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5/12/2017 582776102 1/6 The Inverting Integrator The circuit shown below is the inverting integrator. C i2 (s) vin (s) R v- ideal i1 (s) v+ oc vout (s ) + Since the circuit uses the inverting configuration, we can conclude that the circuit transfer function is: oc vout (s ) G (s ) vin (s ) Z (s ) 2 Z 1 (s ) 1 s C R 1 s RC In other words, the output signal is related to the input as: 5/12/2017 582776102 oc vout (s ) 2/6 1 vin (s ) RC s From our knowledge of Laplace Transforms, we know this means that the output signal is proportional to the integral of the input signal! Taking the inverse Laplace Transform, we find: 1 v (t ) RC oc out t v in (t ) dt 0 For example, if the input is: vin (t ) sinωt then the output is: v oc out 1 (t ) RC t sinωt dt 0 1 1 cosωt RC ω 1 cosωt ω RC We likewise could have determined this result using Fourier Analysis (i.e., frequency domain): 5/12/2017 582776102 G (ω ) oc vout (ω ) vin (ω ) 3/6 Z 2(ω ) Z 1(ω ) 1 j ω C j ω RC R Thus, the magnitude of the transfer function is: G (ω ) j ω RC 1 ω RC And since: j j e 2 cos 2 j sin 2 the phase of the transfer function is: G (ω ) π 2 90 Given that: radians 5/12/2017 582776102 4/6 oc vout (ω ) G (ω ) vin (ω ) and: oc vout (ω ) G (ω ) vin (ω ) we find for the input: vin (t ) sinωt where: vin (ω ) 1 and vin (ω ) 0 that the output of the inverting integrator is: oc vout (ω ) G (ω ) vin (ω ) 1 ω RC and: oc vout (ω ) G (ω ) vin (ω ) 90 0 90 Therefore: 1 sin ωt 90 ω RC 1 cosωt ω RC oc vout (t ) Exactly the same result as before! 5/12/2017 582776102 5/6 If you are still unconvinced that this circuit is an integrator, consider this time-domain analysis. i2 (t) C + vc - vin(t) R v- - i 0 i1 (t) ideal v+ oc vout (t ) + From our elementary circuits knowledge, we know that the voltage across a capacitor is: 1 vc (t ) C t i (t ) dt 2 0 and from the circuit we see that: oc oc vc (t ) v (t ) vout (t ) vout (t ) therefore the output voltage is: 1 v (t ) C oc out t i (t ) dt 2 0 5/12/2017 582776102 6/6 From KCL, we likewise know that: i1(t ) i2(t ) and from Ohm’s Law: i1(t ) Therefore: vin (t ) v (t ) vin (t ) R1 R1 i2 (t ) vin (t ) R1 and thus: t 1 v (t ) i2(t ) dt C 0 oc out 1 RC The same result as before! t v in 0 (t ) dt