Microwave Engineering Assignment-114235


1)Isitpossibletodesignalosslessnetworkwhichmatchesanarbitrary complexloadperfectlyoveragivenfinitebandwidth?Whyorwhynot? (5pts)

2)Whyis 50 Ω chosen as astandard for the characteristicimpedance in mostmicrowave systems?(5pts)

3) Compare thepros and cons ofthefollowingthreeplanar transmissionlines, i.e. microstrip,stripline, and CPW. (10pts)

Analytical Questions[80 pts]:

4) Considerthe two-portSparameter network shown below:

Port1      1.13Ω             0.678Ω               Port2


790Ω             2210Ω

a)Derive [S]ofthe resistive networkat 60 GHz. (25pts)b)Isthenetwork lossless? Please justify.(5pts)

c)Isthenetwork reciprocal? Please justify.(5pts)

d)Calculate the power lossfrom i)port1to port2andii)port2toport1. (Zo= 50 Ω) (10pts)

5) Considerthe wirelessreceiver front-end shownbelow:

MatchingZant          Network








Γopt=0.75       -147°

a)Pleasedeviseasingle-stubmatchingnetworktoachievemaximumpowertransferusingopenshuntstub(pleaseshowyourschematicand leave allunits interms of wavelength).UseZY-Chartonly!(20pts)

b)Whatisthetypeofcomponentassociatedwiththeopenshuntstubbased on your matching network?(5pts)

c)Whatisthevalueofthecomponentthatyouidentifiedin(b)at5GHz (assume striplineimplementationwith ɛr =4)?(10pts)

Microwave engineering

Table of Contents

Answer 1). 2

Answer 2). 2

Answer 3). 2

Answer 4). 3

Answer 5). 5

References. 8

Answer 1)

The loss less network matches the arbitrary load complex perfectly over a finite bandwidth.The loss less reciprocal circuit prototype allows designing of a complex filter that includes improvement in the first phase. The design of loss less network combines frequency technique over finite bandwidth with appropriate replacement techniques and application of the network decomposition(Garg, 2010).

Answer 2)

50 Ohms is used as it is the desired impedance of the respective co-axial cable, except the audiovisual applications which are of 75 Ohms. Therefore, people avoid using the circuit of impedance matching for connecting PCB to the exterior section.Thus, 50 Ohms is chosen to be the standard within the microwave systems(Hibbeler, 2010).

Answer 3)

Micro strip is known to be the most accepted transmission lines of PCB. The easy fabrication, as well as the connection, is little bit dispersive and difficult to analyse.

Strip line is much more enhanced than micro strip,and this is not dispersive and this can be easily analysed. The fabrication and the connection have become more difficult(Garg, 2010).

Coplanar waveguide

The coplanar waveguide is the advanced technology, which can be used to connect as well as fabricate the different components on both the ground as well as the conductor are located on one side of the board.

Answer 4)



= power reflected from network input (port 1) /power incident on input (port 2)

= 9.39475


= power reflected from network output (port 2)/power incident on output (port 2)

= 9.39475


=transmitted power from network output (port 2)/power incident on output (port 1)


= transmitted power from network output (port 1)/power incident on output (port 2)


The network is loss less just because the network cannot absorb more power. Loss less network means that one port has to be accounted by adding the output power at other port with power reflected at the incident port. The S-parameter matrix of the lossless network is the sum of squares of magnitudes(Garg, 2010). The resistive network at 60 GHz is 593.608 Ω.


Linear and passive microwave components would turn to be reciprocal. The reciprocity is a natural effect of linear material such as conductors and dielectrics. The network is reciprocal that greatly simplifies an admittance or impedance matrix.



Power Loss = Volt*Current

                     = V*I                       (V = IR)

                     = IR*I

                     = I2R

                     = I21.808Ω


Power Loss = Volt*Current

                     = V*I                       (V = IR)

                     = IR*I

                     = I2R

                     = I2593.608Ω

Answer 5)


Period Z Y
0 0 0
1 2 0.5
2 4 1
3 6 1.5
4 8 2
5 10 2.5
6 12 3
7 14 3.5
8 16 4
9 18 4.5
10 20 5




The components associated with open shunt tub based on the matching network are LNA, receiver, network to a receiver.


V = IR

    = 0.75*50

     = 37.5V


Garg, M. (2010). Network aware energy-efficient communication (NAEEC) protocol for heterogeneous wireless sensor networks.

Hibbeler, R. (2010). Engineering mechanics. Upper Saddle River, NJ: Prentice Hall.