adder);Second:Integrator;Third:Differentiator;Right:Proportional-integrated-differential circuit(PID). 3) 4)input and output signals are on the circuits.5)summer:the output is the sum of the input voltages.Integrator:produces a ramp of output voltage,which is a linearly increasing or decreasing voltage.Differentiator:produces an output voltage proportional to the instantaneous rate of change of the input voltage.PID:They amplify the beginning and the end of the pulse signal.6)Summer:Uout=-(U1R/R1+U2R/R2+U3R/R3).Integrator:Uout=-1/T int(Uindt);Iin=Uin/R;Uout=-Iint/C = -Uint/T;Uout= -Uin/(4fT).Differentiator:Uout= -T dUin/dt.PID:K=R2/R1;T1=R1C1;T2=R2C2 7)Substracter:The output voltage,which is proportional to the difference of the input voltages when R1=R2 and R=R3. Logarithmic:It is the inverting amplifier with a feedback diode rather than feedback resistor.Once the diode and the resistor positions replace on other, the exponential amplifier appears:
T , ( ) UR(p) exp ( p . ) , .. , . - , T T . exp ( p . ) U R(p) U m. U m. . (1 p. T ) ( 1 p. T ) - UR(p), .. uR(t)=L-1[UR(p)]. - , (3.14). U m. T U1 R ( p ) . ( 1 p. T ) U1R(p) N ( p ) 1 p . T , 1 p 1 . (3.12) (3.14), T U1R(p) Res1 p1 ( t ) U1R(p)ept p1. Mathcad U m. T Res1 p1 ( t ) lim . ep . t . p p 1 d (N(p)) dp 113 assume U m , T U m
/ ,t,-t) / ",1:{. -azt,,.;,fa io,rr).r,ti= ' prt; " r), k ^ n o* - ^ / , -V- -, 7 n a u r : . / - 1-u1r r (e= -t = -.> f t,r-F1..to', L= .46// )r0=0fi).'rv^ .)-l = t -t0,2 . =,42y 'l / ,; ront t + r g 7 $
/ ,t,-t) / ",1:{. -azt,,.;,fa io,rr).r,ti= ' prt; " r), k ^ n o* - ^ / , -V- -, 7 n a u r : . / - 1-u1r r (e= -t = -.> f t,r-F1..to', L= .46// )r0=0fi).'rv^ .)-l = t -t0,2 . =,42y 'l / ,; ront t + r g 7 $
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