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Unformatted text preview: Peterson: Transmission Lines for Electrical and Computer Engineers 2/99 Lecture 13 Frequencydomain Analysis of Transmission Lines Objectives: Introduce the treatment of sinusoidal signals on transmission
lines, through the use of phasors. Deﬁne phase constant, frequency, and
related frequencydomain parameters. Convert the transmission line
equations into their phasor form, and present the general solution. Finally,
derivexthe input impedance of a line with a general load. Previous lectures have focused on transmission lines excited with general timedependent
functions. In many practical applications involving transmission lines, the signals are
sinusoids or modulated sinusoids. The study of a linear system response for sinusoidal
excitation is known as frequencydomain analysis, as opposed to the general timedomain
analysis considered previously. Applications involving modulated sinusoids include cable
TV distribution lines, transmission lines feeding broadcast antennas for radio or television,
and other elements of “wireless” communication systems. In addition, measurements are
often made using sinusoidal signals, since the quantity being measured can be made more
sensitive (and therefore measured more accurately) if the measurement setup incorporates
resonant circuits tuned to that particular frequency. Finally, losses are frequency
dependent, and are usually easier to characterize from a frequencydomain perspective. In
our previous discussions, we considered waves having the general form f(t :t Z/VP).
Beginning in this lecture, we conﬁne our attention to sinusoidal waves such as V*(z,t) = V+ cos[u)(t — z/vp)] ' (13.1)
Equation, (13.1) represents a voltage wave propagating in the +2 direction with velocity VP. The parameter (c in (13.1) is the radian frequency, typically in units of radians/second. The
radian frequency is related to the frequency f, with units of Hertz, by u) = 23rf ‘ (13.2)
Equation (13.1) can be rewritten as V+(Z,t) = V+ cos(wt — ﬁz) (13.3)
where p = m/vp = m sqrt(LC) (13.4) is the phase constant of the transmission line at radian frequency m, with units of
radians/meter. Equivalently, the phase constant is given by B = Zarf/vp = 23:0» (13.5) Peterson: Transmission Lines for Electrical and Computer Engineers 2/99 where 7» is the line wavelength at this frequency, typically in meters. It is apparent that the argument of (13.1) exhibits the functional form of a onedimensional
wave, and therefore the previous aspects of transmission line theory can be applied.
However, there are additional simplifications to the theory that can be realized when all the
signals in a problem are sinusoidal at the same frequency. In a linear electrical system, all the signals arising from a sinusoidal excitation are also
sinusoidal with the same frequency as the excitation, but generally having different
amplitudes and phases. Under these circumstances, it is convenient to use a shorthand
notation that deals directly with the amplitudes and phases of the signals, and suppresses
the sinusoidal time dependence. This notation is called phasor notation. [We mention here
that phasors are directly related to Fourier transform quantities; since we do not assume the
reader has a background in Fourier analysis, we do not pursue this connection] A phasor is a complex—valued quantity, whose magnitude and phase represent the
amplitude and phase angle of the associated sinusoid. Phasors representing waves on
transmission lines are also a function of position on the line (2). The manipulation of
phasors requires a familiarity with complex exponential functions. Most of our readers
should already be familiar with the Taylor series for e", given by the expansion e":1+x+x2/2!+...+x“/n!+... (13.6)
If we consider the exponential function of an imaginary argument, say jB , we obtain
e‘°=1+j6—62/2!—j63/3!+ 94/4! + (13.7)
By separating the real and imaginary parts, the series can be rewritten as
L eie={l—62/2!+64/4!+...}+j{0 —63/3!+05/5!+ ...} (13.8) However, these expansions are readily recognized as the Taylor series for cosine and sine,
respectively. Therefore, we obtain Euler’s identity. ei" = cos(6) +j sin(6) (13.9) Euler’s identity suggests a graphical interpretation (Figure 13.1). Given an angle 6, the
function ei8 is a point on the unit circle in the complex plane corresponding to the angle 6
measured from the positive real axis. The real part of 6‘9 is the projection onto the real axis
(cosB), while the imaginary part of ei9 is the projection onto the imaginary axis (sinB). The
magnitude of the complex exponential function ei9 is always one. Thus, one way of
representing a general sinusoid with amplitude A and phase angle 9 is to express it as FLXM \3.‘ Peterson: Transmission Lines for Electrical and Computer Engineers 2/99 A cos 6 = Re{A cw} (13.10) and work with the complex function {A e59}. The symbol “ e” means to take the real part of the complex quantity within the brackets, or equivalently to project it onto the real axis in
the complex plane. The phasor can be denoted using an arrow in the complex plane pointing from the origin to the point {A 6°}. This arrow indicated the magnitude (A) and
the phase angle (6) of the sinusoid. [Insert Figure 13.1 here] A voltage waveform such as (13.1) that is a function of position can be expressed as a
phasor V(z), where the phasor is deﬁned so that v+ cos(wt— [32) = Re{V(z) em} (13.11)
For this speciﬁc signal, the phasor is given by
V(z) = V+ €531 (13.12) Using Euler’s identity, we observe that (since the coefficient V+ is presumed to be real
valued in this situation) Re{V(z) eimt} = Re{V” ei("’“ﬂz’} = V+ cos((nt — 52) (13.13) Note that the phasor does not include the factor e3“. ' By convention, all the phasor
quantities we work with will be assumed to have the identical frequency and therefore will exhibit the same eimt time dependence, which will be suppressed. For visualization
purposes, we can conceptually think of the phasor in Figure 13.1 rotating around the origin
with radian frequency w as time progresses. To summarize, the realvalued sinusoidal function v(z,t) = V+ cos(wt — 62) has been replaced by a complex—valued phasor V(z) = V+ e’j‘”. We have traded the time dependence
of v(z,t) for the complexvalued nature of V(z). To minimize confusion, we Will usually
denote functions of time with lower—case letters, and phasors with upper—case letters. As an example of constructing phasors, consider the time function
u(z,t) = A sin(u)t — 62 + 113/6) (13.14)
To convert this to a phasor, we rewrite the function as a cosine, to obtain u(z,t) = A cos((nt — $2 + n/6 — at/Z) (13.15) Peterson: Transmission Lines for Electrical and Computer Engineers 2/99 u(z,t) = Re{A em" ““"’3)} (13.16)
and identify the phasor as
U(z) = A 6“" e"""3 (13.17) As an example illustrating the construction of the time function from a phasor, consider the
phasor given by W(z) = —j B 6““ (13.18)
Where B is realvalued. This function is equivalent to W(z) = B either?"2 ' (13.19)
Therefore, the time function is w(z,t) = Re{B elm" 5" m’} ( 13.20)
or w(z,t) = B cos(u)t + Bz  3/2) = B sin(u)t + 52) (13.21)
This function represents a wave propagating in the —z direction. We now focus more speciﬁcally on transmission line problems. For a general line, the
transmission line equations derived in Lecture 1 are 3v ~ ._ a'
53‘ R’ L37: (13.22) 34’. E
Manew c.“ (13.23) To recast these equations into phasor form, we replace the time functions v(z,t) and i(z,t) with phasors V(z) and 1(2), and replace all time derivatives by multiplications with ju). The
phasor equations take the form 1! a —(m3u.\ 12(2) A: (13.24)
£5 . (& ﬁrm) We) (13.25) A: Peterson: Transmission Lines for Electrical and Computer Engineers 2/99 For the moment, we consider only the lossless case, and set R and G equal to zero. The,
lossless equations are $1 _._ 5uL 32(2) as (13.26)
AI . It is a relatively straightforward process to show that these equations have the general
solution V(z) = v; e'sz + V; cm“ (13.28) I(z) = (V0+ 0) e‘jﬁz — (VO‘IZO) 6““31 (13.29) where the phase constant p = w sqrt(LC) and the characteristic impedance Z0 = sqrt(UC)
have been deﬁned previously. We introduce two coefﬁcients, V0+ and Vo', which
represent the complexvalued coefficients associated with the waves traveling in the +2 and
—z directions. [The reader should substitute V(z) and 1(2) back into equations (13.26) and
(13.27) to confirm that they are the solution!] [Insert Figure 13.2 here]
Consider a transmission line driven from the left by a sinusoidal generator and loaded at the right end with an impedance ZL (Figure 13.2). The generator end of the line is located at
z=0 while the load impedance is located at z=L. The line has characteristic impedance Z0 and phase constant 3. It is assumed that the generator has been turned on for a long time, so that the initial transients have died out and the remaining signals are entirely sinusoidal at
a single frequency. The voltage and current waves are given by equations (13.28) and
(13.29). However, at the load VL = V(z=L) and IL = I(z=L) are related by IL = VL / ZL (13.30)
Imposing this constraint yields V(L) = VL = v; ej'3L + V; eff“L (13.31) 1(2) = vL / zL = (v0+ 0) e'i"L — (V0720) eat? (13.32)
These equations can be solved to ﬁnd V0+ and V0" in terms of VL. We determine that +__ ‘ §ﬂL  %
V0 — as (H’f‘: v._ (13.33) Peterson: Transmission Lines for Electriail and Computer Engineers 2/99 Vo= 13" aft—3h __ (13.34) After substituting these coefﬁcients back into (13.28) and (13.29), we construct the
impedance at a location z as , ' (H) _~ (1,9
We) Y§[(le%)edl3 +(l»§—‘i)em I 3““ “‘ = » (13.35)
I“) )3: £2 ' we) . ' (Le)
a2. [("eJedF "(“§;)€,3F ]
which can be rewritten using Euler’s identity as
EL cos/ewe.) + 4%.: sin/5&1) }
20» = Z. 2° MPG,” +3, 2‘ synce) (13.36)
and ﬁnally simplified to
EL + ,3 z. to... fan—2)
zca= 2 2° +3 a Jew “Ti—(L's) (13.37) If we specialize this result to the input end of the line, we obtain the input impedance in the form
V“ Z VZL+5 “Ea '6pr }
Zia: E— — 0 204‘3 JEL tam/5L (13.38) Equation (13.38) is a useful result that we will frequently refer back to. Among other
things, it specifies the form of an equivalent circuit for the line, as depicted in Figure 13.3. [Insert Figure/l3 .3 here] MwQPMAMSJRMWM... Mam“. . , .WUf ( %,A+.\W .=_W A (305(40’100112 tWWe \WW um... 2. .. when... ed....£...m.c a Fad A MA "96%). So‘ujlalq I tWW (wi\ 12)
\r(2t\= Ke{ «£39,130 __}= 2.:{43 6.0 an; ,4 3 ¢._(.§_—nt) _;..;. (4.36; we A , f f ' “:54 gm..) A V ...
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