這是一題單純的R與C並聯，Z是並聯後的阻抗，請問Rx和Cx怎麼推導

(b) Z, remains the same as in Eq. (9.15.4) but R, and X, are in parallel. Assuming an RC parallel combination, Zx = (5.993-j1.356) MQ 1 RK Rx (1-ju = Rx || = jwCx 1 + jwRxCx i+w Rx 5.993 R₁ = By equating the real and imaginary parts, we obtain Real(Z)² + Imag(Zx)² _ 5.993² + 1.356² Real(ZX) Re Zx = 6.3 M2 5.993 Imag(Z) Rx = 5.993² +1.3562 5.993 C₁ = = [Real(Z)² + Imag(Z)²] Cx = - -1356 2Tu₂k[5.993)² + -1.356 2z(2000)(5.9172+1.356²) = = 2.852 μF

無線網路作業，第五題

5. 關於直接序列展頻(Direct Sequence Spread Spectrum, DSSS)的技術是將原始的資料與展頻碼(chipping sequence)作XOR 的運算後,可將原本較高功率且頻寬較窄的訊號,轉換為低功率高頻寬的訊號。如下圖,已知 Data signal=< 0101 > 且由LSB(最左邊位元先輸入),PRN=<10110111001>,請完成Output的解碼過程,需劃出圖形? Data signal Output ·SMS 1-bit 11-bir barker code (PRN): 10710111001 PRN 11 Chips 11 Chips 圖2.22:直接序列展的編碼方式

請問這幾題怎麼解？ 拜託🙏

1. 請以方塊圖簡化方法(聚合點左移、分歧點右移、回授方塊圖公式求轉移函數),求 C(s) 下圖所示方塊圖系統的 轉移函數。 R(s) R(s). C(s) C(s) 2. 請繪出上圖(題1.)之訊號流程圖,並利用梅森增益公式求 轉移函數。 R(s) 3. 已知某線性非時變系統的微分方程式如下,其中u(t)與y(t)分別表輸入及輸 出,試求轉移函數。 dt³ (1) d³y(t) + 5 d² y(t) + 4 dy(t) dr² dt + y(t) = )=2. +3u(t) du(t) dt dt² (2) d²¸y (1) + 3 dy (t) + 2 y (t) + √ " y(t)dt=u(t) dt Y(s) a 4. 某控制系統的方塊圖如下所示,若其具有轉移函數為 R(s) S² + bs + a 則a 及b 之值為何? 4 R(s). Y(s) s+1 5. 求下題的輸出y(t) = ? S 若8(1) 線性 -h(1)=e-31(120) 非時變 線性 則 3e-StuG (1) - -y(1)之值 非時變

請問這題a為何用DTFT再逆轉換算出來y[n]=0與解答用z轉換結果不同

2.5. A causal linear time-invariant system is described by the difference equation y[n]-5y[n-1]+6y[n = 2] = 2x[n— 1]. (a) Determine the homogeneous response of the system, i.e., the possible outputs if x[n] = 0 for all n. (b) Determine the impulse response of the system. (c) Determine the step response of the system. ==

戴維寧等效電路 求解Vth和Rth🙏🏻

2/2 SA ↑ 45 + EUK Vx all a M 252 be b

求救

The shaft shown in the figure is machined from AISI 1040 CD steel. The shaft rotates at 1600 rpm and is supported in rolling bearings at A and B. The applied forces are Fi 10 kN and F2 = 4 kN. Determine the minimum fatigue factor of safety based on achieving infinite life. If infinite life is not predicted, estimate the number of cycles to failure. Also check for yielding. 30 mm- A 192 mm 40 mm 12 mm 240 mm 192 mm -45 mm F₂ 240 mm All fillets 1.5 mm R. 192 mm 42 mm 72 mm 12 mm B 30 mm

想問這兩個式子怎麼知道什麼時候用加的什麼時候用減的啊？

Sidizzo son ei ji povaworl grey line tolera L apapagoslingeri 14.0 V gool sign S qisning nanalintuniendizzo 00 6.00 (+ 21.21. Snus sur le mure on Roomet while keeping it electrically the 4.0 2 es or parallel combinations of 918 b + a oissant 10.0 V sociated with combining resis-mollastunquodh (a) emoved and replaced by a wire qoo circuit.) Because the circuit is his problem is one in which we → I www 6.02 100 9.00 (2 18.00 bus M 2.0 2 abcda: (2) 10.0 V – (6.0 N) Į — (2.0 ₪) I₂ = 0 befcb: - (4.0 N) I2 − 14.0 V + (6.0 M) — 10.0 V = 0 - 1₂ Figure 21.21 (Example 21.4) led in branches. moq as labeled in Figure 21.21.mais12A circuit containing different i svig 19 d od oni bi u bliud tot lo mo battib 210109 26919rw 1+ en slin es (1) 1₁ + 1₂2-1₂ = 01S sugit ni nobı aldı YOU ARE C 15usdwes (2.02) 4₂2 01 Toyons de = L d (3) -24.0 V + (6.0 N) - (4.0 N)₂ = 0lvom snigul grans di anioq gnilise or of am gids de stod bad i mi smo duo d Monosh 10.0 V (6.02), − (2.0 N) (1,₁ + 1₂) = 0 Tom Tom (4) 10.0 V - (8.02) 4₁ 10 0 sob

4.26 4.37 9.66 謝謝🥹🥹11點半以前都可以 非常感恩

*D4.26 Design the common-source circuit in Figure P4.26 using an n-channel MOSFET with &= 0, The quiescent values are to be Ang=6 mA, Vesg = 2.8 V, and Vosg = 10 V. The transconductance is gm = 2.2 mA/V. Let R₁ = 1 ks2, A, = -1, and Rin = 100 k22. Find R₁, R₂, Rs. RD. Kn and VTN. Rm1 Figure P4.26 VDD=18 V R₂ R₂ RD CC2 Ry % 1/₂ Figure P4.27 +9 V 10 持 Co₂ R2 = 500 KSD R₂ -9 V C's K₂ ko

網目電流的題目 想請教詳解

4.11 使用網目電流法求解圖示電路中的網目電流 10 A i。 75 V 292 + Vo (502 答:15 62 滄海圖書/陳在注 總校閱 / 電路學(上冊) (Nilsson 10th edition) 1922 200 5 CH04-62

有人會這些東西嗎🥹感謝

3. 如下圖為一四連桿機構,Q2Q4為固定桿,Q2A 線段為輸入桿,設反時針方向之角速度 為100 rpm, Q.B線段長40cm,試用圖解法求A、B及C三點之速度大小(C為QB之中點), 繪圖比例1cm=10cm。 9₂ B A 60° 55⁰ 75 cm 460⁰ A, B 30 cm C 40cm La Wi=1nonyienc 4. 向量 長 25 公分,向量長 50 公分,二者之間夾60°,求Aa➔BD線段大小及方向。 25 Aa Bb E 54334 2