Problem Set 2 Solutions

 

3.8

 

In this question, we basically use the idea of the probe stub to allow us to design a multidrop bus. A bus must communicate from the bus to the driver, and also from the driver to the bus. In order to do this, we need to terminate the stub at the driver itself, which is why the book refers to using a source-terminated driver.

 

 

There are three different questions that are to be asked. How much voltage appears on the bus created from a signal driven by driver to the bus? How much voltage appears at the driver given a voltage on the bus? How much of this signal is attenuated by the stubs while propagating along the bus?

 

 

 

 

 

 

Lets first look at direction of a signal traveling from the bus to the driver/receiver. Lets first assume that R_at = inf. (infinity). In this situation, R1 = 50 Ohm||(inf.) = 50 Ohm. So, we see no reflection and the initial pulse V1(t) just propagates down the bus, without any power dissipation in the probe stub. However, if R_at = 950 Ohm, then R1 = (950+50)||50 = 47.6 Ohm. Because R1 is not 50 Ohm, we know that there will be a reflection here. Kr = (R1 Z0)/(R1+Z0) = -.024. So, there is a reflected wave of Kr*V1. V2 (voltage that propagates down the line) is (1-Kr)*V1=0.976*V1. So, decreasing R_at increases the amount of attenuation that we see for a signal traveling along the bus.

 

What happens to the signal that now travels through the probe stub? Well, V3 = V2*50/(R_at + 50). So, for R_at = 950 Ohm, V2 ~= 1/20. Hence, the 20:1 attenuation. Here we see that depending on the value of R_at, our attenuation from the one end of the bus to the other end of the bus is inversely proportional to the attenuation from the bus to the driver/receiver stub.

 

Attenuation to the driver/receiver = (R_at + 50)/((1+Kr)*50)

 

Attenuation of bus = V1/(1+Kr)*V1, where Kr = (R1-50)/(R1+50), and R1 = (R_at+50)||50

 

 

 

We also need to determine how much voltage we see on the bus when we apply a voltage V4 from the transceiver. This is akin to a driver driving a value onto the bus. Basically, what is the value of V6 when V4 is applied?

 

Well, V5=V4/2 Rin = R_at + 25 Kr1 = (Rin-50)/(Rin+50). Vi = (1+Kr1)*V5

 

V6 = (50||50)*Vi/(50||50 + R_at) = (50||50)*(1+Kr1)*(V4/2)/(50||50 + R_at)

 

So, by increasing the value of R_at, we also are degrading our signal that is traveling from the driver to the bus.

 

 

 

 

 

3.11

 

a)

 

 

 

Impedance of the circuit for part (a). This graph makes sense at low frequencies the capacitor is an open circuit (so the impedance is high), while it becomes short at high frequencies (so that the total impedance approaches 50 Ohm).

 

 

Waveform received at the right side of the circuit (top) and reflected from the left side (bottom) for rise time of 100 ps. Note the relatively long time it takes for the voltages to reach steady state (explained by our relatively large capacitance).

 

 

 

 

 

 

Similar waveforms but with a rise time of 1 ns. As we would expect, theres pretty much no visible difference from the previous graph.

 

 

b)

 

 

 

The impedance of circuit in part (b). As we would expect, the inductor acts as a short circuit at low frequencies, but as an open circuit at high frequencies.

 

 

 

The voltage at the right side of the circuit (top) and the reflection from the left side of the circuit (bottom) for tr=100ps. Since the inductor acts a an open circuit, the reflected voltage shoots up initially, but eventually goes back down to 0 (when the inductor starts acting as a short circuit).

 

 

 

 

Same two waveforms for tr=1ns. As we would expect, voltage overshoot is much less here (since the incoming edge is less sharp, it generally contains lower frequency components than the 100 ps case).

 

 

c)

 

The impedance of the circuit is the same as in part (b). The waveform reflected from the left side of the circuit is also the same as in part (b) (see bottom graph).

 

 

The top graph is the waveform at the right side of the circuit. In the beginning, when the inductor acts as an open circuit, it does not let all the voltage through hence the voltage at the right is less than one. Eventually, the inductor becomes a short circuit and the voltage goes to 1.

 

 

 

Same two waveforms for tr=1ns. The effect of the inductor is much less prominent.

 

d)

 

Impedance: I have not been able to obtain a satisfactory impedance graph (apparently, Spice does not like AC analysis of transmission lines too much). However, we can predict what the impedance will look like. The impedance of the 200 ps transmission line is going to be purely reactive (for f>0). For some frequencies, it will look like a capacitor, and for some it will look like an inductor (and for others, it will be a combination of the two). Thus, the total impedance of the circuit will oscillate with frequency.

 

 

For tr=100ps:

 

 

 

(Voltage on the right side (end of the 200 ps transmission line) on the top, voltage reflected from the left side of the circuit on the bottom). Initially, the incoming wave sees the impedance of 50||50 = 25 Ohm hence the negative reflection coefficient. However, eventually the reflections from the open-circuited end of the 200 ps line bring the overall reflection back to 0.

 

 

Heres the same waveforms for tr=1ns. As we can see, there is much less ringing in the circuit response.

 

 

 

3.16

 

Here is what our circuit looks like:

 

 

 

The lengths of the transmission line segments can be found from the lengths of the flat regions of the TDR waveform. The first spike is from the capacitor (points down), while the second spike is from the inductor (points up). The 50 Ohm termination is evident from the fact that as time approaches infinity, voltage on the line approaches 1 (thus, the end of the line is matched). The 40 Ohm impedances of the 2nd and 3rd segments are responsible for the waveform not going all the way back to 1 V after encountering the capacitor. The rough values of the capacitor and the inductor can be computed using the formula from the book: deltaV/V = (tau/tr)*(1-exp(-tr/tau)), or, more simply, tau = tr*deltaV/V (for tau << tr). The computed value for the inductor turns out to be not very accurate, since the preceding capacitor reflects all the high-frequency components and degrades the tr; however, it could be used as a starting value to plug into Spice and adjust as needed.