mattyjay7

Have you noticed the number 70 is found throughout the bible in both the old and new testament? Here I unpack some of those and attempt to explain why the number is used and draw connections between it's use in various places in the bible, with a particular focus on its use in the book of Daniel, chapter 9. In a prior video I had discussed the number 12 in the bible It is found most prominently as the number of tribes, which form the basic structure of the Old Testament, And as the number of disciples of Jesus, who founded the early church Well, it turns out the number 70 is involved there too. Recall when the twelve tribes, or at that time, just the sons, of Israel, entered Egypt? As you might imagine, they weren't children at the time, and it wasn't just Israel's immediate family that went into Egypt, it was his extended family including grandchildren, and did you know their were a total of 70 of them at that time? And fast forward a bit to the 12 disciples, or followers of Jesus. It appears Jesus had a sort of inner circle of 3 disciples he was closest to, including James, John and Peter. Well, he also had a sort of outer circle of followers that weren't as close as the 12, and guess how many there were? Yep, 70. So in this way it appears 70 is connected to the number 12 as this size of this next outer ring from both the tribes and the disciples. It's used several other places in the Hebrew scriptures for the number of sons prominent leaders had, or the number of leaders of Israel, and in other symbology, but why 70? One reason is it is described as the life span of a man. Psalm 90:10 The years of our life are seventy Isaiah 23 Seventy years, like the days of one king As a sort of an addendum I also discuss the mathematical or scientific basis or significance for the significance as it relates back to the ancient year of 12 months of 30 days each for 360 days/year. The number is perhaps most prominent in the prophecy recording in the book of Daniel, chapter 9. I describe how the '70 weeks' in this chapter corresponds to 490 years of disobedience by Israel, 70 years of captivity in Babylon, and 70 'weeks' of years until the Messiah comes and sets everything right. I conclude by noting the fulfillment and significance of this prophecy, and about how Isaac Newton wrote about it. Hope the video is helpful, take care, -Matthew

This video follows up on a prior video regarding the number 12 and how we tell time, and is followed by a video on the the number 12 in music. The focus here is on the usage of the number in various religions, including Astrology, Judaism, Islam, Christianity, and Greek Mythology. I argue the prominence is based on 12 lunar cycles in a year. From the Hebrew Scriptures I discuss Abraham and his two sons. Ishmael having 12 sons and Israel having 12 sons, making up the 12 tribes of Israel and being allotted twelve regions of land. I discuss the connection between the numbers 13/12 in our solar system, and various other places the number 12 comes up in the Hebrew Scriptures. I then discuss the number 12 in the New Testament, most prominently with the 12 Disciples, and also including several references to the number 12 in the book of revelation. I finally discuss the number 12 in Greek Mythology including the 12 Titans and 12 Olympians, and how this is similar to the pattern in old and new testament.

The number 12 shows up prominenty in our calendar, how we we tell time, several major religions, and even in music. This first video is on the science behind it, The second one regarding its usage in religions And the third it's usage in music, which gets a little more mathy And what I'd like to argue is that this number showing up is not a coincidence There is a reason behind it, Before we get to the importance of the number 12, let’s start with a bigger number, 360 A number you’re probably familiar with, with regards to the unit circle, represented by 360 degrees around the circle. But why 360 degrees in a circle, well, The ancient Babylonians’ used this and they also considered there to be 360 days in a year, just like the ancient Hebrews did. It’s very interesting they chose this isnt’ it? Because we know that a year is based on earths full orbit, or circle if you will, around the sun. 360 days (off by 5 technically) then the seasons repeat, but it doesn’t seem that our heliocentric understanding of the solar system came until much, much later than the Babylonians, so it’s very interesting they chose this number of days in a year to represent the number of degrees around the unit circle. The Babylonians also used a sort of base 60 number system and 360 was 6x60’s, so it’s a very mathematically convenient number. So 360 can be divided evenly by 60, or 6, or by many other numbers. It is in fact regarded as a ‘highly composite number’ in that it has more divisors than any smaller number 1,2,3,4,5,6,8,9,10,12,15,18,20,24,30,36,40,45,60,72,90,120,180 and 360 All divisors of 360, lots of them One of those very important divisors of 360 is 12, which is also a ‘superior highly composite number’, in that it has more divisors than any number smaller than it. 12 is of course important in the context of the year, because there are 12 months in a year. And where does the definition of month come from, well, think of the origin as ‘moonth’. This is not just true in English but the word for month in Mandarin and many other, if not most, languages comes from their terms for the moon as well. English Moon Moonth Mandarin 月 (yue) 月 (yue) Indonesian Bulan Bulan Turkish Ay ay So we have 12 months in a year in our calendar, and the ancient Babylonian and Hebrew calendars did too, except their months were all 30 days long. Why 30, well, the moon orbits the earth in a little over 27 days. However, in that time it takes to orbit, the earth has gone about 30 degrees around the sun, so the moon needs to orbit a little bit more than one full rotation to go back to a new moon from the earth’s perspective. The new moon, by the way, happens when viewed from the earth, the moon is not reflecting any light from the sun, so it's basically when the moon falls between the earth and sun So you may notice that if in 30 days, the earth goes about 30 degrees around the sun, then each day the earth moves about 1 degree around the sun, with 12 months of 30 days giving 360 days in a year, each day like the earth moving (about) a degree. So the Babylonians choice of 360 degrees in a circle, after which the circle repeats, was quite ingenious. And our use of the number 12 in keeping time doesn’t stop there of course, We use a circular clock to keep track of 12 hours as the hour hand completes one revolution in a half-day. So, we really like this number 12 when it comes to time and circles, and it shows. But these 12 ‘moonths’, if you will, also give rise to elements of human activity beyond just keeping track of time. Consider the Zodiac signs of astrology, which can be traced back all the way to the same Babylonians discussed earlier, and still constitute a sort of religion of astrology today. Perhaps you're familiar with the 12 Zodiac signs in Astrology, which are: Aries Taurus Gemini Cancer Leo Virgo Libra Scorpio Sagittarius Capricorn Aquarius, and Pisces And each represent a span of 30 degrees, or about a month, of the earth’s orbit around the sun. So the prominence of the 12 signs in this religion clearly trace to the 12 lunar cycles in a year, So you can see that the 12 number is integral to the relation of the sun, moon and earth, which forms the basis for it’s importance in other areas of human existence as well, areas like religion, which is what we’ll look into in the next video. So check that out if you're interested. If some other instances of 12 in science come to mind though, let us know in the comments below. But otherwise, have a good one, and take care.

If you'd like to understand how a low pass RL filter works, you have come to the right place. I show how they work using the falstad circuit simulator, and first giving a quick review on how inductors work. By the end of this video you can see how taking the voltage across the resistor instead of across the inductor, we get basically the opposite effect as a high pass RL filter. This same RL circuit that worked as a high pass filter in the previous video now works as a low pass filter, by taking the voltage across the other circuit element. And this is true, basically, because KVL says the voltage drops across both need to add up to the source voltage, so if the voltage across one element is following the source well, the voltage across the other element isn't, and vice versa

Perhaps you know that the D chord consists of the first (tonic) the 3rd (mediant) and 5th (dominant) notes. The first note is D itself, the second most important note is the 5th 'A' note and also important is the 3rd 'F#' note. But why are these notes important, and why do they sound nice when played together, and why do we call this the D chord, and not the A or F# chord, since they are both played along with the D? This video answers these questions by graphing out the time-domain signals for these different notes as well as the lower-octave d notes. The script is loosely transcribed below: So perhaps you know how to play a D chord on the guitar or piano. On the piano, the D scale looks like this This follows the progression of W-W-H-W-W-W-H steps back from D back to a higher-octave D Perhaps you know that the D chord consists of the first (tonic) the 3rd (mediant) and 5th (dominant) notes So our first note is D itself, the second most important note is the 5th 'A' note And also important is the 3rd 'F#' note But why are these notes important, and why do they sound nice when played together, And why do we call this the D chord, and not the A or F# chord, since they are both played along with the D? Let me try and answer those questions by plotting out these notes with respect to time OK, so lets start by plotting out the D note, which is at 293.665 Hz This plot is just showing how the signal oscillates in time Now let's also include the 5th, or A note, which is at 440 Hz You can see that their peaks periodically align Adding these signals together, you can see where their peaks aligns This is why the 5th is an important note, it has a frequency with peaks that periodically match up with the the D note. Now, lets show the 3rd, or F# note You can see the 3 notes all align at another point If we add all these signals up, we get this So you can see they add up constructively at periodic points in time This is why these three notes sound nice together, they constructively combine at periodic points in time, so they sort of match up with each other in time. But if we just considered these points where they added up, what note would this correspond to, with peaks in the same places Well, lets show the D note two octaves down This is two octaves down from d because the frequency is divided by 2^2, or 293.665/2^2 = 73.4 Hz You can see this note also has peaks at the same place So that is why we call this the D chord, because those three notes constructively combine at the peaks of the D note, 2 octaves down. So that is the mathematical basis of the D chord, and why the 3rd and fifth notes in the D scale are especially important, why this particular combination of notes is called the D chord, not the F# or A chord. The same applies to other chords like the E or F chord, where the 3rd and fifth notes of the scales are important of those scales are important for the same reason.

This video introduces the basic concept of analog to digital conversion. I start by providing some motivation for such a conversion. I then plot out an example analog signal in the time domain, then discretize the signal in the dimensions of time (x-axis) and voltage (y-axis). I then show how we can find the closest discrete voltage value for each time interval, and how that value is transcribed into a digital signal on the same time axis.

Basic introduction to transducers and a quick example of amplifying a microphone signal to a computer speaker. Below is the basic format: Sensor to electric signal -- amplify electric signal and change bias – thermometer output thermocouple -- amplify electric signal and change bias – thermometer output 2 to 4 mV corresponds to 0 to 100 degrees Celsius -- amplify electric signal and change bias – thermometer output, 0 to 5 V corresponds to 0 to 100 degrees Celcius Here we can't just multiply by K because, say it's 0 degrees and we needed to translate 2 mV to 0 V, we would have to multiply by 0, but of course then every temperature would give us 0 output voltage. So we have to introduce a bias as well. To do this it may help to create a little table: V_out = K V_in + B So you may notice this looks like an equation for a line, y=mx+b and we can just consider this like a graph with V_in as my x and V_out as my y So we have two points we need to connect And you notice if we a draw a line through these two points we get an intersection on the x axis at our B value. Two find this equation we can first find the slope as the rise/run Rise = (V_out2-V_out1) = (y_2-y_1)= (5-0) = 5 Run = (V_in2 – V_in1) = (x_2-x_1) = 0.004-0.002= 0.002 So our slope will be 5/0.002 = 2500, which will be our b value Then use our point-slope formula Y-y_1= m (x-x_1) Y-0 = 2500 (x-0.002) Y=2500x-5 So our m value of 2500 is our K And our b value of –5 is our big B And our diagram for the transducer would look like this:

A basic introduction to transducers is provided along with solving for the amplification required in a basic transducer between a microphone input and speaker output. Below is basically what's covered. A transducer converts energy from one form to another Or it is "a device that converts variations in a physical quantity, such as pressure or brightness, into an electrical signal, or vice versa" microphone amplification example Sound to electric signal -- amplify electric signal – speaker (12 W, 8 Ohms) (like a computer speaker) Input Voltage range from 0 to 250 mV -- amplify electric signal – speaker (max 10 V) To amplify correctly, we need to multiply 0.25 * K = 10 V So K = V_out/V_in = 10/0.25 = 40 Volage range from 0 to 250 mV -- amplify electric signal by 40 – speaker (max 10 V) In this case, we'd call that amplification value K=40

This video demonstrates a rudimentary circuit animation for converting from 120 Vrms AC to 5 V DC (like AC to DC conversion for a USB charger). The falstad circuit simulator is used to demonstrate. Below is the basic script: o the goal is to get about a constant 5 V DC output here from the usb side, but the challenge is we are plugging this directly into the wall at 120 Vrms. So let's go ahead and do a rudimentary model of this in falstad. So we have our AC source here. Notice it is maxing at 170 V because it is coming in at 120 Vrms Now if we look at the voltage across our load here, it is the same thing, because they are directly connected. This is not quite what we want coming out of the usb side, right, so the first thing we do is add a transformer. Now this transformer has an a value of 29 so it is reducing the voltage across our load down to a little over 5 V, so that's the first step. But we want DC here so the next thing we can do is get rid of the negative voltage by using a 'rectifier', which only allows current to flow in one direction. For this, we'll use a 'diode'. OK, so this diode only allows current to travel in one direction. So it doesn't let the current go backwards through it. So the current is only traveling top to bottom through our load, so there is only a voltage drop from top to bottom across it. So you can see in this plot down here that the diode basically cut off the negative portion of our input AC signal. You may imagine it would be more efficient to invert that negative voltage instead of losing it, and we could do that with a more complicated setup of diodes like this. So you can see this setup inverts those negative voltages and would be more efficient, but let's just stick with our simple setup here for the sake of simplicity. So back to this setup, we want a constant voltage, not something going up periodically like this. So next component we can add is a capacitor. So now we through in a capacitor in parallel, and this helps us out a lot, because you can see the voltage across our load is pretty flat now, just above 5 V. But how did the capacitor do this? Well you can tell by looking at the current. When the diode lets a pulse of current through, that current is going mainly to the capacitor to charge it up. When the diode is not letting current through, the capacitor is discharging into our load. So that capacitor is periodically charging and discharging, allowing a fairly constant current flow through the load, which you can see, stays at a relatively constant voltage. To flatten that further you could use a regulator to sort of shave off the top of the curves. Now this is a very rudimentary AC/DC converter, and there's lots of other components to a real AC/DC converter to make it more effective and efficient, like that more-complicated diode setup.

I discuss the following: why transformers are important, basic physics of transformers, circuit depiction of transformers, how N values affect current and voltage transformation, how transformers work with a DC source, and a simpler depiction of circuits with transformers.

A partially graphical but mainly mathematical explanation of how we get the equation for RMS voltage

This video provides a graphical explanation of what RMS voltage is and why it ends up being basically the peak voltage divided by the square root of 2. The video uses some DESMOS plots to show how a sinusoidal voltage corresponds to power and energy, how that energy compares to the energy generated from a DC voltage source, and backs up to finding a comparable equivalent voltage, or RMS voltage.

This video shows how to use Kirchhoff's Voltage Law (KVL) to find voltages for a simple circuit with 3 loops. Below is the basic script: OK so let's go ahead and apply KVL to this circuit here. So notice this symbol is for a voltage source and these other symbols are for resistors. So if current is going clockwise here, it makes sense that the + and – are on different sides here right, because the voltage source adds voltage, while the voltage will drop across the resistors. So for a simple example, let's say we know V_1, V_2 and V_5 in this case, and we want to find the voltage drop across V_3 and V_4. For this we can use KVL. So first thing we do is draw the loops. There are 3 closed loops here, let's draw them out and call them loop 1, loop 2 and loop 3. For loop 1, we start in the bottom left corner and start adding the voltages, knowing they will sum to 0. Consistency with the +/- signs is very important here, because as discussed, the voltage source behaves differently than the resistors. So for consistency we start with the sign we run into first along our clockwise path. For V_1, we hit the minus sign first, so we start our equation with –V_1. Next we hit V_2 on the plus side, so we add V_2, then we hit V_3 on the plus side, so we add V_3. This gets us back to our starting location, so we know based on KVL these must sum to 0. That's our first equation. We can either use loop 2 or 3 to solve this, so let's just use loop 2 since it maybe helps to clarify a bit more. For loop 2, we start at the bottom left. We first hit V_3 and since we hit on the negative side, we start with –V_3. Now you may be thinking this is a resistor so we should always consider these positive, but this is important, you just have to be consistent with the +/- sides we assigned and with the direction of our loop. So we start with –V_3 first, then +V_4, then +V_5 and we are back to our starting location, knowing these sum to 0. Now this is all we need, 2 equations and 2 unknowns. We first solve for V_3 from our loop 1 equation. Then plugging that V_3 value into our loop 2 equation, we solve for V_4, and we're done. Now if we had 3 unknowns, say we weren't given our source voltage V_1, we could have solved for that too by also using the loop 3 equation, which would have given us 3 equations to solve for 3 unknowns. But that provides a basic example of how to use Kirchoff's Voltage Law, so hope it was helpful, and until next time, take care.

Basic explanation of Kirchoff's Voltage Law by analogy to elevation while taking a hike. Below is the script: Imagine you are hiking this terrain here. Say the trail head is here, you go out to one of the peaks, head back down a different way, then go back down. You could call that trail a 'closed loop'. Something obvious here is that when you get back, you are at the same place you started, which is at the same elevation. This is analogous to electric circuits. Say we have a circuit here. If I make a closed loop around the circle, I come back to the place I started, This place has a certain electric potential, which is of course the same whether you're starting or ending here. This is the basis of Kirchoff's Voltage law, which states: When going around a loop, you have to get back to the same voltage you started from, or Sum of voltage rises = sum of voltage drops, or ∑Vrise= ∑Vdrop By analogy to this trail, the elevation you gain going up the hill, you drop the same elevation when going back down to the hill to your starting place, or another way to say it is: Sum of all voltage differences around any closed loop is zero, or By analogy you can sum up all the elevation changes across your hike, and if going up is positive and going down is negative, they will cancel each other out There are just a few things to be careful about with this analogy. On a hike you might go up a hill and then back the same way you came, but say we are applying KVL in an electric circuit like this, You are not going to go out and back. This is because in a DC circuit, current will be flowing in one direction only. So you can't pass through the same node twice. Current has to go around.

Let's say you want to buy a car and say you know for sure you want a full-on electrical vehicle because gas prices are killing you. You start looking around and some are cheaper, some more efficient, and some more reliable, and ultimately you go with what feels right based on your priorities. That's a typical way to make a decision like that, but there's a more rigorous method you can use for decisions like this to make sure you are getting the car you actually want. This type of process is perhaps more commonly used in organizations where maybe you want to offer your boss a better supported recommendation than just that it feels right to you. For this, we turn to a decision matrix. Now lets' say we looked up four fully electric cars, a Kia Niro, a Volkswagon ID4 which Elon Musk likes to criticize, and of course a Tesla. Now you do some research and drill down to 5 things you really want in your electric car. Good price, good efficiency, good range, good driving and good reliability. So let's draw out a table and list these out: We have the Kia Niro The volkswagen Id4 And the Tesla Model 3 First thing we'll look at is the low-end basic price on each one of these.. So if this was our only factor, we'd simply go with the Kia Niro. We could show this by giving it a 'normalized score'. If smaller is better: Normalized score = (best price)/(actual price) Kia score = 39500/39500 = 1 VW ID4 = 39500/41230=0.95 TM3 = 39500/46990 = 0.83 Or more generically: Normalized score = (best value)/(actual value) If bigger is better: Normalized score = (actual MPGE)/(best MPGE) Kia score = 112/130 = 0.86 VW ID4 = 93/130 = 0.72 TM3 = 130/130 = 1 Or more generically Normalized score = (actual value)/(best value) Say there are a couple other factors you'd like to consider as well. There was range.... where the Tesla Model 3 comes in better at 310 compared to 240 There's reliability... which the Kia Niro is actually the best at with a full 5/5 according to consumer reports, which is where I'm getting all this info And there's basic owner satisfaction, which the Niro's a tad lower in. (I was going to put their road test scores on here also, but they are all basically the same so it's not a big factor) Now we can get normalized scores for each of these other three components, but which of these normalized scoring methods would we use? Bigger is better for these, so we'd use the same formula we used for the overall MPGE And crunching those numbers, here are the scores. Now we could simply add all these normalized scores up and see who wins, right? We do that and we get.. 4.44 3.85, and 4.44 So we have a tie between the Kia and the Tesla. So either seems like a decent option, but notice we allotted an equal 1 point to each of these 5 characteristics. It's likely these aren't all equally important to you though, right? So in order to differentiate between these, we have to consider which of these are more or less important. So perhaps you're thinking these are all important, but that initial price is importanter, and the 'owner satisfaction' is maybe a little less important since it's rather subjective. So we can now assign weights to each of these factors. There are five factors so if lets say the best possible score would be 1, then the default weight for each factor is 1/5 =0.2. So lets assign that default weight to the MPGE, Range, and reliability. All important. But price is more important than satisfcation, so we boost that to 0.3 for price and lower owner satisfaction down to 1. Now we can determine our new weighted score by multiplying each of the normalized score by the associated weight of it's factor. So for the price weighted score, we multipy Price weighted score = (price weight)(normalized score) Kia Niro price weighted score = (0.3) (1) -0.3 VW ID4 price weighted score = 0.3(0.96)=0.29 Tesla Model 3 price weighted score = 0.30(0.84)=0.25 We can do the same for the other weighted scores, so for example, Kia Niro MPGE weighted score = (MPGE weight) (normalized score) = 0.2(0.86)=0.17 We can then fill in all these weigthed scores accordingly And then we can add up all these weighted scores, and we find that the Kia Niro wins out. So maybe you initially really wanted a Tesla, but you crunched the numbers here and find that, based on what's really the most important to you, the Kia's the car for you. That's why a decision matrix like this can be helpful, because it can help make sure you choose the right product based on both the data and your actual, objective priorities. But then again, maybe you just want the Tesla anyone because it's cool or you’re an Elon Musk fan, and you want to show it off to all your friends, so if you do buy it, now you at least now why. Hope this was helpful, and until next time, take care,

Snells Law is derived based on two things: 1. The fact that electromagnetic waves travel at speeds inversely proportional to refractive index, and 2. Geometry A finite difference time domain simulation is also shown to remind viewers of intuition behind Snell's Law.

Let me help provide some intuition behind Snell's Law. If you're not satisfied with just the intuition I'll show in a follow-up video how the geometry works to get Snell's Law First thing to understand with snell's law is what the index of refraction, n, is n is basically a measure of how fast travels through a medium Our baseline is n=1 for light traveling through space If light traveled half that fast through a medium, the medium would have n2=2 so the speed of light through the medium would be v2=cn2 Or half the speed of light in a vacuum Demo: So lets see how this would work for a plane wave So imagine we have a region with n=1 on top and n=2 in the gray region below the green line The plane wave is starting up here, propagating downward You can see when it hits this new medium, the wave slows down It has the same frequency because the phase at the line has to match But it has different wavelength If the wavelength on top is lambda_1 Then the wavelength on bottom is 𝜆2=𝜆1∗n1n2 , or .71 lambda_1 in this case Now imagine if this plane wave was incident at an angle The part of the wavefront that hits the medium first slows down While the part that hasn't hit it yet keeps going at normal speed The side slowing down causes the wavefront angle to shift And this wave front is perpendicular to the direction the beam is traveling So you can see up here the wavefront is like this, And propagation is perpendicular Now here the wavefront angle shifts since one side slowed down first And the propagation direction is perpendicular, so it also shift This gives rise to Snell's law, which states: n_1sin(\theta_1)=n_2sin(\theta_2) Where n_1 and n_2 are the refractive indecis' as discussed And theta_1 and theta_2 are the angles with respect to the normal line to the interface of the mediums So this is the basic intuition of how it works. If you're not completely satisfied with the inuition and want to really understand where this equation comes from, I'll show that in the next video. It's not that bad, just need a little geometry to show how we get this.

Pro-gay Rogan bashes child indoctrination and supports don’t say gay bill

Video reviews where e comes from, how we deal with exponents, and ultimately explains how e^ix yields a unit circle in the complex plane. Before we get into e^ix, let's first discuss e^x, which I trust you are familiar with it. By way of review though, say we are plotting out a function f(x)=a^x in red here. If a=2 we have this red line here, and the derivative of the function is g(x) = ln(2)*2^x or 0.69*2^x, which is going to be less than the initial function as shown by this plot. But if a=4, then the derivative will be ln(4)*4^x, which would be higher than the original function, So e is of course that special number in the number that if fine-tuned to around 2.7, we get the derivative being the same as the original function. So recall what the exponential function means. If I want to find e^3, that is just e*e*e, right? You could also consider this e^1 * e^1 *e^1, and then you add up all the 1's in the exponent to get e^3. But what if I want e^0.5? Well, based on the properties of exponents, we know that e^0.5 * e^ 0.5 would equal e^1 since the exponents add, so (e^0.5)^2 = e, so we know e^0.5 must be the square root of e. What about e^2.5, well that would be e^1*e^1*e^0.5 = e*e*sqrt(e). But how in the world could you find e^2i, for example. If the exponent is the number of times your are multiplying the number e, what does it even mean to multiple something by itself the square root of –1 times? It almost looks like nonsense right? But to see this we need to turn to a more rigourous way to calculate e^x, like how your calculator does it, through it's Taylor Series expansion, which perhaps you recall as So for example if we want to find e^1, we plug in 1 for all those x values and we get... e1= 1 +11+½ +13∗2+14∗3∗2+… = 1 +1 +½ +16+124+.. = 1 + 1 + 0.5 + 0.167 + 0.042 Which gives you 2.709, which is getting close to our initial e value, or e^1. OK, so instead of x=1, lets simple plug in x = 1*I, or x=I. Now we get e^1i = 1 +i + i^2/2! + i^3/3! + i^4/4! = 1 + i + (-1)/2! +(-i)/3! +1/4! = 1 + i –1/2 -i/6 +1/24 = 1+I-0.5-0.167i+0.042 So this is similar to the numbers we had earlier, except now we have some I's and some negatives sprinkled in here, so to see what this means, let's take a look at the complex number plane where the x axis is our real numbers and the y axis is our imaginary numbers. So we start off by adding 1, then we go up i, then we go back –0.5, then we go down 0.167 in the negative I direction, then we finally go up 0.042, and after all that we are around 0.542+0.833i So when we include the I in there, we are not simply adding more and more values like in the case above with e^1. Now each new value adds and subtracts on the real and imaginary lines so we are sort of spiraling around to our value. So this is why e^ix doesn't take off 'exponentially' and get really big like e^x, it's because these terms in the taylor series are cancelling each other out. In fact you, can notice that each time successive term is multiplied by another I, and that effectively causes our addition to rotate 90 degrees on the is complex plane each time, so we are adding smalling values in a spiral. Crazy thing, well, lots of craziness with e^ix, but this spiral keeps going and ends up putting us on the unit circle here. Since we're on the unit circle, where do you imagine the value of pi/2 might get us? If we plug x=pi/2 into our Taylor Series we get: Which spirals around to 0.02+0.92i, which is pretty close to the point (0,1) on the unit circle, but not quite so ideally we would have added a few more values in the series. Then if we plug in x=pi, we need to add several more terms to get this spiral under control, and adding up the first eleven terms we get: -1.002 + 0.007i So you can see the addition and subtraction of the subsequent Taylor series values spirals us into the correct spot on the unit circle. So e^pi*I represents traveling from (1,0) to the point (-1, 0) along the unit circle over a distance of pi. This connection with the unit circle is why we get Euler's formula E^ix = cos(x)+isin(x) Where the real values are reprsented by the cosine function and the imaginary values by the sin function. Often times e^ix is much easier to use for calculations which makes this identity extremely helpful in electrical engineering. There is so much more to be unpacked with this formula, such as insights from the derivatives of the functions, but, while perhaps still baffling, I hope this gives you a little appreciation for how the exponential function can actually represent the unit circle and correspond to cosine and sine values when you throw an I inside the exponent. And if it does make sense a little, this will help you immensely towards understanding how and why we use this formula to model the physics behind so many aspects of electrical engineering, such as AC circuits or electromagnetic waves.

This video shows how we get the Doppler Equation for RADAR to find the new wavelength, new frequency, or the speed of an object. Some animations are used to provide better intuition. Below is a basic script. Let's consider how to tell what exactly these wavelength and frequency changes are Let's look at the wavelength So here is the propagation of a typical wave when the source is not moving It's wavelength, lambda-not, is the distance it travels over one period Now, let's say the source is moving, so that within the timeframe of this period, it moved up to this point Now the wavelength is bunched in, so let's determine that new wavelength It is the same as the old wavelength minus that distance it traveled That distance it traveled is the velocity of the source times the period of time for the one oscillation we're considering That period is simply one over the originial frequency, f_0, so let's replace that And this is our equation for the new wavelength We could also solve for the observed frequency, f, by subbing in here Then taking the inverse of both sides, we can isolate the observed frequency by multiplying by c Now this is a little tricky, but assuming the velocity is small compared to the speed of light, we can adjust this equation to make it easier as follows And simplify with a 1 here. So this would be the observed frequency given the source is traveling towards the observeer, which would be higher than the original frequency since we're adding, as expected Consider though if were are looking at radar, we need to double this, because the radar wave goes to the object and back, so we get this effect twice 1 more thing to consider, and that is the angle. We determined the equation for when the source is going straight towards the observer But we allready noted this would have the opposite effect if the source was going away from the observer, But also, what happens if the source is traveling perpendicular to the observer? We can see that here that an observer on the top would observe the same frequency in either case. So there should be no 'Doppler shift' of the frequency for an observer here. All we need to do to account for this is consider this angle, theta, and throw it into a cosine. So checking this out, if theta is 0, we do get larger f and lower wavelength if theta is 90 degrees, that cosine term goes to 0, and we get no change. If theta is 180 degrees, we get a negative value, corresponding to a smaller f and larger wavelength So this checks out Another way to consider this equation is if you always just consider theta as the angle between the objects velocity and the line between them, then you can consider this should be additive if they are going more toward each other, and negative if the velocity is taking the observer and source further away from each other. And that is the basic intuition and a basic derivation of where we get the Radar Doppler Equation. Let me know if you have anything to add or any questions, and hopefully now you are properly inspired to dress up as the Doppler effect next Halloween. Take care,

This video introduces the Doppler effect, provides some applications and intuition behind it, and finally introduces the governing equation. Below is a loose script: Time to discuss the Doppler Effect This is a fairly well known effect you may have heard of and an effect you have certainly witnessed whether or not you realized it It's very important in applications from the radar gun the police use to discofer your speed, but perhaps the most profound discovery based on the Doppler effect was that the universe had a beginning. The 'red shift' of distant objects in the universe, due to the Doppler effect, is the most commonly understood reason scientists believe the universe had a beginning, the big bang, which is kind of a big deal. But speaking of such, the TV show big bang theory had this little clip on the doppler effect, which is worth checking out: So perhaps you 'get it' and that explanation was sufficient for you, but if not, let's take a closer look at what sheldon was talking about, by first considering what that sound was. Car: So here is car honking its horn whilst driving by an observer Notice the sound changed as the car approached and passed right? It was higher pitched on the approach, then lower as it left Let's look at why that's the case with a little cartoon below it So the car pulls up and begins honking its horn The sound waves travel down And they reach the observer on the side of the road, denoted by the ear. Now as the car approaches, you can see the sound waves get bunched up, which makes the observed frequency higher, hence the higher pitch And as the car leaves, the sound waves get more spread out, leading to a lower perceived frequency, hence the lower pitch. So that's the basic concept of how motion of a source can affect the perceived frequency Let's look at this in 2D now So say we have a source here of electromagnetic radiation This source emit equally in all directions Observers at any place around this will perceive the same emitted frequency Now, say this source is moving In this case you can see the waves bunch up on the side it's moving toward Wavelength gets smaller Frequency gets higher And spread out on the side its moving away from Wavelength gets larger Frequency gets lower Same concept as the car earlier.

This video gives visual demonstrations of how capacitors and low pass filters work. It shows the circuit, the equations, the time-varying inputs and outputs, and the frequency response plot. Start with visual of capacitor and current and voltage and time Show applied DC voltage and voltage across capacitor Note current initially flows until capacitor 'charges up' to source voltage level Capacitor discharge in same time Note that increasing capacitance makes it take longer to charge Increased capacitance corresponds to decreased impedance per the equation Which means it's not 'impeding' as much on stopping the current from flowing through it. Discuss connection between temporal response and cutoff frequency Go back to 10 u Now instead of a DC input hear let's go with a square wave with 20 Hz This is going to cycle between 5 V to –5 V 20 times a second, so one cycle every 50 ms. Notice the voltage across the capacitor takes a little bit of time to follow this, but it catches up and is the same as the square wave most of the time That means this signal at a frequency of 20 Hz would 'pass'. Now let's crank this up to 85 Hz, which happens to be the cutoff frequency of this filter. Notice that as soon as the capacitor 'catches up' to the source voltage, it gets switched. So at this frequency, the source signal might pass, but it will be attenuated since the voltage across the capacitor is spending most of the time 'catching up' to the changing source Now let's crank this up to 500 Hz, which is beyond the cutoff frequency. Now you can see the capacitor voltage never has enough time to 'catch up' with the alternating source voltage. So some of the signal would pass but it would be significantly attenuated If we crank this up even further to 5 kHz you can see there is basically no response from our output voltage. The 5 kHz input square wave is (almost) completely cutoff. Now let's visualize what we've shown on a frequency response graph. So here the x axis represents frequencies in a log scale, and the y axis represents how much of the source signal gets through to the ouput. So if we go on this graph to the frequency we first showed of 20 Hz, you can see the the loss is small, and this is where our output signal was pretty close to the input signal Then we go up to 80 hZ and you can see this curve has gone down a bit, this repsresents the –3dB where our output has lost about half the power of the input signal. Then we go to 500 Hz and you can see it's further down, And 5 kHz is of course further down still representing more attenuation of our source square wave So you can see how this setup yields a low-pass filter, because it will pass the lower-frequency 20 Hz signal pretty closely, but will almost completely reject the 5 kHz squarewave. So this is how a low pass filter works to allow lower frequencies and reject higher frequencies, and as discussed this is due to the time delay capacitors have with storing electrical energy. If you were able to follow this, then you should have a good fundamental understanding of how capacitors and low pass RC filters work, and if you understand this, then it should be relatively straight forward to extend this understanding to how high-pass filters work, and to how filters using inductors instead of capacitors.

The current through an RLC Circuit including a resistor, inductor and capacitor is found by determining the impedance of the elements, the equivalent impedance of the circuit, and then using Ohm's Law. Understand: Here, we have a circuit with a sinusoidal input, a resistor, an inductor, and a capacitor. The end goal is to use Ohm’s Law (with complex numbers) to find the current. Identify Key Information: • Knowns: We know the rectangular form of the input signal and the resistance, inductance, and capacitance of the components. Additionally, we know the frequency, so we can calculate the angular frequency, 𝜔. • Unknowns: The polar and phasor form of the source. • Assumptions: We are looking for the sinusoidal steady state answer. Plan: Although this problem seems daunting at first, if we can transform to the frequency domain, we can simply use equivalent impedances with Ohm’s Law in the same way we used equivalent resistances with Ohm’s Law in previous lessons. In order to transform to the frequency domain, we have to convert the input signal to phasor form, and then transform the components to impedances. From there, we can solve for the current using Ohm’s Law and find the rectangular form of the current, is(t). Solve: First, we convert the input signal to phasor form (ensuring we use the RMS voltage for the magnitude). Recognizing that the frequency of the circuit will not change, we only need to track the amplitude and phase angle; the amplitude and phase can be used to write the output signal in the standard form at the end of the problem: Ṽ = 169.7 V ∠30∘ = 120𝑉 ∠30° s 𝑅𝑀𝑆 √2 𝑅𝑀𝑆 Next, we find the angular frequency. 𝜔 = 2𝜋𝑓 = 2𝜋(3.18 𝐻𝑧) = 20 radians/s We can use this angular frequency to find the impedances. ZR = R = 40Ω 𝑍𝐿 = 𝑗𝜔𝐿 = 𝑗(20radians/s)(3𝐻) = 𝑗60Ω 𝑗 𝑗 𝑍𝐶 = − 𝜔𝐶 = − (20radians/s)(1.667 ∗ 10−3F) = −𝑗30Ω Now, since the components are in series and we have all the values in Ohms, we can simply add them together to get: 𝑍𝑒𝑞 = 𝑍𝑅 + 𝑍𝐿 + 𝑍𝐶 = 40𝛺 + 𝑗60𝛺 – 𝑗30𝛺 = 40𝛺 + 𝑗30𝛺 This leads to the following equivalent circuit: 120V 30o Now, we can use Ohm’s Law to find the current: Ĩs = Ṽ𝑠 Zeq 120VRMS∠30∘ = (40 + j30)Ω 120VRMS∠30° = 50Ω∠36.9° = 2.4Arms∠ − 6.9∘ We can use the magnitude (being sure to convert from RMS value to peak value) and the phase angle of the phasor to convert to rectangular (standard) form of the current. In doing so, we start with the AC equation for current, 𝐼𝑚cos(360°f𝑡 + 𝜙). Then, we set 𝐼𝑚 equal to the RMS value of the phasor multiplied by √2 and 𝜙 equal to the phase angle of the phasor. Note that the frequency did not change from the source frequency. 𝑖𝑠(𝑡) = (2.4 ∗ √2) cos(360°f𝑡 + 𝜙) A = 3.394 cos(360°(3.18 𝐻𝑧)𝑡 − 6.9∘) A Answer: The current is 3.394 cos(360°(3.18 𝐻𝑧)𝑡 − 6.9∘) A Often, when doing analysis on RLC circuits, we want to find the voltage across a specific component – this will be what we refer to as our output voltage. In this case, we can use a voltage divider. To do this, we can use our voltage divider equation, but we replace the “R”s with “Z”s and use the phasor form of the voltage 𝑉𝑡𝑜𝑡𝑎𝑙. Recall that voltage dividers only work with components placed in series, so the impedances must be in series when using a voltage divider.

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