Masalalar — Kompleks Sonlar va Furye

20 ta masala: osondan qiyinga tartiblangan.

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Boshlang'ich (1-8)

Masala 1: Kompleks son arifmetikasi

$z_1 = 3 + 4i$ va $z_2 = 1 - 2i$ uchun $z_1 + z_2$, $z_1 - z_2$, $z_1 \cdot z_2$ ni toping.

Yechim

$z_1 + z_2 = (3+1) + (4-2)i = 4 + 2i$

$z_1 - z_2 = (3-1) + (4-(-2))i = 2 + 6i$

$z_1 \cdot z_2 = (3)(1) + (3)(-2i) + (4i)(1) + (4i)(-2i)$ $= 3 - 6i + 4i - 8i^2 = 3 - 2i + 8 = 11 - 2i$

$\boxed{z_1 + z_2 = 4+2i, \quad z_1 - z_2 = 2+6i, \quad z_1 \cdot z_2 = 11-2i}$

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Masala 2: Modul va argument

$z = -3 + 3i$ uchun modul $|z|$ va argument $\theta$ ni toping.

Yechim

$|z| = \sqrt{(-3)^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$

$\theta = \arctan\left(\frac{3}{-3}\right) = \arctan(-1)$

Lekin $z$ ikkinchi chorakda ($\text{Re} < 0$, $\text{Im} > 0$):

$\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$ (yoki $135°$)

$\boxed{|z| = 3\sqrt{2}, \quad \theta = \frac{3\pi}{4}}$

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Masala 3: Eksponensial ko'rinish

$z = 2 + 2i$ ni eksponensial ko'rinishda yozing.

Yechim

$|z| = \sqrt{4 + 4} = 2\sqrt{2}$

$\theta = \arctan\left(\frac{2}{2}\right) = \arctan(1) = \frac{\pi}{4}$

$\boxed{z = 2\sqrt{2}e^{i\pi/4}}$

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Masala 4: Euler formulasi

$e^{i\pi/3}$ ni algebraik ko'rinishda yozing.

Yechim

$e^{i\pi/3} = \cos\frac{\pi}{3} + i\sin\frac{\pi}{3}$

$= \frac{1}{2} + i\frac{\sqrt{3}}{2}$

$\boxed{e^{i\pi/3} = \frac{1}{2} + \frac{\sqrt{3}}{2}i}$

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Masala 5: Kompleks sonni darajaga ko'tarish

$(1 + i)^4$ ni hisoblang.

Yechim

Avval eksponensial ko'rinishga o'tkazamiz:

$|1+i| = \sqrt{2}$, $\theta = \frac{\pi}{4}$

$1 + i = \sqrt{2}e^{i\pi/4}$

$(1+i)^4 = (\sqrt{2})^4 e^{i \cdot 4 \cdot \pi/4} = 4e^{i\pi} = 4(\cos\pi + i\sin\pi) = -4$

$\boxed{(1+i)^4 = -4}$

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Masala 6: Qo'shma va modul

$z = 5 - 12i$ uchun $\bar{z}$ va $z \cdot \bar{z}$ ni toping.

Yechim

$\bar{z} = 5 + 12i$

$z \cdot \bar{z} = (5-12i)(5+12i) = 25 + 60i - 60i - 144i^2$ $= 25 + 144 = 169 = |z|^2$

Tekshiruv: $|z|^2 = 5^2 + 12^2 = 25 + 144 = 169$ ✓

$\boxed{\bar{z} = 5+12i, \quad z\bar{z} = 169}$

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Masala 7: Kompleks bo'lish

$\frac{3 + 4i}{1 - 2i}$ ni hisoblang.

Yechim

Maxrajni qo'shmasiga ko'paytiramiz:

$\frac{3+4i}{1-2i} \cdot \frac{1+2i}{1+2i} = \frac{(3+4i)(1+2i)}{(1-2i)(1+2i)}$

Surat: $(3+4i)(1+2i) = 3 + 6i + 4i + 8i^2 = 3 + 10i - 8 = -5 + 10i$

Maxraj: $(1-2i)(1+2i) = 1 + 4 = 5$

$\boxed{\frac{3+4i}{1-2i} = \frac{-5+10i}{5} = -1 + 2i}$

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Masala 8: Ildiz chiqarish

$\sqrt{i}$ ni toping (ikkala ildiz).

Yechim

$i = e^{i\pi/2}$ (modul=1, argument=$\pi/2$)

$\sqrt{i} = e^{i\pi/4}$ yoki $e^{i(\pi/4 + \pi)} = e^{i5\pi/4}$

Birinchi ildiz: $e^{i\pi/4} = \cos\frac{\pi}{4} + i\sin\frac{\pi}{4} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i$

Ikkinchi ildiz: $e^{i5\pi/4} = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i$

$\boxed{\sqrt{i} = \pm\frac{\sqrt{2}}{2}(1 + i)}$

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O'rta (9-15)

Masala 9: Furye koeffitsientlari

$f(t)$ to'rtburchak to'lqin: davri $T = 2\pi$, $f(t) = 1$ agar $0 < t < \pi$, $f(t) = -1$ agar $\pi < t < 2\pi$.

$a_0$, $a_1$, $b_1$ koeffitsientlarini toping.

Yechim

$\omega_0 = 2\pi/T = 1$

$a_0 = \frac{2}{T}\int_0^T f(t)\,dt = \frac{1}{\pi}\left[\int_0^\pi 1\,dt + \int_\pi^{2\pi} (-1)\,dt\right]$ $= \frac{1}{\pi}[\pi - \pi] = 0$

$a_1 = \frac{2}{T}\int_0^T f(t)\cos(t)\,dt$ $= \frac{1}{\pi}\left[\int_0^\pi \cos t\,dt - \int_\pi^{2\pi} \cos t\,dt\right]$ $= \frac{1}{\pi}[0 - 0] = 0$

$b_1 = \frac{1}{\pi}\left[\int_0^\pi \sin t\,dt - \int_\pi^{2\pi} \sin t\,dt\right]$ $= \frac{1}{\pi}[[-\cos t]_0^\pi - [-\cos t]_\pi^{2\pi}]$ $= \frac{1}{\pi}[(1+1) - (-1-1)] = \frac{4}{\pi}$

$\boxed{a_0 = 0, \quad a_1 = 0, \quad b_1 = \frac{4}{\pi}}$

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Masala 10: To'rtburchak to'lqin spektri

Yuqoridagi to'rtburchak to'lqin uchun umumiy $b_n$ formulasini toping (faqat toq $n$ uchun).

Yechim

$b_n = \frac{2}{T}\int_0^T f(t)\sin(nt)\,dt$

$= \frac{1}{\pi}\left[\int_0^\pi \sin(nt)\,dt - \int_\pi^{2\pi} \sin(nt)\,dt\right]$

$= \frac{1}{\pi}\left[-\frac{\cos(nt)}{n}\Big|_0^\pi + \frac{\cos(nt)}{n}\Big|_\pi^{2\pi}\right]$

$= \frac{1}{n\pi}[(-\cos(n\pi) + 1) + (\cos(2n\pi) - \cos(n\pi))]$

$= \frac{1}{n\pi}[2 - 2\cos(n\pi)]$

$\cos(n\pi) = (-1)^n$, shuning uchun:

• $n$ toq: $b_n = \frac{1}{n\pi}[2 - 2(-1)] = \frac{4}{n\pi}$
• $n$ juft: $b_n = 0$

$\boxed{b_n = \frac{4}{n\pi} \text{ (faqat toq } n \text{ uchun)}}$

To'rtburchak to'lqin: $f(t) = \frac{4}{\pi}\left[\sin t + \frac{\sin 3t}{3} + \frac{\sin 5t}{5} + \ldots\right]$

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Masala 11: Parsevalning energiya tenglamasi

$f(t) = \sin t$ uchun $\frac{1}{T}\int_0^T |f(t)|^2\,dt$ va Furye koeffitsientlari orqali energiyani tekshiring.

Yechim

Vaqt sohasida: $\frac{1}{2\pi}\int_0^{2\pi} \sin^2 t\,dt = \frac{1}{2\pi} \cdot \pi = \frac{1}{2}$

Furye koeffitsientlari: $a_n = 0$, $b_1 = 1$, boshqalari = 0

$c_1 = \frac{-ib_1}{2} = \frac{-i}{2}$, $c_{-1} = \frac{i}{2}$

Spektral energiya: $|c_1|^2 + |c_{-1}|^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$ ✓

$\boxed{\text{Energiya} = \frac{1}{2}}$

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Masala 12: DFT qo'lda hisoblash

$x = [1, 0, -1, 0]$ uchun 4-nuqtali DFT ni hisoblang.

Yechim

$X[k] = \sum_{n=0}^{3} x[n]e^{-i2\pi kn/4}$

$W = e^{-i\pi/2} = -i$

$X[0] = 1 + 0 + (-1) + 0 = 0$

$X[1] = 1 \cdot W^0 + 0 \cdot W^1 + (-1) \cdot W^2 + 0 \cdot W^3$ $= 1 + 0 + (-1)(-1) + 0 = 2$

$X[2] = 1 \cdot W^0 + 0 + (-1) \cdot W^4 + 0$ $= 1 + (-1)(1) = 0$

$X[3] = 1 \cdot W^0 + 0 + (-1) \cdot W^6 + 0$ $= 1 + (-1)(-1) = 2$

$\boxed{X = [0, 2, 0, 2]}$

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Masala 13: Nyquist chastotasi

Signal maksimal chastotasi 500 Hz. Minimal namunalash chastotasini toping.

Yechim

Nyquist teoremasi: $f_s \geq 2f_{max}$

$f_s \geq 2 \times 500 = 1000$ Hz

$\boxed{f_s \geq 1000 \text{ Hz}}$

Amalda 2-3 marta ko'proq olish tavsiya etiladi (masalan, 2000-3000 Hz).

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Masala 14: Filtr dizayni

Past chastotali RC filtrning kesish chastotasi (cutoff) formulasini chiqaring va $R = 1\,k\Omega$, $C = 100\,nF$ uchun hisoblang.

Yechim

RC filtr uzatish funksiyasi: $H(j\omega) = \frac{1}{1 + j\omega RC}$

Kesish chastotasida $|H| = \frac{1}{\sqrt{2}}$ (-3 dB):

$\frac{1}{\sqrt{1 + (\omega_c RC)^2}} = \frac{1}{\sqrt{2}}$

$\omega_c RC = 1$

$f_c = \frac{1}{2\pi RC}$

$R = 1000\,\Omega$, $C = 100 \times 10^{-9}\,F$

$f_c = \frac{1}{2\pi \times 1000 \times 100 \times 10^{-9}} = \frac{1}{2\pi \times 10^{-4}}$

$\boxed{f_c \approx 1592 \text{ Hz}}$

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Masala 15: Konvolyutsiya teoremasi

$f(t) = e^{-t}u(t)$ va $g(t) = e^{-2t}u(t)$ uchun $(f * g)(t)$ ni toping.

Yechim

Chastota sohasida: $F(\omega) = \frac{1}{1 + i\omega}$, $G(\omega) = \frac{1}{2 + i\omega}$

$Y(\omega) = F(\omega)G(\omega) = \frac{1}{(1+i\omega)(2+i\omega)}$

Parchalab: $\frac{1}{(1+i\omega)(2+i\omega)} = \frac{1}{1+i\omega} - \frac{1}{2+i\omega}$

Teskari transformatsiya: $y(t) = (e^{-t} - e^{-2t})u(t)$

$\boxed{(f * g)(t) = (e^{-t} - e^{-2t})u(t)}$

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Qiyin (16-20)

Masala 16: Kompleks impedans

RLC ketma-ket zanjirda $R = 100\,\Omega$, $L = 0.1\,H$, $C = 10\,\mu F$.

$\omega = 1000\,rad/s$ da umumiy impedans $Z$ ni toping.

Yechim

$Z_R = R = 100\,\Omega$

$Z_L = i\omega L = i \times 1000 \times 0.1 = 100i\,\Omega$

$Z_C = \frac{1}{i\omega C} = \frac{1}{i \times 1000 \times 10^{-5}} = \frac{1}{0.01i} = -100i\,\Omega$

$Z = Z_R + Z_L + Z_C = 100 + 100i - 100i = 100\,\Omega$

Bu rezonans chastotasi! ($\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.1 \times 10^{-5}}} = 1000\,rad/s$)

$\boxed{Z = 100\,\Omega \text{ (sof rezistiv, rezonans)}}$

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Masala 17: Fazor tahlili

AC zanjirda $v(t) = 170\cos(377t + 30°)$ V. Fazor $\tilde{V}$ ni toping va amplituda hamda RMS qiymatlarini hisoblang.

Yechim

$\tilde{V} = V_m e^{i\phi} = 170e^{i \cdot 30°} = 170(\cos 30° + i\sin 30°)$ $= 170(0.866 + 0.5i) = 147.2 + 85i$ V

Amplituda: $V_m = 170$ V

RMS: $V_{rms} = \frac{V_m}{\sqrt{2}} = \frac{170}{\sqrt{2}} \approx 120.2$ V

$\boxed{\tilde{V} = 170\angle 30° \text{ V}, \quad V_{rms} \approx 120 \text{ V}}$

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Masala 18: IMU sensor filtrlash

Akselerometr signali: $a(t) = 9.8 + 0.5\sin(2\pi \cdot 2 \cdot t) + 0.3\sin(2\pi \cdot 50 \cdot t)$

50 Hz shovqinni yo'qotish uchun qanday filtr kerak?

Yechim

Signal tarkibi:

• DC: $9.8$ (gravitatsiya)
• 2 Hz: $0.5\sin(...)$ (real harakat)
• 50 Hz: $0.3\sin(...)$ (shovqin - motor tebranishi yoki elektr shovqin)

Yechim variantlari:

1. Past chastotali filtr (LPF): $f_c \approx 10$ Hz

   • 2 Hz o'tadi, 50 Hz to'siladi
   • Eng oddiy yechim

2. Band-stop filtr: 50 Hz atrofida tor diapazon to'siladi

   • Agar yuqori chastotali ma'lumot ham kerak bo'lsa

$\boxed{\text{LPF, } f_c = 10 \text{ Hz (yoki Band-stop 45-55 Hz)}}$

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Masala 19: Spektral tahlil

Signal: $x(t) = 3\cos(100\pi t) + 2\cos(200\pi t + \pi/4)$

FFT spektrida qaysi chastotalarda piklar bo'ladi va amplitudalari qancha?

Yechim

Chastotalar:

• $\omega_1 = 100\pi \Rightarrow f_1 = 50$ Hz, amplituda = 3
• $\omega_2 = 200\pi \Rightarrow f_2 = 100$ Hz, amplituda = 2

FFT da:

• $f = 50$ Hz da pik, $|X| = 3 \times N/2$ (bir tomonlama)
• $f = 100$ Hz da pik, $|X| = 2 \times N/2$

Normallashtirilgan amplituda spektrida:

$\boxed{f_1 = 50 \text{ Hz} (A=3), \quad f_2 = 100 \text{ Hz} (A=2)}$

Faza spektrida: $f_1$ da $\phi = 0$, $f_2$ da $\phi = 45°$

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Masala 20: Dron tebranish tahlili

Quadcopter propelleri 5000 RPM da aylanadi. Tebranishning asosiy chastotasini va uning garmonikalari chastotalarini toping.

Yechim

Asosiy chastota: $f_0 = \frac{5000}{60} = 83.33$ Hz

Propellerda odatda 2 yoki 3 ta parra bo'ladi. 2 parrali propeller uchun:

Blade Pass Frequency (BPF): $f_{BPF} = 2 \times f_0 = 166.67$ Hz

Garmonikalar:

• 1-garmonika: $f_0 = 83.33$ Hz (aylanish)
• 2-garmonika: $2f_0 = 166.67$ Hz (BPF)
• 3-garmonika: $3f_0 = 250$ Hz
• ...

$\boxed{f_0 = 83.3 \text{ Hz}, \quad f_{BPF} = 166.7 \text{ Hz}}$

Bu chastotalarni filtrlab, motor tebranishlarini kamaytirish mumkin.

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Natijalar Jadvali

Daraja	Masalalar	Mavzular
Boshlang'ich	1-8	Kompleks arifmetika, modul, argument, Euler
O'rta	9-15	Furye qatori, DFT, Nyquist, filtrlar
Qiyin	16-20	Impedans, fazorlar, sensor filtrlash, spektral tahlil

Jami: 20 masala