EE Lecture 7. Finding gamma. Alternate form. I i. Transmission line. Z g I L Z L. V i. V g - Z in Z. z = -l z = 0

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1 Impedance on lossless lines EE - Lecture 7 Impedance on lossless lines Reflection coefficient Impedance equation Shorted line example Assigned reading: Sec.. of Ulaby For lossless lines, γ = jω L C = jβ; we will focus on lossless lines almost always from here on out Thus Ṽ (z) = V e jβz V ejβz () Ĩ(z) = Z ( V e jβz V ejβz) () where V and V are two phasor constants and Z = L/C for a lossless line The impedance on a lossless line is then Z(z) = Ṽ (z) Ĩ(z) = Z which is a function of z! [ V e jβz V ejβz V e jβz V ejβz Thus, the impedance on a transmission line changes depending where you measure it on the line ] Alternate form Note that we can factor out V e jβz on the top and bottom to get: Z(z) = Z V e jβz V e jβz V V = Z ( Γe jβz Γe jβz where we have defined Γ = V V coefficient e jβz V e jβz V ) as a voltage reflection Note that if Γ = (we have no V wave), Z(z) = Z meaning that the impedance is no longer a function of z This is usually the case when we talk about infinitely long lines, where propagating waves can propagate forever without generating a traveling wave Finite length lines terminated in an impedance, however, will typically have Γ as we will see next... ~ V g - Finding gamma Consider a Xmission line of length l terminated in an impedance Z L. A picture is below The two parallel lines in this picture represent a transmission line. It could be any type of line (parallel plate, coax, etc.) but we are told the Z and the µ, ɛ inside the line which is all we need to know to talk about voltages, currents, and impedances Since we ve drawn a circuit impedance Z L at z =, the impedance on the transmission line at z = must be Z L Z g Generator ~ I i ~ V i ~ Z in Z Z L V L - Transmission line z = -l z = - ~ I L Load

2 Using our previous formula for Z(z) and plugging in z = we get Z() = Z Γ Γ = Z L which can be solved to obtain Γ = Z L Z Z L Z Note Γ in general is complex; Ulaby writes Γ = Γ e jθr Knowing Γ we can find the impedance anywhere else on the line, for example at z = l, by using Γe jβz Z(z) = Z Γe jβz since all quantities in this equation are now known (note β = ω LC = ω µɛ) Another way of writing this equation is Γ(z) Z(z) = Z Γ(z) where Γ(z) = Γe jβz Impedance equation without Γ We can also plug in Γ = ZL Z Z LZ so that and simplify some to get ) Z(z) = Z ( ZL jz tan(βz) Z jz L tan(βz) ZL jz tan(βl) Z( l) = Z Z jz L tan(βl) This last formula is probably the most useful of all the equations since it directly gives the impedance at z = l if we know Z, Z L, β, and l Notice the dependence on l occurs only in the tan(βl) terms, which are periodic in βl with period π. Since β = π/λ, the impedance is periodic in l with period λ/. Thus impedances on lossless transmission lines repeat themselves every half wavelength 6 Shorted line example Consider a transmission line which is short circuited (Z L = ) Plugging Z L = into our formula gives [ ] jz tan(βl) Z( l) = Z = jz tan(βl) Z This is purely imaginary since both Z and tan(βl) are real; makes sense because a real impedance allows power dissipation and there is no way to dissipate power in a lossless line terminated in a short! l Shorted line impedance Now let s plot the impedance versus position on the line: Impedance sc Z in jz Z short circuit -λ -λ -λ -λ z 7 8

3 Our plot is just a plot of tan(βl). Note that a shorted line looks like a positive reactance for < l < λ/ but looks like an open circuit when l = λ/. This is not what you expect from circuit theory! Reasonable because transmission line theory only applies when lines become a significant fraction of a wavelength long (i.e. λ/ as above) This rarely happens (except in very long power system transmission lines) at low frequencies (as in circuits) because the wavelengths are so long at low frequencies Transmission line effects however are extremely important when dealing with higher frequency systems where wire lengths can be appreciable fractions of a wavelength Note by choosing an appropriate length of our short circuited line we can get any possible value of reactance we desire, i.e. the line can look like any inductor or capacitor at a specific frequency as long as l is chosen correctly. See Example -7 in book. EE - Lecture 8 More on shorted line Voltage and current on shorted line Standing waves: VSWR Shorted line example Other examples Assigned reading: Sec.. of Ulaby 9 Reduction to circuit theory Note our shorted line can be useful because often lumped element inductors or capacitors are not available at high frequencies and transmission lines represent the only way of obtaining the necessary impedances If we think a little more about the circuit theory limit (where the line is much shorter than a wavelength, so l/λ << and βl is also << ) we can see that transmission line theory gives us the right answer since lim βl Z as βl,tan(βl) = ZL jz tan(βl) = Z L Z jz L tan(βl) Thus the impedance measured on a wire connected to Z L is just Z L as long as βl << (the line is much shorter than a wavelength) Improved circuit theory approximation We can do one step better than this for our shorted line by using a better approximation for the tangent: lim tan(βl) βl βl This gives Z( l) = jz tan(βl) jz βl Recalling Z = L/C and β = ω LC gives Z( l) j L/Cω LCl = jωll, which shows that a short circuited line that is small compared to the wavelength has the same reactance as an inductor with inductance Ll, which is inductance per unit length times length! This makes sense because a shorted line looks like a small turn of wire = an inductor in the circuit theory limit. However, when βl is not <<, the graph of Z( l) shows that the shorted line no longer follows our circuit theory expectations

4 Voltages and currents on a shorted line Voltage and current plots It is also interesting to examine the voltages and currents on a short circuited line. Note for a short, Γ = ZL Z Z LZ = since Z L = This gives Plots of V (z) and I(z) on a shorted line look like: λ/ V V (z) = V V = jv sin(βz) () I(z) = ( V Z e jβz V = V cos(βz) Z () which apply for a shorted line only! If we plot V (z) and I(z) versus z, we observe a standing wave pattern, meaning that V (z) and I(z) oscillate periodically on the line as z is changed z -λ -λ -λ -λ ~ V(z) λ/ V -λ -λ -λ -λ z VSWR Voltage and current plots Note that V (z) and I(z) also repeat themselves every λ/ on the line, which is true for all lossless transmission line voltage and current amplitudes, and for impedances To describe the standing wave voltage pattern on a Xmission line, a term known as the voltage standing wave ratio (VSWR) is used This is defined to be the ratio of the maximum to minimum voltage amplitude on the line The VSWR on a shorted line is infinite since the minimum voltage is zero In general however we can write which can be rewritten as V SW R = Max V e jβz V ejβz Min V e jβz V ejβz V SW R = Max [ V e jβz Γe jβz ] Min [ V e jβz Γe jβz ] Thus we are looking for the maximum and minimum of Γe jβz since the V term cancels. This is the amplitude of a sum of two complex numbers (i.e. a b ) 6

5 VSWR equation Thus, we ll get the largest amplitude when the two numbers are exactly in phase, the smallest amplitude when the two are out of phase. The maximum value of Γe jβz will therefore occur when Γe jβz has phase degrees, giving Γe jβz = Γ. The minimum will occur when Γe jβz = Γ The VSWR can therefore be written in general as V SW R = Γ Γ The same equation for VSWR is obtained if the ratio of the maximum to minimum current on the line is considered Matched line example A matched line is a line terminated in its own characteristic impedance, i.e. Z L = Z. Then, ZL jz tan(βl) Z( l) = Z = Z Z jz L tan(βl) Thus on a matched line the impedance is not a function of z but rather is everywhere Z Note this occurs because Γ = since Z L = Z so V = and there is no reflected wave People working with transmission lines usually try to make all loads matched loads to avoid varying impedances on the line. Unfortunately this is not always possible so we need transmission line theory for the unmatched cases. 7 8 Quarter wave line A Quarter wave line has length l = λ/, so plugging this in to the impedance equation and noting that βl = π λ λ = π, we get ZL jz tan(π/) Z( λ/) = Z = Z Z jz L tan(π/) Z L since tan(π/) = sin(π/) cos(π/) = Thus the impedance on a lossless Xmission line λ/ away from a load is proportional to the inverse of the load impedance. Our short circuited line was an example since a short (Z L = ) became an open (Z( λ/) = ) a quarter wavelength away Recall that two quarter wave lines form a λ/ line which should result in Z( λ/) = Z L EE - Lecture 9 A sample problem Shorted lossy line Power on lines Assigned reading: Secs of Ulaby Let s check: Z A = Z Z L. Using the equation again to find Z B we get Z B = Z Z A = Z ZL = Z Z L, which agrees with our expectations! 9

6 A sample problem For this circuit, how much power is dissipated in the load? Answer: Here we ve got TL and ccts mixed together, so we need to use TL theory to replace the line with an impedance Z in, then use circuit theory to find the power Thus we find Z in from TL theory, replace the right hand side of the picture with Z in, and then use circuit theory to find the power dissipated in Z in Sample problem continued Because the line is lossless, this is the same as the power dissipated in the load Goal: replace previous picture with picture below, then circuit theory tells us the power into the load is Re {V LIL} = { } Re ()Zin Z in Zin = Re {Z in} Z in V G = V f= MHz R G = Ω Z in z= λ/8 Z = Ω Ω z= 6 µ H V G = V f= MHz R G = Ω I L Z in V L Finding Z in To use the transmission line equation to find Z in = Z( λ/8), first we need to find the load impedance Placing the resistance and inductance in parallel we find Z L = Ω jω(6 6 ) j Then = ZL jz tan(βl) Z( λ/8) = Z Z jz L tan(βl) [ j j tan( π λ λ 8 ) ] j( j) tan( π λ λ 8 ) = ( j) Thus the power dissipated in the load is Re { j} =. Watts j Note we could also have found V (z) and I(z) everywhere on the line and then taken Re {V ()I() }, but this would have been a lot harder! Impedance on lossy lines So far we ve been talking about impedance on lossless lines only. We can generate equations to handle lossy lines just by realizing that γ = α jβ for lossy lines instead of just jβ Our old equation for impedance ZL jz tan(βl) Z( l) = Z Z jz L tan(βl) becomes for a lossy line ZL Z tanh(γl) Z( l) = Z Z Z L tanh(γl) Here tanh is the hyperbolic tangent function: ( e γl e γl) / ( e γl e γl) and is taken of a complex argument γl

7 Short circuited lossy line For a shorted lossy line, Z L =, so the general equation becomes Z( l) = Z tanh(γl) which can be expanded as Z( l) = Z sinh(αl) cos(βl) j cosh(αl) sin(βl) cosh(αl) cos(βl) j sinh(αl) sin(βl) where sinh and cosh are hyperbolic sine and cosine functions Short circuited lossy line impedance The plots below compare impedances on lossy and lossless short circuited lines: Lossy line Real Imaginary Lossless line Real Imaginary This is quite complicated compared to the lossless answer jz tan(βl), but we can still find some points of interest For example, if l = nλ, βl = nπ and sin(βl) =. Thus Z( nλ ) = Z tanh(αl) Also if l = nλ for n odd, βl = nπ Z( nλ ) = Z / tanh(αl) for n odd and cos(βl) =. Thus Impedance/Z. Distance along line (λ) 6 Impedance/Z. Distance along line (λ) Power on lines Analysis of impedance Note that the impedance now has both real and imaginary parts since the line is lossy. Also note that the resonance effects at λ/, λ/, etc. get smaller as the line gets longer due to attenuation of the reflected wave Eventually as l, Z( l) should approach Z (which may be complex) on a lossy line because all the reflected wave gets attenuated before returning to z = l This is the extent of our study of lossy lines; we will focus on lossless lines only from here on... We have both voltages and currents on a transmission line, so the line carries power We re working with sinusoidal steady state quantities, so we should use our AC circuit power concepts The instantaneous power is p(z, t) = v(z, t)i(z, t). However as with AC circuits this will contain a term that oscillates rapidly on top of a DC offset. The time average power is what we are interested in. It can be shown that this is given by: P av = Re {Ṽ (z) Ĩ(z) } as in AC circuit theory 7 8

8 Power on terminated lines On a terminated lossless line: Ṽ (z) = V ( e jβz Γe jβz) Ĩ(z) = V /Z ( e jβz Γe jβz) Plugging these in gives a result of the form P av = V Re {(a b) (a b )} Z P av = V Re {( a b ba ab )} Z P av = V ( Γ ) Z The power carried by a lossless line is the same everywhere on the line, even though the voltages and currents may vary. EE - Lecture Gamma plane analysis Voltage and current in the Gamma plane Impedance in the Gamma plane Assigned reading: Secs of Ulaby This is because a lossless line cannot absorb any power. 9 Gamma plane analysis One of our earlier ways of writing Z(z) was Γ(z) Z(z) = Z Γ(z) where Γ(z) = Γe jβz We can also write Complex Gamma plane Let s draw a picture of how Γ(z) varies by sketching the complex Gamma plane If we take z =, then Γ(z) = Γ, which is just some particular complex number (a point in the Gamma plane) Im{Γ(z)} V (z) = V e jβz V ejβz = V e jβz ( Γe jβz) = V e jβz ( Γ(z)) I(z) = ( V Z e jβz V ejβz) = V e jβz Z ( Γ(z)) Taking a closer look at Γ(z) = Γe jβz, we can see that Γ(z) never changes on the line as we vary z, only the phase changes due to the e jβz term Γ( d) Re{Γ(z)}

9 Moving in the Gamma plane Note that we can think either in terms of rectangular, Γ(z) = Γ R (z) jγ I (z), or polar coordinates, Γ(z) = Γ(z) e jθr in the Gamma plane What happens when z? Since Γ(z) = Γe jβz, only the phase of Γ(z) changes, not its magnitude Thus, we just rotate around a circle of radius Γ(z) Note that as z becomes more negative (as in z = l), the angle of Γ(z) decreases, so we rotate clockwise in the Gamma plane As we move down the line, we rotate through an angle corresponding to e jβl which is βl radians clockwise= πl/λ radians. Note this repeats every time l increases by λ/ Interesting things about the Gamma plane:. Since Γ = ZL Z Z LZ, it can be shown that the maximum possible Γ is. Thus all our possible Γ s will fit inside a circle of radius in the Gamma plane. The origin of the Gamma plane means Γ =, so there are no reflections at the load and Z L must be Z. Real axis: we wrote V (z) = V e jβz ( Γ(z)) so V (z) = V Γ(z). Just like our VSWR analysis, the maximum voltage will occur when and Γ(z) are in phase. This is when Γ(z) is on the positive real axis. Minimum voltage amplitude will occur when and Γ(z) are out of phase, this is when Γ(z) is on the negative real axis. Note that the reverse is true for the current since it goes as Γ(z) Visualizing voltage and currents These statements are summarized below. Note we can use our knowledge about the location of maxima and minima to find Γ if we know Γ and the distance l from the load to a voltage max or min. We just spin Γ(z) back from the max or min point thorugh πl/λ radians to get to Γ L A graphical interpretation of voltage variations on line which are proportional to Γ(z) is the paddle wheel diagram Im{Γ(z)} Paddle wheel diagram Note that the length Γ(z) will vary as we move on the line and Γ(z) rotates It is clear from the diagram that we ll get a maximum length of the paddle arm ( Γ(z) ) when Γ(z) is on the positive real axis It is also clear we ll get a minimum length of the paddle arm when Γ(z) is on the negative real axis Maximum Current Minimum Voltage Γ(z) Γ( d) Re{Γ(z)} Also we can see that the maximum possible voltage on the line is V, minimum possible is, both of which can only occur if Γ(z) =, i.e. short or open circuit load Maximum Voltage Minimum Current 6

10 ..8.9 < WAVELENGTHS TOWARD LOAD < EE - Lecture Smith Chart Smith Chart example Properties of Smith Chart Assigned reading: Secs of Ulaby Impedance in the Gamma plane We ve seen that looking in the Gamma plane makes it easier to see the behavior of voltages and currents on lines. What about impedance? We know Z(z) = Z ( Γ(z) Γ(z) ), now define Zn(z) = Z(z) Z = Γ(z) Γ(z) as a normalized impedance, i.e. the impedance relative to the characteristic impedance of the line The above equation provides a map between Γ(z) and Zn(z). A little math shows that a given Γ(z) yields a unique value of Zn(z) and vice-versa Thus we can overlay scales of Zn(z) on our Gamma plane to get impedances graphically if we know Γ(z) 7 Smith Chart The Smith Chart is a Gamma plane with labeled contours of constant Zn(z) What are these contours? To find them solve ΓR(z)jΓI (z) Zn(z) = ΓR(z) jγi (z) by setting the real or imaginary parts of Zn(z) to a constant to find a relationship between ΓR and ΓI It turns out the resulting contours are off-center circles in the Gamma plane. Constant Normalized Resistance Constant Normalized Reactance Im{Γ(z)} Im{Γ(z)} Re{Γ(z)} Re{Γ(z)} INDUCTIVE REACTANCE COMPONENT (jx/zo), ORCAPACITIVESUSCEPTANCE(jB/Yo) ± 8 6 ANGLE OF TRANSMISSION COEFFICIENT IN DEGREES > WAVELENGTHS TOWARD GENERATOR > ANGLE OF REFLECTION COEFFICIENT IN DEGREES 9.. RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) -7 7 CAPACITIVE REACTANCE COMPONENT (-jx/zo), ORINDUCTIVESUSCEPTANCE(-jB/Yo) ATTEN. [db] S.W. LOSS COEFF RFL. LOSS [db] S.W. PEAK (CONST. P) TRANSM. COEFF, P RADIALLY SCALED PARAMETERS TOWARD LOAD > < TOWARD GENERATOR TRANSM. COEFF, E or I SWR dbs RTN. LOSS [db] RFL. COEFF, P RFL. COEFF, E or I CENTER

11 ..8.9 < WAVELENGTHS TOWARD LOAD < Work on example To get into the Gamma plane, we could calculate Γ = ZL Z but ZLZ we don t have to because the chart does it for us! Steps for using chart to find impedance on a terminated line:. Calculate ZnL = ZL = j Z = / j/ in this problem. Plot ZnL on the chart: find circle centered on real axis labeled real part /, circles centered off chart above real axis labeled imaginary part /. See next page for plot on the chart. Draw a line from the center of the chart through this point out to the scales on the outer edge of the chart. Read the number off the wavelengths toward generator scale. Here it is.88. Add l/λ (here.) to this number, and find that number of the wavelengths toward generator scale. We rotated clockwise like we know we are supposed to in the Gamma plane. Here we are finding.88. =. Draw a line from the center of the chart to this new location on the wavelengths toward generator scale Smith Chart example Circles of constant real part of Zn(z) are centered on the real axis, while constant imaginary part circles are centered off the chart (positive above real axis, negative below) Thus, the Smith Chart enables us to find impedances on transmission lines graphically as opposed to using the formulas We can work entirely in terms of impedances without worrying about Γ(z) even though the chart actually is just a labeled Γ(z) plane Repeat our earlier problem but now use the Smith Chart = Ω R G 8 = Ω jω=z L 6 7 Z V G = V f= MHz z= Z in z= λ/8 6. Since we know Γ(z) remains constant, the distance from the center of the chart to the new point should be the same as the old point. Measure and label the location of the new point on the chart. 7. Read off new value of Zn( l) at this point. Multiply by Z to get Z( l) = ZZn( l) Using the chart in this problem we get Zn( λ/8) = j so Z( l) = ( j), exactly the same as before! But using the chart required almost no calculations! INDUCTIVE REACTANCE COMPONENT (jx/zo), ORCAPACITIVESUSCEPTANCE(jB/Yo).88.= ± 8 6 ANGLE OF TRANSMISSION COEFFICIENT IN DEGREES > WAVELENGTHS TOWARD GENERATOR > ANGLE OF REFLECTION COEFFICIENT IN DEGREES 9.. RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) CAPACITIVE REACTANCE COMPONENT (-jx/zo), ORINDUCTIVESUSCEPTANCE(-jB/Yo) ATTEN. [db] S.W. LOSS COEFF RFL. LOSS [db] S.W. PEAK (CONST. P) TRANSM. COEFF, P RADIALLY SCALED PARAMETERS TOWARD LOAD > < TOWARD GENERATOR TRANSM. COEFF, E or I SWR dbs RTN. LOSS [db] RFL. COEFF, P RFL. COEFF, E or I CENTER

12 Things to notice about the Smith Chart. Notice the scales inside the wavelengths toward generator scale. One is just labeled in the opposite direction so it is wavelengths toward load, while the innermost scale is a label of degrees in the Gamma plane, i.e. angle of reflection coefficient Γ(z). Notice the scales in wavelengths only range between and., this is because impedances on lines repeat every. wavelengths. If you add the length of the line to your original number on the scale, then can t find the result, remember that the scale wrapped around at. so you need to keep rotating by the remainder.. Notice the radially scaled parameters on the side of the chart. These all relate to Γ(z) since that is our radius in the Gamma plane. The two of most interest to us are the Refl-vol scale, which is a scale for the magnitude of the reflection coefficient Γ(z), and the standing wave-vol ratio which is the VSWR More things to notice.... Thus if we know Γ(z) on a line (which you can measure just by measuring the distance from your impedance point to the center of the chart) you can just as easily read the VSWR by reading off the VSWR scale, note it is labeled from the center outward. In our example problem, we find Γ(z) = and VSWR=.6 just by using these scales. Notice that purely real Z n (z) is on the real axis of the Gamma plane, purely imaginary Z n (z) is on the Γ(z) = outermost circle 6. Plotting the conjugate of an impedance Z n (z) on the chart flips a point over the real axis since Γ = ZL Z Z L Z = Γ 7. Plotting the reciprocal of an impedance on the chart gives Γ = /Zn /Z = Zn n Z n = Γ. Thus Γ becomes Γ for the reciprocal impedance, which means a reversal in Γ, i.e. a flip through the center of the chart. 6 And more things to notice Thus a normalized admittance (= /Z n ) plotted on the chart is just rotated 8 from the same normalized impedance 9. This fact makes it such that the chart can be used exactly the same way if we work in terms of normalized admittance, Y n = YL Y = Z Y L. This is sometimes easier if we working with circuits or transmission lines connected in parallel, since admittances add in parallel. EE - Lecture Slotted line example Impedance matching networks Assigned reading: Sec.. of Ulaby 7 8

13 V G A slotted line example A slotted line is a transmission line with a slot cut into it to allow measurements of voltage and/or current as a function of z along the line. Such a measurement is made for the following circuit, and the resulting current amplitude plotted I(z) (ma)... R G = Ω. Current amplitude measured along line z= L Z = Ω Air filled z= z 9 (m). Z L Example Question : What is the frequency of operation, f? Question : Find Γ. Question : What is Z L? Question : We know that voltages and current amplitudes, and impedances on a lossless transmission line repeat themselves every half wavelength. By examining the current plot, the distance between two adjacent minima is m. Thus λ/ =, so λ = m. Since the line is air filled, β = π/λ = ω µ ɛ and f = ω/(π) = c = MHz Question : We know that the VSWR on the line is the ratio of the maximum to minimum voltage or current amplitude on the line. Here the maximum is ma while the minimum is ma, giving VSWR=. Since VSWR= Γ Γ, we can solve to find Γ = VSWR VSWR = We can also find Γ using the VSWR= distance on the Standing wave-vol scale measured on the Refl-vol scale Solution for Question We can use the Smith Chart to find Z L. First, we know Z nl must be on a circle in the Gamma plane with radius Γ(z) =.. We can find this distance either using the Refl-vol scale or the VSWR= scale on the side of the chart. We can locate one point on the circle by recognizing that there is a current minimum cm from the load, or cm/ m = λ/8 from the load. Recall that a current minimum is at the same location as a voltage maximum, which is on the positive real axis in the Gamma plane Thus, cm up the line from the load, we are on the positive real axis. To get to Z nl we rotate λ/8 toward the load (counter-clockwise) since we are going back to the load from the current minimum Finish example We can then read off the value of Z nl and un-normalize (i.e. multiply by Z ) to get the load impedance Z L. See the chart on the next page Final answer: Z L = ( j) = j Ohms Notice here that finding the complex valued Z L requires us to know the complex valued Γ. Although we can obtain Γ from knowledge of the VSWR on the line, VSWR alone does not provide the phase of Γ. The additional information regarding the location of a current (or voltage) maximum or minimum is therefore necessary to determine the phase of Γ which then provides Z L The slotted line procedure we have studied is a basic procedure for determining the impedance of an unknown load.

14 ..8.9 < WAVELENGTHS TOWARD LOAD < Impedance matching As mentioned previously, ideally we d like to terminate all our transmission lines in their characteristic impedance so we never get any reflected waves or impedance variations This would be nice because reflected waves carry power back to the source, not to the load where we usually want power to go. However, we can t always get ZL = Z so we have to think of other ways to get as much power into our loads as we can This is the goal of impedance matching networks 9 First let s consider how we can design our systems to provide maximum power transfer by looking back into circuit theory. Remember Xmission line problems are usually worked by using Xmission line theory to replace the lines with circuit impedances, and then using circuit theory from there. Taking the derivatives gives Conjugate matching (RS RL) (XL XS) (RL RS)RL ((RL RS) (XL XS) ) = which can be solved to find RL RS RL =, XL XS = Thus, we find RL = RS and XL = XS will provide maximum power dissipation. In other words, ZL = ZS This is called a conjugate match between a source and a load. We will usually try to design our impedance matching networks to create a conjugate match ~ Vg - Zg Z Zin Generator Transmission line Load M M Matching network A A ZL INDUCTIVE REACTANCE COMPONENT (jx/zo), ORCAPACITIVESUSCEPTANCE(jB/Yo) ± 8 6 ANGLE OF TRANSMISSION COEFFICIENT IN DEGREES > WAVELENGTHS TOWARD GENERATOR > ANGLE OF REFLECTION COEFFICIENT IN DEGREES 9.. RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) CAPACITIVE REACTANCE COMPONENT (-jx/zo), ORINDUCTIVESUSCEPTANCE(-jB/Yo) ATTEN. [db] S.W. LOSS COEFF RFL. LOSS [db] S.W. PEAK (CONST. P) TRANSM. COEFF, P RADIALLY SCALED PARAMETERS TOWARD LOAD > < TOWARD GENERATOR TRANSM. COEFF, E or I SWR dbs RTN. LOSS [db] RFL. COEFF, P RFL. COEFF, E or I CENTER Maximum power transfer Consider the generalized circuit below If ZS and VS are fixed, what value of ZL will result in maximum power dissipation in ZL? Find using calculus. For ZL = RL jxl, Dissipated power = I RL ] [ RL (RS RL) j(xs XL) VS ZS ZL = V S = R L Maximize this by setting = and =. RL XL Z S Z L V S

15 Impedance matching networks We know Xmission lines are very useful for turning one impedance into another, so we can design impedance matching networks using Xmission lines. Note we don t necessarily have to have Z S = Z but this will often be the case In general, we d like to be able to change an arbitrary Z L into some other impedance Z S You can see that we re going to need at least degrees of freedom in our network since we have to match both the real and imaginary parts of the two impedances We also want to use a lossless network because we don t want to dissipate power in the matching network, only in the load. While it is possible to make impedance matching networks out of lossless circuit elements (i.e. inductors and capacitors), circuit elements are often hard to find at high frequencies making Xmission line networks essential 7