TITLE: Waveguide bench experiment
OBJECTIVE:
• Familiarization with the waveguide bench.
• Measuring standing wave distribution on a slotted line with short-circuit terminations.
• Measurement of guide wavelength λg and the frequency f.
• Plot of dispersion diagram (ω-β plot).
• Identification of type of load with help of smith chart.
APPRATUS USED:
• DC Power supply
• Gunn diode
• Waveguide bench
• VSWR meter
• Load
THEORY:
The distance between successive minima of a standing wave pattern determines the half
wavelength λg/2 on the line.
Fig.1 Standing wave distribution inside a waveguide with short-circuit termination
For the TE10 mode, the guide wavelength λg is related to the free space wavelength λ0 by:
where λ0 is given by,
𝑐
and 𝜔 = 2𝜋𝑓𝑜
𝜆𝑜 = 𝑓𝑜
where a is the larger internal dimension of the guide and c is the velocity of light in free
space. For a lossless line, the phase constant β is
2𝜋
𝛽=
𝜆𝑔
For finding type of load, we use smith chart. If you are moving towards load, we have to move
anticlockwise direction and if we are moving towards source we have to move clockwise
direction. When going towards load if first Vmin comes then the load is inductive and if Vmax
comes first then load is capacitive.
Fig.2 Smith chart
BLOCK DIAGRAM:
Fig.3 Block Diagram
Fig. 4 Experimental setup
OBSERVATIONS:
S.No. fo (λg)obs. a β/k
1. f1 = 9.89 GHz λg1 =4.1 cm a1 = 2.2 cm β1/k1 = 0.739
2. f2 = 9.995 GHz λg2 = 4cm a2 = 2.26 cm β2/k2 = 0.751
3. f3 = 10.065 GHz λg3 = 3.9cm a3 = 2.309 cm β3/k3 = 0.764
By using smith chart, we find that:
S.No. fo Type of load
1. f1 = 9.89 GHz Capacitive
2. f2 = 9.995 GHz Capacitive
3. f3 = 10.065 GHz Capacitive
CALCULATIONS AND FORMULA USED:
1) For f1 = 9.89 GHz:
λo1 = c/f1 = 3.03 cm
Consecutive minima at x=10.9, 13, 15,17
Average λg1 = 4.1 cm;
By using the above formula: a=2.2 cm
2) For f1 = 9.995 GHz:
λo2 = c/f2 = 3.001 cm
Consecutive minima at x=9.8,
11.3,13.7,15.8,17.8
Average λg2 = 4 cm;
By using the above formula: a=2.26 cm
3) For f1 = 10.065 GHz:
λo3 = c/f3 = 2.98 cm
Consecutive minima at x=9.4, 11.3,
13.1,15.2,17.2
Average λg3 = 3.9 cm;
By using the above formula: a=2.309 cm
CONCLUSION:
Thus, we successfully formed the dispersion diagram for the frequencies taken as well as we
learnt to find the nature of load with the help of smith chart.