II Transmitter and Receiver Design

Transcription

1 8/3/6 transmission lines 1/7 II Transmitter and Receiver Design We design radio systems using RF/microwave components. Q: Why don t we use the usual circuit components (e.g., resistors, capacitors, op-amps, transistors)?? A: We do use these! But we require new devices because: 1.. A. Microwave Components et s carefully examine each of the microwave devices that are useful for radio design: 1) ) 3) 4) 5) 6) 7) 8)

2 8/3/6 transmission lines /7 1. Transmission ines Q: So just what is a transmission line? A: Q: Oh, so it s simply a conducting wire, right? A: HO: The Telegraphers Equations HO: Time-Harmonic Solutions for inear Circuits a) Basic Transmission ine Theory Q: So, what complex functions I(z) and (z) do satisfy both telegrapher equations? A: HO: The Transmission ine Wave Equations Q: Are the solutions for I(z) and (z) completely independent, or are they related in any way?

3 8/3/6 transmission lines 3/7 A: HO: The Transmission ine Characteristic Impedance Q: So what is the significance of the constant β? What does it tell us? A: HO: The Propagation Constant Q: Is characteristic impedance the same as the concept of impedance I learned about in circuits class? A: HO: ine Impedance Q: These wave functions ( z ) and ( z ) important. How are they related? seem to be A: HO: The Reflection Coefficient

4 8/8/6 transmission lines 4/7 HO:, I, or, -, Γ?? b) The Terminated, ossless Transmission ine We now know that a lossless transmission line is completely characterized by real constants and β. ikewise, the waves propagating on a transmission line are completely characterized by complex constants and. Q: and β are determined from, C, and ω. How do we find and? A: Every transmission line has boundaries 1) ) Typically, there is a source at one end of the line, and a load at the other. et s apply the load boundary condition!

5 8/8/6 transmission lines 5/7 HO: The Terminated, ossless Transmission ine HO: Special alues of oad Impedance Q: So what is the significance of the constant β? What does it tell us? A: HO: The Propagation Constant Q: So the line impedance at the end of a line must be load z z ); what is the line impedance (i.e., ( ) impedance at the beginning of the line (i.e., z z? )? ( ) A: HO: Transmission ine Input Impedance Q: You said the purpose of the transmission line is to transfer E.M. energy from the source to the load. Exactly how much power is flowing in the transmission line, and how much is delivered to the load? A: HO: Power Flow and Return oss Note that we can specify a load with:

6 9/1/6 transmission lines 6/1 1) ) 3) A fourth alternative is SWR. HO: SWR c) A second boundary condition: Applying a generator to the transmission line Q: A passive load specifies (z) and Γ(z), but we still don t explicitly know (z), I(z), (z), or - (z). How are these functions determined? A: HO: A Transmission ine Connecting Source and oad Q: OK, we can finally ask the question that we have been concerned with since the very beginning: How much power is delivered to the load by the source? A: HO: Delivered Power

7 9/1/6 transmission lines 7/1 Q: So the power transferred depends on the source, the transmission line, and the load. What combination of these devices will result in maximum power transfer? A: HO: Special Cases of Source and Input Impedances Q: Yikes! The signal source is generally a Thevenin s equivalent of the output of some useful device, while the load impedance is generally the input impedance of some other useful device. I do not want to nor typically can I change these devices or alter their characteristics. Must I then just accept the fact that I will achieve suboptimum power transfer? A: HO: Matching Networks Q: But in microwave circuits, a source and load are connected by a transmission line. Can we implement matching networks in transmission line circuits? A: HO: Matching Networks and Transmission ines Q: Matching networks seem almost too good to be true; can we really design and construct them to provide a perfect match?

8 9/1/6 transmission lines 8/1 A: It is relatively easy to provide a near perfect match at precisely one frequency! But, since lossless matching networks are made entirely of reactive elements (not to mention the reactive components of the source and load impedance), we find that changing the signal frequency will typically mismatch our circuit! Thus a difficult challenge for any microwave component designer is to provide a wideband match to a transmission line with characteristic impedance. d) Scattering Parameters Note that a passive load is a one-port device a device that can be characterized (at one frequency) by impedance or load reflection coefficient Γ. However, many microwave devices have multiple ports! Most common are two-port devices (e.g., amplifiers and filters), devices with both a gozenta and a gozouta. gozenta gozouta

9 9/1/6 transmission lines 9/1 Note that a transmission line is also two-port device! Q: Are there any known ways to characterize a multi-port device? A: Yes! Two methods are: 1.. HO: The Impedance Matrix Q: You say that the impedance matrix characterizes a multiport device. But is this characterization helpful? Can we actually use it to solve real problems? A: Example: Using the Impedance Matrix Q: The impedance matrix relates the quantities (z) and I(z), is there an equivalent matrix that relates (z) and - (z)? A: HO: The Scattering Matrix Q: Can the scattering matrix likewise be used to solve real problems?

10 9/1/6 transmission lines 1/1 A: Of course! Example: The Scattering Matrix Example: Scattering Parameters Q: But, can the scattering matrix by itself tell us anything about the device it characterizes? A: Yes! It can tell us if the device is matched, or lossless, or reciprocal. HO: Matched, ossless, Reciprocal e) Types of Transmission ines Perhaps the most common transmission line structure is coaxial transmission line. HO:Coaxial Transmission ines Coaxial transmission lines are used with connectorized devices. HO: Coax Connectors We can also construct transmission lines on printed circuit boards. HO: Printed Circuit Board Transmission ines

11 8/3/6 The Telegrapher Equations 1/3 The Telegrapher Equations Consider a section of wire : i (z,t) i (z z,t) v (z,t) - v (z z,t) - z Q: Huh?! Current i and voltage v are a function of position z?? Shouldn t i ( z, t) i ( z z, t) and v ( zt, ) v( z zt, )? A: NO! Because a wire is never a perfect conductor. A wire will have: 1) Inductance ) Resistance 3) Capacitance 4) Conductance

12 8/3/6 The Telegrapher Equations /3 i.e., i (z,t) R z z i (z z,t) v (z,t) - G z C z v (z z,t) - z Where: R resistance/unit length inductance/unit length C capacitance/unit length G conductance/unit length resistance of wire length z is R z. Using K, we find: i ( z, t) v ( z z, t) v ( z, t) R z i( z, t) z t and from KC: v ( z, t) i( z zt, ) i( zt, ) G zv( zt, ) C z t

13 8/3/6 The Telegrapher Equations 3/3 Dividing the first equation by z, and then taking the limit as z : v( z zt, ) v( zt, ) i( zt, ) lim Ri( zt, ) z z t which, by definition of the derivative, becomes: v ( z, t) i ( z, t) Ri( zt, ) z t Similarly, the KC equation becomes: i ( z, t) v ( z, t) Gv( z, t) C z t These equations are known as the telegrapher s equations! v ( z, t) i ( z, t) Ri( zt, ) z t i ( z, t) v ( z, t) Gv( z, t) C z t

14 8/3/6 Sinusoidal wave propagation 1/4 Time-Harmonic Solutions for inear Circuits There are an unaccountably infinite number of solutions v ( z,t ) and i ( z,t ) for the telegrapher s equations! However, we can simplify the problem by assuming that the function of time is time harmonic (i.e., sinusoidal), oscillating at some radial frequencyω (e.g.,cos ωt ). Q: Why on earth would we assume a sinusoidal function of time? Why not a square wave, or triangle wave, or a sawtooth function? A: We assume sinusoids because they have a very special property! Sinusoidal time functions and only a sinusoidal time functions are the eigen functions of linear, time-invariant systems. Q:??? A: If a sinusoidal voltage source with frequency ω is used to excite a linear, time-invariant circuit (and a transmission line is both linear and time invariant!), then the voltage at each and every point with the circuit will likewise vary sinusoidally at the same frequency ω!

15 8/3/6 Sinusoidal wave propagation /4 Q: So what? Isn t that obvious? A: Not at all! If you were to excite a linear circuit with a square wave, or triangle wave, or sawtooth, you would find that generally speaking nowhere else in the circuit is the voltage a perfect square wave, triangle wave, or sawtooth. The linear circuit will effectively distort the input signal into something else! Q: Into what function will the input signal be distorted? A: It depends both on the original form of the input signal, and the parameters of the linear circuit. At different points within the circuit we will discover different functions of time unless, of course, we use a sinusoidal input. Again, for a sinusoidal excitation, we find at every point within circuit an undistorted sinusoidal function! Q: So, the sinusoidal function at every point in the circuit is exactly the same as the input sinusoid? A: Not quite exactly the same. Although at every point within the circuit the voltage will be precisely sinusoidal (with frequency ω ), the magnitude and relative phase of the sinusoid will generally be different at each and every point within the circuit. Thus, the voltage along a transmission line when excited by a sinusoidal source must have the form:

16 8/3/6 Sinusoidal wave propagation 3/4 ( ) v ( z,t) v( z) cos ωt φ ( z ) Thus, at some arbitrary location z along the transmission line, v z we must find a time-harmonic oscillation of magnitude ( ) and relative phase φ ( z ). Now, consider Euler s equation, which states: jψ e cosψ jsinψ Thus, it is apparent that: jψ Re { e } cos ψ and so we conclude that the voltage on a transmission line can be expressed as: ( ) v ( z,t) v( z) cos ωt φ ( z ) j ( ωt φ( z )) Re { v ( z ) e } jφ( z ) jωt Re v ( z ) e e { } Thus, we can specify the time-harmonic voltage at each an every location z along a transmission line with the complex function ( z ): ( ) ( z ) v ( z ) e jφ z

17 8/3/6 Sinusoidal wave propagation 4/4 where the magnitude of the complex function is the magnitude of the sinusoid: v ( z ) ( z ) and the phase of the complex function is the relative phase of the sinusoid : φ z arg z { } ( ) ( ) Q: Hey wait a minute! What happened to the time-harmonic jωt function e?? A: There really is no reason to explicitly write the complex jωt function e, since we know in fact (being the eigen function of linear systems and all) that if this is the time function at any one location (such as qt the excitation source) then this must be time function at all transmission line locations z! The only unknown is the complex function ( z ). Once we determine ( z ), we can always (if we so desire) recover the real function v ( z,t ) as: jωt { } v ( z,t) Re ( z) e Thus, if we assume a time-harmonic source, finding the v z,t reduces to solving for the transmission line solution ( ) complex function ( z ).

18 8/3/6 The Transmission ine Wave Equation 1/8 The Transmission ine Wave Equation et s assume that v ( zt, ) and i( zt, ) each have the timeharmonic form: j t { ze ω j t } and i ( z, t) Re { I ( z) e ω } v ( zt, ) Re ( ) The time-derivative of these functions are: v ( z, t) t i ( z, t) t jωt e Re ( z) Re ( ) t jωt { jω z e } jωt e Re I ( z) Re j I ( z) e t jωt { ω } The telegrapher s equations thus become: ( z) jωt Re e Re R j I z e z jωt { ( ω ) ( ) } I( z) jωt Re e Re G j C z e z jωt { ( ω ) ( ) } And then simplifying, we have the complex form of telegrapher s equations:

19 8/3/6 The Transmission ine Wave Equation /8 ( z) z ( R jω) I ( z) I( z) z ( G jωc ) ( z) Note that these complex differential equations are not a function of time t! * The functions I(z) and (z) are complex, where the magnitude and phase of the complex functions describe the j t magnitude and phase of the sinusoidal time function e ω. * Thus, I(z) and (z) describe the current and voltage along the transmission line, as a function as position z. * Remember, not just any function I(z) and (z) can exist on a transmission line, but rather only those functions that satisfy the telegraphers equations. Our task, therefore, is to solve the telegrapher equations and find all solutions I (z) and (z)!

20 8/3/6 The Transmission ine Wave Equation 3/8 Q: So, what functions I (z) and (z) do satisfy both telegrapher s equations?? A: To make this easier, we will combine the telegrapher equations to form one differential equation for (z) and another for I(z). First, take the derivative with respect to z of the first telegrapher equation: ( z) ( R jω) I ( z) z z ( z) I ( z) ( R jω) z z Note that the second telegrapher equation expresses the derivative of I(z) in terms of (z): I ( z) z ( G jωc ) ( z) Combining these two equations, we get an equation involving (z) only: ( z) ( R jω)( G jω C ) ( z) z Now, we find at high frequencies that: R jω and G jω C

21 8/3/6 The Transmission ine Wave Equation 4/8 and so we can approximate the differential equation as: ( z) z ( )( ) jω jωc ( z) ω C ( z) β ( z) where it is apparent that: β ω C In a similar manner (i.e., begin by taking the derivative of the second telegrapher equation), we can derive the differential equation: I ( z) β I ( z) z We have decoupled the telegrapher s equations, such that we now have two equations involving one function only: ( z) z I ( z) z β ( z) β I ( z) These are known as the transmission line wave equations.

22 8/3/6 The Transmission ine Wave Equation 5/8 Note only special functions satisfy these equations: if we take the double derivative of the function, the result is the original function (to within a constant)! Q: Yeah right! Every function that I know is changed after a double differentiation. What kind of magical function could possibly satisfy this differential equation? A: Such functions do exist! j z For example, the functions ( z) e β j and ( z) e β z each satisfy this transmission line wave equation (insert these into the differential equation and see for yourself!). ikewise, since the transmission line wave equation is a linear differential equation, a weighted superposition of the two solutions is also a solution (again, insert this solution to and see for yourself!): β ( z) e e j z jβz In fact, it turns out that any and all possible solutions to the differential equations can be expressed in this simple form! Therefore, the general solution to these wave equations (and thus the telegrapher equations) are:

23 8/3/6 The Transmission ine Wave Equation 6/8 β ( z) e e j z jβz β I ( z) I e I e j z jβz where,,i, and I are complex constants. It is unfathomably important that you understand what this result means! It means that the functions (z) and I(z), describing the current and voltage at all points z along a transmission line, can always be completely specified with just four complex constants (,,I,I )!! We can alternatively write these solutions as: ( z) ( z) ( z) where: I ( z) I ( z) I ( z) β ( z) e ( z) e j z jβz β I ( z) I e I ( z) I e j z jβz

24 8/3/6 The Transmission ine Wave Equation 7/8 The two terms in each solution describe two waves propagating in the transmission line, one wave ( (z) or I (z) ) propagating in one direction (z) and the other wave ( - (z) or I - (z) ) propagating in the opposite direction (-z). ( z ) e j β z ( z ) e j β z z Q: So just what are the complex values,,i,i? A: Consider the wave solutions at one specific point on the transmission line the point z. For example, we find that: ( ) z e e ( 1) j β ( z ) In other words, is simply the complex value of the wave function (z) at the point z on the transmission line! ( )

25 8/3/6 The Transmission ine Wave Equation 8/8 ikewise, we find: ( ) z ( ) I I z ( ) I I z Again, the four complex values,i,,i are all that is needed to determine the voltage and current at any and all points on the transmission line. More specifically, each of these four complex constants completely specifies one of the four transmission line wave z I z z I z. functions ( ), ( ), ( ), ( ) Q: But what determines these wave functions? How do we find the values of constants,i,,i? A: As you might expect, the voltage and current on a transmission line is determined by the devices attached to it on either end (e.g., active sources and/or passive loads)! The precise values of,i,,i are therefore determined by satisfying the boundary conditions applied at each end of the transmission line much more on this later!

26 8/3/6 The Characteristic Impedance of a Transmission ine 1/4 The Characteristic Impedance of a Transmission ine So, from the telegrapher s differential equations, we know that the complex current I(z) and voltage (z) must have the form: (z) e e j β z jβ z I(z) I e I e j β z jβ z et s insert the expression for (z) into the first telegrapher s equation, and see what happens! d ( z ) j z j z j β e j β β β e jω I(z) dz Therefore, rearranging, I (z) must be: β j βz jβz I(z) ( e e ) ω

27 8/3/6 The Characteristic Impedance of a Transmission ine /4 Q: But wait! I thought we already knew current I(z). Isn t it: I(z) I e I e?? j βz jβ z How can both expressions for I(z) be true?? A: Easy! Both expressions for current are equal to each other. j βz j βz β j βz j βz I(z) I e I e ( e e ) ω For the above equation to be true for all z, I and must be related as: β β I e e and I e e ω ω γ z γz γz γz j z Or recalling that β e ( z ) in terms of the two propagating waves: (etc.) we can express this β β I ( z ) ( z ) and I ( z ) ( z ) ω ω Now, we note that since: β ω C

28 8/3/6 The Characteristic Impedance of a Transmission ine 3/4 We find that: β ω C C ω ω Thus, we come to the startling conclusion that: ( z ) ( z ) ( z ) ( z ) and I C I C Q: What s so startling about this conclusion? A: Note that although the magnitude and phase of each propagating wave is a function of transmission line position z (e.g., ( z ) and I ( z )), the ratio of the voltage and current of each wave is independent of position a constant with respect to position z! ± ± Although and I are determined by boundary conditions (i.e., what s connected to either end of the transmission line), ± the ratio I ± is determined by the parameters of the transmission line only (R,, G, C). This ratio is an important characteristic of a transmission line, called its Characteristic Impedance.

29 8/3/6 The Characteristic Impedance of a Transmission ine 4/4 I I C We can therefore describe the current and voltage along a transmission line as: (z) e e j βz jβz I(z) e e j βz jβz or equivalently: (z) I e I e j βz jβz I(z) I e I e jβz jβz Note that instead of characterizing a transmission line with real parameters and C, we can (and typically do!) describe a lossless transmission line using real parameters and β.

30 8/5/6 ine Impedance 1/3 ine Impedance Now let s define line impedance ( z ), a complex function which is simply the ratio of the complex line voltage and complex line current: ( z ) ( z ) I ( z ) Q: Hey! I know what this is! The ratio of the voltage to current is simply the characteristic impedance, right??? A: NO! The line impedance ( z ) is (generally speaking) NOT the transmission line characteristic impedance!!! It is unfathomably important that you understand this!!!! To see why, recall that: ( z ) ( z ) ( z )

31 8/5/6 ine Impedance /3 And that: Therefore: I ( z ) ( z ) ( z ) ( z ) ( z ) ( z ) ( z ) I ( z ) ( z ) ( z ) Or, more specifically, we can write: j z jβz jβz jβ z β e e ( z ) e e Q: I m confused! Isn t: ( z ) I ( z )??? A: Yes! That is true! The ratio of the voltage to current for each of the two propagating waves is ±. However, the ratio of the sum of the two voltages to the sum of the two currents is not equal to (generally speaking)! This is actually confirmed by the equation above. Say that ( z ), so that only one wave ( ( z ) ) is propagating on the line.

32 8/5/6 ine Impedance 3/3 In this case, the ratio of the total voltage to the total current is simply the ratio of the voltage and current of the one remaining wave the characteristic impedance! ( z ) ( z ) ( z ) ( z ) ( z ) I ( z ) ( z ) I ( z ) (when ) Q: So, it appears to me that characteristic impedance is a transmission line parameter, depending only on the transmission line values and C. Whereas line impedance is ( z ) depends the magnitude and phase of the two propagating waves ( z ) and ( z )--values that depend not only on the transmission line, but also on the two things attached to either end of the transmission line! Right!? A: Exactly! Moreover, note that characteristic impedance z is a function is simply a number, whereas line impedance ( ) of position (z ) on the transmission line.

33 8/5/6 The Reflection Coefficient 1/4 The Reflection Coefficient So, we know that the transmission line voltage ( z ) and the transmission line current I ( z ) can be related by the line impedance ( z ): ( z ) ( z ) I ( z) or equivalently: I ( z ) ( z ) ( z ) Q: Piece of cake! I fully understand the concepts of voltage, current and impedance from my circuits classes. et s move on to something more important (or, at the very least, more interesting). Expressing the activity on a transmission line in terms of voltage, current and impedance is of course perfectly valid. However, let us look closer at the expression for each of these quantities:

34 8/5/6 The Reflection Coefficient /4 ( z ) ( z ) ( z ) I ( z ) ( z ) ( z ) ( z ) ( z ) ( z ) ( z ) ( z ) It is evident that we can alternatively express all activity on the transmission line in terms of the two transmission line z z. waves ( ) and ( ) j ( z ) e β z j ( z) e β z z Q: I know ( z ) and I ( z ) are related by line impedance ( z ): ( z ) But how are ( z ) ( z ) I ( z ) and ( z ) related?

35 8/5/6 The Reflection Coefficient 3/4 A: Similar to line impedance, we can define a new parameter Γ z as the ratio of the two the reflection coefficient ( ) quantities: Γ ( z ) ( z ) ( z ) Γ( z ) ( z) ( z ) More specifically, we can express Γ ( z ) as: e Γ ( z ) e j βz jβz e j βz Note then, the value of the reflection coefficient at z is: ( ) ( ) z Γ ( z ) z We define this value as Γ, where: j β ( ) e Γ Γ ( z ) Note then that we can alternatively write Γ ( z ) as: ( z ) Γ Γ e j βz

36 8/5/6 The Reflection Coefficient 4/4 So now we have two different but equivalent ways to describe transmission line activity! We can use (total) voltage and current, related by line impedance: ( z ) ( z ) ( z ) ( z ) I ( z ) I ( z ) Or, we can use the two propagating voltage waves, related by the reflection coefficient: ( z ) ( z ) ( z ) ( z ) ( z ) ( z ) Γ Γ These are equivalent relationships we can use either when describing a transmission line. Based on your circuits experience, you might well be tempted to always use the first relationship. However, we will find it useful (as well as simple) indeed to describe activity on a transmission line in terms of the second relationship in terms of the two propagating transmission line waves!

37 8/8/6 I or 1/7,I, or, -,Γ? Q: How do I choose which relationship to use when describing/analyzing transmission line activity? What if I make the wrong choice? How will I know if my analysis is correct? A: Remember, the two relationships are equivalent. There is no explicitly wrong or right choice both will provide you with precisely the same correct answer! For example, we know that the total voltage and current can be determined from knowledge wave representation: ( z ) ( z ) ( z ) ( z ) 1 Γ( z ) ( ) I ( z ) ( z ) ( z ) ( ) ( z ) 1 Γ( z )

38 8/8/6 I or /7 j z Or explicitly using the wave solutions ( z) e β and j ( z) e β z : β ( z ) e e j z jβz jβz jβz ( ) e Γ e I ( z ) e e jβz jβz j βz jβz ( Γ ) e e More importantly, we find that line impedance z z I z can be expressed as: ( ) ( ) ( ) ( z ) ( z ) ( z ) ( z ) ( z ) 1 Γ ( z ) ( z ) 1 Γ ook what happened the line impedance can be completely and unambiguously expressed in terms of reflection Γ z! coefficient ( ) More explicitly:

39 8/8/6 I or 3/7 β e e ( z ) e β e j z jβz j z jβz 1 Γ e jβz jβz 1 Γ e With a little algebra, we find likewise that the wave functions can be determined from ( z ),I ( z) and ( z ): ( z ) ( z ) ( z ) I ( z ) ( z ) ( z ) ( z ) ( z ) I ( z) ( z ) ( z ) ( z ) From this result we easily find that the reflection coefficient Γ z can likewise be written directly in terms of line ( ) impedance: ( z ) Γ ( z ) ( z )

40 8/8/6 I or 4/7 Thus, the values Γ ( z ) and ( ) z are equivalent parameters if we know one, then we can directly determine the other! Q: So, if they are equivalent, why wouldn t I always use the current, voltage, line impedance representation? After all, I am more familiar and more confident those quantities. The wave representation sort of scares me! A: Perhaps I can convince you of the value of the wave representation. Remember, the time-harmonic solution to the telegraphers equation simply boils down to two complex constants and. Once these complex values have been determined, we can describe completely the activity all points along our transmission line. For the wave representation we find: ( z ) e ( z ) e j βz j βz Γ ( z ) e j βz

41 8/8/6 I or 5/7 Note that the magnitudes of the complex functions are in fact constants (with respect to position z): ( z ) ( z ) ( z ) Γ While the relative phase of these complex functions are expressed as a simple linear relationship with respect to z : arg { ( z )} βz arg { ( z )} βz arg { Γ ( z )} βz Now, contrast this with the complex current, voltage, impedance functions:

42 8/8/6 I or 6/7 β ( z ) e e j z jβz I ( z ) e e jβz jβz With magnitude: β e e ( z ) e β e j z jβz j z jβz ( z ) β e e j z jβz?? I ( z ) jβz jβz e e?? j z jβz β e e ( z ) e β e j z jβz?? and phase: arg arg jβz jβz { ( z )} arg{ e e } jβz jβz { I ( z )} arg{ e e }???? arg jβz jβz { ( z )} arg{ e e } arg?? jβz jβz { e e }

43 8/8/6 I or 7/7 Q: It appears to me that when attempting to describe the activity along a transmission line as a function of position z it is much easier and more straightforward to use the wave representation. Is my insightful conclusion correct (nyuck, nyuck, nyuck)? A: Yes it is! However, this does not mean that we never determine (z), I(z), or (z); these quantities are still fundamental and very important particularly at each end of the transmission line!

44 1/8/5 The Terminated ossless Transmission.doc 1/8 The Terminated, ossless Transmission ine Now let s attach something to our transmission line. Consider a lossless line, length, terminated with a load. I(z) I (z) -, β z z z z - z Q: What is the current and voltage at each and every point on the transmission line (i.e., what is I ( z ) and ( z ) for all points z where z z z?)? A: To find out, we must apply boundary conditions! In other words, at the end of the transmission line ( z z) where the load is attached we have many requirements that all must be satisfied!

45 1/8/5 The Terminated ossless Transmission.doc /8 1. To begin with, the voltage and current ( I ( z z ) and ( z z )) must be consistent with a valid transmission line solution: ( ) ( ) ( ) z z z z z z e e jβz jβz ( ) ( ) z z z z I ( z z ) jβz jβz e e. ikewise, the load voltage and current must be related by Ohm s law: I 3. Most importantly, we recognize that the values I ( z z ), ( z z ) and I, are not independent, but in fact are strictly related by Kirchoff s aws! I(zz ) I, β (zz ) - - z z z z

46 1/8/5 The Terminated ossless Transmission.doc 3/8 From K and KC we find these requirements: ( ) z z ( ) I z z I These are the boundary conditions for this particular problem. Careful! Different transmission line problems lead to different boundary conditions you must access each problem individually and independently! Combining these equations and boundary conditions, we find that: I ( ) ( ) z z I z z z z z z z z z z ( ) ( ) ( ) ( ) ( ) Rearranging, we can conclude: ( ) ( ) z z z z

47 1/8/5 The Terminated ossless Transmission.doc 4/8 Q: Hey wait as second! We earlier defined ( z ) ( z ) as reflection coefficient Γ ( z ). How does this relate to the expression above? A: Recall that Γ ( z ) is a function of transmission line position z. The value ( z z ) ( z z ) ( z ) is simply the value of function Γ evaluated at z z (i.e., evaluated at the end of the line): ( ) ( ) z z Γ z z ( z z ) This value is of fundamental importance for the terminated transmission line problem, so we provide it with its own special symbol ( Γ )! Γ Γ ( z z ) Q: Wait! We earlier determined that: so it would seem that: ( z ) Γ ( z ) ( z ) ( z z ) Γ Γ ( ) ( ) z z z z Which expression is correct??

48 1/8/5 The Terminated ossless Transmission.doc 5/8 A: They both are! It is evident that the two expressions: Γ and Γ ( ) ( ) z z z z are equal if: ( ) z z And since we know that from Ohm s aw: and from Kirchoff s aws: I ( ) ( ) z z I I z z and that line impedance is: ( ) ( z ) z z I z ( ) z z we find it apparent that the line impedance at the end of the transmission line is equal to the load impedance: ( ) z z The above expression is essentially another expression of the boundary condition applied at the end of the transmission line.

49 1/8/5 The Terminated ossless Transmission.doc 6/8 Q: I m confused! Just what are were we trying to accomplish in this handout? A: We are trying to find (z) and I(z) when a lossless transmission line is terminated by a load! We can now determine the value of in terms of. Since: We find: ( ) ( ) z z e Γ j βz j βz z z e e Γ j βz And therefore we find: β ( ) ( Γ ) z e e j z jβz ( ) ( Γ ) z e e e jβz jβz jβz where: I ( z ) e ( e Γ ) e j βz jβz jβz Γ

50 1/8/5 The Terminated ossless Transmission.doc 7/8 z Now, we can further simplify our analysis by arbitrarily assigning the end point z a zero value (i.e., z ): I(z) I (z) -, β z z - z If the load is located at z (i.e., if z ), we find that: ( ) ( ) ( ) z z z ( ) jβ( ) e e jβ I ( z ) ( ) ( ) z z e jβ ( ) jβ( ) e ( ) z

51 1/8/5 The Terminated ossless Transmission.doc 8/8 ikewise, it is apparent that if z, Γ and Γ are the same: ( ) ( ) z Γ Γ ( z z) Γ z Therefore: Γ Γ Thus, we can write the line current and voltage simply as: β ( z) e Γ e β I ( z) e Γ e j z jβz j z jβz for z Q: But, how do we determine?? A: We require a second boundary condition to determine. The only boundary left is at the other end of the transmission line. Typically, a source of some sort is located there. This makes physical sense, as something must generate the incident wave!

52 8/3/6 Special alues of oad Impedance 1/11 Special alues of oad Impedance It s interesting to note that the load enforces a boundary condition that explicitly determines neither (z) nor I(z) but completely specifies line impedance (z)! j β z jβ z e Γe cos βz j sin βz ( z) jβz jβz e Γe cos βz j sin βz β ( z) e e Γ Γ j z jβz ikewise, the load boundary condition leaves ( z ) and ( z ) undetermined, but completely determines reflection Γ z! coefficient function ( ) et s look at some specific values of load impedance R jx Γ z result! and see what functions (z) and ( ) 1. In this case, the load impedance is numerically equal to the characteristic impedance of the transmission line. Assuming the line is lossless, then is real, and thus:

53 8/3/6 Special alues of oad Impedance /11 R and X It is evident that the resulting load reflection coefficient is zero: Γ This result is very interesting, as it means that there is no z! reflected wave ( ) Thus, the total voltage and current along the transmission line is simply voltage and current of the incident wave: ( z ) ( z ) e j βz I ( z ) I ( z ) e j βz Meaning that the line impedance is likewise numerically equal to the characteristic impedance of the transmission line for all line position z: j βz ( z ) e ( z ) jβz I ( z ) e And likewise, the reflection coefficient is zero at all points along the line:

54 8/3/6 Special alues of oad Impedance 3/11 ( z ) ( z) ( z ) ( z ) Γ We call this condition (when ) the matched condition, and the load a matched load.. jx For this case, the load impedance is purely reactive (e.g. a capacitor of inductor), the real (resistive) portion of the load is zero: R The resulting load reflection coefficient is: jx Γ jx Given that is real (i.e., the line is lossless), we find that this load reflection coefficient is generally some complex number. We can rewrite this value explicitly in terms of its real and imaginary part as: jx X X Γ j jx X X Yuck! This isn t much help!

55 8/3/6 Special alues of oad Impedance 4/11 et s instead write this complex valueγ in terms of its magnitude and phase. For magnitude we find a much more straightforward result! jx X jx X Γ 1 Its magnitude is one! Thus, we find that for reactive loads, the reflection coefficient can be simply expressed as: where θ Γ Γ j e θ Γ 1 X tan X We can therefore conclude that for a reactive load: θ e Γ j As a result, the total voltage and current along the transmission line is simply (assuming z ): ( ) ( j β z j θ j β z ) z e e e j( βz θγ ) j( βz θγ ) ( ) j θγ e e e ( β θ ) e z j θγ cos Γ

56 8/3/6 Special alues of oad Impedance 5/11 I z e e jβz jβz ( ) ( ) j( βz θ ) j( βz θ ) ( ) e j θ e e j θ j e sin z ( β θ ) Meaning that the line impedance can be written in terms of a trigonometric function: ( z ) ( z ) j βz I ( z ) ( ) cot θ Γ Note that this impedance is purely reactive (z) and I(z) are 9 out of phase! We also note that the line impedance at the end of the transmission line is: ( ) ( ) z j cot θ Γ With a little trigonometry, we can show (trust me!) that: and therefore: cot ( ) θ Γ X ( ) ( ) z j cot j X θ Γ

57 8/3/6 Special alues of oad Impedance 6/11 Just as we expected (and our boundary condition demanded)! Finally, the reflection coefficient function is: jθγ jβz ( z ) e e Γ ( z ) e jβz ( z ) e ( βz θ ) j Γ Meaning that for purely reactive loads: ( z ) ( βz ) j θ Γ Γ e 1 In other words, the magnitude reflection coefficient function is equal to one at each and every point on the transmission line. 3. R For this case, the load impedance is purely real (e.g. a resistor), and thus there is no reactive component: X The resulting load reflection coefficient is: R Γ R

58 8/3/6 Special alues of oad Impedance 7/11 Given that is real (i.e., the line is lossless), we find that this load reflection coefficient must be a purely real value! In other words: Re { Γ } R R { } Im Γ So a real-valued load results in a real valued load reflection coefficient G. Now let s consider the line impedance ( z ) and reflection coefficient function Γ ( z ). Q: I bet I know the answer to this one! We know that a purely imaginary (i.e., reactive) load results in a purely reactive line impedance. Thus, a purely real (i.e., resistive) load will result in a purely resistive line impedance, right?? A: NOPE! The line impedance resulting from a real load is complex it has both real and imaginary components! Thus the line impedance, as well as reflection coefficient function, cannot be further simplified for the case where R.

59 8/3/6 Special alues of oad Impedance 8/11 Q: Why is that? A: Remember, a lossless transmission line has series inductance and shunt capacitance only. In other words, a length of lossless transmission line is a purely reactive device (it absorbs no energy!). * If we attach a purely reactive load at the end of the transmission line, we still have a completely reactive system (load and transmission line). Because this system has no resistive (i.e., real) component, the general expressions for line impedance, line voltage, etc. can be significantly simplified. * However, if we attach a purely real load to our reactive transmission line, we now have a complex system, with both real and imaginary (i.e., resistive and reactive) components. This complex case is exactly what our general expressions already describes no further simplification is possible! 4. R jx Now, let s look at the general case, where the load has both a real (resitive) and imaginary (reactive) component. Q: Haven t we already determined all the general expressions (e.g., Γ, ( z ),I ( z ), ( z ), Γ ( z )) for this general case? Is there anything else left to be determined?

60 8/3/6 Special alues of oad Impedance 9/11 A: There is one last thing we need to discuss. It seems trivial, but its ramifications are very important! For you see, the general case is not, in reality, quite so general. Although the reactive component of the load can be either positive or negative ( < X < ), the resistive component of a passive load must be positive ( R > ) there s no such thing as negative resistor! This leads to one very important and useful result. Consider the load reflection coefficient: Γ ( ) ( ) ( ) ( ) R jx R jx R jx R jx Now let s look at the magnitude of this value:

61 8/3/6 Special alues of oad Impedance 1/11 ( ) ( ) Γ R jx R jx ( ) ( ) R X R X ( ) ( ) ( ) ( ) R R X R R X R X R R X R It is apparent that since both R and are positive, the numerator of the above expression must be less than (or equal to) the denominator of the above expression. In other words, the magnitude of the load reflection coefficient is always less than or equal to one! Γ 1 (for R ) Moreover, we find that this means the reflection coefficient function likewise always has a magnitude less than or equal to one, for all values of position z. Γ( z ) 1 (for all z)

62 8/3/6 Special alues of oad Impedance 11/11 Which means, of course, that the reflected wave will always have a magnitude less than that of the incident wave magnitude: ( z ) ( z ) (for all z) We will find out later that this result is consistent with conservation of energy the reflected wave from a passive load cannot be larger than the wave incident on it.

63 8/3/6 The Propagation Constant B 1/5 The Propagation Constant β Recall that the activity along a transmission line can be expressed in terms of two functions, functions that we have described as wave functions: ( z) e ( z) e j β z j β z where β is a real constant with value: β ω Q: What is this constant β? What does it physically represent? C A: Remember, a complex function can be expressed in terms of its magnitude and phase: Thus: j f f ( z ) f ( z ) e φ ( z ) ( z) φ ( z) βz φ ( z) φ ( z) βz φ

64 8/3/6 The Propagation Constant B /5 Therefore, βz φ represents the relative phase of wave z ; a function of transmission line position z. Since phase φ ( ) is expressed in radians, and z is distance (in meters), the value β must have units of: β φ z radians meter The wavelength λ of the propagating wave is defined as the distance over which the relative phase changes by π radians. So: z π π φ( z z )- φ( z) β z βλ π π or, rearranging: β π λ Thus, the value β is thus essentially a spatial frequency, in the same way that ω is a temporal frequency: ω π T where T is the time required for the phase of the oscillating signal to change by a value of π radians, i.e.: ωt π

65 8/3/6 The Propagation Constant B 3/5 Note that this time is the period of a sinewave, and related to its frequency in Hertz (cycles/second) as: π 1 T ω f Q: So, just how fast does this wave propagate down a transmission line? We describe wave velocity in terms of its phase velocity in other words, how fast does a specific value of absolute phase φ seem to propagate down the transmission line. Since velocity is change in distance with respect to time, we need to first express our propagating wave in its real form: jωt { } ( ω β φ ) v ( z,t) Re ( z) e cos t z Thus, the absolute phase is a function of both time and frequency: φ z,t ωt βz φ ( ) Now let s set this phase to some arbitrary value of φ c radians. ωt βz φ φ c For every time t, there is some location z on a transmission line that has this phase value φ c. That location is evidently:

66 8/3/6 The Propagation Constant B 4/5 ωt φ φ z β c Note as time increases, so too does the location z on the line where φ ( z,t) φc. The velocity v p at which this phase point moves down the line can be determined as: ωt φ φc d dz β ω v p dt dt β This wave velocity is the velocity of the propagating wave! Note that the value: v p ω β ω f λ β π π and thus we can conclude that: v p f λ as well as: β ω v p Q: But these results were derived for the ( z ) about the other wave ( ( z ))? wave; what

67 8/3/6 The Propagation Constant B 5/5 A: The results are essentially the same, as each wave depends on the same value β. The only subtle difference comes when we evaluate the phase z, we find: velocity. For the wave ( ) φ ( z,t) ωt βz φ Note the plus sign associated with βz! We thus find that some arbitrary phase value will be located at location: φ φc ωt z β Note now that an increasing time will result in a decreasing value of position z. In other words this wave is propagating in the direction of decreasing position z in the opposite direction of the ( z ) wave! This is further verified by the derivative: v p φ φc ωt d dz β ω dt dt β Where the minus sign merely means that the wave propagates in the z direction. Otherwise, the wavelength and velocity of the two waves are precisely the same!

68 8/3/6 Transmission ine Input Impedance 1/3 Transmission ine Input Impedance Consider a lossless line, length, terminated with a load. I(z) I (z) -, β - z z et s determine the input impedance of this line! Q: Just what do you mean by input impedance? A: The input impedance is simply the line impedance seen at the beginning (z ) of the transmission line, i.e.: ( z ) in ( z ) I ( z ) Note in equal to neither the load impedance nor the characteristic impedance! and in in

69 8/3/6 Transmission ine Input Impedance /3 To determine exactly what in is, we first must determine the voltage and current at the beginning of the transmission line (z ). jβ jβ (z ) e Γ e Therefore: jβ jβ I(z ) e e Γ ( ) ( ) z e Γ e jβ jβ in jβ jβ I z e Γe We can explicitly write in in terms of load using the previously determined relationship: Γ Combining these two expressions, we get: j β j β ( ) e ( ) e β β ( ) ( ) jβ jβ jβ jβ ( e e ) ( e e ) jβ jβ jβ jβ ( ) ( ) e e in j j e e e e Now, recall Euler s equations: e e j β j β cos β j sin β cos β j sin β

70 8/3/6 Transmission ine Input Impedance 3/3 Using Euler s relationships, we can likewise write the input impedance without the complex exponentials: cos β j j in cos β sin β j tan β tan j β sin β Note that depending on the values of β, and, the input impedance can be radically different from the load impedance! Q: So is there a similar concept of input reflection coefficient? A: There sure is! As you might expect, it is simply the value of reflection coefficient function Γ ( z ) evaluated at the beginning of the transmission line (i.e., at z ): in ( z ) e Γ Γ Γ j β Note that the input impedance and input reflection coefficient are related in the same way as and Γ are at every other point on the transmission line: Γ in in in

71 8/3/6 Power Flow and Return oss 1/5 Power Flow and Return oss We have discovered that two waves propagate along a z and z transmission line, one in each direction ( ( ) ( ) ). j βz j βz I( z) e e Γ I j βz jβz ( z) e Γ e - - z z The result is that electromagnetic energy flows along the transmission line at a given rate (i.e., power). Q: How much power flows along a transmission line, and where does that power go? A: We can answer that question by determining the power absorbed by the load!

72 8/3/6 Power Flow and Return oss /5 The time average power absorbed by an impedance is: P 1 Re{ I } 1 Re{ ( z ) I( z ) } 1 Re Γ Γ abs { ( ) ( ) } jβ jβ jβ jβ e e e e { 1 ( ) } Re Γ Γ Γ ( 1 ) Γ The significance of this result can be seen by rewriting the expression as: P abs ( 1 ) Γ Γ The two terms in above expression have a very definite physical meaning. The first term is the time-averaged power of the wave propagating along the transmission line toward the load.

73 8/3/6 Power Flow and Return oss 3/5 We say that this wave is incident on the load: P inc P ikewise, the second term of the P abs equation describes the power of the wave moving in the other direction (away from the load). We refer to this as the wave reflected from the load: Γ Pref P Γ P inc Thus, the power absorbed by the load is simply: Pabs Pinc Pref or, rearranging, we find: Pinc Pabs Pref This equation is simply an expression of the conservation of energy! It says that power flowing toward the load (P inc ) is either absorbed by the load (P abs ) or reflected back from the load (P ref ).

74 8/3/6 Power Flow and Return oss 4/5 P abs P inc P ref Note that if Γ 1, then P inc P ref, and therefore no power is absorbed by the load. This of course makes sense! The magnitude of the reflection coefficient ( Γ ) is equal to one only when the load impedance is purely reactive (i.e., purely imaginary). Of course, a purely reactive element (e.g., capacitor or inductor) cannot absorb any power all the power must be reflected! Return oss The ratio of the reflected power to the incident power is known as return loss. Typically, return loss is expressed in db: R.. 1 log P 1 log ref 1 1 Pinc Γ

75 8/3/6 Power Flow and Return oss 5/5 For example, if the return loss is 1dB, then 1% of the incident power is reflected at the load, with the remaining 9% being absorbed by the load we lose 1% of the incident power ikewise, if the return loss is 3dB, then.1 % of the incident power is reflected at the load, with the remaining 99.9% being absorbed by the load we lose.1% of the incident power. Thus, a larger numeric value for return loss actually indicates less lost power! An ideal return loss would be db, whereas a return loss of db indicates that Γ 1--the load is reactive!

76 9/1/6 SWR 1/3 SWR Consider again the voltage along a terminated transmission line, as a function of position z : β ( z) e Γ e j z jβz Recall this is a complex function, the magnitude of which expresses the magnitude of the sinusoidal signal at position z, while the phase of the complex value represents the relative phase of the sinusoidal signal. et s look at the magnitude only: β ( z) e Γ e j z jβz Γ j βz jβz e 1 e Γ jβ z 1 e ICBST the largest value of (z) occurs at the location z where: j βz Γ e Γ j while the smallest value of (z) occurs at the location z where: j βz Γ e Γ j

77 9/1/6 SWR /3 As a result we can conclude that: ( ) ( 1 Γ ) z max ( ) ( 1 Γ ) z min The ratio of ( z ) to ( z ) max Standing Wave Ratio (SWR): min is known as the oltage ( z ) ( z ) 1 Γ max SWR 1 SWR 1 Γ min Note if Γ (i.e., ), then SWR 1. We find for this case: ( z) ( z) max min In other words, the voltage magnitude is a constant with respect to position z. Conversely, if Γ 1 (i.e., jx ), then SWR. We find for this case: ( z) and ( z) min max In other words, the voltage magnitude varies greatly with respect to position z.

78 9/1/6 SWR 3/3 As with return loss, SWR is dependent on the magnitude of Γ (i.e, Γ ) only! (z) z λ (z) max (z) min z

79 9/1/6 A Transmission ine Connecting Source 1/5 A Transmission ine Connecting Source & oad We can think of a transmission line as a conduit that allows power to flow from an output of one device/network to an input of another. To simplify our analysis, we can model the input of the device receiving the power with it input impedance (e.g., ), while we can model the device output delivering the power with its Thevenin s or Norton s equivalent circuit. g I i I i - g i I g g i - - g i g I i i Ig Ii g

80 9/1/6 A Transmission ine Connecting Source /5 Typically, the power source is modeled with its Thevenin s equivalent; however, we will find that the Norton s equivalent circuit is useful if we express the remainder of the circuit in terms of its admittance values (e.g., Y, Y, Y( z )). I i g - g i - z z Recall from the telegrapher s equations that the current and voltage along the transmission line are: (z) e e j βz jβz I(z) e e j β z jβz At z, we enforced the boundary condition resulting from Ohm s aw: ( ) ( z ) I I( z )

81 9/1/6 A Transmission ine Connecting Source 3/5 Which resulted in: Γ So therefore: (z) e Γ e jβz jβz j βz jβz I(z) e e Γ We are left with the question: just what is the value of complex constant?!? This constant depends on the signal source! To determine its exact value, we must now apply boundary conditions at z. We know that at the beginning of the transmission line: (z ) e Γ e jβ jβ jβ jβ I(z ) e e Γ ikewise, we know that the source must satisfy: g i g Ii

82 9/1/6 A Transmission ine Connecting Source 4/5 To relate these three expressions, we need to apply boundary conditions at z : I i I ( z ) g - g i ( z ) z z From K we find: ( z ) i And from KC: Ii I ( z ) Combining these equations, we find: I g i g i e e e e jβ jβ jβ jβ g Γ g Γ One equation one unknown ( )!! Solving, we find the value of :

83 9/1/6 A Transmission ine Connecting Source 5/5 e g j β 1 1 ( Γ ) ( Γ ) in g in where: in ( z ) Γ Γ Γ e j β There is one very important point that must be made about the result: j β g e ( 1Γ ) ( 1Γ ) And that is the wave ( z ) dependent on the value of load!!!!! in g in incident on the load is actually Remember: in ( z ) Γ Γ Γ e j β We tend to think of the incident wave ( z ) the source, and it is certainly true that ( z ) source after all, ( z ) if being caused by depends on the. However, we find from the equation above that it likewise depends on the value of the load! Thus we cannot in general consider the incident wave to be the cause and the reflected wave the effect. Instead, each wave must obtain the proper amplitude (e.g.,, ) so that the boundary conditions are satisfied at both the beginning and end of the transmission line. g

84 9/1/6 Delivered Power 1/3 Delivered Power Q: If the purpose of a transmission line is to transfer power from a source to a load, then exactly how much power is delivered to for the circuit shown below?? I ( z ) g - g in ( z ) z z A: We of course could determine and, and then determine the power absorbed by the load (P abs ) as: P 1 Re abs z I z { ( ) ( )} However, if the transmission line is lossless, then we know that the power delivered to the load must be equal to the power delivered to the input (P in ) of the transmission line: 1 Re Pabs Pin { ( z ) I ( z )}

85 9/1/6 Delivered Power /3 However, we can determine this power without having to solve for and (i.e., ( z ) and I ( z )). We can simply use our knowledge of circuit theory! We can transform load to the beginning of the transmission line, so that we can replace the transmission line with its input impedance in : I ( z ) g - g ( z ) ( z ) in Note by voltage division we can determine: in ( z ) g g in And from Ohm s aw we conclude: I ( z ) g g in And thus, the power P in delivered to in (and thus the power P abs delivered to the load ) is:

86 9/1/6 Delivered Power 3/3 1 Re 1 g Re in g g in ( g in) Pabs Pin { ( z ) I ( z ) } 1 g g in Re { } in 1 g g in in Re { Y } in Note that we could also determine P abs from our earlier expression: P abs ( 1 ) Γ But we would of course have to first determine (! ): e g j β 1 1 ( Γ ) ( Γ ) in g in

87 9/6/6 Special Cases of Source and oad Impedance 1/8 Special Cases of Source and oad Impedance et s look at specific cases of g and, and determine how they affect and P abs. g For this case, we find that simplifies greatly: e g e g e g j β j β j β 1 g e ( 1Γ ) ( 1Γ ) ( 1Γ ) ( 1Γ ) in 1 1Γ 1Γ j β in in g in in in ook at what this says! It says that the incident wave in this case is independent of the load attached at the other end!

88 9/6/6 Special Cases of Source and oad Impedance /8 Thus, for the one case g, we in fact can consider ( z ) as being the source wave, and then the reflected wave ( z ) as being the result of this stimulus. Remember, the complex value is the value of the incident wave evaluated at the end of the transmission line ( ( z ) ). We can likewise determine the value of the incident wave at the beginning of the transmission line (i.e., ( z )). For this case, where g, we find that this value can be very simply stated (!): ( z ) e jβ ( z ) 1 jβ g e e g ikewise, we find that the delivered power for this case can be simply stated as: P abs ( 1 ) Γ ( 1 ) g Γ 8 jβ

89 9/6/6 Special Cases of Source and oad Impedance 3/8 in g For this case, we find takes on whatever value required to make in g. This is a very important case! First, using the fact that: in g Γ in in g We can show that (trust me!): j g β g e 4Re { g } Not a particularly interesting result, but let s look at the absorbed power.

90 9/6/6 Special Cases of Source and oad Impedance 4/8 P abs 1 g g in Re { } in 1 g g g Re { g } 1 Re g { g } 1 1 g Re { g } { g } P 4Re Although this result does not look particularly interesting either, we find the result is very important! It can be shown that for a given g and g the value of input impedance in that will absorb the largest possible amount of power is the value in g. avl This case is known as the conjugate match, and is essentially the goal of every transmission line problem to deliver the largest possible power to in, and thus to as well! This maximum delivered power is known as the available power (P avl )of the source. There are two very important things to understand about this result!

91 9/6/6 Special Cases of Source and oad Impedance 5/8 ery Important Thing #1 Consider again the terminated transmission line: g - g in z z Recall that if, the reflected wave will be zero, and the absorbed power will be: P abs g g But note if, the input impedance in but then * in g (generally)! In other words, does not (generally) result in a conjugate match, and thus setting does not result in maximum power absorption! Q: Huh!? This makes no sense! A load value of will minimize the reflected wave ( P ) all of the incident power will be absorbed. Any other value of will result in some of the incident wave being reflected how in the world could this increase absorbed power?

92 9/6/6 Special Cases of Source and oad Impedance 6/8 After all, just look at the expression for absorbed power: P abs ( 1 ) Γ Clearly, this value is maximized when Γ (i.e., when )!!! A: You are forgetting one very important fact! Although it is true that the load impedance affects the reflected wave powerp, the value of as we have shown in this handout likewise helps determine the value of the incident wave (i.e., the value of P ) as well. Thus, the value of that minimizes P will not generally maximize P, nor will the value of that maximizes P likewise minimize P. Instead, the value of that maximizes the absorbed power is, by definition, the value that maximizes the difference P P. We find that this value of close as possible to the ideal case of in g. is the value that makes in as Q: Yes, but what about the case where g? For that case, we determined that the incident wave is independent of. Thus, it would seem that at least for that case, the

93 9/6/6 Special Cases of Source and oad Impedance 7/8 delivered power would be maximized when the reflected power was minimized (i.e., ). A: True! But think about what the input impedance would be in that case. Oh by the way, that provides a in conjugate match ( in g )! Thus, in some ways, the case g (i.e., both source and load impedances are numerically equal to ) is ideal. A conjugate match occurs, the incident wave is independent of, there is no reflected wave, and all the math simplifies quite nicely: 1 j g e β P abs g 8 ery Important Thing # Note the conjugate match criteria says: Given g and g, maximum power transfer occurs when. in g It does NOT say: Given g and in, maximum power transfer occurs when. g in

94 9/6/6 Special Cases of Source and oad Impedance 8/8 This last statement is in fact false! A factual statement is this: Given g and in, maximum power transfer occurs when. g A fact that is evident when observing the expression for available power: P available 1 1 g g 4Re 8R { g } In other words, given a choice, use a source with the smallest possible output resistance (given that g remains constant). This will maximize the available power from your source! g

95 9/6/6 Matching Networks 1/1 Matching Networks Consider again the problem where a passive load is attached to an active source: g g R jx The load will absorb power power that is delivered to it by the source. 1 P Re{ I} 1 g Re g g g 1 Re { } g 1 g g g R Recall that the power delivered to the load will be maximized (for a given g and ) if the load impedance is equal to the g complex conjugate of the source impedance ( ). g

96 9/6/6 Matching Networks /1 We call this maximum power the available power P avl of the source it is, after all, the largest amount of power that the source can ever deliver! P max P avl g 8 R g * Note the available power of the source is dependent on source parameters only (i.e., g and R ). This makes sense! g Do you see why? * Thus, we can say that to take full advantage of all the available power of the source, we must to make the load impedance the complex conjugate of the source impedance. * Otherwise, the power delivered to the load will be less than power made available by the source! In other words : P P avl Q: But, you said that the load impedance typically models the input impedance of some useful device. We don t typically get to select or adjust this impedance it is what it is. Must we then simply accept the fact that the delivered power will be less than the available power?

97 9/6/6 Matching Networks 3/1 A: NO! We can in fact modify our circuit such that all available source power is delivered to the load without in any way altering the impedance value of that load! To accomplish this, we must insert a matching network between the source and the load: R jx g g g I in g I in in Matching Network R jx The sole purpose of this matching network is to transform the load impedance into an input impedance that is conjugate matched to the source! I.E.: in * in g I in in * g Matching Network R jx

98 9/6/6 Matching Networks 4/1 Because of this, all available source power is delivered to the input of the matching network (i.e., delivered to ): in P in P avl Q: Wait just one second! The matching network ensures that all available power is delivered to the input of the matching network, but that does not mean (necessarily) that this power will be delivered to the load. The power delivered to the load could still be much less than the available power! A: True! To ensure that the available power delivered to the input of the matching network is entirely delivered to the load, we must construct our matching network such that it cannot absorb any power the matching network must be lossless! We must construct our matching network entirely with reactive elements! Examples of reactive elements include inductors, capacitors, transformers, as well as lengths of lossless transmission lines.

99 9/6/6 Matching Networks 5/1 Thus, constructing a proper lossless matching network will lead to the happy condition where: P Pin Pavl * Note that the design and construction of this lossless network will depend on both the value of source impedance and load impedance. g * However, the matching network does not physically alter the values of either of these two quantities the source and load are left physically unchanged! Now, let s consider the matching network from a different perspective. Instead of defining it in terms of its input impedance when attached the load, let s describe it in terms of its output impedance when attached to the source: R jx g g g I out g Matching Network out

100 9/6/6 Matching Networks 6/1 This new source (i.e., the original source with the matching network attached) can be expressed in terms of its Thevenin s equivalent circuit: R jx out out out s This equivalent circuit can be determined by first evaluating oc (or measuring) the open-circuit output voltage : out R jx g g g I out g Matching Network oc out And likewise evaluating (or measuring) the short-circuit output current I sc : out R jx g g g sc I out g Matching Network out

101 9/6/6 Matching Networks 7/1 oc sc From these two values ( and I ) we can determine the out Thevenin s equivalent source: out oc oc out s out out oc Iout Note that in general that s g and the matching out g network transforms both the values of both the impedance and the voltage source. Q: Arrrgg! Doesn t that mean that the available power of this transformed source will be different from the original? A: Nope. If the matching network is lossless, the available power of this equivalent source is identical to the available power of the original source the lossless matching network does not alter the available power! P avl g 8R 8R g s out

102 9/6/6 Matching Networks 8/1 Now, for a properly designed, lossless matching network, it turns out that (as you might have expected!) the output impedance is equal to the complex conjugate of the load out impedance. I.E.: out The source and load are again matched! Thus, we can look at the matching network in two equivalent ways: R jx g g g g Matching Network R jx 1. As a network attached to a load, one that transforms its impedance to in a value matched to the source impedance g : g g in g

103 9/6/6 Matching Networks 9/1. Or, as network attached to a source, one that transforms its impedance to out a value matched to the load impedance : g out g s Either way, the source and load impedance are conjugate matched all the available power is delivered to the load! Recall that a primary purpose of a transmission line is to allow the transfer of power from a source to a load. g g

104 9/6/6 Matching Networks 1/1 Recall that the efficacy of this power transfer depends on: 1. the source impedance. g. load impedance. 3. the transmission line characteristic impedance. 4. the transmission line length.

105 9/6/6 Matching Networks and Transmission ines 1/7 Matching Networks and Transmission ines Recall that a primary purpose of a transmission line is to allow the transfer of power from a source to a load. g g in Q: So, say we directly connect an arbitrary source to an arbitrary load via a length of transmission line. Will the power delivered to the load be equal to the available power of the source? A: Not likely! Remember we determined earlier that the efficacy of power transfer depends on: 1. the source impedance. g. load impedance. 3. the transmission line characteristic impedance.

106 9/6/6 Matching Networks and Transmission ines /7 4. the transmission line length. Recall that maximum power transfer occurred only when these four parameters resulted in the input impedance of the transmission line being equal to the complex conjugate of the source impedance (i.e., in g ). It is of course unlikely that the very specific conditions of a conjugate match will occur if we simply connect a length of transmission line between an arbitrary source and load, and thus the power delivered to the load will generally be less than the available power of the source. Q: Is there any way to use a matching network to fix this problem? Can the power delivered to the load be increased to equal the available power of the source if there is a transmission line connecting them? A: There sure is! We can likewise construct a matching network for the case where the source and load are connected by a transmission line. For example, we can construct a network to transform the input impedance of the transmission line into the complex conjugate of the source impedance.

107 9/6/6 Matching Networks and Transmission ines 3/7 g g g Matching Network Q: But, do we have to place the matching network between the source and the transmission line? A: Nope! We could also place a (different) matching network between the transmission line and the load. g g g Matching Network In either case, we find that at any and all points along this matched circuit, the output impedance of the equivalent source (i.e., looking left) will be equal to the complex conjugate of the input impedance (i.e., looking right).

108 9/6/6 Matching Networks and Transmission ines 4/7 g out in g s in out Q: So which method should we chose? Do engineers typically place the matching network between the source and the transmission line, or place it between the transmission line and the load? A: Actually, the typical solution is to do both! We find that often there is a matching network between the a source and the transmission line, and between the line and the load. g g Matching Network Matching Network

109 9/6/6 Matching Networks and Transmission ines 5/7 The first network matches the source to the transmission line in other words, it transforms the output impedance of the equivalent source to a value numerically equal to characteristic impedance : g out g s The second network matches the load to the transmission line in other words it transforms the load impedance to a value numerically equal to characteristic impedance : in Q: Yikes! Why would we want to build two separate matching networks, instead of just one?

110 9/6/6 Matching Networks and Transmission ines 6/7 A: By using two separate matching networks, we can decouple the design problem. Recall again that the design of a single matching network solution would depend on four separate parameters: 1. the source impedance. g. load impedance. 3. the transmission line characteristic impedance. 4. the transmission line length. Alternatively, the design of the network matching the source and transmission line depends on only: 1. the source impedance. g. the transmission line characteristic impedance. Whereas, the design of the network matching the load and transmission line depends on only: 1. the source impedance.. the transmission line characteristic impedance. Note that neither design depends on the transmission line length!

111 9/6/6 Matching Networks and Transmission ines 7/7 Q: How is that possible? A: Remember the case where. For that special g case, we found that a conjugate match was the result regardless of the transmission line length. Thus, by matching the source to line impedance and likewise matching the load to the line impedance, a conjugate match is assured but the length of the transmission line does not matter! In fact, the typically problem for microwave engineers is to match a load (e.g., device input impedance) to a standard transmission line impedance (typically 5Ω); or to independently match a source (e.g., device output impedance) to a standard line impedance. A conjugate match is thus obtained by connecting the two with a transmission line of any length! s

112 9/6/6 The Impedance Matrix 1/7 The Impedance Matrix Consider the 4-port microwave device shown below: I ( z ) ( z ) I1 ( z 1) z ( ) 1 1 port 1 z z port 4-port microwave device 1 1P port 4 z z z P z z 4 4P port 3 z 3 3P I3 ( z 3) z ( ) 3 3 I4 ( z 4) ( z ) 4 4 Note in this example, there are four identical transmission lines connected to the same box. Inside this box there may be a very simple linear device/circuit, or it might contain a very large and complex linear microwave system.

113 9/6/6 The Impedance Matrix /7 Either way, the box can be fully characterized by its impedance matrix! First, note that each transmission line has a specific location that effectively defines the input to the device (i.e., z 1P, z P, z 3P, z 4P ). These often arbitrary positions are known as the port locations, or port planes of the device. Thus, the voltage and current at port n is: ( ) I ( z z ) z z n n np n n np We can simplify this cumbersome notation by simply defining port n current and voltage as I n and n : ( ) I I ( z z ) z z n n n np n n n np For example, the current at port 3 would be I I ( z z ) P Now, say there exists a non-zero current at port 1 (i.e., I1 ), while the current at all other ports are known to be zero (i.e., I I I ). 3 4 Say we measure/determine the current at port 1 (i.e., determine I 1 ), and we then measure/determine the voltage at the port plane (i.e., determine ).

114 9/6/6 The Impedance Matrix 3/7 The complex ratio between and I 1 is know as the transimpedance parameter 1 : 1 I 1 ikewise, the trans-impedance parameters 31 and 41 are: and I I1 1 We of course could also define, say, trans-impedance parameter 34 as the ratio between the complex values I 4 (the current into port 4) and 3 (the voltage at port 3), given that the current at all other ports (1,, and 3) are zero. Thus, more generally, the ratio of the current into port n and the voltage at port m is: m mn (given that Ik for all k n) I n

115 9/6/6 The Impedance Matrix 4/7 Q: But how do we ensure that all but one port current is zero? A: Place an open circuit at those ports! I I port microwave device I 3 3 I 4 4 Placing an open at a port (and it must be at the port!) enforces the condition that I.

116 9/6/6 The Impedance Matrix 5/7 Now, we can thus equivalently state the definition of transimpedance as: m mn (given that all ports k n are open ) I n Q: As impossible as it sounds, this handout is even more boring and pointless than any of your previous efforts. Why are we studying this? After all, what is the likelihood that a device will have an open circuit on all but one of its ports?! A: OK, say that none of our ports are open-circuited, such that we have currents simultaneously on each of the four ports of our device. Since the device is linear, the voltage at any one port due to all the port currents is simply the coherent sum of the voltage at that port due to each of the currents! For example, the voltage at port 3 can be determined by: I I I I

117 9/6/6 The Impedance Matrix 6/7 More generally, the voltage at port m of an N-port device is: N I m mn n n 1 This expression can be written in matrix form as: I Where I is the vector: [ ] I I,I,I,,I N 1 3 T and is the vector: T,,, 1 3, N And the matrix is called the impedance matrix: 11 1 n m1 mn The impedance matrix is a N by N matrix that completely characterizes a linear, N -port device. Effectively, the impedance matrix describes a multi-port device the way that describes a single-port device (e.g., a load)!

118 9/6/6 The Impedance Matrix 7/7 But beware! The values of the impedance matrix for a particular device or network, just like, are frequency dependent! Thus, it may be more instructive to explicitly write: ( ω) 11 ( ω) 1 n ( ω) m1 ( ω) mn ( ω)

119 9/6/6 Example Using the Impedance Matrix 1/3 Example: Using the Impedance Matrix Consider the following circuit: 1 I 1 I 16-1 I Where the 3-port device is characterized by the impedance matrix: et s now determine all port voltages 1,, 3 and all currents I,I,I 1 3.

120 9/6/6 Example Using the Impedance Matrix /3 Q: How can we do that we don t know what the device is made of! What s inside that box? A: We don t need to know what s inside that box! We know its impedance matrix, and that completely characterizes the device (or, at least, characterizes it at one frequency). Thus, we have enough information to solve this problem. From the impedance matrix we know: I I I I I 4I 1 3 I 4I I Q: Wait! There are only 3 equations here, yet there are 6 unknowns!? A: True! The impedance matrix describes the device in the box, but it does not describe the devices attached to it. We require more equations to describe them.

121 9/6/6 Example Using the Impedance Matrix 3/3 1. The source at port 1 is described by the equation: ( ) I 1 1. The short circuit on port means that: 3. While the load on port 3 leads to: ( 1) I (note the minus sign!) 3 3 Now we have 6 equations and 6 unknowns! Combining equations, we find: 16 I I I I I I I 1 3 I I 4I 1 3 I I 4I 1 3 I I 4I I I 4I I 1 3 Solving, we find (I ll let you do the algebraic details!): I 1 7. I 3. I

122 9/6/6 The Scattering Matrix 1/1 The Scattering Matrix At low frequencies, we can completely characterize a linear device or network using an impedance matrix, which relates the currents and voltages at each device terminal to the currents and voltages at all other terminals. But, at microwave frequencies, it is difficult to measure total currents and voltages! * Instead, we can measure the magnitude and phase of each of the two transmission line waves ( z) and ( z). * In other words, we can determine the relationship between the incident and reflected wave at each device terminal to the incident and reflected waves at all other terminals. These relationships are completely represented by the scattering matrix. It completely describes the behavior of a linear, multi-port device at a given frequency ω.

123 9/6/6 The Scattering Matrix /1 Consider the 4-port microwave device shown below: ( z ) ( z ) ( z ) 1 1 ( z ) 1 1 port 1 port port 3 4-port microwave device z z 1 1P port 4 z z P z z z 4 4P z 3 3P ( z ) 3 3 ( z ) 3 3 ( z ) 4 4 ( z ) 4 4 Note in this example, there are four identical transmission lines connected to the same box. Inside this box there may be a very simple linear device/circuit, or it might contain a very large and complex linear microwave system. Either way, the box can be fully characterized by its scattering parameters!

124 9/6/6 The Scattering Matrix 3/1 First, note that each transmission line has a specific location that effectively defines the input to the device (i.e., z 1P, z P, z 3P, z 4P ). These often arbitrary positions are known as the port locations, or port planes of the device. Say there exists an incident wave on port 1 (i.e., ( z ) 1 1 ), while the incident waves on all other ports are known to be zero z z z ). (i.e., ( ) ( ) ( ) Say we measure/determine the voltage of the wave flowing into z z ). port 1, at the port 1 plane (i.e., determine ( ) 1 1 1P Say we then measure/determine the voltage of the wave flowing z z ). out of port, at the port plane (i.e., determine ( ) P The complex ratio between 1 ( z 1 z 1 P) and ( z z P) is know as the scattering parameter S 1 : S 1 j βzp ( z z) e jβz1p ( z z ) e e ( ) jβ z z P 1P ikewise, the scattering parameters S 31 and S 41 are: S ( z z ) ( z z ) ( ) ( ) 3 3 3P 4 4 4P 31 and S41 1 z1 z1p 1 z1 z1p

125 9/6/6 The Scattering Matrix 4/1 We of course could also define, say, scattering parameter S 34 as the ratio between the complex values 4 ( z 4 z 4 P ) (the wave into port 4) and 3 ( z3 z3p ) (the wave out of port 3), given that the input to all other ports (1,, and 3) are zero. Thus, more generally, the ratio of the wave incident on port n to the wave emerging from port m is: m ( zm zmp) Smn (given that k ( zk ) for all k n) ( z z ) n n np Note that frequently the port positions are assigned a zero value (e.g., z1, z ). This of course simplifies the P scattering parameter calculation: P S mn ( z ) e z j β m m m m j β n ( n ) n e n We will generally assume that the port locations are defined as z np, and thus use the above notation. But remember where this expression came from!

126 9/6/6 The Scattering Matrix 5/1 Q: But how do we ensure that only one incident wave is non-zero? A: Terminate all other ports with a matched load! Γ ( z ) ( z ) ( z ) 1 1 ( z ) 1 1 ( z ) port microwave device ( z ) 3 3 Γ 3 ( z ) 3 3 ( z ) 4 4 Γ 4

127 9/6/6 The Scattering Matrix 6/1 Note that if the ports are terminated in a matched load (i.e., ), then Γ and therefore: n n n ( z ) In other words, terminating a port ensures that there will be no signal incident on that port! Q: Just between you and me, I think you ve messed this up! In all previous handouts you said that if Γ, the wave in the minus direction would be zero: ( z ) if Γ but just now you said that the wave in the positive direction would be zero: ( z ) if Γ Of course, there is no way that both statements can be correct! A: Actually, both statements are correct! You must be careful to understand the physical definitions of the plus and minus directions in other words, the propagation directions of waves ( z ) and ( z ) n n n n!

128 9/6/6 The Scattering Matrix 7/1 For example, we originally analyzed this case: ( z ) Γ ( z ) if Γ ( z ) In this original case, the wave incident on the load is ( z ) (plus direction), while the reflected wave is ( z ) (minus direction). Contrast this with the case we are now considering: port n n ( z ) n N-port Microwave Network Γ n n ( z ) n For this current case, the situation is reversed. The wave incident on the load is now denoted as ( z ) (coming out of port n), while the wave reflected off the load is now denoted as ( z ) (going into port n ). n n As a result, ( z ) when Γ! n n n n n

129 9/6/6 The Scattering Matrix 8/1 Perhaps we could more generally state that: reflected incident ( z z ) Γ ( z z ) For each case, you must be able to correctly identify the mathematical statement describing the wave incident on, and reflected from, some passive load. ike most equations in engineering, the variable names can change, but the physics described by the mathematics will not! Now, back to our discussion of S-parameters. We found that if z np for all ports n, the scattering parameters could be directly written in terms of wave amplitudes n and m. S (given that z for all k n) m mn k k n ( ) Which we can now equivalently state as: S mn (given that all ports, except port n, are matched ) m n

130 9/6/6 The Scattering Matrix 9/1 Q: As impossible as it sounds, this handout is even more boring and pointless than any of your previous efforts. Why are we studying this? After all, what is the likelihood that a microwave network will have only one incident wave that all of the ports will be matched?! A: OK, say that our ports are not matched, such that we have waves simultaneously incident on each of the four ports of our device. Since the device is linear, the output at any port due to all the incident waves is simply the coherent sum of the output at that port due to each input wave! For example, the output wave at port 3 can be determined by (assuming z np ): S S S S More generally, the output wave voltage at port m of an N-port device is: N m mn n np n 1 ( ) S z

131 9/6/6 The Scattering Matrix 1/1 This expression can be written in matrix form as: Where is the vector: S T 1,, 3,, N and is the vector: T 1,, 3,, N Therefore S is the scattering matrix: S S11 S1 n Sm1 S mn The scattering matrix is a N by N matrix that completely characterizes a linear, N-port device. Effectively, the scattering matrix describes a multi-port device the way that Γ describes a single-port device (e.g., a load)! But beware! The values of the scattering matrix for a particular device or network, just like Γ, are frequency dependent! Thus, it may be more instructive to explicitly write: S ( ω ) S11 ( ω ) S1 n ( ω ) Sm1 ( ω ) Smn ( ω )

132 9/6/6 Example The Scattering Matrix 1/6 Example: The Scattering Matrix Say we have a 3-port network that is completely characterized at some frequency ω by the scattering matrix:...5 S A matched load is attached to port, while a short circuit has been placed at port 3: z P 1 1 (z) (z) port 1 port 3 1 (z) (z) port 3-port microwave device z P (z) 3 (z) 3 z P 3

133 9/6/6 Example The Scattering Matrix /6 Because of the matched load at port (i.e., Γ ), we know that: ( z ) ( z ) and therefore: You ve made a terrible mistake! Fortunately, I was here to correct it for you since Γ, the constant (not ) is equal to zero. NO!! Remember, the signal ( z ) is incident on the matched load, and ( z ) is the reflected wave from the load (i.e., ( z ) is incident on port ). Therefore, is correct! ikewise, because of the short circuit at port 3 ( Γ 1): ( z ) 1 z ( 3 ) 3 and therefore: 3 3

134 9/6/6 Example The Scattering Matrix 3/6 Problem: a) Find the reflection coefficient at port 1, i.e.: Γ1 1 1 b) Find the transmission coefficient from port 1 to port, i.e., T 1 1 I am amused by the trivial problems that you apparently find so difficult. I know that: and Γ 1 1 S T.5 1 S 1 1 NO!!! The above statement is not correct! Remember, 1 1 S 11 only if ports and 3 are terminated in matched loads! In this problem port 3 is terminated with a short circuit.

135 9/6/6 Example The Scattering Matrix 4/6 Therefore: and similarly: Γ 1 1 S 11 1 T 1 S 1 1 To determine the values T 1 and Γ 1, we must start with the three equations provided by the scattering matrix: and the two equations provided by the attached loads: 3 3

136 9/6/6 Example The Scattering Matrix 5/6 We can divide all of these equations by 1, resulting in: Γ T ook what we have 5 equations and 5 unknowns! Inserting equations 4 and 5 into equations 1 through 3, we get: Γ T

137 9/6/6 Example The Scattering Matrix 6/6 Solving, we find: Γ 1 ( ) ( ) T

138 9/6/6 Example Scattering Parameters 1/4 Example: Scattering Parameters Consider a two-port device with a scattering matrix (at some specific frequency ω ): and 5Ω. S ( ω ω ) 1. j 7. j 7.. Say that the transmission line connected to port of this device is terminated in a matched load, and that the wave incident on port 1 is: where z1 z. P P j z1 ( ) z j e β 1 1 Determine: 1. the port voltages ( z z ) and ( z z ) 1 1 1P. P. the port currents I ( z z ) and I ( z z ) 1 1 1P. P 3. the net power flowing into port 1

139 9/6/6 Example Scattering Parameters /4 1. Since the incident wave on port 1 is: j z1 ( ) z j e β 1 1 we can conclude (since z 1P ): ( ) z z j e 1 1 1P je j j βz j β 1P ( ) and since port is matched (and only because its matched!), we find: The voltage at port 1 is thus: ( P ) ( P ) 1. ( j) z z S z z j. ( ) ( ) ( ) z z z z z z 1 1 1P 1 1 1P 1 1 1P j. j. j. j. e π ikewise, since port is matched: ( ) z z P

140 9/6/6 Example Scattering Parameters 3/4 And also: Therefore: ( P ) ( P ) j 7. ( j) z z S z z ( ) ( ) ( ) z z z z z z P P P e j. The port currents can be easily determined from the results of the previous section. ( P ) ( P ) ( P ) ( z z ) ( z z ) I z z I z z I z z P 1 1 1P.. j j j 5 j 36. j 36. e π and:

141 9/6/6 Example Scattering Parameters 4/4 ( P ) ( P ) ( P ) ( z z ) ( z z ) I z z I z z I z z P P j 8. e π 3. The net power flowing into port 1 is: P P P ( ) (. ) 5 ( ) 396. Watts

142 9/8/6 Matched reciprocal lossless 1/9 Matched, ossless, Reciprocal Devices Often, we describe a device or network as matched, lossless, or reciprocal. Q: What do these three terms mean?? A: et s explain each of them one at a time! Matched A matched device is another way of saying that the input impedance at each port is numerically equal to when all other ports are terminated in matched loads. As a result, the input reflection coefficient of each port is zero no signal will come out of a port when a signal is incident on that port (and only that port!). In other words, we want: m mm m S for all m a result that occurs when: S for all m if matched. mm

143 9/8/6 Matched reciprocal lossless /9 We find therefore that a matched device will exhibit a scattering matrix where all diagonal elements are zero. Therefore: S.1 j..1.3 j..3 is an example of a scattering matrix for a matched, three port device. ossless For a lossless device, all of the power that delivered to each device port must eventually find its way out! In other words, power is not absorbed by the network no power to be converted to heat! Recall the power incident on some port m is related to the amplitude of the incident wave ( m ) as: m P m While power of the wave exiting the port is: m P m

144 9/8/6 Matched reciprocal lossless 3/9 Thus, the power delivered to that port is the difference of the two: m m Pm Pm Pm Thus, the total power incident on an N-port device is: N N 1 m m m 1 m 1 P P Note that: N H m m 1 where operator H indicates the conjugate transpose (i.e., Hermetian transpose) operation, so that is the inner product (i.e., dot product, or scalar product) of complex vector with itself. H Thus, we can write the total power incident on the device as: P N H 1 m m 1 Similarly, we can express the total power of the waves exiting our M-port network to be: P N H m m 1 1

145 9/8/6 Matched reciprocal lossless 4/9 Now, recalling that the incident and exiting wave amplitudes are related by the scattering matrix of the device: S Thus we find: H H H P S S Now, the total power delivered to the network is: Or explicitly: M P P P P m 1 P P P H H H S S 1 H H I S S where I is the identity matrix. Q: Is there actually some point to this long, rambling, complex presentation? A: Absolutely! If our M-port device is lossless then the total power exiting the device must always be equal to the total power incident on it.

146 9/8/6 Matched reciprocal lossless 5/9 If network is lossless, then P P. Or stated another way, the total power delivered to the device (i.e., the power absorbed by the device) must always be zero if the device is lossless! If network is lossless, then P Thus, we can conclude from our math that for a lossless device: 1 H H P for all I S S This is true only if: H I S S S S I H Thus, we can conclude that the scattering matrix of a lossless device has the characteristic: If a network is H lossless, then S S I Q: Huh? What exactly is this supposed to tell us? A: A matrix that satisfies known as a unitary matrix. H S S I is a special kind of matrix

147 9/8/6 Matched reciprocal lossless 6/9 If a network is lossless, then its scattering matrix S is unitary. Q: How do I recognize a unitary matrix if I see one? A: The columns of a unitary matrix form an orthonormal set! S S11 S1 S31 S 41 S S S S S S S S S S S S matrix columns In other words, each column of the scattering matrix will have a magnitude equal to one: N m 1 S mn 1 for all n while the inner product (i.e., dot product) of dissimilar columns must be zero. N SniS nj S1 is 1j S is j SNiS Nj for all i j n 1 In other words, dissimilar columns are orthogonal.

148 9/8/6 Matched reciprocal lossless 7/9 Consider, for example, a lossless three-port device. Say a signal is incident on port 1, and that all other ports are terminated. The power incident on port 1 is therefore: 1 P 1 while the power exiting the device at each port is: m Sm1 1 m m1 1 P S P The total power exiting the device is therefore: 1 3 P P P P S P S P S P ( ) S S S P Since this device is lossless, then the incident power (only on port 1) is equal to exiting power (i.e, P P1 ). This is true only if: S S S Of course, this will likewise be true if the incident wave is placed on any of the other ports of this lossless device: S S S S 1 S S S S S S S S 1

149 9/8/6 Matched reciprocal lossless 8/9 We can state in general then that: 3 m 1 S mn 1 for all n In other words, the columns of the scattering matrix must have unit magnitude (a requirement of all unitary matrices). It is apparent that this must be true for energy to be conserved. An example of a (unitary) scattering matrix for a lossless device is: 1 3 j 1 3 j S 3 j 1 3 j 1 Reciprocal Reciprocity results when we build a passive (i.e., unpowered) device with simple materials. For a reciprocal network, we find that the elements of the scattering matrix are related as: S mn S nm

150 9/8/6 Matched reciprocal lossless 9/9 For example, a reciprocal device will have S1 S1 or S S. We can write reciprocity in matrix form as: 3 3 T S S if reciprocal where T indicates (non-conjugate) transpose. An example of a scattering matrix describing a reciprocal, but lossy and non-matched device is:.1.4 j..5.4 j. j.1 S j..1 j j.1.1

151 3/1/5 Coaxial Transmission ines.doc 1/4 Coaxial Transmission ines The most common type of transmission line! Outer Conductor a b Inner Conductor ε Coax Cross-Section - The electric field ( the direction â ρ. )points in The magnetic field ( in the direction â φ. )points E. M. Power flows in the direction a. ˆz A TEM wave!

152 3/1/5 Coaxial Transmission ines.doc /4 Recall from EECS that the capacitance per/unit length of a coaxial transmission line is: π ε farads C ln b/a meter And that the inductance per unit length is : µ b Henries ln π a m Where of course the characteristic impedance is: o C 1 π µ ε b ln a and: β ω C ω µ ε Therefore the propagation velocity of each TEM wave within a coaxial transmission line is: v p ω c β µ ε µ ε ε ε r r where ε εε is the relative dielectric constant, and c is the r 8 speed of light ( c 3 1 m/ s).

153 3/1/5 Coaxial Transmission ines.doc 3/4 Note then that we can likewise express β in terms ε r : ω β ω µ ε ω µ ε ε ε c r r Now, the size of the coaxial line (a and b) determines more than simply and β ( and C) of the transmission line. Additionally, the line radius determines the weight and bulk of the line, as well as its power handling capabilities. Unfortunately, these two characteristics conflict with each other! 1. Obviously, to minimize the weight and bulk of a coaxial transmission line, we should make a and b as small as possible.. However, for a given line voltage, reducing a and b causes the electric field within the coaxial line to increase (recall the units of electric field are /m). A higher electric field causes two problems: first, it results in greater line attenuation (larger α); second, it can result in dielectric breakdown. Dielectric breakdown results when the electric field within the transmission line becomes so large that the dielectric material is ionized. Suddenly, the dielectric becomes a conductor, and the value G gets very large!

154 3/1/5 Coaxial Transmission ines.doc 4/4 This generally results in the destruction of the coax line, and thus must be avoided. Thus, large coaxial lines are required when extremely low-loss is required (i.e., line length is large), or the delivered power is large. Otherwise, we try to keep our coax lines as small as possible!

155 3/1/5 Coaxial Connectors.doc 1/ Coaxial Connectors There are many types of connectors that are used to connect coaxial lines to RF/microwave devices. They include: SMA The workhorse microwave connector. Small size, but works well to > GHz. By microwave standards, moderately priced. BNC The workhorse RF connector. Relatively small and cheap, and easy to connect. Don t use this connector past GHz! F A poorman s BNC. The RF connector used on most consumer products such as Ts. Cheap, but difficult to connect and not reliable.

156 3/1/5 Coaxial Connectors.doc / N The original microwave connector. Good performance (up to 18GHz), and moderate cost, but large (about cm in diameter)! However, can handle greater power than SMA. UHF The poorman s N. About the same size, although reduced reliability and performance. RCA Not really an RF connector. Used primarily in consumer application for video and audio signals (i.e., < MHz). Cheap and easy to connect. APC-7 and APC-3.5 The top of the line connector. Best performance, but cost big $$$. Used primarily in test equipment (e.g., network analyzers). 3.5 can work to nearly 4 GHz.

157 9/8/6 Printed Circuit Board Transmission ines 1/ Printed Circuit Board Transmission ines ε r Microstrip Probably most popular PCB transmission line. Easy fabrication and connection, yet is slightly dispersive, lossy, and difficult to analyze. ε r Stripline Better than microstrip in that it is not dispersive, and is more easily analyzed. However, fabrication and connection is more difficult. εε r r Coplanar Waveguide The newest technology. Perhaps easiest to fabricate and connect components, as both ground and conductor are on one side of the board. ε r Slotline Essentially, a dual wire tranmission line. Best for balanced applications. Not used much.

158 9/8/6 Printed Circuit Board Transmission ines / An antenna array feed, constructed using microstrip transmission lines and circuits. A wideband microstrip coupler.