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### Section 1: Motivation

Before delving into simulations, several useful RF concepts will be presented. In this lesson, the difference between DC (steady state) analysis and RF analysis are described.

### Section 2: Waves

Much of RF design is the study of how energy goes into and out of a system. It turns out that mechanical waves (sound and vibration) and RF waves share a lot in common- these fields use the same equations to define a propagating (traveling) wave. This section will not cover propagation in general, but instead will try to provide some analogies to help visualize what's going on in later sections. Using a simple plastic megaphone (search your favorite online retailer for "plastic megaphone"), you can do these experiments at home or in the class..

The Source and the Load

Consider a talking head. The blue man is our “source”, and the sound coming from his mouth is the signal. The green man's ear is the “load”, and we care about how much energy goes to that load. Notice that audio energy goes all over the place: the energy that doesn't reach the load (the green man's ear) from the source (the blue man's mouth) is considered “lost”.

Figure 1.2.1: Energy travels from the source (a mouth) to the load (an ear)

Transmission, Reflection, and Gain

We can introduce a device between the source and the load that can change how much energy is lost, and how much is transmitted. For example, we can use a megaphone. With a megaphone, we can see most of the energy is transmitted to the load, so much less is considered loss. The sound that goes to the load is the “transmitted” energy, the sound that goes back towards the source is the “reflected” energy. The difference between how much energy gets to the load with and without the megaphone is the “gain” of the megaphone.

Figure 1.2.2: Gain allows for more energy to be transmitted to the destination (the load)

Insertion Loss and Return Loss

The energy that does not reach the load is considered “lost”. If the megaphone is turned backwards, more energy is actually reflected back to the source than is transmitted to the load. The amount of energy that is reflected back, and does not reach the load, is often referred to as the “return loss”. The energy has been reflected back to the source.

Figure 1.2.3: Sound can be "lost" when reflected (returned back) to the source

If a large piece of cotton is put in the megaphone, a lot of the sound energy will be absorbed by the cotton- it will be neither reflected nor transmitted. This is the “insertion loss”; generally this is the audio energy that is turned from sound into heat (or RF into heat). So energy may be transmitted, reflected, or lost in the device... energy will always be conserved.

Figure 1.2.4: Sound can be lost due to lossy materials

Ports

These examples have had two openings, or “ports”. A port is a place that energy can enter and leave the system. It is possible to consider a megaphone with three ports, such that some of the energy is reflected, some goes towards one port (whose load is the green man's ear), and some is diverted to a third port (whose load is the orange man's ear). This third port is also considered a load, the energy that goes to the third port is not considered loss.

Figure 1.2.5: Sound with three "ports": 1) blue mouth, 2) green ear, 3) orange ear

Propagation Delay

Keep in mind that it takes time for sound to travel from the source (the blue man's mouth) to the load (the green man's ear). We normally don't notice the difference in time between when a sound is made and when it is heard, but this effect is very important in RF analysis of circuits. The next section will look into how propagation delay effects the voltage waves bouncing back and forth through a device.

Sound when first spoken

Figure 1.2.6: Sound after a few milliseconds of propagation time

**Main Point**

This section covers basic terminology and concepts for how waves travel. Energy can be transmitted from a source to a load, it can be reflected back to the source, and it can also be lost by being transformed into heat (which is what happens when sound is muffled).

### Section 3: Voltage, Current, Resistance, Power, and Voltage Division

This section will describe some of the ways that main concepts in electronics, using mechanical analogies.

**Voltage**

While the water analogy isn't perfect, the first analogy I'll provide does help many with understanding these concepts- consider a water tank and a water tower. The water represents electrons. Water in a tank on the ground can provide a certain amount of water pressure at ground level. Move the same amount of water up to to top of the tower, and we have increased the pressure that can be provided at ground level. The increase in water pressure is caused by the increase in potential energy. We did work on that water, so we can get more work out of the water. Voltage is the level of pressure (relative to ground) per the amount of water... that is, the amount of energy per unit of charge.

**Current**

If we let water flow through a pipe, the amount of water going through the pipe at any given cross-section is the current. At your faucet, if you open your tap a little, you get a weak current. If you open your tap a lot, you get a strong current. It's one of those happy times the words and the concepts match up.

**Resistance**

If you think about how much pressure is in your pipes, a lot more water could flow than your tap is letting. Your tap is putting up a resistance to the current flow. In electronics, resistors resist the flow of current- bigger resistors allow less current to flow.

**Power**

Let's say you have a pipe coming out of your water tower, and it is going to power a turbine. You would like to know how much power you could reasonably get. Intuitively, it should be a function of the pressure (our ability to do work), and it should have something to do with how how fast the water can flow... a trickle won't provide a lot of power. It turns out that:

Mechanical: Power = (Force from pressure) * Velocity

Electrical: Power = Voltage * Current

**Voltage Dividers**

If our pipe is fairly big, we would expect a large current- the pipe has a low resistance to the flow of water. If the pipe is narrow, we would expect a small amount of current- the pipe has a high resistance to the flow of water. But, what if the pipe isn't the same size all the way down? If we have a wide pipe with a nozzle, we would expect more water to come out that if the whole pipe was narrow- we use this effect in garden hoses to spray a lot of water just in a focused beam. If the whole garden hose was that size, we wouldn't get much flow at all. There isn't much drop in our water pressure through the garden hose because it has low resistance, and there isn't much drop in our water pressure through the nozzle because it's so short. We can measure the force at the tap where the garden hose is, and we can measure the force where the spray comes out... but what if we want to calculate the force between the hose and the nozzle, where we can't directly measure?

In electronics, sometimes we want to know what the voltage potential is in between two resistors, as in Figure 1.3.1. We have an input voltage potential (Vin) and an output voltage potential (Vout). Since water pressure levels are relative, we can define a zero potential (ground) wherever is convenient. For the garden hose, it would be on the output of the spray. If we know the resistance R1 (above, the garden hose) and the resistance R2 (above, the nozzle), then:

Vout = Vin * R2 / (R1 + R2)

Figure 1.3.1: Resistor divider

**Electrical and Mechanical Analogues**

The second analogy is more exact, if perhaps less straightforward. The mechanical (vibration and sound) forces describes above are familiar to us because most of us experience them everyday. Hopefully, it should not be too difficult to understand how sound or waves propagate, and translate that understanding to how voltage waves propagate. Since all of the math behind these mechanical and electrical forces are differential equations, it shouldn't be too surprising that the equations behind both are similar. The mechanical waves described above have a force associated with them. This force is what is translated into voltage in our ear and is how most people hear. This force is analogous to an electrical voltage. Objects resist changes in motion by this mechanical force, and we call this friction. In electronics, friction is called resistance. The mechanical force travels at a certain velocity, and the velocity is analogous to current, in that:

Mechanical: Force = Velocity * Friction

Electrical: Voltage = Current * Resistance

Mechanical: Power = Force * Velocity

Electrical: Power = Voltage * Current

Further analogies exist between the mechanical and electrical. In electronics, inductors are devices which resist changes in current; mechanically, a mass will resist changes in velocity. In electronics, capacitors are devices which hold voltage potentials which can be used later; mechanically, a string can store mechanical force. If the reader is interested, I recommend Googling "mechanical and electrical system analogies" for more details.

### Section 4: Voltage waves and Propagation Delay

The "resistance" of a resistor is defined as the relationship between the voltage and the current. The impedance of a transmission line is the relationship between the propagating voltage and current waves. Any voltage wave traveling down the transmission line will have a corresponding current wave, maintaining a relationship between the potential and the actual electrons flowing down the line.

Figure 1.4.1: Impedance of a transmission line

Consider the transmission line shown in Figure 1.4.2: a switch connects a DC voltage source with a 50Ώ source impedance to a 1 ft transmission line terminated with a 50Ώ resistor. As long as the switch is open, the voltage between P1/P1' is 0V. At the moment the switch is closed, the voltage source (which includes the 50Ώ resistor) sees the 50Ώ transmission line, forming a voltage divider. Therefore, the voltage at P1 at the time the switch closes is 0.5V.

DC Voltage Wave Bounce, Matched

Since it takes time for energy to propagate, there is a delay between the switch closing and the voltage at P2 changing: at the exact moment the switch closes, the voltage at P2 is still 0V. In air, RF waves propagate at 299792458 m/s (approximately 1ft/ns). Therefore, the voltage at P2 will change approximately 1ns after the switch is closed.

Figure 1.4.2: Voltage wave traveling down a matched transmission line

**Main Point**

The normal steady state analysis for a resistor divider (the two 50Ώ resistors connected by a foot of wires) ignores the propagation delay for voltages to set up along the line. After 1ns, the voltage wave analysis converges to the same answer as the simple steady state analysis.

### Section 5: Wave Bounce Diagram for DC voltages

This analysis is more complicated if the resistances are not all equal, as they are in Figure 1.4.2. Consider the circuit shown in Figure 1.5.1. At the moment the switch is closed, the voltage at P1 will be 0.586V (a 50Ώ source resistor into a 70.7Ώ transmission line). But we also know that over time, the steady state voltage at P2 will be 0.667V. How do we resolve this apparent conflict? The key is to consider what happens when the voltage wave hits P2.

Figure 1.5.1: Transmission line used for wave bounce

Just before the voltage wave hits P2, the voltage wave is 0.586V and has an associated current wave of 8.29mA (for a 70.7Ώ). As soon as the voltage wave hits P2, there is an apparent conflict: a voltage wave of 0.586V across a 100Ώ resistor implies a current of 5.86mA. It is impossible for the current going into the terminating resistor to differ from the current leaving the transmission line (Kirchhoff's current law), but it is entirely possible for the voltage waves to differ. So a voltage wave bounces back off of the terminating resistor towards the source resistor.

DC Voltage Wave Bounce with Mismatch

The amount of reflection (Γ, defined below) off of the resistors is a relationship between where the resistances, and depends on the direction of the wave. When the wave bounces off of the terminating resistor, Z2 = 70.7 and Z1 = 100; when the wave is bouncing off of the source resistor, Z2 = 70.7 and Z1 = 50.

Equation 1.5.1

Therefore, the amount of reflected voltage is the reflection coefficient times the voltage wave (so the voltage wave going back to the source is 0.1V), and the voltage at P2 is the sum of the incoming and outgoing voltage waves (0.586 in plus 0.1V out = 0.686V). But this is still not 0.667V as expected for the steady state.

You can see the effects of the voltages bouncing off of these discontinuities if you look at the finite difference time domain section of the video above. Transitioning from high speed to low speed dielectric is just like going from a high impedance to a low impedance line- we see similar reflections. Here, however, we don't have a short traveling pulse wave. We have a constant DC voltage which is trying to figure out to what value it should settle.

The Wave Bounce Diagram is a useful tool which helps track the voltages at P1 and P2 over time. It uses a vertical axis for time, and the voltage nodes (P1 and P2) go along the horizontal axis. The voltage at a point is equal to the sum of the existing voltage, plus the incoming and outgoing voltages. Therefore, for the 2^{nd} bounce at P1, the voltage is equal to the existing voltage (5.86V) plus the incoming voltage wave (0.1V) plus the outgoing voltage wave (-0.0171V). One can see that after a few bounces, the voltage converges to the expected value.

Figure 1.3.2: Wave bounce diagram for two nodes (P1 and P2)

**Main Point**

The normal steady state analysis for a resistor divider (a 50Ώ resistor and a 100Ώ resistor connected by a foot of wires) ignores the propagation delay for voltages to set up along the line. After around 3ns (three bounces down the line), the voltage wave analysis converges to the same answer as the simple steady state analysis (within 0.1%).

### Section 6: Wave Bounce Diagrams for Digital signals

Now imagine if the switch in the previous figures, rather than staying closed, opens and closes at a regular interval. Instead of a steady state voltage, voltage waves would continue to bounce around back and forth along the transmission line. The reflections caused by this square wave traveling along a transmission line is a real problem with high speed digital lines. When everything is matched (Figure 1.4.2) there are no voltage bounces. When impedances are mismatched (Figure 1.5.1), voltage waves bounce back and forth. Some of the energy is transmitted to the load, and some of the energy is reflected back to the source.

If the transmission line is short enough, this problem is mitigated: the voltage settles more quickly if it doesn't have a long distance to bounce around.

Digital Signal Voltage Wave Bounce with Mismatch

The energy that is transmitted to the load is (generally) useful energy. The energy that is reflected back is (generally) wasted energy. How much of the energy is wasted? According to the Wave Bounce Diagram 0.1V was reflected off of the first bounce, but some portion of that was reflected back to the source, so the actual amount of energy wasted is proportional to the voltage actually received by the source, 0.0828V from the first bounce plus 0.0024V from the third bounce, seen in Figure 1.6.1.

Figure 1.6.1: Wave Bounce Diagram showing wasted energy

**Main Point**

Because the waves are constantly varying, tracking how voltages are transmitted and reflected are much more difficult for digital analysis compared to what goes on in the steady state. Because the voltages keep changing, if there is a mismatch some amount of energy is periodically being reflected back to the source. Not only is this wasted power, the reflected voltages interfere with the digital output- potentially corrupting the signal. This is exactly why USB cables cannot be more than 5 meters long.

### Section 7: Wave Bounce Diagrams for RF signals

Instead of considering the “step function” voltage wave in Figure 1.4.2, consider the “sine wave” voltage wave of Figure 1.7.1. We no longer can track a discrete bounce of a voltage wave, instead we will have a reflection that constantly varies over time. The simple Wave Bounce Diagram tracking voltage amplitudes no longer applies. Additionally, since the voltage never settles to a steady state, some amount of reflected energy is continuously being directed back to the source.

An RF signal is a constantly varying signal, and so it is no longer possible to think of discrete voltage waves bouncing along the transmission line. However, is is possible to treat each discrete frequency as a single voltage signal using complex numbers. Looking at Figure 1.7.1, a sine wave can be seen traveling down the transmission line. We can observe that *at this frequency*, the transmission line is quarter of a wavelength long. A quarter of a wavelength corresponds to 90º (a half wavelength is 180º, a full wavelength is 360º). At 1ns, when the sine wave hits the end of the transmission line, the voltage wave will reflect, and travel another 90º (for a total phase shift of 180º).

Here is the interesting thing... the voltage wave begins to reflect at 1ns. From 0.5ns to 1ns, the transmitted signal travels 90º down the transmission line, where it is reflected (let's say the end of the transmission line is an open: all of the voltage is reflected). From 1ns to 1.5ns, the reflected voltage travels another 90º, for a total phase shift of 180º. It took 1ns for the wave to travel from the beginning of the transmission line and reflect back to the starting point. In that same 1ns, the transmitted voltage did not stay still, though, it too has changed by 180º. If we sum the transmitted voltage with the reflected voltage, we see that they cancel- the voltage is zero at the start of the voltage line. It is as if the start of the transmission line is a short... even though the end has been left open. So, a quarter-wavelength line has the interesting property that it makes an open look like a short, and it also makes a short look like an open.

Figure 1.7.1: Sine wave voltage signal on a transmission line (click to enlarge)

The effective length (in terms of wavelengths) of the transmission line changes over frequency. It is possible to track how the voltage waves will bounce along the transmission line, for a single frequency at a time, using the phase information of the line length, since:

Exactly the same type of voltage wave bounces will occur as described above

Because the voltage is constantly changing, just tracking the amplitudes isn't enough

Since the phase length changes with frequency, the Wave Bounce Diagram has to be solved for each frequency

Figure 1.7.2: Wave bounce diagram using complex numbers for the RF Analysis

Equations 1.7.1a

Equations 1.7.1b

Equations 1.7.1c

Equations 1.7.1d

Equations 1.7.1e

Equations 1.7.1f

Rather than trying to graphically solve the problem (which is useful only to introduce the concepts), we can use a numerical version of the wave bounce diagram (which is also really only useful to introduce concepts), using the diagram in Figure 1.7.2 and the equations above. As a practicing engineer, I can say that I have not used wave bounce diagrams between the time I learned “The Theory of Small Reflections” and the time I wrote this guide.

**Main Point**

Because the waves are constantly varying, tracking how voltages are transmitted and reflected are much more difficult for RF analysis compared to what goes on in the steady state. Because the voltages are constantly varying, if there is a mismatch some amount of energy is constantly being reflected back to the source. In most cases, this is wasted power, however sometimes we can use the reflected energy to our advantage, like making an open circuit look like a short circuit (or more commonly useful, a short circuit look like an open circuit).

Copyright 2010, Gregory Kiesel