If abatteryandswitchwere connected to one end of a 10-mile long cable, and two oscilloscopes were used to record voltage at either end of the cable, how far apart in time would those two pulses be, assuming a propagation velocity equal to the speed of light (in other words, the cable has a速度因子equal to 1.0)?
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53.68 microseconds
学生应该意识到问题的措辞tion that the voltage signal probably does not arrive at the far end of the cable instantaneously after the switch is closed. Although the speed of light is very, very fast, it is not instant: there will be a measurable time delay.
If a battery and switch were connected to one end of a 10-mile long cable, and two oscilloscopes were used to record voltage at either end of the cable, how far apart in time would those two pulses be, assuming a propagation velocity equal to 69% the speed of light (in other words, the cable has a速度因子equal to 0.69)?
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77.80微秒
学生应该意识到问题的措辞tion that the voltage signal probably does not arrive at the far end of the cable instantaneously after the switch is closed. Although 69% of the speed of light is still very, very fast, it is not instant: there will be a measurable time delay.
What does the “50 ohm” rating of an RG-58/U coaxial cable represent? Explain how a simple cable, with no continuity between its two conductors, could possibly be rated inohms。
Hint: this “50 ohm” rating is commonly referred to as thecharacteristic impedance电缆。此参数的另一个术语是涌impedance，我认为这更具描述性。
A cable with acharacteristic, or涌, impedance of 50 ohms behaves as a 50-ohmresistorto any voltage surges impressed at either end, at least until the surge has had enough time to propagate down the cable’s full length and back again.
This concept will seem very strange to students who are only familiar with resistance in the context of resistors and other simple electrical components, where resistance does not change appreciably over time. In this example, though, the “resistance” of the cable is extremely time-dependent, and the time spans involved are typically very short - so short that measurements made with ohmmeters will not reveal it at all!
鉴于以下测试电路，示波器用于记录currentfrom the battery to the cable (measuring voltage dropped across a shunt resistor), what sort of waveform or pulse would the oscilloscope register after switch closure?
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The oscilloscope would register a square-edged pulse of voltage approximately equal to 480 μV, which of course corresponds to a current of approximately 480 mA:
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The pulse duration should range somewhere between 162.67 microseconds and 170.42 microseconds (based on two different figures I obtained for RG-58/U cable velocity factors).
Answering this question requires several steps, and the combining of multiple concepts. It should be apparent from the answer thatOhm’s Law(I =E/r）足以计算脉冲电流，但是答案中给出的时间延迟数字可能会使某些学生感到困惑。对于那些计算时间数字的学生是答案中给出的时间的一半，请鼓励他们思考为什么（错误）答案可能会减少50％。这样的2：1比率的存在意味着一个简单的概念误解。
For the RG-58/U cable velocity factor, I obtained two different figures: 0.63 and 0.66, which accounts for the two time delay answers given.
鉴于以下测试电路，示波器用于记录currentfrom the battery to the cable (measuringvoltage droppedacross a shunt resistor), what sort of waveform or pulse would the oscilloscope register after switch closure?
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The oscilloscope would register a continuous current of 480 mA any time the switch is closed.
Challenge your students to think of another electrical component (besides an RG-58/U cable of infinite length) that would behave like this, drawing 480 mA of current from a 24 volt source any time the switch is closed. Hint: you don’t have to think very hard!
Suppose this 10-mile-long RG-58/U cable were “terminated” by a resistor with a resistance equal to the cable’s owncharacteristic impedance:
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What sort of waveform or pulse would the oscilloscope register after switch closure?
The oscilloscope would register a continuous current of 480 mA any time the switch is closed.
问问你的学生比较的行为circuit with that of an unterminated RG-58/U cable. How does this circuit’s behavior differ? Why is that?
为了以不同的方式表达问题，终止电阻的包含对明显的作用是什么length电缆？换句话说，RG-58/U电缆的长度与此电路完全相同？
When a pulse propagates down an “unterminated” cable and reaches an open-circuit, what does it do? Does it simply vanish, or does it go some place else?
A voltage pulse, upon reaching the open end of a cable, will be “reflected” back in the direction from which it came, its polarity being maintained while the current moves in the opposite direction.
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反射脉冲到达源后，电源端子处将有最大电压，电缆中的电流为零。
To help answer this question, it is helpful to ask students howvoltage and current在开路条件下彼此相关（最大电压，零电流）。
当脉冲向下传播由短路终止的电缆时，它做了什么？它只是消失了，还是去其他地方？
到达电缆的短端时，电压脉冲将“反映”到其到达的方向上，其极性在朝着相同方向移动时逆转。
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After the reflected pulse reaches the source, there will be minimum voltage at the source terminals and maximum current in the cable.
To help answer this question, it is helpful to ask students how voltage and current relate to each other in a short-circuit condition (minimum voltage, maximum current).
如果电缆被不正确值的电阻终止（不等于电缆的特性阻抗）会发生什么？
与电缆的特性电阻不等于的任何终止电阻（太小或太大）都会导致反射波，尽管幅度要小于电缆未终止或直接终止的电缆。
Answering this question is an exercise in qualitative thinking: compare the results of termination with the proper amount of resistance, versus termination with infinite or zero resistance. A terminating resistor of improper value will produce an effect somewhere between these extreme cases.
例如，与正确终止的电缆相比，将电缆阻抗（如电压源“看到”），而不是开放式或缩短的电缆。在传播延迟时间过去后，通过不正确值电阻“外观”终止的电缆将是什么？
A two-conductor cable of uniform construction will exhibit a uniformcharacteristic impedance（z0) due to its intrinsic, distributed inductance and capacitance:
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如果我们要使电缆更窄，以使导体更加紧密，那么所有其他维度都保持不变，将会发生什么？
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Z0woulddecrease。我将留给您解释为什么会发生这种情况。
Be sure to ask your students to explain why the characteristic impedance will change in the direction it does, based on the known changes to both capacitance and inductance throughout the cable. It should fairly simple for students to explain why capacitance will increase as the two conductors are brought closer together, but it may not be as apparent why the inductance will decrease. A good “Socratic” question to ask is about magnetic field strength, assuming one end of the cable were shorted, and a DC current source connected to the other end. Be sure to remind them to discuss the right-hand corkscrew rule for current and magnetic fields in their answer to this follow-up question!
A two-conductor cable of uniform construction will exhibit a uniformcharacteristic impedance（z0) due to its intrinsic, distributed inductance and capacitance:
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What would happen to the value of this characteristic impedance if we were to make the cable wider, so that the conductors were further apart, all other dimensions remaining the same?
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Z0wouldincrease。我将留给您解释为什么会发生这种情况。
Be sure to ask your students to explain why the characteristic impedance will change in the direction it does, based on the known changes to both capacitance and inductance throughout the cable. It should fairly simple for students to explain why capacitance will increase as the two conductors are brought closer together, but it may not be as apparent why the inductance will decrease. A good “Socratic” question to ask is about magnetic field strength, assuming one end of the cable were shorted, and a DC current source connected to the other end. Be sure to remind them to discuss the right-hand corkscrew rule for current and magnetic fields in their answer to this follow-up question!
A two-conductor cable of uniform construction will exhibit a uniformcharacteristic impedance（z0) due to its intrinsic, distributed inductance and capacitance:
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What would happen to the value of this characteristic impedance if we were to shorten the cable’s length, all other dimensions remaining the same?
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Z0would remain exactly the same!
后续问题：什么电气特性wouldchange for this shortened cable?
This is sort of a “trick” question, designed to make studentsthinkabout characteristic impedance, and to test their real comprehension of it. If a student properly understands the physics resulting in characteristic impedance, they will realize length has nothing whatsoever to do with it. Although the cable’s total capacitance will change as a result of shortening the cable’s length, and the cable’s total inductance will likewise decrease for the same reason, these electrical changes should not present a conceptual difficulty to students unless they are modeling the cable in terms ofonelumpedcapacitanceandone(or two) lumpedinductance(s). If they are thinking in these terms, they have not yet fully grasped the reason why characteristic impedance exists at all.
Suppose we were designing a pair of BJT放大器circuits to connect to either end of a long two-conductor cable:
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How would we choose the component values in each transistor amplifier circuit to naturally terminate both ends of the 75 Ω cable?
rCof the transmitting amplifier should be 75 Ω, as should the parallel equivalent resistance RB1||rB2接收放大器。
This question is really a review ofThévenin’s theorem适用于common-emitter，分隔偏置的BJT放大器电路。
In case anyone asks, the “zig-zags” in the four lines for the cable represent an unspecified distance between those points. In other words, the cable islonger比可能比例代表the schematic diagram.
Suppose we were designing a pair ofBJT放大器circuits to connect to either end of a long two-conductor cable, each end coupled to its respective amplifier through a transformer:
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How would we choose the component values in each transistor amplifier circuit to naturally terminate both ends of the 75 Ω cable?
rCof the transmitting amplifier should be 1.875 kΩ, as should the parallel equivalent resistance RB1||rB2接收放大器。
这个问题实际上是对Thévenin定理的评论，因为它适用于公共发射极，偏见的BJT放大器电路，以及适用于加速和降落变压器的阻抗变换。
In case anyone asks, the “zig-zags” in the four lines for the cable represent an unspecified distance between those points. In other words, the cable islonger比可能比例代表the schematic diagram.
Some communications networks use cables to not only provide a path for data transmission, but also DC power to energize the circuits connected to the cable.
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However, if we were to terminate the cable as shown, the termination resistor would dissipate a substantial amount of power. This is wasted energy, and would unnecessarily burden the power supply providing DC power to the network cable.
How can we eliminate the problem of power dissipated by the termination resistor in a DC power/signal cable and yet still maintain proper termination to avoid reflected signals?
A capacitor must be connected in series with the termination resistance to prevent the resistance from acting as a DC load on the network:
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Understanding this answer requires that students recall the filtering behavior of aseries capacitor在交流电路中。
找到一长长的同轴电缆，并将其带上您的课堂进行讨论。在讨论之前，请尽可能多地了解有关您的电缆的信息：
如果可能的话，请查找组件的制造商的数据表（或至少与类似组件的数据表）一起讨论。
The purpose of this question is to get students to kinesthetically interact with the subject matter. It may seem silly to have students engage in a “show and tell” exercise, but I have found that activities such as this greatly help some students. For those learners who are kinesthetic in nature, it is a great help to actually触碰real components while they’re learning about their function. Of course, this question also provides an excellent opportunity for them to practice interpreting component markings, use amultimeter, access datasheets, etc.
