FWLec14 (1) - Peterson Transmission Lines for Electrical...

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Unformatted text preview: Peterson: Transmission Lines for Electrical and Computer Engineers 2/99 Lecture 14 Short-circuited and Open-circuited Lines Objectives: Examine the input impedance of lines with short-circuit and open-circuit loads, and quarter—wave and half-wave lines with general loads. Investigate the voltage and current standing waves on short-circuited and open-circuited lines. In the last lecture, the input impedance of a lossless transmission line of length L was determined in the form Zia = Eu» a in “t0“ [5(— where 20 denotes the characteristic impedance, ZL is the load impedance, and B is the phase constant. In this lecture, we specialize this result to several situations to illustrate how the frequency—domain voltage and current phasors are related on transmission lines. Since we are working with frequency-domain parameters, it will be understood that the excitation and responses are sinusoidal at the same frequency and that all the initial transients have died down. Short-circuited line First, consider a short-circuited line of length L. Equation (14.1) can be specialized using ZL = 0 to obtain 21.. =1 20 tan(fiL) (14.2) We immediately make two observations. First, the lossless, short-circuited line has a purely imaginary input impedance. Equivalently, it is purely reactive. This is consistent with the physical nature of a system that contains no energy-dissipating mechanism and therefore cannot absorb energy. The second observation is that because of the tangent function in (14.2), the input reactance of the line can assume any value from —00 to +00. Figure 14.1 shows a plot of the input reactance as a function of its length L. [Insert Figure 14.1 here] By adjusting the length of a short—circuited line, it can be made to mimic any value of capacitance or inductance. (Capacitors exhibit negative values of reactance, while inductors have positive reactance.) This property is useful at high frequencies, where traditional inductors and capacitors have unreliable values due to parasitic reactances. Consequently, in high frequency circuits, short-circuited transmission lines can be used in place of these components. (Such lines are sometimes referred to as stubs.) From an inspection of Figure 14.1, we also observe that the short-circuited transmission line appears to be an open-circuit when its length is M4, where 7» is the line wavelength 53m 14.! Imfwt Maud—i «SAN-(:— Peterson: Transmission Lines for Electrical and Computer Engineers 2/99 A = 2m (14.3) The line also acts as an open circuit whenever its length is an odd multiple of M4. On the other hand, whenever the line’s length is M2 or a multiple of M2, it appears to be a short circuit. The explanation for this behavior lies with the voltage and current waves on the line, a topic to be discussed shortly. Open-circuited line As a second special case, consider a line terminated in an open circuit (ZL -> 00). A specialization of Equation (14.1) to an open circuit yields Zn. = —j Z0 cot(BL) (14.4) In common with the short-circuited line, the open-circuited line has a purely imaginary input impedance, and can assume any value of reactance from —00 to +00. Consequently, by adjusting its length, it can also be made to mimic any value of capacitance or inductance. We observe that the open-circuited line appears to be a short circuit whenever its length is an odd multiple of M4, and it appears to be an open circuit whenever its length is a multiple of M2. Figure 14.2 shows a plot of the input reactance as a function of its length L. [Insert Figure 14.2 here] Matched line As a third special case, we consider a matched termination (ZL = Zn). In this situation, Equation (14.1) simplifies to 2.520 (14.5) Thus, a matched line always has input impedance equal to its characteristic impedance. This is usually the desired situation, since it eliminates reflections from the load and ensures complete power transfer to the load. In practice transmission lines are matched as closely as possible, keeping in mind that in many situations systems must be used over a range of frequencies and it is generally not possible to realize a perfect match over a range of frequencies. ’x/ Peterson: Transmission Lines for Electrical and Computer Engineers 2/99 Quarter-wave line In the previous discussion, we observed that something special seems to happen when the lines are a multiple of M4 or M2 in length. To evaluate Equation (14.1) for the case of a quarter-wave line, we observe that if L = M4, BL = ammo/4) = n/2 (14.6) Since tan(Jr/2) -—> 00, Equation (14. 1) reduces to 2,, = (2.)” 2L (14.7) As observed previously, a quarter-wave line converts a short-circuit load into an open circuit, and vice versa. Equation (14.7) indicates that it also converts any real-valued load impedance (any resistance) into another real—valued impedance. This property is sometimes useful for matching a load resistance RL to a line with characteristic impedance Z01, by using a quarter-wave section of a line with a different characteristic impedance given by 202 :J Z01 RL (14.8) The quarter-wave section has input impedance equal to 201, by (14.7), and will therefore accept all the incident energy and deliver it all to the load resistor RL. Quarter-wave matching will be discussed in more detail in Lecture 19. Half-wave line The input impedance of transmission lines with length L = M2 can be investigated using {3L = (Zn/KXMZ) = at (14.9) Since tan(n) = 0, Equation (14. 1) reduces to zin = zL (14.10) Thus, a line that is a multiple of a half wavelength has input impedance equal to its load impedance, regardless of the value of characteristic impedance on the line. There is also a general property, suggested in Figures 14.1 and 14.2, that the input impedance repeats as the line length changes by half-wavelength multiples. This property will be illustrated in the following discussion. EVOH'l-ac nu“: wk“ . ‘ V(e)' (5(i'L\ I: -n‘“’ av (£43: - 2i). 1- t i n I K(%) I I ‘ ‘ I ' l ? Apat- h L w {the-19 M U ”(7044, l’fi‘)’ #0457. WIN (MWW‘VV Peterson: Transmission Lines for Electrical and Computer Engineers 2/99 Voltages and currents on a short-circuited line [Insert Figure 14.3 here] Figure 14.3 shows a short-circuited line of length L, characteristic impedance Z”, and phase constant B. The general solution for voltage and current waves on the line is repeated from Equations (13.28) and (13.29): V(z) = v; e-ifil + V; eat“ (14.11) 1(2) = (V0720) e‘jB‘ — (VO‘ 0) efl'tl (14.12) Since the load is a short circuit, it follows that V(L) = 0, and therefore V0” = — V0+ e‘m‘L ' (14.13) We can write the voltage as V(z) = V0+ {e’j‘h — efl'WU} = V0+ e—ifiL {e—jfl(z-L) _ e+j8(z—L)} = V0+ e‘j"L {—2j sinfi(z—L)} = K1 sinB(z—L) (14.14) where K1 is a complex constant. Similarly, the current can be written as 1(2) = (Vo+ 0) e—lflL {e-isz-L) + CHM-14)} = K2 cosB(z—L) (14.15) where K2 is another constant. These functions, and the imaginary part of the impedance 2(2) = V(z)/1(1), are plotted in Figure 14.4. It is apparent that the impedance varies widely due to the alternating zeros in the numerator (voltage) and the denominator‘(current). The fact that the voltage or current can vanish at points along the line explains the possibility that the line input impedance mimics a short circuit or open circuit. [Insert Figure 14.4 here] The nulls in the voltage and current plots are spaced at half-wavelength intervals on the line, as are maximum values for voltage and current. For a short-circuited line, the voltage is always zero at the load (the short circuit), While the current is maximum there. Thus, the Peterson: Transmission Lines for Electrical and Computer Engineers 2/99 voltage maximum and current minimum occur together, as do the voltage minimum and current maximum. The voltage and current behavior repeats, with the inclusion of a 180° phase shift, every half wavelength. Since the phase shift cancels in the impedance, it repeats exactly every half wavelength. It is important to keep in mind that the voltage V(z) is a phasor quantity. To visualize the voltage behavior with respect to time, we must convert from a phasor to a time function. Suppose the complex constant K1 in equation (14.14) is written as K: IKJ e” (14.16) Then, the time function can be constructed using v(z,t)= Re{V(z) em} = Re{|K,l sin[B(z—L)] elm eie} =|K1lsin[B(z—L)] cos((nt + 9) (14. 17) and is sketched in Figure 14.5. It is apparent that the functional form of (14.17) is not that of a traveling wave, since the dependence on 2 and t are separated. Instead, the behavior illustrated in Figure 14.5 is that of a standing wave. A standing wave is produced by the interference pattern of the forward and reflected waves in Equations (14.11) and (14.12). The nulls in a standing wave are fixed in location, while the amplitude oscillates up and down as a function of time. In contrast, a traveling wave (Figure 14.6) shifts its location as time progresses, nulls and all. [Insert Figure 14.5 here] [Insert Figure 14.6 here] A similar behavior can be observed for the open-circuited transmission line. The condition at the load end of the line is I(L) = O, and therefore V0“ = V; e‘jz“L ' (14.18) The voltage and current can be written as V(z) = v; {e‘it‘z + e+j5‘"m} = Vo+ e-ifiL {e-iflflrL) + e+ifi(Ir-L)} = V0+ e‘j"L {2 cosfi(z—L)} Raw-L ULS' VW o‘c- A. SM mm one. ~Fun~étho£ ‘KW ' Raw 14.4, VWtM/ LWWZ‘? wow u «WIQVAW. I ‘f . wag, IQ) I . l X(z) I 1 EM . é". .z.:l. - (lac-L) I?“ “6-? V0457” W) Mud/(WW MM W'WCLLLM “601% Peterson: Transmission Lines for Electrical and Computer Engineers 2/99 = K, cos[3(z—L) (14.19) and 1(2) = (v0+ 0) e‘it‘L {e‘ifiw — e+j“(‘"”} = K4 sinB(z—L) (14.20) where K3 and K4 are complex-valued constants. Figure 14.7 shows plots of the voltage, current, and reactance. The behavior is again that of a standing wave created by the perfect interference of the forward and reflected waves on the line. The current vanishes at the open-circuit end of the line, while the voltage exhibits a maximum there. Voltage minima and maxima alternate, separated by quarter wavelength distances. Current minima and maxima occur at the locations of voltage maxima and minima, respectively. [Insert Figure 14.7 here] Application As an example illustrating the above theory, consider the following. A transmission line used in a CATV distribution system is damaged, resulting in an open circuit somewhere on the line. Assume that you, as the system operator, have access to the input end of the line. Assume also that you can send a sinusoidal signal at any frequency into this transmission line, and determine the input impedance at that frequency. Can you determine the location of the break? Or must you physically inspect every foot of cable? The solution process begins with equation (14.4), which can be rewritten as Xin = — Z0 cot(2:rLf/vp) (14.21) Figure 14.8 shows a plot of this function versus frequency f. Two successive frequencies where the input impedance vanishes are related by f, — f1 = vp/2L (14.22) Thus, assuming that you can determine two successive frequencies where the input reactance vanishes, and that you also know the propagation velocity vp for your line, equation (14.22) can be solved for the distance L to the break. We note that a similar approach can be used if a short-circuit condition exists somewhere on the line. This procedure can also be used to locate faults in wiring boards, packages, and integrated circuits, which is an important aspect of the manufacture of these components. Example. A lossless transmission line of length blm and characteristic impedance 20:50!) is short-circuited at the load end. The phase velocity of the line is vp=108 m/s. 2"“: 4. z-W/SL : ' a“ 17741 1‘ ( mug b) Find the input impedance of the line at f = 70 MHz. fl... whom: um .zmhssotmxnwwj IS‘I JL —'——-——. Io" £.=l c) Is it possible to achieve the 24., of part (b) with a shorter length of the same short-circuited transmission line? If yes, provide the length of the shorter line. thAmeso! s{%(0.‘¢'fl) —» 1:4, m (51‘: 0.4T!" 572° 1‘ = 0.290 M -—--______-:_""..—""'————- 0444A. «remade—’3 0‘91“»?! M 2i: Ofli‘i M Sm 3;: '\$l¢$$+€-M'Ml.h~,wst\.mft¢4~ lime/L3 2" we! 3.; IN 2|“, L "2mm A lossless air-filled transmission line has charactetistic impedance 20 = 300 Q. a) A2511) length of this lincisterminatcdinanopcn circuit. Findtheinputimpedanceof the lincatf=100Ml-Iz. 32.5 m = 0.8333 ) -— 522.30 vacLiavw A ‘ “if“ 5‘3: 20 cot fl! = +4l73'1 J]. b) Is there a shorter length of this transmission line that has the same input impedance as the 2.5 m section? If so, what is its length? Sam. M > “7.3% ‘13: m\ma%t= (5 23¢; —"V)W= 2 o‘iLt méw - = o 3'53 wavdma‘WS = I. 0 mi“ c) Sketch the voltage standing wave pattern on the 2.5m line: = l, 5' M N 2‘ lv (21 l INPUT é____.> imp O. 5‘ m d) If the inductance per unit length of this transmission line is 1.0 uH/m, what is the capacitance per unit length? L. \ -“ = ____= =1,m>uo E- n. F C m - H F /.M b) d) PM‘J\&W\ {'4 .i A lossless transmission line filled with a material having e=4.5eo and u=uo has characteristic impedame 20 = 100 0. Find the inductance per unit length. Find the capacitance per unit length. Al.6(-m)lengthofthislineis temiinatedinanopen circuit. Findtheinputimpedanceof thelineatf=75MHz. ' Isthereashorter length ofthis transmission linethat has the same input 'nnpedance as the 1.6 (m) section? If so, what is its length? Sketch the voltage standing wave pattern on the 1.6 meter line, clearly showing the location ofminima and maxima: (We) ' b) d) th‘w/x Hal A lossless transmission line is constructed with a uniform dielectric material in such a manner to produce inductance per unit length L = 1><10—6 (H/m) and capacitance per unit length C = 1x10-10 (F/m). Find the relative permittivity 81- of the uniform dielectric material filling the line interior, assuming that the permeability is HO = 41tx10-7 (H/m). Find the characteristic impedance 20. A 15 (cm) length of this line is terminated in a short circuit. Find the input impedance of the line at f=500 MHz. Is there a shorter length of this transmission line that has the same input impedance as the 15 (cm) section? If so, what is its length? Sketch the Mag; standing wave pattern on the 15 cm line, clearly showing the location of minima and maxima: [le Pwiqiwn (LL 3 W The lossless transmission line system shown below is driven with a signal VG(t) = 10 cos(21t x 109 t) — 5 cos(61c x 109 t) (V) 20:75.9. Volt.) VP = 1.5' moguls S.— L=7.§-cu ——> (a) Determine the input impedance Zm at radian frequency 21: x 10’. (b) Determine the input impedance Zin at radian frequency 6m x 109. (c) Find Vin(t). (d) Find vLa). Pwuw HA SW M a. \osskss WMKSSIM KM 04 \«z‘fl. Mg «£qu km [Ellyl = Ea :Dti'um M Mai? ‘LWW 04‘ (4.3 a. skew-7t“ Mull-eaa “8 M ) M (L) and Dw-Qwi-d x18 “3&1.“ Pmuwx “LS- Pmuew. W come A M m, L.== if). ...
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