Determine its output voltage a) -13 v b) 13 v c) 13 v d) None of the mentioned View Answer Answer: a Explanation: The differential output voltage V id V in1 V in2 3-7v -4v.The output of the op-amp in this circuit depends on polarity of differential voltage V 0 -V sat -V ee -13 v.
Integral Of A Square Wave Generator Is TView Answer Answer: b Explanation: The time period of the output waveform for a square wave generator is T 2RC ln(2R 1 R 2 )( R 2 ).
Assume the resistor value to be 10k and find the capacitor value a) 3.9 F b) 0.3 F c) 2 F d) 0.05F View Answer Answer: d Explanation: Lets take R 2 1.16 R 2, therefore the output frequency f o 12RC C 12Rf o 1 (210k1khz) 0.05F. Integral Of A Square Wave Free Certificate OfParticipate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs. He is Linux Kernel Developer SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals Storage Protocols such as iSCSI Fiber Channel. ![]() Integral Of A Square Wave Series Of WeightedVideo transcript so we started with a square wave that had a period of 2pi then we said hmm can we represent it as an infinite series of weighted signs and co-signs and then working from that that idea we were actually able to find expressions for the coefficients for a sub zero and a sub n when n does not equal zero and the B sub ends and evaluating it for this particular square wave we were able to get that a sub n is going to be equal or a sub zero is going to be three halves that a sub n is going to be equal to zero for any n other than zero and that B sub n is going to be equal to zero if n is even and six over n pi if n is odd so one way to think about it youre going to get your a sub youre gonna get your a sub zero youre not going to have any of the cosine terms and youre only going to have the odd sine terms and if you think about it just visually if you look at the square wave it makes sense that youre going to have the signs and not the cosines because a sine function is going to look something like this so a sine function is going to look something like this while a cosine function looks something like looks something like let me make it a little bit neater a cosine function would look something like something like that and so cosine and multiples of cosine of X so cosine of 2x cosine of 3x is going to be out of phase while the sine of X or I should say cosine of Ts and the sines of Ts sine - 2 T sine 3t is going to be more in phase with the way this function just happened to be so it makes sense that our a sub NS were all 0 for n not equaling 0 and so based on what we found for our a sub 0 and our a sub ends and our B sub ends we could expand out this actual and we did in the previous video what does this Fourier series actually look like so 3 halves plus 6 over pi sine of T plus 6 over 3 pi sine of 3t plus 6 over 5 pi sine of 5t and so on and so forth and so a lot of you might be curious what does this actually look like and so I actually just you can type these things into Google and it will just graph it for you and so this right over here is just the first two terms this is three-halves plus six over pi sine of T I notice is starting to look right because our art square wave looks something like it goes it looks something like this where so its going to go like that and then its going to go down to zero and then its going to go and its going to go up it looks something like that doesnt have the pies in the two pies marked off clean these because its going to look something like that so even just the two terms its kind of a decent approximation for even two terms but then as soon as you get to three terms if you add the six the six over three pi sine of 3t to the first turn two terms so if you look at these first three terms and thats looking a lot more like a square wave and then if you add the next term well looks like even more like a square wave and then if you add to that to what we already wrote down here if you were to add to that six over seven pi times sine of 70 it looks even more like a square wave so this is pretty neat we get you can visually see that we were actually able to do it and all kind of just fell out from the mathematics Finding Fourier coefficients for square wave Our mission is to provide a free, world-class education to anyone, anywhere. I know integrating gave us the sum of the differentiated energies, but why the Energy is the square of the signal. In the former case, the total energy stored in EM fields is the integrated modulus of E and B, as described at scienceworld.wolfram.comphysicsElectromagneticField.html. ![]() Charges are generators of a continuous symmetry via Noethers theorem and exist whenever there is a conserved current, e.g. PVI is a statement about the proportionality of current to power. Although voltage can take a constant value, it can also vary as a function independent of the current. Another way to conceptualize it is to think in terms of accounting. If I have a budget and I use a double entry accounting process, debits and credits should sum to zero. Some things I use my money for give me greater marginal utility, analogous to voltage, making Utility analogous to power. For example, energy stored in a capacitor is proportional to square of its voltage (frac12CU2). If that is so, it is not true that this square of signal always gives value of some physically sensible energy. Formally, if S(t) is quadratically integrable function of time, one can always define energy as the above square of the signal. Physically, this is because the square of electric field in the Poynting theorem can be interpreted as EM energy density. When on the path of a signal we place a detector, what impinges on it, is the flux (defined as quantity of incoming energyunit surface in the unit time). So, in order to get what amount of energy impinges on the detector in an interval of time, e.g. Now the things becomes simple because for the plane wave the direction of the product E x H is along the direction of propagation of the wave, and H is proportional to E. This is why, up to some constant factor, we need E2 - of course, for E dependent on t we need E(t)2 - and why we integrate over this. Now, why there appears the absolute square is because sometimes we work with E represented as a complex number. Perhaps you can modify your post by adding the relevant portions of the link. Provide details and share your research But avoid Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. MathJax reference. To learn more, see our tips on writing great answers. ![]()
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