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- a) Obtain the open-loop transfer function. Go to page: 12 he closed-loop transfer function.. c) Find the value of gain and closed-loop poles at the imaginary axis crossing:... d) Write the range of k for which the closed system is stable.......….. e) Write the value of k that makes the system marginally stable:... f) What would be the period of oscillation. g) Find %OS, Tp, Ts, atk = 15. h) Find the steady-state errors when the input is r(t)= 0.62 u(t) step at k=15:.. H(s)G(s): k (s + 7)(s +1-j)(s +1+j)Given the feedback circuit below and assuming the voltage amplifier has gain A₂, input resistance R₁ and output resistance Ro, answer the following questions: + a. Loop gain b. Closed loop gain Re C. Input resistance d. Output resistance Rin м + RE A) Sketch the feedback small signal model of this circuit. (Hint: You may convert the input VIN and source resistance Rs to its Norton current equivalent if needed) B) Find the following in terms of the amplifier parameters and resistances R, and RF: RoutThe figure below shows a block diagram with multiple nested feedback loops which needs to be analyzed the functions Fi shown in the blocks are simplified representations of Fi(s). (a) determine a canonical feedback configuration with one feedforward and one feedback path, (b) determine the overall transfer function Y(s)/X(s). The individual steps are to be explained by redrawing necessary figures. Show your solution steps.
- Question 5. A unity feedback system has the overall transfer function Y(s) R(s) 24 (5s + 3)(2s + 8) Determine the system type and the corresponding error constants (Kp, K, and K.) for tracking step, ramp and parabolic reference inputs in terms of ( and wn.Q1 Design an operational-amplifier circuit using two inverting configurations to produce the output vo = +201 + 0.5v2 – 0.2v3 – 1v4. Choose feedback resistor Rf = 220kN for each amplifier. %3D | -Given the following values and shunt-series feedback circuit below, answer the following question: The open loop input resistance of the circuit is: A. 441.18 Ω B. 3035.71 Ω C. 3125 Ω D. 500 Ω
- An uncompensated system shown in Figure 3(a) has forward transfer function G(s) with a unity feedback. Summarize the various performance parameters in tabular form and comment on the design results. Mention about the speed of response and stability.of tion A voltage-voltage feedback amplifier employs a basic amplifier with input resistance of 20 MOhms, output resistance of 2kOhms and gain A = 1500. The feedback factor k= 0.2. The closed loop input resistance Rinct, and the closed loop output resistance RocLof the closed-loop amplifier are: RinCL= GO and RoutCL= S2the fystlen A A unity feedback Susten has a forrard traf trandsfer functin of 10Es+20)(5t 30) 52(5+25)(5+35)CH50) is as state the fystem ype. step, Cu) fuid the stahi emr Coustants for Stepe, raup and farabolic inputs.
- a- Zero. b- Slightly different from zero. O Maximum positive or negative. d- An amplified sin wave. 9-Negative feedback reduces @ The feedback fraction. b- Distortion. c- The input offset voltage. d- The open-loop gain. 10-The input impedance of a current-voltage converter is Small. b- Large. c- Ideally zero. d- Ideally infinite. a- 11-In a linear op-amp circuit, the @ Signal are always sin wave. b- Op-amp does not go into saturation. c- Input impedance is ideally infinite. d- Gain-bandwidth product is constant.A voltage-voltage feedback amplifier employs a basic amplifier with input resistance of 20 MOhms, output resistance of 2kOhms and gain A = 2000. The feedback factor k = 0.1. The closed loop input resistance Rincu, and the closed loop output resistance RocLof the closed-loop amplifier. Select one: O a. Rinci= 4.02GO, Rautc= 402ko Ob. none of these O c. RinCl= 4.02GN, RoutCL= 9.950 O d. Rinci= 99.5k2 and RoutCL= 402kOQ) In the negative feedback the low frequency and the high frequency * :with the feedback are (a) flf= fl /(1+AB) and fhf= fh /(1+Aß), where flf = low frequency with feedback (b) fhf= fh /(1+AB) and flf= fl (1+AB), where fhf = high frequency with feedback (c) flf= fl (1+AB) and fhf= fh (1+AB) (d) fhf= fh (1+AB) and flf= fl /(1+AB) O