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ECE 450: Spring 2009 Homework Set 3 Solutions Due on Friday, Feb 6, 2009 ECE 450: Spring 2009 Homework Set 3 Solutions Due on Friday, Feb 6, 2009 1. A loss-less transmission line of characteristic impedance of 50 Ohm is to be terminated with a real 150-Ohm load. Between the load and the transmission line, a quarter-wave impedance transformer will be introduced to match the load to the line. The transmission line is a coaxial cable filled with insulating material of relative dielectric constant of 2.25. The operating frequency at which perfect matching is desired is 1GHz. a. Given the constraint that the impedance transformer cannot be placed closer than 0.25 m from the load, yet it should be inserted as close as possible to the load, design the matching network. Give explicit values for the distance, d, of the quarter-wave transformer from the load, the characteristic impedance of the transformer, and its physical length, L. c 31× 08 λ == =0.2m 9 f ε r 10 × 2.25 Since the load resistance is real, the quarter wave transformer can be placed at any multiple of λ/4 from the load, which in this case is any multiple of 0.05m from the load. Thus, given the constraint of the problem, the first place we can put the transformer is at d=0.25m At d = 0.25m, the load has effectively undergone one quarter wave transformation, by a 50Ω line. Thus, Zo2250 Zeff ===16.67Ω ZL 150 We can place our quarter wave transformer (of length 0.05m) here, to transform this impedance to 50Ω. ZZtransformer =×eff Zo=16.67*50 =28.87Ω b. Plot the voltage standing wave pattern on the section of the line between the load and the transformer. ZZ− 150 − 50 Γ= L 0 = =0.5 and Γ(de) = Γ -2 jdβ =0.5e− j20π d ZZL ++0 150 50 − jd20π Vd()=+V++(1 Γ()d)=V(1+0.5e ) At the output of the transformer/input of the line (d = 0.25m), V(d) = 0.5V+ ECE 450: Spring 2009 Homework Set 3 Solutions Due on Friday, Feb 6, 2009 V d 1.4H L 1.2 1.0 0.8 0.6 0.4 0.2 d 0.05 0.10 0.15 0.20 0.25 c. Plot the voltage standing wave pattern on the transformer. ZZ− 16.67 − 28.87 Γ= L 0 = =−0.268 and Γ(de) = Γ -2 jdβ =−0.268e− j20π d ZZL ++0 16.67 28.87 − jd20π Vd( )=+V'(++1 Γ(d))=V'(1−0.268e ) Continuity between the transformer and the 50Ω line requires V(0) = 0.5V+ − jd20π Thus, 0.732V '++==0.5VV and (d) 0.683V+(1−0.268e) V d 0.8 H L 0.6 0.4 0.2 d 0.01 0.02 0.03 0.04 0.05 ECE 450: Spring 2009 Homework Set 3 Solutions Due on Friday, Feb 6, 2009 d. Plot the magnitude of the reflection coefficient, |Γ|, at the input of the transformer as a function of βL from 0 to π. 1− 0.268e− jd20π − jd20π 28.87 − 50 1+Γ(de) 1−0.268 ZZ− − jd20π ZZ=Γ=28.87 and =in 0 =1+ 0.268e in transformer 1−Γ(de) 1+0.268 − jd20π in Z+Z1− 0.268e− jd20π in 0 28.87 + 50 1+ 0.268e− jd20π Gin 0.5 0.4 0.3 0.2 0.1 bd 0.5 1.0 1.5 2.0 2.5 3.0 ECE 450: Spring 2009 Homework Set 3 Solutions Due on Friday, Feb 6, 2009 e. If the maximum acceptable value of reflection is |Γ|=0.2, estimate the bandwidth of the matching network. For this question, we must examine the complete effects of the matching network. The change in frequency will lead to a change in the lengths of both the transformer and the subsequent line section. 22π fd −−10 π fd 9 βπdftrans ==×5×10 and βdline ==πf×2.5×10 vvpp −9 ⎛⎞⎛⎞10+ .5e22××jf×.5×10 ⎜⎟50×−28.87 ⎜⎟22××jf×.5×10−9 ⎜⎟⎜⎟10− .5e 25××jf××10−10 1+ −9 e ⎜⎟⎜⎟10+ .5e22××jf×.5×10 50×+28.87 ⎜⎟⎜⎟22××jf×.5×10−9 ⎜⎟⎝⎠10− .5e 28.87× −9 − 50 ⎜⎟⎛⎞10+ .5e22××jf×.5×10 50×−28.87 ⎜⎟⎜⎟22××jf×.5×10−9 ⎜⎟⎜⎟10− .5e 25××jf××10−10 1− −9 e ⎜⎟⎜⎟10+ .5e22××jf×.5×10 ⎜⎟⎜⎟50×+−9 28.87 ⎝⎠10− .5e22××jf×.5×10 Γ= ⎝⎠ −9 ⎛⎞⎛⎞10+ .5e22××jf×.5×10 ⎜⎟50× − 28.87 ⎜⎟22××jf×.5×10−9 ⎜⎟⎜⎟10− .5e 25××jf××10−10 1+ −9 e ⎜⎟⎜⎟10+ .5e22××jf×.5×10 50×+28.87 ⎜⎟⎜⎟22××jf×.5×10−9 ⎜⎟⎝⎠10− .5e 28.87× −9 + 50 ⎜⎟⎛⎞10+ .5e22××jf×.5×10 50×−28.87 ⎜⎟⎜⎟22××jf×.5×10−9 ⎜⎟⎜⎟10− .5e 2××jf×51×0−10 1− −9 e ⎜⎟⎜⎟10+ .5e22××jf×.5×10 ⎜⎟⎜⎟50×+−9 28.87 ⎝⎠⎝⎠10− .5e22××jf×.5×10 ECE 450: Spring 2009 Homework Set 3 Solutions Due on Friday, Feb 6, 2009 We can plot Г and obtain: G f 0.8 H L 0.6 0.4 0.2 f 8 8 9 9 9 6. µ10 8. µ 10 1. µ10 1.2 µ10 1.4 µ 10 And we can zoom in around the operating frequency to get: G f 0.30 H L 0.25 0.20 0.15 0.10 0.05 f 8 8 9 9 9 9 9.8 µ10 9.9 µ10 1. µ10 1.01 µ10 1.02 µ 10 1.03 µ 10 Thus, we can see that the bandwidth of this network is 36 MHz, which can be easily confirmed numerically. ECE 450: Spring 2009 Homework Set 3 Solutions Due on Friday, Feb 6, 2009 2. Consider the transmission line circuit shown below. All transmission lines are lossless. The quarter-wave transformer section is inserted at the position of the first voltage minimum from the load. Given that the voltage standing wave ratio on the transmission line to the left of the transformer is 1.0, calculate the following: a. The characteristic impedance of the quarter-wave transformer. As we are given that the VSWR left of the transformer is unity, we know that the input impedance at the transformer must be 50Ω. We can calculate the impedance at the voltage minimum and then find the impedance of the transformer. ZZ− Γ= R 0 =0.5 −0.5 j ∴| Γ|=0.707 ZZR + 0 1|−Γ| ZZ==8.58Ω min 0 1|+Γ| ZZtransformer =×min Z0 =20.71Ω b. The distance of the quarter-wave transformer from the load. We have a voltage minimum whenever we have the following condition: − jd4π dn3 Γ=ee−−2/jdβπ−| Γ| ⇒j 4×eλ =−1 ⇒=+ λ 16 2 Thus d = 3λ/16 c. The time-average power that gets dissipated at the load. As a result of the presence of the transformer, the impedance seen by the source is simply 50Ω, and no load reflections propagate back to the source. Thus, the power dissipated by the load is simply the power accepted by a 50Ω load connected in series with the source and its impedance of 50 + j50Ω. ZV50 VV==IN 100 =40 −j20V and I=G =0.8 −j0.4A IN ZZ++G 100 j50 IN Z+Z IN G IN G 11 PV=×Re{ I**} =Re{(40 −j20)×(0.8 −j0.4) } =20W IN 22IN IN ECE 450: Spring 2009 Homework Set 3 Solutions Due on Friday, Feb 6, 2009 3. (You must hand in a Smith Chart showing all work for this problem) In the network shown below the VSWR on Line 1 is 1. Both lines are lossless. The characteristic impedance of Line 1 is 100 Ohm, while the characteristic impedance of Line 2 is 50 Ohm. Use the Smith Chart to calculate the unknown load impedance. Since the VSWR on Line 1 is 1, we know that the input impedance, Zin = 100Ω. Transmission Line 2 has Z0 = 50Ω, hence the normalized input impedance, zin = Zin/Z0=2. yin = 1/zin=0.5, also Yb=-j0.04 Ω, thus yb = Yb/Y0 = Yb*Z0 = -j2 Thus, yin - yb = 0.5 + j2 The wavelength on the Smith Chart at that point is 0.321λ. Moving 0.211λ towards the load brings us to a wavelength of 0.532 λ. = 0.032 λ This corresponds to yR = 0.11 - j0.19, and zR = 2.2 + j4 Thus the load impedance is 110 + j200 Ω. 4. (You must hand in a Smith Chart showing all work for this problem) A loss-less transmission line with characteristic impedance Z0 = 75 Ohm is terminated by a load ZR = 150 - j 37.5 Ohm. Use the Smith Chart to calculate: a. The VSWR on the line; If ZR = 150 - j37.5Ω, zR = ZR/Z0 = 2-j0.5; From the Smith Chart, we obtain VSWR = 2.2 b. The distance in wavelengths of the first current maximum from the load; Moving clockwise, toward the generator, d = 0.225λ c. The distance in wavelengths of the first voltage maximum from the load; Moving clockwise, toward the generator, d = 0.475λ d. The input impedance of a segment of line of length 2.5 m operating at a frequency of 800 MHz. Assume that the phase velocity on the line is 2.0 x 108 m/s. ECE 450: Spring 2009 Homework Set 3 Solutions Due on Friday, Feb 6, 2009 8 6 λ = vp/f = 2.0 x 10 /800 x 10 = 0.25m. Thus the line is 10λ in length, and its input impedance is the same as the load impedance = 150 - j37.5Ω ECE 450: Spring 2009 Homework Set 3 Solutions Due on Friday, Feb 6, 2009 5.
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