# Tutor profile: Serge K.

## Questions

### Subject: Discrete Math

How many binary sequences of length $$26$$ have equal numbers of $$0$$'s and $$1$$'s?

The number of binary sequences of length $$26$$ that have equal numbers of 0's and 1's is determined by the number of possible ways to choose $$13$$ positions for the zeros out of the $$26$$ available positions (the other $$13$$ positions are set to $$1$$'s). Therefore, the answer is: $$C(26,13)=\binom{26}{13}=10,400,600$$ sequences.

### Subject: Electrical Engineering

Consider the signal $$x(t)=e^{-2t}u(t)$$, where $$u(t)$$ is the unit step function. Determine the energy spectral density of $$x(t)$$, and then deduce the energy of $$x(t)$$.

Let $$X(f)$$ be the fourier transform of $$x(t)$$. The energy spectral density of $$x(t)$$ is $$S_{xx}(f)=|X(f)|^2$$. $$X(f)=\int_{-\infty}^{+\infty} x(t) e^{-i2\pi ft} dt= \int_0^{+\infty} e^{-2t} e^{-i2\pi ft} dt = \int_0^{+\infty} e^{-2(i\pi f+1)t} dt=\frac{e^{-2(i\pi f+1)t}}{-2(i\pi f+1)}\bigg\vert^{+\infty}_{0}$$. Therefore, $$X(f)=\frac{e^{-2(i\pi f+1)\infty}}{-2(i\pi f+1)} - \frac{e^{-2(i\pi f+1)0}}{-2(i\pi f+1)}=0-\frac{1}{-2(i\pi f+1)} =\frac{1}{2(i\pi f+1)}.$$ Hence, $$S_{xx}(f)=\left \lvert \frac{1}{2(i\pi f+1)} \right \rvert^2 =\frac{1}{\left \lvert 2(1+i\pi f) \right \rvert^2}=\frac{1}{4(1+\pi^2f^2)}. $$ Hence, the energy of $$x(t)$$ is $$E=\int_{-\infty}^{+\infty} S_{xx}(f) df= \int_{-\infty}^{+\infty} \frac{1}{4(1+\pi^2f^2)} df .$$ Let $$u=\pi f$$, then $$du = \pi df$$. We apply the substitution rule and get $$E=\frac{1}{4\pi}\int_{-\infty}^{+\infty} \frac{1}{1+u^2} du =\frac{1}{4\pi}\arctan(u)\big\vert^{+\infty}_{-\infty}=\frac{1}{4\pi} [\arctan(+\infty)-\arctan(-\infty)]=\frac{1}{4\pi}\left[\frac{\pi}{2}-(-\frac{\pi}{2})\right].$$ Therefore, $$E=\frac{1}{4\pi}[\pi]=\frac{1}{4}.$$

### Subject: Calculus

Compute the following definite integral: $$\int_{1}^{e} \frac{1}{x(1+\ln^2 x)} dx,$$ where $$\ln x$$ is the natural logarithm of $$x$$, and $$e$$ is Euler's number.

We solve this problem by applying the substitution rule. Let $$u=\ln x$$, then we have $$du=\frac{1}{x}dx$$. Therefore, by applying the substitution we get: $$\int_{1}^{e} \frac{1}{x(1+\ln^2 x)} dx=\int_{\ln(1)}^{\ln(e)} \frac{1}{1+u^2} du=\int_{0}^{1} \frac{1}{1+u^2} du=\arctan(u)\vert^{1}_{0}=\arctan(1)-\arctan(0)=\frac{\pi}{4}-0=\frac{\pi}{4}.$$

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