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# Tutor profile: Ohmeko O.

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Ohmeko O.
Electrical Engineering and Computer Science Tutor
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## Questions

### Subject:HTML Programming

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Question:

What are the three ways you can apply CSS to your HTML website?

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Ohmeko O.

### Subject:Electrical Engineering

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Question:

A 75 $$\Omega$$ coaxial transmission line has a length 2.0 cm and is terminated with a load impedance of $$37.5 + j75 \Omega$$. If the dielectric constant of the line is $$\epsilon_r = 2.56$$ and the frequency is 3.0 GHz, find the input impedance to the line, the reflection coefficient at the load, the reflection coefficient at the input, and the SWR on the line.

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Ohmeko O.

Step 1) Write down what we are given: $$Z_o = 75 \Omega$$ coaxial line, Length of Line is 2.0 cm, $$Z_L = 37.5+j75 \Omega$$, $$\epsilon_r = 2.56$$, $$f_c = 3.0$$ GHz. Step 2) Find the reflection coefficient at the load. The reflection coefficient $$\Gamma_L$$ is defined as: $$Gamma_L = \frac{Z_L - Z_o}{Z_L + Z_o} = \frac{37.5+j75 -75}{37.5+j75 +75} = 0.078+j0.615$$ Step 3) Find the SWR of the line. The SWR is defined by the following equation: $$SWR = \frac{V_{max}}{V_{min}} = \frac{Vp(1+|\Gamma_L|)}{Vp(1+|\Gamma_L|)} = \frac{1+0.6202}{1-0.6202} = 4.266$$ Step 4) Find the reflection coefficient at the input. The reflection at the input can be obtained by multiplying the reflection at the load by the delay introduced to the reflection by the length of the line: $$\Gamma_{in} = \Gamma(d) = \Gamma_L e^{-2j\beta d} = 0.602\angle(1.446-2*\frac{2\pi}{\lambda}*2*10^{-2})$$ We have to convert 82.875 into radians for the calculation which equates to: $$82.875^{\circ} * \frac{\pi}{180^{\circ}} = 1.446$$ Thus we get the reflection at the input to be: $$\Gamma_{in} = 0.6202\angle 212.08$$ Step 5) Find the input impedance of the line. The input impedance of the line could be found with this formula: $$Z_{in}(d) = \frac{Z_{L} + Z_{o}\tan(\beta*l)}{Z_{o} + Z_{L}\tan(\beta*l)} = 18.95 - j20.30$$

### Subject:Algebra

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Question:

Find the solutions to $$|3x + 9| > -2$$

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Ohmeko O.

The solutions to the above inequality is: Step 1) You have to realize that you have an absolute value so you have two cases. Step 2) You have two equations to solve: $$3x+9 > -2$$ and $$-3x-9 > -2$$ Step 3) First equation solution to solve is $$3x + 9 > -2$$ $$3x > -11$$ $$x > \frac{-11}{3}$$ Step 4) Second equation solution to solve is $$-3x-9 > -2$$ $$-3x > 7$$ $$x > \frac{-7}{3}$$ Step 5) The solution to the inequality is thus $$x > \frac{-7}{3}$$ and $$x > \frac{-11}{3}$$

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