EEC353 Practice Problem Set #6. The value of load impedance mut be found by meaurement, at 200 MHz. An engineer meaure the tanding-wave pattern a hown in the figure above. The figure how the amplitude of the voltage on the tranmiion line in volt RMS a a function of ditance along the tranmiion line in millimetre. Marker # i poitioned at a voltage maximum, of 6.370 volt at z303.423 mm. Marker #2 i poitioned at a voltage minimum, of 3.633 volt at z675.727 mm. The characteritic reitance of the tranmiion line i Rc 50 ohm and the peed of propagation i u 300 meter per microecond. Define Rc Γ. + Rc (i) What i the value of the tanding-wave ratio on the tranmiion line (ii) What i the magnitude of the reflection coefficient Γ (iii) What i the angle of the reflection coefficient Γ (iv) What i the value of the load impedance 2. The antenna in problem # of aignment 5 mut radiate a much power a poible at 2450 MHz, and the bandwidth mut be a wide a poible. The generator i modified o that the internal reitance i R 50 ohm, and it ha an open-circuit voltage V 0 volt RMS. The antenna input impedance i 20 j56 ohm. To increae the power radiated by the antenna, a matching circuit i ued. The matching circuit conit of a length t of tranmiion line of characteritic reitance the load R ct in erie with of, and a tuning tub connected in parallel with the load. The tuning tub i a length tranmiion line of characteritic reitance R c, terminated with a hort circuit. The line of length t i
a quarter-wave tranformer terminated with the tub/load parallel combination. The 8.7-cm line and the tuning tub have characteritic reitance R c 50 ohm. All three tranmiion line have peed of travel u 4 cm/n. (i) Chooe the length of the tub o that the admittance Y BB Y + of the tub admittancey in parallel with the load admittance i real, with zero imaginary part. The input admittance of the tub i Y j cot β. Rc (ii) The parallel combination of the tub impedance and the antenna impedance i BB RBB +j0. What i the value of reitance R BB (iii) Deign the quarter-wave tranformer by chooing the characteritic reitance to be R ct RcRBB, and the length t to be a quarter-wavelength at 2450 MHz. (iv) Find the power delivered to the antenna at 2450 MHz. (v) Ue TRINE to find the bandwidth for a return lo of 20 db or better. (Hint: Ue the power plitter circuit template. Make the length of line #4 equal to zero, and make load 2. ine #3 i the tub and 0 i the hort circuit.) 3. Your bo i not atified with the deign in problem #2 becaue the bandwidth i not wide enough. The bo think that all engineer are the ame o he intruct you, a computer engineer, to deign a ingletub matching circuit. A before, the load i 20 j56 ohm. The generator ha open-circuit voltage V 0 volt RMS and internal reitance R 50 ohm, at 2450 MHz. The cable length i 8.7 cm. All the cable have charcteriteric reitance R 50 ohm and peed of travel u 4 cm/n. (i) Chooe o that the input admittance Y G + jb of the line of length and the load ha real part G 20 ms. Note that there are two poible value for. 50 (ii) Chooe the tub length uch that the input admittance Y in of the tub in parallel with Y i equal to 20 ms. For each value of there i a correponding value of. c
(iii) (iv) Ue TRINE to find the bandwidth of each of the two deign, for a return lo of 20 db or better. Which ha a wider bandwidth Doe the ingle-tub deign have a wider bandwidth than the tub-plu-quarter-wavetranformer deign
Solution to Practice Problem Set #6. The value of load impedance mut be found by meaurement, at 200 MHz. An engineer meaure the tanding-wave pattern a hown in the figure above. The figure how the amplitude of the voltage on the tranmiion line in volt RMS a a function of ditance along the tranmiion line in millimetre. Marker # i poitioned at a voltage maximum, of 6.370 volt at z303.423 mm. Marker #2 i poitioned at a voltage minimum, of 3.633 volt at z675.727 mm. The characteritic reitance of the tranmiion line i Rc 50 ohm and the peed of propagation i u 300 meter per microecond. Define Rc Γ. + Rc (v) What i the value of the tanding-wave ratio on the tranmiion line (vi) What i the magnitude of the reflection coefficient Γ (vii) What i the angle of the reflection coefficient Γ (viii) What i the value of the load impedance Solution (Thank to Armin Para) Solution : (i) V VSWR V (ii) Γ min VSWR VSWR + 6.37 3.633 max 0.27362 (iii) u 300 λ.5 m f 200.75337
min 0.6757 φ.720 0.25.720 24. 336 λ 4.5 (iv) Γ Γ 0.27362 24. 336 φ + Γ Γ o + 0.27362 24.336 50 0.27362 24.336 80.273 - j9.5670 ohm 2. The antenna in problem # of aignment 5 mut radiate a much power a poible at 2450 MHz, and the bandwidth mut be a wide a poible. The generator i repaired o that the internal reitance i R 50 ohm, and it ha an open-circuit voltage V 0 volt RMS. The antenna input impedance i 20 j56 ohm. To increae the power radiated by the antenna, an impedance-matching circuit i inerted at the antenna terminal. The circuit conit of a tuning tub connected in parallel with, and a quarter-wave tranformer between the 8.7 cm tranmiion line and the tub/load parallel combination. The 8.7 cm line and the tuning tub have characteritic reitance R c 50 ohm. All three tranmiion line have peed of travel u 4 cm/n. (vi) Chooe the length of the tub o that the admittance Y BB Y + of the tub admittancey in parallel with the load admittance i real, with zero imaginary part. The input admittance of the tub i Y j cot β. Rc (vii) The parallel combination of the tub impedance and the antenna impedance i BB RBB +j0. What i the value of reitance R BB
(viii) Deign the quarter-wave tranformer by chooing the characteritic reitance to be R ct RcRBB, and the length t to be a quarter-wavelength at 2450 MHz. (ix) Find the power delivered to the antenna at 2450 MHz. (x) Ue TRINE to find the bandwidth for a return lo of 20 db or better. Solution (thank to Guilin Sun) Known: V 0 volt RMS, R 47 ohm, f2450 MHz, 8.7 cm, u 4 cm/n40m/u, Rc 50 ohm, 20 j56 ohm The following formulae will be ued: Y j R ct Rc RBB Rc tan β the following parameter will be ued: 7 u 4 *0 4 λ 0.05743m5.743cm f 6 2450 *0 245 olve: i) the admittance of the load i 20 + j56 3 3 G + jb 4.4457 *0 + j.855* 0 () 2 2 20 j56 20 + 56 the admittance at BB i YBB Y + G + jb we want to have a real admittance at BB, o it imaginary part i zero, that i Y S + jb 0 therefore we have Y j jb Rc tan β 3 β tan (/ B * R c ) tan (/(.855*0 *50)) 86.6 the length of tuning tub i 86.6 86.6 400 λ *.3748 cm 360 360 245 the admittance of the deigned tuning tub i about 3 Y.840*0 j j () Rc tan β 50 tan(360 *.3748/ 5.743) which i quite cloe to.855(m). ii) obviouly the reitance at BB i 000 R BB 224.94 ohm G 4.4457 iii) the characteritic reitance of the quarter-wave tranformer i R ct Rc RBB 50 * 224.94 06.05 ohm 5.743 the length i quarter wavelength, that i t λ. 4286 cm 4 4 iv) ince the T i lole and now the load i matched, o 2 2 V V P av I * V 25 / 50 0.5 Watt. Rc 2 Rc Note the problem give u the RMS voltage, there i no need to divide the factor of 2.
Compared to the problem # of aignment 5, the radiated power i much higher : without match, it i 0.3W. With a quarter-wave tranformer and tuning tub, the power i increaed to 0.5 W, an increae of about.7x. The anwer: i) ength of the tuning tub i.3748cm ii) R i 224.94 ohm iii) the characteritic impedance i 06.05 ohm iv) the power delivered i 500 mw. v) The bandwidth i 04.7MHz (ee below) Note that if the calculation i approximate with limited data digit, the real T may not be matched very well. For example, there maybe a reidual uceptance at BB, and the length of the quarter-wave tranformer i not exactly quarter wavelength. Then your reult may be not a good a thi reult. In addition, the T length may not be controlled at the micrometer preciion a we did in calculation in reality, it may introduce ome error. v) TRINE imulation. The tranmiion line, pleae chooe Two tranmiion line in erie, with hunt load at the main manu and input the parameter.
2. Chooe Find the voltage, current and power to get the power. To check our calculation, we can view all the parameter at the load, at the T #2 (tuning tub),#(the quarter-wave tranformer). Firt, view the load by clicking the load # we can ee that the power i exactly 500 mw. Second, view the tuning tub by click line #2
ince there i no reitance for load #2, there hould be no power conumption. The power diplayed above i negative and i due to computation error. Third, view the tranformer by click ine# Note the power i the ame a in the load, which hould be true ince it i lole. You can ee that the parallel impedance at BB i 224.93 ohm, which agree with our calculation. Alo it ha a reidual reactance of 0.02 ohm at it input (ource end). Finally, we can ee the Smith chart to check the matching.
the matching i perfect! The bandwidth can be found by clicking the plot a parameter a a function of frequency in the main manu, then chooe return o and ue port #, pecify frequency range from 2300 to 2600 MHz, and calculate it by click calculate the frequency repone and graph the reult. With the help of nap manu and et the nap value 20dB, you can mark the curve and get the bandwidth. It how the bandwidth i 04.7MHz. 3.
Your bo i not atified with the deign in problem #2 becaue the bandwidth i not wide enough. The bo think that all engineer are the ame o he intruct you, a computer engineer, to deign a ingletub matching circuit. A before, the load i 20 j56 ohm. The generator ha open-circuit voltage V 0 volt RMS and internal reitance R 50 ohm, at 2450 MHz. The cable length i 8.7 cm. All the cable have charcteriteric reitance R 50 ohm and peed of travel u 4 cm/n. (v) Chooe o that the input admittance Y G + jb of the line of length and the load ha real part G 20 ms. Note that there are two poible value for. 50 (vi) Chooe the tub length uch that the input admittance Y in of the tub in parallel with Y i equal to 20 ms. For each value of there i a correponding value of. (vii) Ue TRINE to find the bandwidth of each of the two deign, for a return lo of 20 db or better. Which ha a wider bandwidth (viii) Doe the ingle-tub deign have a wider bandwidth than the tub-plu-quarter-wavetranformer deign c Solution (Thank to Ibrahem Abdalla) λ u / f β 2π / λ Γ Γ 0.64 φ 7.4 R + R o 4/ 2.45 C C 5.7 360/ 2*0.057 cm 0.64 7.4 6300 o deg/ m i- To find the length of the line :
θ co( Γ o θ 29.8, θ 230.2 φ θ 2β.09 cm ) co( 0.64) 2 2.884 cm Compenating for the ve ign by adding λ/2: λ / 2 5.7/ 2 2.855cm.09 + 2.855.77cm.884 + 2.855 0.97cm 2 ii- To find the length of the tub : tan β β β S S 2 S 0.492cm S 2 S ± 30.98 49.02 2.37cm Γ 2 Γ o o ± 0.6 cae (cm) Y S(cm) YS BW(MHz).77 20.2-j33.22 2.370 j33.0 59.66 2 0.97 9.98+j33.07 0.492 -j33.0 6.882
The mith chart for Smith Chart for Return o The mith chart for 2
The mith chart for 2 Retrun lo