Facebook
Google
GitHub
LinkedinMathematics for Electronics
Calculus for Electric Circuits
30 questions By Tony R. Kuphaldt
-
Question 4 of 30
∫f(x) dx Calculus alert!
One of the fundamental principles of calculus is a process called integration. This principle is important to understand because it is manifested in the behavior of capacitance. Thankfully, there are more familiar physical systems which also manifest the process of integration, making it easier to comprehend.
If we introduce a constant flow of water into a cylindrical tank with water, the water level inside that tank will rise at a constant rate over time:
In calculus terms, we would say that the tank integrates water flow into water height. That is, one quantity (flow) dictates the rate-of-change over time of another quantity (height).
Like the water tank, electrical capacitance also exhibits the phenomenon of integration with respect to time. Which electrical quantity (voltage or current) dictates the rate-of-change over time of which other quantity (voltage or current) in a capacitance? Or, to re-phrase the question, which quantity (voltage or current), when maintained at a constant value, results in which other quantity (current or voltage) steadily ramping either up or down over time?
Reveal answerIn a capacitance, voltage is the time-integral of current. That is, the applied current “through” the capacitor dictates the rate-of-change of voltage across the capacitor over time.
Challenge question: can you think of a way we could exploit the similarity of capacitive voltage/current integration to simulate the behavior of a water tank’s filling, or any other physical process described by the same mathematical relationship?
Notes:The concept of integration doesn’t have to be overwhelmingly complex. Electrical phenomena such as capacitance and inductance may serve as excellent contexts in which students may explore and comprehend the abstract principles of calculus. The amount of time you choose to devote to a discussion of this question will depend on how mathematically adept your students are.
Hopefully, the challenge question will stir your students’ imaginations, as they realize the usefulness of electrical components as analogues for other types of physical systems.
-
Question 5 of 30
∫f(x) dx Calculus alert!
One of the fundamental principles of calculus is a process called integration. This principle is important to understand because it is manifested in the behavior of inductance. Thankfully, there are more familiar physical systems which also manifest the process of integration, making it easier to comprehend.
If we introduce a constant flow of water into a cylindrical tank with water, the water level inside that tank will rise at a constant rate over time:
In calculus terms, we would say that the tank integrates water flow into water height. That is, one quantity (flow) dictates the rate-of-change over time of another quantity (height).
Like the water tank, electrical inductance also exhibits the phenomenon of integration with respect to time. Which electrical quantity (voltage or current) dictates the rate-of-change over time of which other quantity (voltage or current) in an inductance? Or, to re-phrase the question, which quantity (voltage or current), when maintained at a constant value, results in which other quantity (current or voltage) steadily ramping either up or down over time?
Reveal answerIn an inductance, current is the time-integral of voltage. That is, the applied voltage across the inductor dictates the rate-of-change of current through the inductor over time.
Challenge question: can you think of a way we could exploit the similarity of inductive voltage/current integration to simulate the behavior of a water tank’s filling, or any other physical process described by the same mathematical relationship?
Notes:The concept of integration doesn’t have to be overwhelmingly complex. Electrical phenomena such as capacitance and inductance may serve as excellent contexts in which students may explore and comprehend the abstract principles of calculus. The amount of time you choose to devote to a discussion of this question will depend on how mathematically adept your students are.
-
Question 6 of 30
∫f(x) dx Calculus alert!
Both the input and the output of this circuit are square waves, although the output waveform is slightly distorted and also has much less amplitude:
You recognize one of the RC networks as a passive integrator, and the other as a passive differentiator. What does the likeness of the output waveform compared to the input waveform indicate to you about differentiation and integration as functions applied to waveforms?
Reveal answerDifferentiation and integration are mathematically inverse functions of one another. With regard to waveshape, either function is reversible by subsequently applying the other function.
Follow-up question: this circuit will not work as shown if both R values are the same, and both C values are the same as well. Explain why, and also describe what value(s) would have to be different to allow the original square-waveshape to be recovered at the final output terminals.
Notes:That integration and differentiation are inverse functions will probably be obvious already to your more mathematically inclined students. To others, it may be a revelation.
If time permits, you might want to elaborate on the limits of this complementarity. As anyone with calculus background knows, integration introduces an arbitrary constant of integration. So, if the integrator stage follows the differentiator stage, there may be a DC bias added to the output that is not present in the input (or visa-versa!).
⌠ ⌡ d dx[ f(x) ] dx = f(x) C In a circuit such as this where integration precedes differentiation, ideally there is no DC bias (constant) loss:
d dx ⌠ ⌡ f(x) dx = f(x) However, since these are actually first-order “lag” and “lead” networks rather than true integration and differentiation stages, respectively, a DC bias applied to the input will not be faithfully reproduced on the output. Whereas a true integrator would take a DC bias input and produce an output with a linearly ramping bias, a passive integrator will assume an output bias equal to the input bias.
fTherefore, the subsequent differentiation stage, perfect or not, has no slope to differentiate, and thus there will be no DC bias on the output.
Incidentally, the following values work well for a demonstration circuit:
Footnotes:
If this is not apparent to you, I suggest performing Superposition analysis on a passive integrator (consider AC, then consider DC separately), and verify that VDC(out) = VDC(in). A passive differentiator circuit would have to possess an infinite time constant (τ = ∞) in order to generate this ramping output bias
!
