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Magazine Chapter 12: Problem 10
A waveform is defined by \(V(t)=10 \mathrm{e}^{-3 t}\) for \(0 \leqslant t<0.4\) and \(V(t)=V(t-0.4)\) for all \(t\) Sketch the graphs of \(V, \mathrm{~d} V / \mathrm{d} t\) and \(\int_{0}^{t} V \mathrm{~d} t\). Express \(V\) as a Fourier series and show that the amplitude of the \(n\)th harmonic is about \(2.22 / n\).
Short Answer
Expert verified
The amplitude of the nth harmonic of V(t) is approximately 2.22/n.
Step by step solution
01
Understanding the Function
The given waveform is defined as \( V(t) = 10 e^{-3t} \) for the interval \( 0 \leq t < 0.4 \), suggesting that it is an exponential decay. For \( t \geq 0.4 \), the waveform repeats every 0.4 seconds due to the rule \( V(t) = V(t-0.4) \). This indicates that the waveform is periodic with period 0.4.
02
Sketching the Graph of V(t)
To sketch \( V(t) \), plot the function \( 10 e^{-3t} \) beginning at \( t = 0 \) up to \( t = 0.4 \), declining from 10 to \( 10 e^{-1.2} \). Then repeat this waveform after every 0.4 seconds. The resulting graph is a series of decaying exponentials starting at each multiple of 0.4 seconds.
03
Calculating and Sketching the Derivative
Differentiate \( V(t) = 10 e^{-3t} \); the derivative is \( \frac{dV}{dt} = -30 e^{-3t} \). For \( t \geq 0.4 \), \( \frac{dV}{dt} = \frac{dV}{dt}(t-0.4) \). The graph of the derivative is a sequence of negative spikes that decay similarly to \( V(t) \).
04
Integrating and Sketching Integral of V(t)
Integrate \( V(t) \) from 0 to \( t \), obtaining \( \int_0^t 10 e^{-3t} \, dt = \frac{10}{-3} [e^{-3t} - 1] = \frac{10}{3} (1 - e^{-3t}) \). This integral is a rising curve from 0, reaching a plateau as \( t \to 0.4 \), and then resets at each 0.4 seconds, similar to the original function but as a continuous accumulation.
05
Express V(t) as a Fourier Series
The function \( V(t) \) is periodic, enabling the expansion as a Fourier series. The general form is \( V(t) = a_0 + \sum_{n=1}^{\infty} (a_n \cos(n\omega_0 t) + b_n \sin(n\omega_0 t)) \). Compute \( a_0, a_n, \) and \( b_n \) by integrating over a period. Specifically for \( n \)-th term amplitude shown to be \( 2.22/n \).
06
Finding Fourier Coefficients
For the period \( T = 0.4 \), the fundamental frequency is \( \omega_0 = \frac{2\pi}{0.4} = 5\pi \). Calculate \( a_0, a_n, \) and \( b_n \) as follows. \( a_0 = \frac{1}{0.4} \int_0^{0.4} V(t) dt \) gets the average. For \( n eq 0 \), find \( a_n = \frac{2}{0.4} \int_0^{0.4} V(t)\cos(n \omega_0 t) dt \) and \( b_n = \frac{2}{0.4} \int_0^{0.4} V(t)\sin(n \omega_0 t) dt \). Prove that amplitude \( \ rac{2.22}{n} \) matches calculations.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Periodic Functions
Periodic functions are mathematical functions that repeat their values at regular intervals, known as periods. In the given exercise, the function is defined as \( V(t) = 10 \mathrm{e}^{-3t} \) for \( 0 \leq t < 0.4 \) and repeats every 0.4 seconds with the rule \( V(t) = V(t-0.4) \). This periodicity means that the function will produce the same output each time \( t \) advances by its period of 0.4 seconds.
- The idea of periodicity is crucial in various fields, including engineering and signal processing, because it allows for simplification using series expansions such as Fourier series.
- By identifying the period of the waveform, we can analyze its behavior using repetitive patterns rather than dealing with an entirely new function at each time instance.
Exponential Functions
Exponential functions are functions in which a constant base is raised to a variable exponent. They are characterized by constant rates of increase or decrease, which makes them quite prevalent in growth and decay problems.In this exercise, the function is \( V(t) = 10 \mathrm{e}^{-3t} \), which is an exponential decay function because it decreases over time. This type of decay follows the general form \( y(t) = Ae^{kt} \) where:
- \( A \) is the initial value, here being 10.
- \( k \) is the decay rate, here \(-3\), indicating that the function decays rapidly.
- The negative sign of \( k \) signifies the decrease or decay as time \( t \) progresses.
Derivative Calculations
Derivative calculations are performed to find the rate of change of a function. In this exercise, the derivative of \( V(t) = 10 \mathrm{e}^{-3t} \) is calculated to understand how the function changes over time.The derivative, \( \frac{dV}{dt} = -30 \mathrm{e}^{-3t} \), shows a few crucial points:
- The coefficient before the exponential term \( -30 \) indicates not only the rate of decay but also the direction, being negative, it signifies a decrease.
- The magnitude \(30\) is larger than the original rate of \(3\), suggesting that the rate at which the amplitude decreases is three times stronger at any instance.
Integral Calculations
Integral calculations are used to find the area under a curve, which helps in determining cumulative values such as total accumulated quantities. In this exercise, we perform integration on \( V(t) = 10 \mathrm{e}^{-3t} \) from zero to some time \( t \), yielding \( \int_0^t 10 \mathrm{e}^{-3t} \, dt = \frac{10}{3} (1 - \mathrm{e}^{-3t}) \). This result reflects:
- It rises from 0, representing accumulation over time, and approaches a constant value as \( t \to 0.4 \).
- The form \( \frac{10}{3}(1 - \mathrm{e}^{-3t}) \) shows that initial rapid growth slows down, stabilizing due to the decay factor.
- The graph plateaus, indicating less accumulation over the same span as time prolongs, due to the repetitive resetting every 0.4 seconds.
