# Using reverse fourier transformation we have vouttf 1

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Using reverse Fourier transformation, we have: Vout(t)=F -1 (Vout(s))=0.04e - ω 0 t * ε (t) Obviously, the time constant is 1/ ω 0 =10ns.

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Problem 4 : Ignore DC bias analysis. You don’t need it. The transistor has transconductance gm. Its output resistance Rds is infinity. (a) Draw a small-signal equivalent circuit of the circuit. (b) gm=10mS, C=10pF. Find, by nodal analysis, a small-signal expression for Vout(s)/Vin(s) (c) Find any/all pole and zero frequencies of the transfer function, in Hz: Draw a clean Bode Plot on semilog paper of Vout/Vin, LABEL AXES, LABEL all relevant gains and pole or zero frequencies, Label Slopes (d) Vin(t) is a 100 mV amplitude step-function Find Vout(t), and plot it below. Label axes, show initial and final values, show time constants V in V out Q 1 C Solution: a) This s mall-signal equivalent circuit is: b) We can easily derive two functions from this small-signal equivalent circuit Step1: current passing through C gives voltage drop of Vout (Here we neglect Cgs1). sC*Vout= I1= gm1*(Vin-Vout) Step2: we got: H(s)=Vout(s)/Vin(s)= 1/(s*C/gm+1) Step3: We can rewrite the ω 0 as the position of pole, we have: H(s)=Vout(s)/Vin(s)= 1/(s*1ns+1) Where ω 0 =gm/C=10 9 rad/s which is 0.159GHz (gm= 10mS, C=10 pF) c) From discussion above, we know that there are One Zero and One Pole in this system, the frequency of Zero is at 0Hz; the frequency of Pole is ω 0 = 10 9 rad/s. The Bode plot of H(s) is shown below:
d) When the input is 100mV step-function, the input signal is Vin(t) = 0.1* ε (t) Hence, Vin(s)=0.1/s. Therefore, Vout(s)=H(s)*Vin(s)=0.1/s(s*1ns+1). Using reverse Fourier transformation, we have: 1 1 1 . 0 1 . 0 ) 1 1 ( 1 . 0 ) ( + + = + = ns s ns s ns s s s Vout Therefore, ] 1 [ ) ( 1 . 0 ) ( 9 10 t e t t Vout = ε , and we can draw the response as follow: Obviously, the time constant is 1/ ω 0 =1ns. Problem 5 : Ignore DC bias analysis. You don’t need it. The two transistors have transconductance gm1 and gm2 respectively. Their drain-source resistances Rds1 and Rds2 are both infinity. (a) Draw a small-signal equivalent circuit of the circuit

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(b) Find, by nodal analysis, a small-signal expression for Vout/Vin V in V out Q 1 Q 2 R 1 R 2 Solution: a) the small-signal equivalent circuit is: We can see that here are three nodes in this circuit, which can be marked as Nvin, Nvx and Nvout, and their voltages are Vin, Vx, and Vout, respectly. b) We can easily derive two function from this small-signal equivalent circuit Step 1 : the current passing through R1 and Cgs2 cause a voltage drop of Vx, and this current I1=gm 1 *(Vout-Vin). Vx= gm 1 *(Vout-Vin)*(R1//Cgs2) Here R1//Cgs2 stands for the impedance of parallel of R1 and Cgs2, which is R1/(sCgs2*R1+1) Step2: the current flowing into node Nvout should be 0, we have: I1+I2+sCgs1*(Vout-Vin)+Vout/R2 = 0 Step3: we have the transfer function like: H(s)=Vout(s)/Vin(s)=H num (s)/H den (s) Here H num (s)= gm 1 R2(1+gm 2 R1)+sR2(Cgs 2 R1gm 1 +Cgs 1 )+s 2 Cgs 1 Cgs 2 R1R2 H den (s) = 1+gm 1 R2(1+gm 2 R1)+s[R2(Cgs

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