A unity feedback system, having an open loop gain G(s)H(s) = K(1-s) (1+s) A IKI > 1 B K> 1 C|K| < 1 D K<-1 becomes stable when
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- 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: RoutQ2: For the unity feedback second order system shown below, if R(s) is a unit step, trace R(s), E(s) and C(s). Take a. K=50 and kl=10 b. K=k1=10 c. K=25 and kl=10. Comment on the results in each case. R(s) E(s) K. C(s) (s +. kl)sA unity feedback system has the characteristic equation shown below. For some gain K the unity feedback system enters into a state of self-sustained oscillations. Use the Routh - Hurwitz Stability Criterion to find the frequency of this self-sustained oscillation in rad/ sec. Enter your answer in decimal form to two decimal places. 4 + 65³ +12s² + 10s + (3+K)
- 2-) The AC equivalent of a feedback amplifier circuit is given in the figure on the right. (Hfe100, Va = ∞, Ic1 = 15 mA, Ic2 = 5mA and Ic3 = 5 mA) a) State the type of feedback used in the circuit, explaining the reason. b) Draw the small signal equivalent of the amplifier circuit. c) Calculate the value of β for the feedback by drawing the β circuit. d) Find the Avf = Vo / Vs closed loop gain of this circuit. e) Find the Rif and Rof values.A feedback system root locus graph shown in figure below. For damped frequency 8.04 Hz, the gain K = 10 Imaginary Axis (seconds) ÖAN÷O-NWAGALBE -10 -20 -18 -16 Root Locus -14 -12 -10 -8 -6 -4 Real Axis (seconds) Ņ 0Consider the unity negative feedback system depicted in the figure below. The design specification requires a location of dominant poles to yield a 1.2 second settling time and an overshoot of 15%. If a compensator with a zero at -0.5 is used to achieve the specification, find the approximate location of compensator pole. K G(s) (s + 5)³ R(s) C(s) G(s)
- Construct the root locus for a unity-feedback control system having the open-loop transfer function given as 10 s(s – 1)(2s + 3)* Comment on the stability of the system.the 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 feedback amplifier employing series-series feedback is shown in the figure below, where the feedback network is contained inside 200m S, and R, = 3k. The other component values are RA = 1kQ, 2kQ, Rs = 5kQ. and R₁ = 10k2. Express your answers up to 3 decimal places. the dashed box. For the open loop amplifier, Ri = 7kQ, Gm RB Vs = Rs Ri Vin www R₁ Gm. Vin RB = Ro i out RL
- Q) In the negative feedback the low frequency and the high frequency with the feedback are: Note: flf = Low frequency with feedback and fhf = High Frequency with feedback * (a) flf= fl /(1+AB) and fhf= fh /(1+AB) O (b) fhf= fh /(1+AB) and flf= fl (1+AB) O (c) flf= fl (1+AB) and fhf= fh (1+AB) O (d) fhf= fh (1+AB) and flf= fl /(1+AB)Q1: design an inverting op-amp circuit to form the weighted sum Vo = -(v1 + 4v2) .choose the values for R1, R2, and Rf so that for a maximum output voltage of 10 volt the current in the feedback resistor will not exceed 1 ma.Feedback:2. An ideal series-shunt feedback amplifier has Rs negligibly small. If Vi = 100mV, Vfb =99mV, and Vo= 5V determine:a. Av , β, Avf (including the units)b. Rif, Rof if Ri = 5 Kohm & Ro = 4 Kohm.