EECS 117 Homework Assignment 3 Spring ω ω. ω ω. ω ω. Using the values of the inductance and capacitance, the length of 2 cm corresponds 1.5π.
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1 EES 7 Homework Assignment Sprg 4. Suppose the resonant frequency is equa to ( -.5. The oad impedance is If, is equa to ( ( The ast equaity hods because ( -.5. Furthermore, ( Usg the vaues of the ductance and capacitance, the ength of cm corresponds s m cm v v In genera, ( ( v. Thus is ( ( ( ( tan(.5 ( tan(.5 ( tan(.5 ( tan(.5 The approximation hods because (.5 tan( >>,,. This expression has the same form as a parae circuit, with eq, eq, eq
2 Therefore, the put impedance is that of a second order circuit. Aso, eq eq -, so our assumption is correct, i.e., ( -.5. The Q factor is 94 Q eq eq The equivaent circuit is eq 5 µf, eq.64 fh, and eq 5 parae.. For a ossy e, the put impedance has a form tanh( tanh( where. Aso, cosh( cos( s( cosh( cos( sh( tanh( *( cos( s( ( s( s( *( s( cos( cos(. ikewise,. So, cosh( cosh( sh( tanh( The put impedance is then equa to ( sh( (cosh( sh( (cosh( cosh( cosh( sh( cosh( sh( cosh( cosh( cosh( sh( cosh( cosh( sh( The third term the denomator has dependence and is thus negigibe.
3
4 4. This probem is simiar to the exampe -6 the textbook. For matchg of the oad with either a short or an open shunt stub, we have the foowg circuit: tot, A M Where M is either or fity for a short or an open stub, respectivey. Sce we are deag with shunt stub, admittance woud simpify the cacuation. The equivaent admittance at pot A (excudg the stub for a moment is given by tan( tan( where [ ] - 5, and [ ] -. Separatg the expression to the rea and imagary parts gives where x tan(. x x x 5x ( 5x 5x ( 5x Sce a short or an open stub can behave as a purey reactive eement, the rea part of above shoud be equa to order for the matchg to take pace. Thus, we have, x ( 5x 5x 5
5 5 5 The soution of this equation is: tan( or. Both of them are 4 4 negative. This means that is arger than. The more negative vaue represents a pot coser to the oad and this vaue wi be used the foowg cacuation. tan( Substitute this vaue to the imagary part of and get x x ( 5x 5x x The stub needs to have an impedance opposite to the vaue above order to cance the reactive part of the for matchg. For a short stub, cot( '. Therefore, we want.447 cot( '.454. The put impedance at pot A cudg the stub is tot, ( 5 tan(.5 tan tan (.5 5tan (.5 (.5 tan(.5 ( 5 tan(.5 5tan (.5 tan(.454 The imagary part is equa to zero. The SW is given by S In genera, for, i.e., *n, where n is the ratio of to, the put impedance shown above is
6 tot, ( n ( 5 tan(.5n tan tan (.5n 5tan (.5n (.5n tan(.5n ( 5 tan(.5n 5tan (.5n tan(.454* n and the SW is equa to S ( n ( n ( n ( n SW can be potted as a function of n, representg the amount of shift the signa frequency phase constant from the ones used cacuatg the numbers above..4 S The vaues of n where S. are.989 and., for a short stub. Foowg the same procedure, we have these for an open stub n.447 tan( ' tan( '.99.
7 tot, ( n ( 5 tan(.5n tan tan (.5n 5tan (.5n (.5n tan(.5n tan(.99* n ( 5 tan(.5n 5tan (.5n The pot of S vs. n:.7 S The vaues of n where S. are.99 and.. For impedance matchg with a umped eement, we have the foowg circuit: n M To keep the probem simpe, et s make the umped eement M a purey reactive component. With this constrat, the rea part of needs to match the characteristic impedance of the transmission e, ust ike the cases of short and open stubs above. The
8 previous cacuation gives.5 or Just as before, we pick the shortest distance. 5. At this ength, is equa to The positive imagary part impies that the umped eement needs to be an ductor order to make the impedance matchg work. So, M M ( ( v where v is the propagation veocity of the transmission e. The veocity is property of the transmission e, and does not depend on the signa frequency. In genera, tot, is tot, ( n ( 5 tan(.5n tan (.5n 5tan (.5n tan (.5n tan(.5n.4474 ( 5 tan(.5n 5tan (.5n n The pot of S vs. n:.4 S The vaues of n where S. are.989 and.. n
9 In summary, Bandwidth for S. Short.989 to. Open.99 to. umped.989 to.
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