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Chapter 11: Problem 4

Show that \(3\left(\frac{\mathrm{e}^{j \omega}-\mathrm{e}^{-j \omega}}{j \omega}\right)=\frac{6 \sin \omega}{\omega}\)

Short Answer

Expert verified
Question: Prove that \(3\left(\frac{e^{j \omega}-e^{-j \omega}}{j\omega}\right) = \frac{6 \sin \omega}{\omega}\). Answer: To prove the given expression, we applied Euler's formula to convert the complex exponentials into their equivalent trigonometric functions. After simplification, we found that the expression on the left side is equal to the expression on the right side, thus establishing the equality.

Step by step solution

01

Apply Euler's formula to the complex exponentials

Euler's formula states that \(e^{j \omega} = \cos(\omega) + j\sin(\omega)\) and \(e^{-j \omega} = \cos(\omega) - j\sin(\omega)\). Thus, we can replace complex exponentials with their trigonometric equivalents. So, the expression now looks like: \(3\left(\frac{(\cos(\omega) + j\sin(\omega)) - (\cos(\omega) - j\sin(\omega))}{j\omega}\right)\)
02

Simplify the expression

Now, we will simplify the expression by carrying out the subtraction and division: \(3\left(\frac{(\cancel{\cos(\omega)} + j\sin(\omega)) - (\cancel{\cos(\omega)} - j\sin(\omega))}{j\omega}\right)\) \(3\left(\frac{2j\sin(\omega)}{j\omega}\right)\)
03

Cancel out the j terms and simplify further

Now, cancel out the \(j\) term in the numerator and the denominator and simplify: \(3\left(\frac{2\cancel{j}\sin(\omega)}{\cancel{j}\omega}\right)\) \(\frac{6\sin(\omega)}{\omega}\) Now, we see that the expression on the left side has been simplified and is equal to the expression on the right side: \(3\left(\frac{e^{j \omega}-e^{-j \omega}}{j\omega}\right) = \frac{6 \sin \omega}{\omega}\) Thus, the equality is established, and the exercise is solved.

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Most popular questions from this chapter

Use De Moivre's theorem to show that if \(z=\cos \theta+\mathrm{j} \sin \theta\) then (a) \(z^{n}=\cos n \theta+j \sin n \theta\) (b) \(z^{-n}=\cos n \theta-\mathrm{j} \sin n \theta\) Deduce that $$ z^{n}+\frac{1}{z^{n}}=2 \cos n \theta $$ and $$ z^{n}-\frac{1}{z^{n}}=2 \mathrm{j} \sin n \theta $$

Solve each of the following equations leaving (a) \(z^{3}=-1\) (b) \(z^{3}=1\) (c) \(z^{3}-6 \mathrm{j}=0\) your answers in polar form: (a) \(z^{2}=1+\mathrm{j}\) (b) \(z^{2}=1-\mathrm{j}\) (c) \(z^{3}=-2+3 \mathrm{j}\)

Express \(\cos \omega t\) in terms of exponential functions.

Express each of the following in terms of

Consider the complex number \(z=\cos \theta+\mathrm{j} \sin \theta\) (a) Write down the exponential form of this number. (b) By raising the exponential form to the power \(n\), and using one of the laws of indices, deduce De Moivre's theorem.
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