5/10/16 - Introduction to Phasors

Today we start off with a sine function and are given some points. We are asked to find the function of the curve.

Then we give it an offset and find the function with the offset.

We're given our first introduction to the phasor domain here. Since we know that the sine and cosine functions differ by a 90 degree phase shift, we can set up our phasor domain axes as such and then combine the two vectors. We know that their combined magnitude is the root sum square and the angle is arc tangent of the vertical vector over the horizontal vector.

We have the formula for addition of angles in a sine function (which didn't help us for this problem) and voltages given as sine functions. We are told to find the resultant vector and draw it out on the board.

Here we define a phasor, we can say that its components are a real and jimanginary number. The phasor magnitude is simply the root sum square of the real and jimanginary magnitudes and the angle is the arch tangent of the jimaginary over real.

General rules for phasor operations.

Here we attempt our first problem with phasors, we know it's easier to convert them into rectangular and then add, then convert back into polar. As noted by our comment, it only took three tries to attain the correct answer.

There is also a small drawing of a guillotine.

Now applying this knowledge to circuit analysis, we can convert a current or voltage function of time into the phasor domain, but we lose the frequency in doing so.

Another example, we can see that multiplying by the jimaginary results in a 90 degree phase shift and we calculated a phasor from the current given in complex form. We almost had the answer, but didn't account for what region the phasor was in. As you can see by our facial reaction inscribed with detail, we were fairly disgruntled.

Another addition problem, making note of the sin, we modify our functions accordingly and convert to add.

Last problem, we are asked to find the the current across and inductor and are introduce the concept of impedance. We're told that the impedance of an inductor is defined as the jimanginary factors multiplied by the angular frequency and inductance.

5/5/16 - RLC Circuits Cont.

Today, we start off with a RLC circuit and solve for the current across the inductor.

First, we know the initial current and voltage at t = 0, and use it to solve our current equation to find the first constant. 

We take the derivative of the function and set it equal to the voltage over the inductance to solve for our second constant.

Since there is no resistance, out nepper frequency is zero and the response frequency is purely the resonance frequency. We plug it in and then solve for our equation for current across the inductor.

We also have a rude comment from our friendly neighbors.

Now we have another example but the RLC components are now in parallel. We solve for our nepper and resonance frequencies and find our initial conditions. Then we use those values to solve our general differential equation. Finally, since it's being switched on, we say that the value approaches the value at t = ∞.

Lastly, for our lab activity, we set up the circuit depicted on the board on the breadboard. We're observing the current over time with a the alternating signal shown below.

We calculated the first zero of the function's derivative to be about half a mini second.

The input function.

The output on the oscilloscope.

A close up of the function, we can see the first zero of the function occurs around the calculated time.

Here we solve the equation to find zeros with a computer.

5/3/16 - Solving RLC Circuits

Today, we learn about solving RLC circuits, which are second order differential equations.

Here we derive the formula for the output voltage and we can see that it forms a second order differential equation with all three RLC values.

Here's a problem where we are asked to find the voltage across the capacitor. We know that the instantaneous voltage is going to be the same as the resistor and the current will be the voltage over the total resistance.

We find the voltage drop and current drop per second due to the inductor and capactior.

And then finally, we know that at t = ∞, there will be no current across the capacitor and the voltage will be the same as the source.

Here we are to find the characteristic roots of the function. Since alpha is greater than omega_0, we can say that the circuit is overdamped. We then use the equations to find the first and second constants for the solution. 

Lastly, we are to solve for when all three elements are in parallel.

Alpha and omega_0 can be solve for and then used to solve for the constants.

Lastly, the lab of the day was to hook up all three elements in parallel. And since the nepper frequency is less than the resonant frequency, the circuit is underdampped. We can observe the voltage become a sinusoidal function with a exponentially declining amplitude.

4/26/16 - Integrating Op Amp Circuits

Today we take a twist on our standard inverting amplifier, instead of a resistor, we can add a capacitor instead. Solving for the ideal case nodal analysis, we can see the output voltage is equal to a constant times an time integral. If we have a small value for 1/RC, the gain will be very high and quickly saturate.

Given some input signals, we predict the output of the functions by taking their integral.

Now we set up an actual circuit and hook up the oscilloscope to measure our op amp.

Here we can see we are putting in a 100mV sine wave at 1kHz.

Our measured values for the circuit elements.

Here we see the there is a difference in the amplitude of the output voltage as well as there being a phase shift of about 90 degrees from the integral.

Here we learn about the impulse function. First, we take a unit step function and then find its derivative. We can see that it has a constant value of zero, except at the point where it switches. We define the infinite slope to be the unit impulse function and define it to have an area of 1.

Our board's version of the above graph.

Here we are given a problem where a capacitor has an input of a step function and rising function. We can define the derivative of the rising function to be the step function and write the current across the capacitor. We then draw the graphs of the functions.

There is also a peanutt to the side of the board.

Next, we have a similar problem where the voltage is given and we are to find the current functions.

Next, we have a circuit to work with. We are told that the voltage at t < 0 is zero and at t = ∞ is 5/4V. We find the time constant and use it to solve for the voltage function. We then use that to find the current function.

Lastly, we have another problem except with an inductor. We are given the current and find the time constant. Solving for the current as a function of time and then multiplying it by the resistance to find the voltage. Unfortunately, we obtained the wrong answer as denoted by the sad face.

Also, we have an unknown drawing on our board at this time.

4/14/16 - Inductor and Capacitor First Order Cicruits

We are told that inductors follow the same rules to combine as resistors.

Parallel : Add in inverse

Series: Add normally

He is the equivalent inductance from the above circuit.

Next we look at the equations for the voltage and current across inductors and capaticors, as well as their time constants.

Here we can say that the voltage across the capacitor has to be equal to the negative voltage across the resistor, and simplifying yields a first order differential equation.

Here we solve for the voltage across an capacitor as a function of time. We find that the voltage depends an exponential function with a time constant of 1/RC.

We are told that a general rule of thumb for a capacitor to be completely depleted in about five time constants. Solving for 1% of the potential left on a capacitor, we find that it is roughly 4.6 time constants. Rounding up to 5, we can say that after 5 time constants, there will be less than 1% of the original potential left on the capacitor.

Since we define power in terms of Watts, which is volts times amperes, we can find the rate of energy leaving the capacitor as a function of time.

Given a simple circuit, we calculate the voltage and current as a function of time based on the resistance and capacitance values.

Given a range of values for resistance and capacitance values, we calculate the range of time constants that can be produced from it.

Here we hook up the Analog Discovery to a simple circuit with two resistors and a capacitor, we hook up the oscilloscope probes across the capacitor to measure the potential.

We set up Waveforms' oscilloscope function to have a trigger of about 1V and record the climb in potential across the capacitor leads. We can see that it is an exponential function.

Same deal with letting it discharge, an exponential function.

Next, we set up a square wave as an input to the circuit, we find the time constant to be roughly 15ms.

There is also a Kirby Eric in the corner.

Here is the output of the square function as an input.

An attempted close up of the graph, didn't turn out too well from a different phone.

Here we expanded the curve to measure approximately how long it took for the capacitor to fully charge.

We measured that it took approximately 75ms, which is just about 5 of the time constants we calculated.

The zoomed in discharage graph.

Next, we tackle the current across an inductor. We can use our methods to solve a first order differential equation and find the current as a function of time across the inductor. As all general solutions to first order differential solutions, the solution follows a exponential format similar to the capacitor.

Next, we given a basic circuit with a resistor in series with a rather large inductor of 5H, we calculate the time constant and then right the current as a function of time. We then integrate the current function, and multiply by some constants to find the voltage as a function of time.

Now a more complicated circuit, we see the DC voltage source is disconnected at t > 0. We find the equivalent resistance for the resistors and then calculate the time constant. We use a basic current divider equation to find the initial current and solve for our current function.

So we pluck a inductor in the same spot as the capacitor and now measure the voltage across the inductor.

We input a square wave with a frequency of 100kHz and observe the outcome.

We can see the the voltage drops fast across the inductor, which follows our exponential equation.

The charge and dischrage of the inductor due to the wavegen.

Here we calculated our time constant for the circuit and measured the actual time it took. We found it to be roughly 5 time constants like predicted.