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Copy of gate solved paper 2012 ec

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GATE EC 2012 Q. 1 - Q. 25 carry one mark each. MCQ 1.1 The current ib through the base of a silicon npn transistor is 1 + 0.1 cos (10000 t) mA At 300 K , the r in the small signal model of the transistor is (A) 250 (C) 25 SOL 1.1 (B) 27.5 (D) 22.5 Option (C) is correct. Given ib = 1 + 0.1 cos (1000 t) mA So, IB = DC component of ib = 1 mA In small signal model of the transistor VT r = IC = VT = VT IB IC / VT " Thermal voltage IC = I B = VT IB r = 25 mV = 25 VT = 25 mV, IB = 1 mA 1 mA The power spectral density of a real process X (t) for positive frequencies is shown below. The values of E [X 2 (t)] and E [X (t)] , respectively, are So, MCQ 1.2 Brought to you by: Nodia and Company PUBLISHING FOR GATE Visit us at: www.nodia.co.in

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