imperfections can be substantial at high frequencies. ~ + matching network 0 d l Short-circuited stub in parallel ELEC3115 - part B - Andrea Morello 70 When inserting components in parallel, it’s easier to use the admittances, since they simply add: = + In normalized units: = 1 = + So the steps to match the load are: Step 1: calculate d such that = 1 Step 2: calculate l such that = − with these steps we obtain a matched load = 1 Note: since the stub is terminated by a short circuit, the admittance you see upstream from it is always imaginary, i.e. = 0 (think about it…) Matching procedure ELEC3115 - part B - Andrea Morello 71 For shunt short-circuited stub: 1. Calculate the normalized load impedance = /0 and locate it on the Smith chart. This is point P1. 2. Draw the S circle through P1. 3. “Flip the chart”, by drawing a line from P, through O, until it intersects the S circle on the opposite side. This is point P2 and represents the normalized admittance of the load. From now on, the chart will indicate admittances. 4. Extend the line P1O P2 to the periphery of the Smith chart, and mark the point P2’ where the line crosses the “Wavelengths toward generator” scale. Write down the value of 2 at P2’. Remember: the WTG scale is relative. You should think of P2’ as the origin of that scale from now on, because that’s where the load admittance is. Matching procedure ELEC3115 - part B - Andrea Morello 72 5. Mark the points P3 and P4 where the S circles intersects the circle = 1. Note again: the circle = 1 is what “used to be” = 1 but now represents admittances because you’ve flipped the chart once. Notice that 3,4 = 1 ± , i.e. the susceptances differ only by the sign. 6. Extend the lines OP3 and OP4 to the periphery of the chart, and mark the points P3’ and P4’ on the WTG scale. Write down the values of the corresponding 3 and 4. 7. Calculate 3 = 3 − 2 and 4 = 4 − 2. and (relative to the wavelength ) are two possible position where = 1. You have now matched the real part of the admittance (Step 1). 8. Now you can attach a shunt stub at point or to cancel out the imaginary part of the admittance. Since we are considering a short- circuited stub, we start from point Psc which, for admittances, is on the right edge of the chart ( = ∞). Turning clockwise, calculate the lengths 3 and 4 that take you from Psc to P3”, where = −3 and P4”, where = −4. P3,4” are on the outer edge of the chart (Step 2). Remarks on matching ELEC3115 - part B - Andrea Morello 73 The single-stub matching technique we have discussed (as well as most other methods) is frequency-dependent. It relies upon a stub being of a certain length with respect to the wavelength. At a different frequency, is different so the stub has the wrong length! For the same reason, the analysis we did was all based upon phasors, i.e. we considered steady-state sinusoidal excitation at a single frequency. What happens if we apply a step? A big mess. A step function contains a broad spectrum of frequencies, and the response of the circuit isn’t matched at all. Stub-based matching presents a matched impedance to the generator at a specific frequency, but it does so by causing reflections between the stub and the load. Amplitude of incident waves ELEC3115 - part B - Andrea Morello 74 In the preceding lectures, we have seen how to treat the propagation of wave with voltage amplitude 0 +. Consider the general circuit below: Γ~ + 0 = 0 = − Γ How do you calculate 0 +? Amplitude of incident waves ELEC3115 - part B - Andrea Morello 75 In the case of a transient in the time domain (like when using time-domain reflectometry): 0 += 0 0 + This is because, at the instant it turns on, the generator only “sees” the impedance of the cable. In the case of steady-steady harmonic signals (which we describe with phasors), it’s not so simple!! There is still an incident and a reflected wave if there is some mismatch, but finding the correct value of 0 + requires accounting for all reflections between generator, line and load. Amplitude of incident waves ELEC3115 - part B - Andrea Morello 76 Using Eqns. (18a) and (24): = 0 + − + Γ + ; = 0 1 + Γ −2 1 − Γ −2 At the generator ( = −) the total voltage must satisfy the voltage division rule: − = + = 0 + + Γ − From this we extract the value of 0 +: 0 + = + 1 β + Γ −β (33) Or, using Eq. (24): 0 + = 0 0 + 0 −β 1 − ΓΓ −2β (34) For steady-state harmonic signals, 0 + is not a trivial quantity! Waveguides ELEC3115 - part B - Andrea Morello 77 Guided propagation ELEC3115 - part B - Andrea Morello 78 Wave propagation in coaxial cables is easy to understand, and can be derived from simple circuit laws (see Telegrapher’s equations). For very high frequencies, coaxial cables become impractical due to high losses and difficulty in constructing circuits with well-defined dimensions. In this section we study the general theory of wave propagation, including the case where there is only one conductor in the problem. Maxwell’s equations ELEC3115 - part B - Andrea Morello 79 Consider a source-free region of space, i.e. with no charges ( = 0) and no currents = 0 , filled by a non-conducting medium ( = 0) having permittivity and permeability . Maxwell’s equations are: × = − (35) × = (35) ∙ = 0 (35) ∙ = 0 (35) Combine (35a) and (35b) by taking × × = − × = − 2 2 Since × × = ∙ − = −, the above becomes: = 2 2 Wave equations ELEC3115 - part B - Andrea Morello 80 Calling = 1 the speed of light in the medium, we obtain: = 1 2 2 2 (36) Analogously: = 1 2 2 2 (36) These are wave equations. Their solution has the form of a plane wave ; = 0cos ± ∙ (37) Where = (38) is the wave vector. u is a unit vector that indicates the direction of propagation. Eqns. (36) can also be written for the individual Cartesian components , , Time-harmonic fields ELEC3115 - part B - Andrea Morello 81 Use phasors for time-harmonic (steady-state sinusoidal) fields, e.g.: ; = Maxwell’s equations become, for source-free regions: × = − (39) × = (39) Which yield wave equations for the phasors: + 2 = 0 (40a) + 2 = 0 (40b) These are called Helmholtz equations. Guided propagation ELEC3115 - part B - Andrea Morello 82 Let us now assume that there is a structure (e.g. a tube) that guides electromagnetic waves in one direction, z. We write: , , ; = , (−) (41) i.e. we describe a wave where the electric field has some dependence on x and y (to be calculated), and depends on z according to a phase factor (also to be determined) Note: is not necessarily the same as k. The presence of the guiding structure can allow for the two to be different. The way to think of them is: • is only a property of the medium. It’s just 2/ for a wave that propagates freely, at speed = 1 in a space filled by the medium having and . • is a property of the medium + the guiding structure. We need to solve the Helmholtz and Maxwell equations for the specific structure to find out the relation between
and k. Transverse and longitudinal components ELEC3115 - part B - Andrea Morello 83 The phasors , , and , , can be divided into transverse and longitudinal components: , , = ⊥ , + (, ) − (42) , , = ⊥ , + (, ) − (42) Substituting (42a,b) into the Maxwell equations (39a,b) we obtain: + = − (43) − − = − (43) − = − (43) + = (43) − − = (43) − = (43) Transverse components ELEC3115 - part B - Andrea Morello 84 We can solve for the transverse components as a function of the longitudinal ones. For example, can be written as a function of and by replacing, in (43a), obtained from (43e): = 2 − (44) = − 2 + (44) = − 2 + (44) = 2 − + (44) Where = 2 − 2 (45) is the cutoff wavenumber TEM waves ELEC3115 - part B - Andrea Morello 85 Is there a solution of Eqns. (44) where = = 0 ? Yes, if = 0, i.e. = . From Eqns. (44) we would get an undetermined result, but we can go back to Eqns. (43) and, imposing = = 0, find for example: 2 = 2 Which means = = as anticipated The Helmholtz equation (40a) becomes (take the example for ): 2 2 + 2 2 + 2 2 + 2 = 0 (46) However we know from (42a) that , , = , −, so 2 2 = − 2 = − 2 and Eq. (46) becomes: 2 2 + 2 2 = 0 (47) TEM waves ELEC3115 - part B - Andrea Morello 86 The same holds of all other transverse components, so we can write ⊥ 2⊥ , = 0 ⊥ 2⊥ , = 0 Where ⊥ 2 = 2 2 + 2 2 Thus the transverse components of the fields obey Laplace equation, i.e. they are the same as the fields between the conductors. TEM waves cannot exist in single-conductor waveguides. This is because, in the perpendicular plane, ׯ⊥ , ∙ = × ⊥ , = + Τ But, without an inner conductor, = 0. Moreover, since the TEM mode has by definition = 0, then also = 0. You need a centre conductor to produce a current density ≠ 0 so that a nonzero component of ⊥ can be produced around it (Ampere law) TE waves ELEC3115 - part B - Andrea Morello 87 TE stands for “transverse electric”. These are the solutions of Eqns. (44) where = 0 whereas ≠ 0. Eqns. (44) thus become: = − 2 (48) = − 2 (48) = − 2 (48) = 2 (48) Now we have solutions where ≠ 0. To apply Eqns. (48) we first solve the Helmholtz equation for : 2 2 + 2 2 + 2 2 + 2 = 0 → 2 2 + 2 2 + 2 = 0 because 2 2 = − 2 and 2 = 2 − 2 TM waves ELEC3115 - part B - Andrea Morello 88 TM stands for “transverse magnetic”. These are the solutions of Eqns. (44) where = 0 whereas ≠ 0. Eqns. (44) thus become: = 2 (49) = − 2 (49) = − 2 (49) = − 2 (49) The Helmholtz equation for becomes: 2 2 + 2 2 + 2 2 + 2 = 0 → 2 2 + 2 2 + 2 = 0 Wave impedance ELEC3115 - part B - Andrea Morello 89 In the “circuit style” description of transmission lines, we defined the characteristic impedance as 0 = 0 + 0 + = With single-conductor waveguides this becomes meaningless. We therefore use the wave impedance, defined as the ratio between the transverse components of the electric and magnetic field. It applies to any transmission line. • TEM modes: using (43d): TEM = = = ≡ 0 (50) where we have used the fact that, for TEM modes, = = This is the same as 0 for a coax (see Tutorial 1). If and are frequency-independent (non-dispersive medium), then also is frequency-independent. Wave impedance ELEC3115 - part B - Andrea Morello 90 For non-TEM modes, = 2 − 2 ≠ . We define the cutoff (angular) frequency as: = = (51) • TE modes: taking the ratio of (48c) and (48b), and recalling that = : TE = = = 2 − 2 = 2 − 2 = 1 2 − 2 TE = 0 1 − 2 (52) is frequency-dependent and always higher than 0 Wave impedance ELEC3115 - part B - Andrea Morello 91 • TM modes: taking the ratio of (49c) and (49b), and recalling that = : TM = = = 2 − 2 = 2 − 2 = 1 2 − 2 TM = 0 1 − 2 (53) is frequency-dependent and always lower than 0 Parallel plate waveguide ELEC3115 - part B - Andrea Morello 92 Consider two metal plates facing each other, separated by a distance d, filled by a medium described by and . The plates have width ≫ such that fringing fields can be ignored. This parallel plate waveguide is an abstract idealization, but a simple one to analyse. , y z x d W TM modes ELEC3115 - part B - Andrea Morello 93 TM modes have = 0 and ≠ 0. Here, in addition, since ≫ we can consider the waveguide to be invariant along the x direction, thus = 0. The Helmholtz equation is: 2 2 + 2 = 0 (54) and , , = , −. The solution to Eq. (54) is of the form: , = sin + cos (55) where the constants A and B are determined by imposing the boundary conditions , = 0 = 0 and , = = 0 (no electric field along a metal plate). Therefore = 0 and = for = 0,1,2,3, … or: = 56 = 0,1,2,3, … TM modes ELEC3115 - part B - Andrea Morello 94 From Eq. (56): = 2 − 2 → = 2 − Τ 2 (57) , = sin , , = sin − (58) From Eqns. (49) we find the transverse components: = cos − (59) = − cos − 59 = = 0 (59) From Eq. (51), the cutoff frequency for the n-th mode is: = 2 = 2 (60) TM modes ELEC3115 - part B - Andrea Morello 95 We can also define a cutoff wavelength for mode n: = = Τ1 Τ 2 = 2 (61) Or: = 2 (62) This means that, in order to start propagating the n-th mode, you must be able to fit at least n half-wavelengths between the plates. Bouncing plane waves interpretation ELEC3115 - part B - Andrea Morello 96 Take for example the TM1 mode ( = 1): 1 = 2 − Τ 2 = 2 − 2 Τ 1 2 , , , = 1 sin −. This can be rewritten as: , , = 1 2 Τ −1 − − Τ −1 Interpretation: two plane waves travelling along a oblique directions −,+ and +,+ , with wave vector = Τ 2 + 1 2 y z d 0 sinθ = cosθ = 1 Phase velocity ELEC3115 - part B - Andrea Morello 97 The phase velocity of each plane wave along its direction of propagation () is simply Τ = Τ1 which is just the speed of light in the material filling the guide. BUT, the phase velocity along z is: = 1 = 1 cosθ (63) which is always greater than the speed of light in the material!! This already tells you that the phase velocity is not the quantity describing the speed at which information travels along a transmission line. TE modes ELEC3115 - part B - Andrea Morello 98 The TE modes in a parallel plate waveguide are dual to the TM modes: , = sin + cos (64) From (48c): , , = − cos − sin − with = 0 at = 0, and ≡ 0 = 0 , , = cos − (65) = sin − (66) = cos − 66 = = 0 (66) and are the same as for the TM mode. TEM mode ELEC3115 - part B - Andrea Morello 99 Since the parallel plate waveguide has two separate conductors, it can actually sustain a TEM mode, which is just the TM mode with = 0. 0 = ; 0 = 0; = 0 −0; = − 0 −0 TEM TM TE Rectangular waveguides ELEC3115 - part B - Andrea Morello 100 Rectangular waveguides are among the most commonly used in high-frequency microwave applications. Since there is now only one conductor, the TEM mode does not exist. y z x b a , By convention > TM modes ELEC3115 – part B – Andrea Morello 101 Helmholtz equation: (now Τ ≠ 0 because it’s not invariant along x) 2 2 + 2 2 + 2 , = 0; , , = , − This partial differential equation can be solved by separating the variables: , = () Substituting into the Helmholtz equation: 2 2 + 2 2 + 2 = 0 → 1 2 2 + 1 2 2 + 2 = 0 Since the variables are independent, each term must be equal to a constant. We can define the constants and such that: 2 2 + 2 = 0 (67) 2 2 + 2 = 0 (67) 2 + 2 = 2 (67) TM modes ELEC3115 – part B – Andrea Morello 102 The general solution for can be written as: , = cos + sin cos + sin Boundary conditions: , = 0 at = 0, , = 0 at = 0, They impose: = 0; = = 1,2,3, … = 0; = = 1,2,3, … , , = sin sin − (68) The propagation constant is: = 2 − 2 = 2 − 2 − 2 (69) TM modes – cutoff frequency ELEC3115 – part B – Andrea Morello 103 F
rom (69) we see that the propagation constant is real if > , where the cutoff frequency of the TMmn mode is: = 2 = 1 2 2 + 2 (70) TM modes: transverse components ELEC3115 – part B – Andrea Morello 104 From (69) and (49), the transverse components of the fields in the TMmn mode are: = − 2 cos sin − 71 = − 2 sin cos − 71 = 2 sin cos − 71 = − 2 cos sin − 71 Notice that all fields are identically zero unless both and are ≠ 0. Therefore, the lowest propagating TM mode is the TM11, with cutoff 11 = 1 2 1 2 + 1 2 TM modes: field patterns ELEC3115 – part B – Andrea Morello 105 TE modes ELEC3115 – part B – Andrea Morello 106 The solution is obtained similarly to the TM modes: , , = cos cos − (72) = 2 cos sin − 73 = − 2 sin cos − 73 = 2 sin cos − 73 = 2 cos sin − 73 Now the fields are nonzero as soon as either or is ≠ 0. Therefore, the lowest propagating TE mode is the TE10, which has: 10 = 1 2 (74) TE modes: field patterns ELEC3115 – part B – Andrea Morello 107 Waveguide modes: choice ELEC3115 – part B – Andrea Morello 108 The “dominant” (i.e. starting at lowest frequency) mode is always the TE10. Other modes become accessible at higher frequencies. If you want to transmit a frequency , you will normally choose a waveguide such that 10 < < 01 < 11… because that’s the smallest waveguide you can use for the purpose. Waveguides are bulky and heavy, so you don’t normally want to make them bigger than necessary just to propagate higher modes! Dispersion ELEC3115 - part B - Andrea Morello 109 In empty space or in a uniform medium, electromagnetic waves propagate with: • velocity = Τ1 • propagation constant = Τ • wavelength = Τ2 = Τ If and are independent of frequency, then is constant, and and are simply proportional to each other The function is called dispersion relation If it’s linear, the medium is called non-dispersive, because it means that the velocity of propagation is the same for signals of any frequency. Dispersion ELEC3115 - part B - Andrea Morello 110 Consider the TM modes in a parallel-plate waveguide: • TM0 TEM: 0 = = Τ , = Τ1 , 0 = 0, • TM1: 1 = 2 − 1 2 = 1 − 1 2 (1 = 1, 1 = Τ ) • TM2: 2 = 2 − 2 2 = 1 − 2 2 (2 = 2, 2 = Τ2 ) For ≥ 1 the modes are dispersive: the relation is not linear 1 TEM 2 TM1 TM2 Phase velocity ELEC3115 - part B - Andrea Morello 111 The phase velocity is, in general, given by: = (75) 1 2 Τ1 TEM TM1 TM2 For ≥ 1 the phase velocity is infinite at the cutoff frequency, and tends asymptotically to the speed of light in the medium as → ∞ Time delay ELEC3115 - part B - Andrea Morello 112 A transmission line can be treated like a two-port network, with transfer function . We apply a signal () to the input of the line. It Fourier transform is: = −∞ +∞ () − and the inverse is = 1 2 −∞ +∞ If the TL is matched and lossless, then = − = − (76) = () = 1 2 න −∞ +∞ − = 1 2 න −∞ +∞ − = − If is constant (independent of ), the line puts out an exact copy of the input signal, delayed by a time Τ Narrow-band signal ELEC3115 - part B - Andrea Morello 113 Consider a signal () having a narrow band, i.e. its Fourier transform has significant intensity only between ±. t () () +− The signal () then modulates a carrier of the form cos 0 = cos 0 = 0 t () () 0 + 0 0 − Narrow-band signal ELEC3115 - part B - Andrea Morello 114 The Fourier transform of the modulated signal is: = න −∞ +∞ () 0− = − 0 (I have taken only the positive frequencies part. When I go back to the time domain, I will take the real part of the inverse Fourier transform of the above) Now pass this signal through a TL with transfer function: = − i.e. I am now assuming a dispersive TL. = − 0 − Back to the time domain: = 1 2 න −∞ +∞ = = 1 2 න 0− 0+ − 0 − (77) Narrow-band signal ELEC3115 - part B - Andrea Morello 115 Although is no longer constant, I can take a Taylor expansion of it around 0 because the signal is narrow-band around it ≈ 0 + ′ − 0 + … where 0 = 0 , ′ = ቚ =0 Substituting into eq. (77) and making a change of variable = − 0: = 1 2 න 0− 0+ 0+−0− ′ = = 1 2 0−0 න 0− 0+ − ′ = = cos 0 − 0 ∙ − ′ (78) (1) (2) Group velocity ELEC3115 - part B - Andrea Morello 116 0 = 0 = 0 (0) Term (1) becomes cos 0 − (0) Let us define a group velocity 0 = 1 ′ = ቮ −1 =0 (79) Term (2) becomes − (0) The carrier “propagates” at the phase velocity The envelope propagates at the group velocity Group velocity in waveguides ELEC3115 - part B - Andrea Morello 117 Consider an air-filled waveguide, where = Τ ; 0 = Τ0 : = 2 − 2 = Τ 2 − 2 ቤ =0 = Τ0 2 Τ0 2 − 2 = 0 0 = ቮ −1 =0 = 0 (80) The phase velocity is 0 = 0 = 0 (81) Since < (always!) we have Τ 0 < 1 and Τ0 > 1, and therefore: < < The wave packet always propagates slower than the speed of light. Group velocity in waveguides ELEC3115 - part B - Andrea Morello 118 Graphical interpretation: 1 TEM TM1 0 The group velocity is the slope of the dispersion relation. Notice that → 0 as you approach the cutoff frequency (the dispersion relation tends to horizontal) Electromagnetic power ELEC3115 - part B - Andrea Morello 119 Calculate the divergence of the vector product of and . A useful vector identity allows us to write: ∙ × = ∙ × − ∙ × (82) The first term on the right-hand side is, using Faraday’s law: ∙ × = − ∙ (83) The second term is, using Ampere’s law: ∙ × = ∙ + ∙ (84) Eq. (80) is only a function of the magnetic fields, and (84) only a function of the electric fields. Electromagnetic power ELEC3115 - part B - Andrea Morello 120 In electrical circuits, the dissipated power is expressed by Joule’s law: = ∙ (85) Therefore, in microscopic terms, the power density is: = ∙ (86) measured in W/m3. V is the volume. Since ∙ is a power density, so are the terms in Eqns. (83) and (84). The total power is obtained by integrating (82) over the volume V: ම ∙ × = −ම ∙ + ∙ −ම ∙ (87) Eq. (84) can be rewritten using Gauss’ theorem, and the identities ∙ = ∙ 2 ; ∙ = ∙ 2 where ∙ = 2 and ∙ = 2 Poynting theorem ELEC3115 - part B - Andrea Morello 121 × = − ම 2 2 + 2 2 −ම ∙ (88) where S is the surface of volume V. Eq. (88) is in units of W. Eq. (88) is called Poynting theorem. Let us define the Poynting vector: (in units W/m2) = × (89) From Eq. (88) we can identify the various terms: • װ × is the flux of the Poynting vector out of the volume V • 2 2 is the density of magnetic power • 2 2 is the density of electric power • ∙ is the density of power dissipated by Joule heating Poynting theorem ELEC3115 - part B - Andrea Morello 122 The Poynting theorem says that: “the outward flux of the Poynting vector equals the negative of the time change of electric and magnetic power densities, and the Joule dissipation.” The Poynting vector indicates the direction of propagation of the power associated with an electromagnetic wave 2 2 + 2 2 Time-averaged and complex Poynting vector ELEC3115 - part B - Andrea Morello 123 Typically we deal with time-varying harmonic signals, so it’s more meaningful to talk about the Poynting vector averaged over one period of the wave: = 1 න 0 Using the phasor description of and one can demonstrate that = 1 2 × ∗ (90) (it’s the same reason why the power in a circuit is 1 2 ∗ ) We can define a complex Poynting vector as = × ∗ (91) Power in waveguides ELEC3115 - part B - Andrea Morello 124 From the equations of the fields in waveguides (59), (66), (71), (73) we can calculate the Poynting vector for any mode, and integrate it over the cross-section of the waveguide to obtain the tota
l power flowing through. Example: TE modes in rectangular waveguide , , = 2 cos sin − 73 , , = − 2 sin cos − 73 , , = 2 sin cos − 73 , , = 2 cos sin − 73 Power in waveguides ELEC3115 - part B - Andrea Morello 125 The power is the sum of the contributions from ∗/2 and − ∗/2. The − sign is because ො × ො = −ො , = ො 1 2 , ∗ , − , ∗ , Notice, however, that has itself a − sign [Eq. (73b)] so all terms give a positive contribution. The power flow is then װ , ∙ , where = ො For example, for the TE10 mode ( = 1, = 0) one obtains: 10 = 3 10 2 42 (92) Note: we used a generic amplitude . Specifically, one finds that • ≡ for TE modes • ≡ for TM modes Evanescent waves ELEC3115 - part B - Andrea Morello 126 In a lossless waveguide, the propagation constant is = , where = ± 2 − 2 = = = 2 + 2 ≡ Therefore = ± 1 − 2 (93) If > then is real, is imaginary, and the wave propagates normally, accumulating a phase factor. Evanescent waves ELEC3115 – part B – Andrea Morello 127 If < then is imaginary, is real, and the wave decays exponentially as it goes. We can define an attenuation constant below cutoff as: = 2 − 1 (94) Therefore, the electric and magnetic fields in a waveguide below cutoff propagate as (take for example the x-component of the electric field): , , = , , 0 − (95) An electromagnetic wave that decays exponentially due to being below the frequency cutoff of a guiding structure is called evanescent wave. Note that the attenuation constant does not arise from dissipation! The waveguide was assumed lossless, so there is no heating taking place anywhere. The power is reflected back at the point where the waveguide goes below cutoff, but it’s not sudden. Units for propagation constant ELEC3115 - part B - Andrea Morello 128 In general, the propagation constant has a real and an imaginary part: = + Their units are: = rad/m = Np/m Np stands for “Neper”. The Neper is a dimensionless unit that describes by how many multiples of (the Neper constant) the wave gets attenuated. = 1 Np/m means that the field (electric or magnetic) has amplitude reduced by Τ1 after 1 m. Since 20Log10 = 8.69 we can also write: 1 Np/m = 8.69 dB/m Waveguides as high-pass filters ELEC3115 - part B - Andrea Morello 129 Consider for example a waveguide having cutoff for the lowest mode (TE10) 10 = 10 GHz. Calculate the attenuation of a signal at 8 GHz. = 2 10 2 − 1 = 2×8×109 3×108 10 8 2 − 1 = 125.66 Np/m Or 125.66 × 8.69 = 1092 dB/m This is an extremely effective high-pass filter!!
