By Professor Syed Idris Syed HassanSch of Elect. Electron .ppt

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1、By Professor Syed Idris Syed Hassan Sch of Elect. & Electron Eng Engineering Campus USM Nibong Tebal 14300 SPS Penang,Microwave Circuit Design,Introduction,10 weeks lecture + 4 weeks ADS simulation Assessments :8 tests + 2 ADS assignments + 1 final examination Class : 9.00- 10.30 lecture10.30-11.00

2、rest (tea break)11.00-12.30 lecture12.30- 1.00 test,Dates,06/04/02 Morning 20/04/02 Morning 27/04/02 Morning 04/05/02 Morning 11/05/02 Morning 18/05/02 Morning 25/05/02 Morning,08/06/02 Morning 15/06/02 Morning 22/06/02 Morning 29/06/02 morning 06/07/02 Morning 20/07/02 Morning 27/07/02 Morning,Syll

3、abus,Transmission lines Network parameters Matching techniques Power dividers and combiners Diode circuits Microwave amplifiers Oscillators Filters design Applications Miscellaneous,References,David M Pozar ,Microwave Engineering- 2nd Ed., John Wiley , 1998 E.H.Fooks & R.A.Zakarevicius, Microwave En

4、gineering using microstrip circuits, Prentice Hall,1989. G. D. Vendelin, A.M.Pavio &U.L.Rohde, Microwave circuit design-using linear and Nonlinear Techniques, John Wiley, 1990. W.H.Hayward, Introduction to Radio Frequency Design, Prentice Hall, 1982.,Transmission Line,Equivalent Circuit,R,R,L,L,C,G,

5、Lossy line,Lossless line,Analysis,From Kirchoff Voltage Law,Kirchoff current law,(a),(b),Analysis,Lets V=Voejwt , I = Ioejwt,Therefore,then,a,b,Differentiate with respect to z,Analysis,The solution of V and I can be written in the form of,where,Let say at z=0 , V=VL , I=IL and Z=ZL,Therefore,and,e,f

6、,c,d,and,Analysis,Solve simultaneous equations ( e ) and (f ),Inserting in equations ( c) and (d) we have,Analysis,But,and,Then, we have,and,*,*,Analysis,Or further reduce,or,For lossless transmission line , g= jb since a=0,Analysis,Standing Wave Ratio (SWR),node,antinode,Ae-gz,Begz,Reflection coeff

7、icient,Voltage and current in term of reflection coefficient,or,Analysis,For loss-less transmission line,g = jb,By substituting in * and * ,voltage and current amplitude are,Voltage at maximum and minimum points are,and,Therefore,For purely resistive load,g,h,Analysis,Other related equations,From eq

8、uations (g) and (h), we can find the max and min points,Maximum,Minimum,Important Transmission line equations,Zo,ZL,Zin,Various forms of Transmission Lines,Parallel wire cable,Where a = radius of conductor d = separation between conductors,Coaxial cable,Where a = radius of inner conductorb = radius

9、of outer conductorc = 3 x 108 m/s,a,b,Micro strip,w,he,er,t,t=thickness of conductor,Substrate,Conducted strip,Ground,Characteristic impedance of Microstrip line,Where,w=width of strip h=height and t=thickness,Microstrip width,For A1.52,For A1.52,Simple Calculation,Approximation only,Microstrip comp

10、onents,Capacitance Inductance Short/Open stub Open stub Transformer Resonator,Capacitance,Zo,Zo,Zoc,For,For,Inductance,Zo,Zo,ZoL,For,For,Short Stub,Zo,Z,Zo,Zo,ZL,Open stub,Zo,Z,Zo,Zo,ZL,Quarter-wave transformer,Zo,Zo,ZT,l/4,Zmx/min,ZL,x,q in radian,At maximum point,Quarter-wave transformer,q in radi

11、an,at minimum point,Resonator,Circular microstrip disk Circular ring Short-circuited l/2 lossy line Open-circuited l/2 lossy line Short-circuited l/4 lossy line,Circular disk/ring,a,feeding,a,* These components usually use for resonators,Short-circuited l/2 lossy line,=nl/2,Zin,Zo,a,b,where,= series

12、 RLC resonant cct,Open-circuited l/2 lossy line,=nl/2,Zin,Zo,a,b,= parallel RLC resonant cct,where,Short-circuited l/4 lossy line,=l/4,Zin,Zo,a,b,= parallel RLC resonant cct,where,Rectangular waveguide,a,b,Cut-off frequency of TE or TM mode,Conductor attenuation for TE10,Example,Given that a= 2.286c

13、m , b=1.016cm and s=5.8 x 107S/m. What are the mode and attenuation for 10GHz?,Using this equation to calculate cutoff frequency of each mode,Calculation,TE10,a=2.286mm, b=1.016mm, m=1 and n=0 ,thus we have,Similarly we can calculate for other modes,Example,TE10,TE20,TE01,TE11,6.562GHz,13.123GHz,14.

14、764GHz,16.156GHz,Frequency 10Ghz is propagating in TE10.mode since this frequency is below the 13.123GHz (TE20) and above 6.561GHz (TE10),continue,or,Evanescent mode,Mode that propagates below cutoff frequency of a wave guide is called evanescent mode,Wave propagation constant is,Where kc is referre

15、d to cutoff frequency, g is referred to propagation in waveguide and b is in space,g = a +jb,a=attenuation,b=phase constant,When f0 fc ,But,Since no propagation then,The wave guide become attenuator,Cylindrical waveguide,a,TE mode,Dominant mode is TE11,continue,a,TM mode,TM01 is preferable for long

16、haultransmission,Example,Find the cutoff wavelength of the first four modes of a circular waveguide of radius 1cm,Refer to tables,TE modes,TM modes,1st mode,2nd mode,3rd &4th modes,3rd &4th modes,Calculation,1st mode Pnm= 1.841, TE11,2nd mode Pnm= 2.405, TM01,1st mode Pnm= 3.832, TE01 and TM11,Stripline,w,b,Continue,On the other hand we can calculate the width of stripline for a given characteristic impedance,Continue,Where,t =thickness of the strip,