Unformatted text preview: 1 + Γ2 + 2Γ cos(θr − 2βd) (203) and, • We can see from gd that if we want to make that 1 we have to
have,
cos(θr − 2βd) = −Γ
(204)
• That will give us a denominator,
1 + Γ2 − 2Γ2 = 1 − Γ2
Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. (205) Electromagnetics I: Transmission lines 111 • So that, we would have,
gd = 1+ Γ2 1 − Γ2
1 − Γ2
=1
=
+ 2Γ cos(θr − 2βd)
1 + Γ2 − 2Γ2
(206) • So, for the condition,
cos(θr − 2βd) = −Γ (207) we have to adjust d since that is all we have control over. So,
we select a d to do the above.
• Once we ﬁx d, then we have to calculate what bd is and then
add our shunt element so that bs = −bd .
• Many diﬀerent ways to accomplish matching. We could add a
capacitor or inductor, or we canl use transmission lines in an
arrangement called “single stub”, illustrated in Fig. 29. (Remember a transmission line can be the same as an inductor or
capacitor).
Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 112 • To satisfy both our two degrees of freedom we have two length
of d and the length of a “stub” l placed in shunt. Stub is either
SC or OC.
• The procedure:
1. Convert ZL to YL
2. Select a distance d so as to transform the load admittance
into YL into Yd = Y0 + jB (i.e. so that yL = 1 looking into
M M before adding the stub).
3. Add shunt stub (OC or SC) with Ys = −jB so that total
admittance looking into M M (with stub) is Yin = Y0 +
jB − jB = Y0 , as needed! Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines Feedline
Y0 Yin 113 d M
Yd
M' Ys Y0 YL
Load Y0
l Feedline Shorted
stub M
Yd Yin Ys M' Figure 229 Figure 29: Impedance matching using shortedstub. Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 114 1.12. Transients on Transmission lines
Everything we’ve done so far was frequencycentric. Also, largely applicable only to narrowband signals. What do we do with wideband
applications? For that we need the transient response of the transmission line network.
• Start with a single pulse of amplitude V0 and duration τ .
• Decompose it into two step functions:
V (t) = V1 (t) + V2 (t) = V0 U (t) − V0 U (t − τ ) (208) where U (t) is a unit step function. Illustrated in ﬁg. 30.
• Why bother? If we know what happens to a step response we
can ﬁgure out pulse responses! Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 115 V(t) V(t) V0 V1(t) = V0 U(t) V0
τ t τ t V2(t) = V0 U(t  τ)
(a) Pulse of duration τ (b) V(t) = V1(t) + V2(t) Figure 30: Rectangular pulse as a sum of two step functions.
• Transient response
Figure 232
For a transient response we need to include...
View
Full
Document
This note was uploaded on 09/25/2013 for the course ECE 331 taught by Professor Martinsiderious during the Fall '12 term at Portland State.
 Fall '12
 MartinSiderious
 Electromagnet

Click to edit the document details