(2t+2)8(t-1) =. cos (t) 6(t+) S (e*) o(t+1)dt S (?-2) 8(2-t) dt
Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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Three-segment transformers are a type of transformer used to transform voltages of electrical systems into three ranges. Two type transformers are shell-type transformer and core type transformer. In brief, it could be described because of the exquisite kinds of configurations.
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Ever since electricity has been created, people have started using it in its entirety. We see many types of Transformers in the neighborhoods. Some are smaller in size and some are very large. They are used according to their requirements. Many of us have seen the electrical transformer but they do not know what work they are engaged in.
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![Certainly! Here is the transcription and explanation of each item in the image as it would appear on an educational website:
---
### Mathematical Expressions Involving the Dirac Delta Function
1. **Expression 1:**
\[
(2t + 2) \delta(t - 1) = \underline{\hspace{20em}}
\]
This expression involves the Dirac delta function, \(\delta(t - 1)\), which is non-zero only when \(t = 1\). The term \(2t + 2\) is evaluated at this point.
2. **Expression 2:**
\[
\cos(t) \delta \left( t + \frac{\pi}{2} \right) = \underline{\hspace{20em}}
\]
Here, the Dirac delta function \(\delta \left( t + \frac{\pi}{2} \right)\) indicates that the cosine function should be evaluated at \(t = -\frac{\pi}{2}\).
3. **Integral 1:**
\[
\int_{-\infty}^{\infty} \left( e^{t+1} \right) \delta(t + 1) \, dt = \underline{\hspace{20em}}
\]
This integral signifies that the exponential function \(e^{t+1}\) should be evaluated where the delta function \(\delta(t + 1)\) is non-zero, i.e., at \(t = -1\).
4. **Integral 2:**
\[
\int_{-\infty}^{\infty} \left( t^2 - 2 \right) \delta(2 - t) \, dt = \underline{\hspace{20em}}
\]
In this integral, the expression \(t^2 - 2\) is evaluated at the point where the delta function \(\delta(2 - t)\) is active, which is at \(t = 2\).
These expressions and integrals are useful in various applications of signal processing and systems analysis, where the Dirac delta function is often used to model impulse inputs.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fad16e748-6a7d-4448-b034-0f7e391345fb%2F96519ba3-0bb3-481d-b1bf-2f068c2efb50%2F0fpkmf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Certainly! Here is the transcription and explanation of each item in the image as it would appear on an educational website:
---
### Mathematical Expressions Involving the Dirac Delta Function
1. **Expression 1:**
\[
(2t + 2) \delta(t - 1) = \underline{\hspace{20em}}
\]
This expression involves the Dirac delta function, \(\delta(t - 1)\), which is non-zero only when \(t = 1\). The term \(2t + 2\) is evaluated at this point.
2. **Expression 2:**
\[
\cos(t) \delta \left( t + \frac{\pi}{2} \right) = \underline{\hspace{20em}}
\]
Here, the Dirac delta function \(\delta \left( t + \frac{\pi}{2} \right)\) indicates that the cosine function should be evaluated at \(t = -\frac{\pi}{2}\).
3. **Integral 1:**
\[
\int_{-\infty}^{\infty} \left( e^{t+1} \right) \delta(t + 1) \, dt = \underline{\hspace{20em}}
\]
This integral signifies that the exponential function \(e^{t+1}\) should be evaluated where the delta function \(\delta(t + 1)\) is non-zero, i.e., at \(t = -1\).
4. **Integral 2:**
\[
\int_{-\infty}^{\infty} \left( t^2 - 2 \right) \delta(2 - t) \, dt = \underline{\hspace{20em}}
\]
In this integral, the expression \(t^2 - 2\) is evaluated at the point where the delta function \(\delta(2 - t)\) is active, which is at \(t = 2\).
These expressions and integrals are useful in various applications of signal processing and systems analysis, where the Dirac delta function is often used to model impulse inputs.
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