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)

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).

System Equation:

For example purposes, I will solve a system equation with complex roots.  A system equation with complex roots as a function of will appear in the following format (if it does not, you need to manipulate your equation to be in the form):

Roots:

So we have

which also equals

so your first step is to look at your equation and determine your roots, then write out your equation with constants.

Example with initial conditions and

so now we can write our equation as follows:

In order to solve for and we need to use our initial conditions.  To evaluate the first derivative initial condition, we must first take the derivative of our that we just found.

evaluating this equation with t = 0 and the response equal to -4v, we get this:

evaluating our equation with t = 0 and the response equal to 3v, we calculate

Using these two equations, we calculate our constants:

and

Fill these into our equation to determine the final result.

Now you know how to solve this common differential equations and linear systems problem, determine characteristic roots and modes, and write system equations.

Total response = zero-input response + zero-state response.

Our system:

Where represents our input and represents our output.

The zero-input response, which we will be solving for here, is the system response when the input f(t) = 0 so that is the result of the internal system conditions.  It is independent of the external input f(t).  Therefore, in order to solve a differential equation that represents the zero-input response component of a system, we will need to have the initial conditions of the system, or solve for them.  Keep this in mind while working your problem.

Okay, let’s go ahead and solve a problem shall we?

Our first step is to find a differential equation that will represent the system for .  This is because we only care about the behavior after t = 0 for a zero-input response problem.  We can notice that when our switch opens, this is a natural response.  In other words, we do not switch into a circuit that has a current or voltage source in it, just a capacitor, inductor and two resistors.  In order to solve our ZIR problem, we also need initial conditions do we not?  We will worry about those after we get our differential equation.

The problem tells us that we need to find , so lets look at this circuit after the switch opens and decide what kind of equation we need to write.  Can you see that once the switch opens we have a capacitor, an inductor, and two resistors in series?  If not, you need to study up on parallel and series circuits.  After we combine these two resistors, we have three elements in series, a series RLC circuit.

The sum of voltages around this loop should equal zero by KVL

We should know the following relationships to help us with our substitutions:

;   ;   ; 

Because we are solving for , we will leave this term alone and replace as many of the others as we can.  Note that all four of the elements are in series and therefore share the same current (i).

We need this equation in terms of constants and so we need to perform substitutions on all terms that involve the current (i).  From the relationship we note that and can perform the substitutions that we need:

We divide all terms by the amount in front of our highest order term which in this case is LC.

and our differential equation becomes:

By using the quadratic equation, we can form our characteristic equation

Roots:

Now we have some work to do, and a little bit of thinking.  Using the roots that we just derived, we can now write a general equation for this system, remembering that our roots are complex:

Evaluating this at t = 0

We know that is 2V because after time has elapsed prior to the switch opening the capacitor has become an open and assumes all available potential across it.  When the switch opens, voltage can’t change instantaneously across a capacitor so it initially remains at the same potential as it was prior to the switch opening.

We should also note that the current is -1 Amp.  Why is that?  The initial voltage across the inductor (which is a short prior to switch opening) is 2 V, leaving all of this potential for the 2 ohm resistor (remember that the cap is an open and no current flows in that branch).  Therefore there is a 1 Amp current flowing in the clockwise direction when the switch opens.  CURRENT CAN CHANGE INSTANTANEOUSLY IN A CAPACITOR.  Therefore, immediately after the switch opens, the 1 Amp current flows through the capacitor in the direction opposite it’s polarity, resulting in -1 Amp.

We will need to evaluate the derivative of this function in order to get a second equation (need 2 equations to solve for 2 unknowns and ).  We now know that .

(I have removed the SINE terms because at 0 they evaluate to 0, in this case.)

Therefore,

Now that we have both of our constants, we can insert them into our equation which represents the zero-input response of our system where .

This example was a little complex for teaching a new concept, but I believe it covers a wide variety of important concepts and steps that you will likely see.  If you can do this whole problem, not only do you understand how to evaluate the ZIR of a system, but you also understand characterisic roots and modes for complex roots, series RLC circuit analysis, initial conditions, basic inductor and capacitor physics concepts, as well as frequency and time domain analysis for such circuits.  I hope this helped!

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 .