Before we start let’s think a little bit about what we’re doing:
Our goal is going to be to find the pole(s) of the circuit.
Okay? What is a pole and why do I care where it is?
A pole is a frequency at which the gain of the device rolls off. (remember that when it rolls off , it will be at the -3dB frequency with a slope of -20dB/decade)

We care because if the gain of a device rolls off at a certain frequency, then we won’t be able to amplify a signal above that frequency very well because the gain will be decreasing by 20dB/decade.

The procedure is nearly identical whether we are using a BJT of a MOSFET, but we will work each of them side by side just in case there might be any confusion, and we’ll follow these steps as we go through.  (we will also use some values that came from the output file when running a simulation of this circuit in Cadence (or PSPICE) )

1. Take a look at one of the circuits and see what you notice, how about the MOSFET.  This step is just to help us with our knowledge understanding of the circuit.
- At a glance it just looks just like another MOSFET right? Sure is, but let’s take a look at a few things just for kicks. Notice that it is using a bypass capacitor at the source so we don’t have to worry about (at when working with high frequency).  Since the capacitor bypasses to ground, you should notice that this is a common-source amplifier.  You could notice the Values for and and start to think about what the Gate voltage is and how that may affect the circuit.
2. We are talking about frequency response so that means we are probably going to want to draw the small signal equivalent circuit.
Remember that the capacitors and will act like short circuits at high frequencies so we will ignore them, but we will have to account for some of the capacitance internal to the device.

Both devices have internal capacitances that are very similar.  As you can see from the small signal models for a MOSFET (above) and BJT (below), the only significant difference is that the BJT has an additional resistance Rpi between the Base and Emitter.

Most of the analysis we will do is based on the small signal model. Note that small signal models are not typically used in PSPICE so this picture may look a bit odd, especially the controlled source but for our purpose it is good to have a visual reference. To start we will point out what everything is. Cgs is an internal capacitance betwe

en the gate and source. The

values for Cgs was similar to one the a PSPICE simulation may give.  CM1 and CM2 are Miller capacitances which we will find values for later.  ro is a Norton equivalent resistance that makes the model more ideal.  And just pretend that the G2 looks more like a voltage controlled current source and that their gains are gm*Vgs and gm*Vpi. For the BJT CM1 and CM2 are both Miller capacitances, Cpi is similar to Cgs and Rpi the additional component used for BJTs but not MOSFETs. The other part should look familiar from the other figures.

ON TO THE ANALYSIS!!!

We will find the device gain, overall gain, equivalent input and output capacitances, and the input and output poles. The process for both is essentially the same.

Device Gain: This is the gain from the control source to the output so we are looking for Vout/Vgs (or Vout/Vpi for a BJT). We will ignore CM2 for this process. Notice the resistances ro, RD, and RL are in parallel. Vout should be given by that equivalent resistance times the current though it which is gm*Vgs from the control source. So the equation for device gain is,

   (MOSFET)

  (BJT)

Overall Gain: This will be the gain from the source (Vs) to the output (Vout). We already know what Vout/Vgs is so if we find Vgs/Vs, we can multiply them to get Vout/Vs = (Vout/Vgs) * (Vgs/Vs).  Vgs/Vs is a simple voltage divider. Hopefully you can see this from the small signal model (remember that we are ignoring the capacitors for now but they will play a part later).  The equations we will get for Vgs/Vs and the overall gain are.

  (MOSFET)

Overall Gain:   (MOSFET)

  (BJT)

Overall Gain:   (BJT)

Now we will find the input and output poles.  For this we will need to look at the capacitances and use a formula to find the Miller capacitances, CM1 and CM2.  Any explanation for the miller capacitance will have to wait for another post or check out your Electronics Book, Wikipedia, Google, etc. but we will need to use a couple of special equations.  Overall we will need to find the input resistance and input capacitance for the input pole and the output resistance and output capacitance for the output pole.

Each pole will be at a frequency w=1/RC where the R and C are the equivalent R and C at that point, so to find the input pole, we will need to find the input resistance and the input capacitance.  These are found by looking into the input (the left side of the small signal model).  The voltage source will  act like a short so we see Rs in parallel with R1//R2 for the MOSFET (the BJT will have Rpi in parallel also).  The input capacitance will be Cgs in parallel with CM1 (the BJT will be the same).  The output resistance and capacitance are found the same way only looking in from the output (the right side of the small signal model).

      (MOSFET or BJT)

So the input pole will be: (MOSFET)

  =  950                                     =

  =                                               =

=                                                                          =

(BJT)

and the output pole will be: (MOSFET)

(BJT)

=                                  =

=                                                 =

=                                        =

To Do:

finish input & ouput R, input C, Pole (& calculate answers)

npn or pnp?

When looking at a BJT, the easiest way to decide whether it is npn or pnp is to look at the emitter, which is always modeled as the arrow. If you remember that the arrow tail is always at a ‘p’ node and the tip is at an ‘n’ node, you can easily decide whether the BJT is npn or pnp. Remember that the collector and emitter are always either both ‘n’ or both ‘p’.

Voltage Polarities

It is important to know which direction the voltage’s will appear positive when we begin using nodal analysis to solve BJT circuits. Typically, there will be a voltage drop of .7 V over the nodes that will be used in these calculations. Whether or is positive is decided by the type of BJT. The voltage polarities are flipped between pnp and npn BJTs. Obviously, the only difference in the symbols between the two types of BJTS is the arrow, which is the emitter. If we remember the tip of the arrow is the lower voltage, we are able to deduce that for an npn BJT and for pnp.

To be in the active mode, a BJT’s collector-emitter voltage must be above approximately .3 V. As above, this voltage polarity is reversed between npn and pnp BJTs. To determine, whether or should be positive, we can use our deduction of the base-emitter voltage polarity. The voltages, in active mode, drop from collector to base to emitter in npn BJTs and from emitter to base to collector in pnp BJTs. So, if we have figured out that we are using an npn BJT because was a positive .7V, we know that the base voltage is higher than the emitter voltage. From here we know the collector must be higher than the base, and therefore, higher than the emitter. We have just figured out that must be greater than the .3V to be working in active mode. Using the same logic, must be greater than .3V for a pnp BJT to remain in active mode.

Current Directions

Current directions are very simple to figure out. Just use the arrow. The collector and emitter currents always go in the direction of the arrow in active mode. The base current is a little more tricky to figure out, but is also fairly obvious when using the arrow as a reference. As you can see in the above npn circuit, where the arrow is ‘pointing’ away from the base, the base current flows towards the BJT, in the direction the arrow is pointing. Oppositely in the pnp circuit, the base current flows away from the BJT, in the direction the arrow is pointing. There is a table of basic equations listed in my post titled “BJT Transistor Nodal Analysis” which would allow us to calculate each current using a different current, but using Kirchhoff’s Current Law, knowing two currents, we could calculate the third. For a npn BJT, and for a pnp BJT . Note that both of these equations evaluate to .

It is also important to know that can be modeled as .

These equations are not very informative by themselves so a few examples are demonstrated below. In both examples we will assume is very large. What this means for our calculations is . Since we also assume that .

Finding missing voltages in a BJT circuit

Example 1. Solve for V3:

There are several ways to find . The more “difficult” way is to first find the emitter current, , and then use Ohm’s Law. Since we know , we can find the collector current, , and then solve for .

The easier way to find is to recall that behaves like a diode. For this pnp BJT: .

We know that so

may not always be a very large number. Had that been the case here, we would have started by finding the collector current (since it’s voltage drop and resistance are given) and since anymore, we would use the formulas above to the find the base and collector current.

Finding BJT Bias Voltages and Currents

Example 2 Solve for V2 and I1:

Here we will want to start by finding . also equals which approximately equals and this collector current will allow us to find .

Notice that was given as .  If this had not been given, we would have been able to find it because and .

Similar to the previous example, if was not a very large number. We would first find the emitter current and then use the equations in the table to find the other branch currents.

Note that both of these examples used pnp BJTs. The difference in an npn BJT is the base-emitter voltage is reveresed. You would use .

In General:

Most of these problems are very simple to solve. Typically is given and you will need to use Ohm’s Law to identify one of the currents. After one of the currents is found you will be able to solve for the other currents using the basic equations listed above. If one of the currents is not immediately obvious, the base-emitter voltage is likely needed. Most problems have you deduce the emitter voltage from the base, but it is easily possible to find the base voltage from the emitter voltage and then use that to find the base current.

Next, a time shift is demonstrated. Here our function is changed from f(t) to f(t-2). Notice that subtracting 2 from t in the function results in a positive shift of the graph.

In the next two graphs t will be scaled. Scaling t is not quite as intuitive as we may have expected. When we multiply t by 2, corresponding points of the function now occur at 1/2 the time they previously had. When we divide t by 2, each corresponding time on the graph occurs at a t that is now multiplied by 2. Notice that each of these factors directly affects the duration of the signal.

Scaling the amplitude has more intuitive results. If we multiply f(t) by 2, the amplitude of 1 is changed to 2. Multiplying f(t) by 1/2 results in an amplitude of 1/2.

Finally, multiplying t by -1 mirrors our function over the y-axis. Each time now occurs at its negative.

Example:
Here we will attempt to convert f(t) into 2*f(.5t+3). The graph of f(t) is shown below.

The easiest way to handle this type of problem without error is to manipulate the function one step at a time. First, I have converted f(t) into 2*f(t). Only the peaks are changed here (by a factor of 2).

Next, I convert 2*f(t) into 2*f( t). Notice how the actually expands our graph duration by a factor of 2 (from a 6 sec duration to a 12 sec duration).

Finally, we move from 2*f( t) to 2*f( t + 3). As shown in the discussion above, this is a time shift. Time shifts can be a little confusing because adding results in a negative shift of our graph. Try to think of it as our signal occurring 3 seconds earlier than before, reading from left to right on the graph. The easiest way to do this part is shift each x-intercept by 3 seconds (to the left, of course).

To begin I will introduce our variables:

concentration of free electrons (donors)
concentration of holes (acceptors)
= number of free electrons and holes in a unit volume

In thermal equilibrium(or no doping):

and, therefore

However, doping is common in most examples. To increase the concentration of free electrons, an element with 5 valence electrons is used (i.e. Phosphorous). The resultant material is said to be n-type. To increase the number of holes, an element with 3 valence electrons is used (i.e. Boron). The resultant material is said to be p-type.

This introduces subscript n’s and p’s along with our concentration of free electron and hole variables.

In n-type silicon:

concentration of free electrons (in n-type silicon)
concentration of holes (in n-type silicon)

In p-type silicon:

concentration of free electrons (in p-type silicon)
concentration of holes (in p-type silicon)

Note: The subscript indicates whether the material is n-type or p-type.

Calculations

Typically you first want to identify whether the material you are working with is p-type or n-type. This introduces two new variables. which refers to the concentration of donor atoms and which refers to the concentration of acceptor atoms.

In n-type silicon:

Here you will use the variables , , , and .

In p-type silicon:

Here you will use the variables , , , and .

In most cases and or will be given and you will be able to find or . Then you will find or from the equations above.

Isolation

• For the biological subject, against electric shock from the amplifier’s power source(s)–or, how to not kill your patient while taking measurements!

• For the signal-to-noise ratio (SNR): isolation keeps the (60-Hz mains and other) noise out of the sensor pickup–and out of the amplifier input.

One simple way to isolate the input is to use an LED/phototransistor pair:

The sensor is connected to an LED, which outputs light of intensity proportional to signal voltage.

The light from the LED falls on a photodiode or phototransistor, which is biased in such a way that current only flows when light hits the device, and the current (and thus measured voltage) is proportional to light intensity.

Phototransistor output is to the amplifier.

The power circuitry for the amplifier is completely isolated from the sensor (electrodes etc.) Power supplies for the sensor use a transformer or battery, and the “ground” for the sensor side floats relative to the amplifier side (which should have ground connected to earth eventually.)

These devices are manufactured in one piece, so the frequency of light output from the LED is matched to the ideal absorption frequency of the phototransistor. However, there must be no electrical connection between the two sides of the circuit. More exactly, the impedance between the two circuits should be near a Teraohm ().

Single-Ended vs. Differential Input

• Referenced Single-Ended: two leads of sensor are + signal and earth (computer) ground. Note: to avoid ground loops, it is best to avoid connecting sensor earth to computer earth.

• Non-Referenced Single-Ended: two leads of sensor are + signal and – signal, which is connected through a bias resistor to earth ground.

• Differential: three sensor leads are available: + and – signal and (earth) ground.

The 4-BNC -to- DAQ-card lead you’ve probably used in a lab before is referenced single-ended: each of 4 sensors has available a signal wire, and the 4 sensors share a common ground wire. If the DAQ card is configured in Differential mode using this input lead, the signal gradually rises to a saturated level, as the DAQ card assumes a signal on the – differential input, which is in reality floating.

Problems with Single-Ended Input:

• Cross-talk: Because multiple signals share a ground wire, they also share some portion of the signal, so if only one input is connected that signal shows up, somewhat weakened, on the second input trace.

• Lack of isolation: using the ground wire as the negative signal lead means the subject and the amplifier are strongly coupled–and strong electric signals from the amplifier may travel to the subject. Also, stray noise (such as from the power mains) easily couples into the input.

Differential Input:

I’m talking a lot about DAQ cards, these are very important when working with A/D systems and analog and digital filters.  If you are not familiar with these, go do a little light reading.  Hopefully this is making enough sense to get the concepts across.  The DAQ card comes with a finite number (16, in our case) of analog input ports (channels), each with paired input pins: a signal (ACH#) and a ground (AIGND). All the analog input ground pins are tied together on the DAQ card. In differential mode, the analog input ports are paired, so channel 1 uses AICH0 as +, AICH8 as -, and the ground wires for the pair are tied together as the reference ground. Thus a 16-channel DAQ card has only 8 channels in differential mode.

• Because no channels share leads, cross-talk is reduced. The DAQ card has high common-mode rejection, so if the – input leads are tied to ground (simplest configuration), channels will not interfere because the cross-talk is common to both + and – inputs.

• The ground lead cannot be isolated from the input electrodes without causing the DAQ circuitry to saturate, so the DAQ card itself does not provide isolation.

Wiring a DAQ card in Differential Mode:

The figure is adapted from Fig. 4.6, p. 4-15, of the NI 6024E (DAQ card) User Manual, and illustrates the appropriate connections for differential input to the card. Recommended values of R+ and R- depend on the impedance Rs (in series with Vs) and coupling of the source signal Vs:

R+ and R- provide bias current return paths. Bias currents result from not-quite infinite input impedance to the DAQ, and if not balanced the noise they represent will not be common to both + and – inputs, and thus won’t be rejected. However, R+ and R- load down the source with an equivalent 2R+, which will decrease gain if R+, R- are too low. If R+, R- are too large, they will produce a DC offset at the DAQ input.

Bandwidth

Bandwidth is a critical parameter in determining the type of amplifier needed. Often a bioamp has adjustable bandwidth; typical applications have both low-pass and high-pass adjustable filters.

Typical Bandwidths of Biological Signals: (Table from Webster Fig 6.16, p. 259.)

(Electro-oculogram: signal from electrodes placed either side or above and below the eye. Linearly proportional to angle of gaze; DC signal)

Noise Reduction

Can be either or both analog and/or digital. If digital, it is done by post-processing collected data. If analog, wiring of the printed-circuit board, handling of wires on the bench, and internal circuitry are all details to consider carefully.

Protective Shielding

For equipment, against transient large signal sources–includes grounding the outer case of the equipment and surge protection, as well as isolation.

Power Supplies

AC (mains) with rectification and usually transformer isolation. Subject to power failures, risk of electrocution.
Battery: provides its own isolation. Limited lifetime and undesirable behavior just before failure.

Specialized Amplifiers

Lock-In Amplifier Block Diagram

Lock-in Amplifier: (www.lockin.de) If the signal source is mostly at a single frequency but is very weak and/or subject to a great deal of noise, a lock-in amplifier can be used to extract the signal. In biological cases the single-frequency property of the signal is most-often externally generated, by applying a single-frequency excitation in one way or another.

Theory:

Signal source S(t) = A cos(ω1t + θ1) + B cos(ω2t + θ2)

Reference signal R(t) = C cos(ω1t)

Product S(t)R(t) = AC cos(ω1t) cos(ω1t + θ1) + other terms in cos(ω1t) and cos(ω2t)

= AC/2 (cos(2ω1t + θ1) + cos(-θ1) ) + other terms in cos(ω1t) and cos(ω2t)

Integrating over an even number of cycles of the reference signal reduces all terms in ω1 to zero, so the displayed signal is proportional only to the amplitude of the component of the signal source at the reference frequency.

To get the reference frequency into the signal source various means are employed: one may have to excite the subject (nerve, membrane…) at the reference frequency, or one may already know the source has a dominant resonance; alternatively one may look at each of several spectral components in the source signal piecemeal, by tuning the reference frequency.

Pre-Amplifier:  high input impedance, moderate gain, high CMRR. Often differential input and isolation. May include DC offset control, gain control switches, and/or calibration signal.

Chopper-stabilized Amplifier:  removes thermal DC drift from a (very-low-frequency or DC) signal by using negative feedback and chopping the low-frequency signal at a frequency above the amplifier’s high-pass limit (effectively, this is frequency modulation). The signal can be reconstructed by demodulation after amplification; noise signals at both high frequencies and those below the chopper frequency are rejected.

This post was made using some old class notes, a DAQ card user manual and just some good old knowledge.  I know this is not a traditional engineersphere.com lesson, it is more of an ‘informative read’ for the avid electrical engineer interested in biomedical applications.  Enjoy

Amplifiers

Definition (for Bioinstrumentation): Circuit that makes a small signal, usually voltage but occasionally current or power, big enough to do something useful–including excite an output mechanism.

Common uses:

• Biological measurements of small signals
• Audio engineering: a large current is needed to drive speakers
• Wireless communications: far from originating antenna, signal is very weak and must be
amplified to be useful.

Secondary applications:

Many amplifiers are also filters, preferentially amplifying some frequencies over others

• Don’t want to amplify noise along with signal
• Only interested in low- or high-frequency portion of signal
• Active filter provides amplification as added bonus

General Amplifier Characteristics

Common Mode Rejection Ratio (CMRR)–ratio (usually in dB) of the amplifier’s common-mode gain to its differential-mode gain. Common-mode signals are input signals common to both + and – inputs and are usually unwanted noise–60-Hz, thermal, etc; differential signals are applied to only one input.

Gain–voltage out over voltage in, or current out over current in. May be given in dB.  Bioamp requirement: often adjustable; should be 1000 or greater, should be calibrated.

Input Impedance–what the input source sees as its load working into the amplifier: if the entire amplifier circuit were modelled as a resistor, what would be the value of the resistor? Typical bioamp: Rin = 10MΩ–signal source need not provide much current.

Output Impedance–same as input impedance but from the output end: model the entire amplifier as a source, and this is its internal impedance. Bioamp requirement: Ro << Rload.

Frequency Response–over what range of frequencies is the gain constant? Graphically illustrated with a Bode plot of gain vs. frequency.

DC offset–usually an amplifier has an operator-adjustable DC offset knob, to null out any offset associated with non-ideal amplifier or sensor behavior. A DC offset signal results in an incorrect reading unless removed or filtered out (high-pass filter).

Operational Amplifier:

basis for most instrumentation-related amplifiers, cheap, readily available, easy to work with. “Operational” = good for mathematical operations (+, -, log, …)

Meanings and advantages:

• equal input voltages –> within the limits of external power supplies, an op amp outputs whatever current is needed to drive the two input voltages equal. Result is that the output voltage follows the input, scaled by a large gain.
• infinite input resistance means the op amp never loads down the source, even if the source cannot supply much power.
• zero output resistance means the op amp is an ideal voltage source, with output voltage independent of whatever load impedance it must work into.
• infinite open-loop gain means the amplification properties of a circuit containing an op amp are independent of the op amp internal properties.

Carr and Brown go through several common op-amp configurations and show how to derive their voltage gains. Suffice it for now to know that if you want to build an amplifier, an op amp is a good place to start.

Common op-amp circuit configurations:

• Inverting and non-inverting amplifiers
• Summing and difference amplifiers
• Integrating and differentiating amplifiers
• Log and anti-log amplifiers
• Instrumentation amplifier
• Low-pass filter
• High-pass filter
• Band-pass and notch filters
• Buffer (voltage follower, or unity-gain buffer)

Op Amp Equivalent Circuit

This schematic illustrates the important properties of the op amp, and of any amplifier. It can also make it easier to understand circuit operation.

Note the open-circuit inputs– Rin = infinity. The
output voltage supply is a dependent voltage
source. Also, since the gain A is infinite, v2 – v1
must be zero to get a finite output.

Difference (Differential) Amplifier

Example: derive the gain relationship for the basic differential amplifier shown, assuming U1 is ideal and Vin = V2 – V1.

To get equal gain of both V1 and V2, set R2/R1 = R4/R3. Then Vo = R2/R1(V2-V1).

To get a high gain, R2 >> R1, but to get high input impedance R1 (and/or R3) should be large, making R2 and R4 even larger…Result: high gain and high input impedance are difficult to achieve together.

Instrumentation Amplifier

A difference amp with input buffer/gain stages to increase input impedance and gain. To analyze, realize that the same current must flow in R5, R6 and R5 (since no current flows into the op amps). Set R1=R3, R2 = R4; then Vo = G1* (v3(U1) -v2(U1)), where G1 = R2/R1 = gain of second (differential) stage.

Gain of input stage is 1 + 2*R5/R6 = G2. Overall gain is G1*G2. Making R6 a potentiometer allows compensation for inequalities in the two R5s, as well as for variable gain of the entire circuit.

Overall Gain: A practical difference amp can have a gain of 100, so it is not hard to get an overall gain of 10,000 from an instrumentation amp.
Input Impedance: equal to that of the op amps U1 and U2–very large. Use FET-based amps for extremely high input impedance
Output Impedance: close to that of the op amp U1–very small: the amp will provide whatever current is needed to maintain the output voltage regardless of load impedance.

Equal resistors: in practice one cannot buy matched discrete resistors; however it is fairly easy to manufacture them within an integrated circuit. Monolithic diff-amps are available.

Non-idealities of amplifiers

Gain:  TANSTAAFL–you cannot have gain without a power supply to provide it. Real gain is limited by the external power  supplies (+/- 12 or 15 V, for op amp circuits) Exceeding the limits of the power supply results in Saturation, or “hitting the rail”.

Output impedance: a zero output impedance means the circuit will provide whatever current is needed to maintain the requested output voltage. Practically, however, an op amp can only provide some 20mA, meaning RO is negligible only for RL>>15V/20mA = 750 Ω.

Frequency dependence: to avoid oscillation or saturation, circuitry must often be added that limits the bandwidth of an amplifier.
• To keep DC offset signals (from polarizing electrodes, for example) out of the amplifier, a high-pass filter is used to cut off DC (and lower-frequency ac) signals.
• If the load to be driven contains substantial capacitance, the current output limit again becomes a problem, limiting gain at high frequencies, where capacitors look like shorts.

Input bias current: real op amps do have non-zero input currents, which produce voltage drops at the input–another source of DC offset. This source can be minimized by using FET op amps.

Impedance Bridge

Often the measurand is the relation between voltage and current (one applied, the other a response) rather than a biologically generated source. An example in Carr and Brown uses a wire heated by an applied current as an airflow sensor:  air flow from a breathing patient cools the wire, changing its resistance. Similarly, a voltage applied to a membrane induces a current flow; the ratio of voltage to current is a resistance. Such relations are best measured using a Bridge, and if the bridge is made solely of resistors it is called a Wheatstone Bridge.

Usually drawn as a diamond, this configuration of resistors is “balanced” when V+ – V- = 0. If Rtest then varies a little, a differential amplifier across V+ and V- will register a potential difference proportional to
the change in Rtest.

The impedances can have capacitance and/or inductance associated with them, in which case the bridge can measure both energy storage and  resistive loss in an element.

A return path to ground for (DC) bias currents is automatically provided by this circuit to prevent saturation.

Well there you have it, a few common amplifier configurations and some useful terms pertaining to them.  Remember important concepts such as amplifier saturation, Input Impedance, Output Impedance, and Gain.  A solid understanding of these concepts is sure to impress somebody!  Amplifiers  part II will continue to elaborate on more fun amplifier concepts.

References: References: Carr and Brown ch. 7; Webster chs. 3, 6; Neamen, Electronic Circuit Analysis and
Design (McGraw Hill, 2001) ch. 9

Cartesian:

Cylindrical:

Spherical:

In most cases you will only need to work from Cartesian to Cylindrical or Spherical OR back, so I will only supply those equations. If you need to work between Cylindrical and Spherical, it would be one more simple step working from one of those, to Cartesian, then on to the other.

Cartesian Cylindrical:

Cartesian Cylindrical:

Cartesian Spherical:

Cartesian Spherical:

Plug the values from any given points into the correct equation to convert to a different type of coordinate system.

You would begin by finding the vector between these two points. The direction of this vector may be important so look for key words such as . Once we have established the direction we’re going in, to in this case, we subtract the beginning point from the end point. . This will give us the vector we are looking for. The next step would be to convert this vector into a unit vector, by dividing it by it’s magnitude.

These are the two formulas we are looking at:

Note:
Here I use as my vector from to and denotes the unit vector from to .

Implementing these formulas:

Hmm.. The End

Since these integrals can be tedious and certain functions tend to reoccur, a table of Laplace Transforms has been linked:

Laplace Transforms Table

This table can be a little complex to use at first so an example is provided below to get you started. In this problem we implement the Laplace Transform and Inverse Laplace Transform to solve for .

The first step is to take the Laplace Transform of both sides of the equation. Use element 1 of our table for the right side and element 18 for the left side. Note that the initial conditions are necessary to take the Laplace Transform of the left side.

Inputting our initial conditions:

Assuming you are an engineering student and can do a little alegebra, our next step is to find the terms that have in common and factor it out. Our goal is to find because, after all, :

After solving for and factoring the denominator:

Now we arrive at the trickier part of this procedure. We must take the Inverse Laplace of to find . If our function does not match anything in the table, such as this case, factoring is a good place to start. This problem can easily be factored using the function on your TI-89. Just go to and enter your function in parenthesis. Using this function:

Noting and , we can take the Laplace Transform of each term independently and also manipulate the constant terms if necessary or just pull them out. Using the 2nd property in our Laplace Transform table:

To check your work you can plug into the original, differential equation and at we find that .