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19 days ago

Filters are an important component of electronics and communications systems. Chapter 21 presented a full discussion on passive and active filters. Here, we investigate how to design filters to select the fundamental component (or any desired harmonic) of the input signal and reject other harmonics. This filtering process cannot be accomplished without the Fourier series expansion of the input signal. For the purpose of illustration, we will consider two cases, a lowpass filter and a bandpass filter.
The output of a lowpass filter depends on the input signal, the transfer function $ H(\omega) $ of the filter, and the

19 days ago

The Fourier series provides the spectrum of a signal. As we have seen, the spectrum consists of the amplitudes and phases of the harmonics versus frequency. By providing the spectrum of a signal $ f(t) $, the Fourier series helps us identify the pertinent features of the signal.
It demonstrates which frequencies are playing an important role in the shape of the output and which ones are not. For example, audible sounds have significant components in the frequency range of $ 20 \mathrm{~Hz} $ to $ 15 \mathrm{kHz} $, while visible light signals range from $ 10^{5} \mathrm{GHz}

19 days ago

A compact way of expressing the Fourier series in Eq. (A) is to put it in exponential form.
$$f(t)=a_{0}+\sum_{n=1}^{\infty}\left(a_{n} \cos n \omega_{0} t+b_{n} \sin n \omega_{0} t\right) \tag{A}$$
This requires that we represent the sine and cosine functions in the exponential form using Euler's identity:
$$\begin{aligned}\cos n \omega_{0} t &=\frac{1}{2}\left[e^{j n \omega_{0} t}+e^{-j n \omega_{0} t}\right] \\
\sin n \omega_{0} t &=\frac{1}{2 j}\left[e^{j n \omega_{0} t}-e^{-j n \omega_{0} t}\right]\end{aligned} \tag{1}$$
Substituting Eq. (1) into Eq. (A) and collecting terms, we obtain
$$f(t)=a_{0}+\frac{1}{2} \sum_{n=1}^{\infty}\left[\left(a_{n}-j b_{n}\right) e^{j n \omega_{0} t}+\left(a_{n}+j b_{n}\right) e^{-j n \omega_{0} t}\right] \tag{2}$$
If we define a new coefficient $ c_{n} $ so that
$$c_{0}=a_{0}, \quad

19 days ago

To find the average power absorbed by a circuit due to a periodic excitation, we write the voltage and current in amplitude-phase form as
$$v(t)=V_{\mathrm{dc}}+\sum_{n=1}^{\infty} V_{n} \cos \left(n \omega_{0} t-\theta_{n}\right) \tag{1}$$
$$ i(t)=I_{\mathrm{dc}}+\sum_{m=1}^{\infty} I_{m} \cos \left(m \omega_{0} t-\phi_{m}\right) \tag{2}$$
Following the passive sign convention (Fig. 1), the average power is
$$P=\frac{1}{T} \int_{0}^{T} \text { vi } d t \tag{3}$$
Substituting Eqs. (1) and (1) into Eq. (3) gives
$$\begin{aligned}P=& \frac{1}{T} \int_{0}^{T} V_{\mathrm{dc}} I_{\mathrm{dc}} d t+\sum_{m=1}^{\infty} \frac{I_{m} V_{\mathrm{dc}}}{T} \int_{0}^{T} \cos \left(m \omega_{0} t-\phi_{m}\right) d t \\&+\sum_{n=1}^{\infty} \frac{V_{n} I_{\mathrm{dc}}}{T} \int_{0}^{T} \cos \left(n \omega_{0} t-\theta_{n}\right) d t \\&+\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{V_{n} I_{m}}{T} \int_{0}^{T} \cos \left(n \omega_{0} t-\theta_{n}\right) \cos \left(m \omega_{0}

19 days ago

We find that in practice, many circuits are driven by nonsinusoidal periodic functions. To find the steady-state response of a circuit to a nonsinusoidal periodic excitation requires the application of a Fourier series, ac phasor analysis, and the superposition principle. The procedure usually involves three steps.
Steps for Applying Fourier Series:
Express the excitation as a Fourier series.
Find the response of each term in the Fourier series.
Add the individual responses using the superposition principle.
The first step is to determine the Fourier series expansion of the excitation. For the periodic voltage source shown in Fig. 1(a), for example, the Fourier series is expressed

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America, India, Bangladesh

Chapter 11 discussed the self-inductance of a coil. We shall now examine the mutual inductance that exists between coils of the same or different dimensions. Mutual inductance is a phenomenon basic to the
operation of the transformer, an electrical device used today in almost every field of electrical engineering. This device plays an integral part in power distribution systems and can be found in many electronic circuits and measuring instruments.
In this chapter, we will discuss three of the basic applications of a transformer:
to build up or step down the
voltage or current,
to act as an impedance matching device,
and to isolate

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Electrical Engineering and Telecommunications is arguably the origin of most high technology as we know it today. Based on fundamental principles from mathematics and physics, electrical engineering covers but not limited to the following fields:

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America, Pakistan, Canada

A transformer is constructed of two coils placed so that the changing flux developed by one will link the other, as shown in Fig. 1.
This will result in an induced voltage across each coil. To distinguish between the coils, we will apply the transformer convention that
the coil to which the source is applied is called the primary, and the
coil to which the load is applied is called the secondary.
For the primary of the transformer of Fig. 1, an application of Faraday's law will result in
$$ \bbox[10px,border:1px solid grey]{e_p = N_p { d \phi \over dt}} \,\, \text{(volts, V)} \tag{1}$$
revealing

Last 28 days

124

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United States, Philippines, India

We found that voltage sources of different terminal voltages cannot be
placed in parallel because of a violation of Kirchhoff's voltage law.
Similarly,
Current sources of different values cannot be placed in series due to a
violation of Kirchhoff's current law.
However, current sources can be placed in parallel just as voltage
sources can be placed in series. In general,
Two or more current sources in parallel can be replaced by a single
current source having a magnitude determined by the difference of
the sum of the currents in one direction and the sum in the opposite
direction. The new parallel internal resistance is the total resistance
of the resulting parallel

Last 28 days

108

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Kenya, India, United States

Recalling the equation introduced for Ohm's law for electric circuits.
$$ \text{Effect} = {\text{cause} \over \text{opposition}} $$
The same equation can be applied for magnetic circuits. For magnetic circuits, the effect desired is the flux $\Phi$. The cause is the magnetomotive force (mmf) , which is the external force (or "pressure") required to set up the magnetic flux lines within the magnetic material. The opposition to the setting up of the flux $\Phi$ is the reluctance $S$.
Substituting, we have
$$\bbox[10px,border:1px solid grey]{\Phi = {m.m.f \over S}} \tag{1}$$
The magnetomotive force is proportional to the product of the number of turns around the core