1uH C2. 1uH C3. 390pF L8. 1uH C6. 1uH C8. 1uH C10. 1uH C7. 1uH C9. 390pF L pF L pF L pF L pF L12. 1uH C12.

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Laurence Andrews
6 years ago
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1 Lumped Model of Transmission Line In the Instrumentation Studio, we use both a coil of coax (around 8-1 meters long) and an equivalent lumped model of the coil of coax to study transmission lines. The lumped circuit is configured in the following manner: R1 L1 L2 L3 L4 L5 V1 5 C1 C2 C3 C4 C5 L1 C1 L9 C9 L8 C8 L7 C7 L6 C6 L11 L12 L13 L14 L15 C11 C12 C13 C14 C15 L2 L19 L18 L17 L16 R2 93 C2 C19 C18 C17 C16 The physical circuit has one BNC connector at the input end and a BNC connector and two banana plug connector posts at the output end. This circuit can be modeled using PSpice to determine its frequency response. The input voltage as a function of frequency looks like: 1.V.8V Lumped Transmission Line.6V.4V.2V Hz 5MHz 1MHz 15MHz V(R1:2) where the frequency has been plotted on a linear scale to show the details better. The input voltage can also easily be calculated and plotted using ideal formulas and Matlab. In this case the frequency response is more regular:

2 .7 Ideal Transmission Line Volts x 1 6 Note that the response is similar at lower frequencies, but begins to diverge significantly at higher frequencies. Also, no resistance has been included in either model. These two frequency plots are more conventionally plotted using a semilog scale and are: 8mV Lumped Transmission Line 6mV 4mV 2mV 1.KHz 1KHz 1KHz 1.MHz 1MHz 1MHz V(R1:2) and

3 .7 Ideal Transmission Line Volts respectively In Project 2 (Spring 2) we use the lumped line as the line we short and connect as a stub into the line that runs from the function generator to the matched load at the scope. It turns out that this line does not work the way we would like it to work because of the way it is constructed. A simple simulation using PSpice shows why this is the case. First, we can simulate ideal, lossless transmission lines with some selected length in the following configuration: R1 T1 T2 R2 V1 5 5 T3 With a voltage source frequency selected to be 1MHz, the voltages at the input and output look like

4 5mV V -5mV s.2us.4us.6us.8us 1.us V(R1:2) V(T2:B+) where the smaller voltage is the output signal. Disregard the variations at the beginning, since the signal does not reach steady-state until about.2us. Assume that we decide to use two sections of the lumped line to represent an 8 meter cable. We have a circuit that looks like R4 T4 T5 R3 V2 5 5 L1 C1 39uF L2 C2 39uF Note that the short at the end of the lumped line shorts out the capacitor and thus the lumped line is missing a component. The lumped representation for each section of the line must contain both an inductor and a capacitor, but the second section here contains only an inductor. Thus, the line looks like a 4 meter long transmission line with an inductor as a load rather than an 8 meter long cable with a short at the end. The voltages observed at the input and output do not look like the real line. Rather they look like

5 8mV 4mV V -4mV -8mV s.2us.4us.6us.8us 1.us V(R4:2) V(T5:B+) We can fix this problem by constructing our lumped transmission line a little differently. Each section should be represented by a TEE configuration with an inductor on either side of a capacitor so that the 8 meter long line now looks like R5 T6 T7 R6 V3 5 5 L3 C3.5uH 39uF L4 C4 39uF L5.5uH Notice that the short circuit now does not eliminate any of the components. Unfortunately, there is no good way to build a lumped model of a transmission line that works exactly right for both a short and an open circuit load. With our new model, the last inductor would be ignored if the load was an open circuit. For the new model the voltages at the input and output now look much more like the ideal line case. The voltages are still not exactly the same.

6 8mV 4mV V -4mV -8mV s.2us.4us.6us.8us 1.us V(R5:2) V(R6:1) To make the comparison easier to see, we can plot just the input and the output in the three cases and note that the output signals are the most similar for the first and third cases: 8mV 4mV V -4mV -8mV s.2us.4us.6us.8us 1.us V(R5:2) V(R1:2) V(R4:2) 5mV V -5mV s.2us.4us.6us.8us 1.us V(T2:B+) V(T5:B+) V(R6:1) However, the input signals are the most similar for the second and third cases. This is probably due to the inadequacies of a lumped transmission line model with only two

7 sections in it. It is unrealistic to assume that only two sections can represent the full transmission line with any real accuracy. The frequency of 1 MHz was chosen arbitrarily. Let us select a frequency that should be blocked. For blocking we need a frequency for which the stub is a half wavelength long. If we assume a propagation speed 2/3 of the speed of light, then 8 meters of cable will be a half wavelength at 8 u u 2 1 f = = = = 12. 5MHz. The output voltages for this frequency show that λ 2d 2 8 neither lumped line works perfectly, but that the last model is a little better. Note that the ideal line does exactly what it is supposed to do..5v -.V -.5V -1.V s.2us.4us.6us.8us 1.us V(T2:B+) V(R6:1) V(T5:B+) When using the lumped line, most groups found that the frequency for blocking was a bit lower than the ideal model would predict. Let us try such a lower frequency.

Fields and Waves I Spring 2008 Homework 1

Fields and Waves I Spring 2008 Homework 1 Fields and Waves I Spring 28 Due 23 January 28 at : pm Some of the solution is found in the text below, some is attached at the end. 1. Waves and Phasor Notation Be sure that you read the following questions