DAY 16: 1st Order Circuits Step Response/ Inverting Differentiator Lab

Part 1: First order op amp circuit

Today, we discussed the 1st order circuits step response and the inverting differentiator. We also did a experiment to verify the theory of the inverting differentiator. We also discussed a few basic knowledge of the unit step function, the unit impulse function, and the ramp function. 

The deduction for this op amp circuit, the inverting differentiator. By applying ideal op amp assumption and the KCL, we deduced vo(t)= V0 -1/(RC) * integral(vi.dt).

With different vin, we deduce the vo based on the relationship between them. 

vo(t)= V0 -1/(RC) * integral(vi.dt). 

Part 2: Inverting Differentiator Op Amp Lab
Purpose: The lab aims to verify the theory of inverting differentiator. We will observe the vin, vout of an inverting differentiator on the oscilloscope and expect to gain two sinusoidal graphs that are greater or less pi/2.

Pre-lab:

Based on the given values, we calculate the equation between vin and vo, vo=(ARCw).sin(wt) = 6.28*sin (wt), with f=500Hz. Similarly for f=1000Hz, vo=1000*sin(wt); and f=2000Hz, vo=2500*sin(wt)

The set-up of a inverting differentiator

500Hz

Expected output voltage = 500, measured out voltage = 628.3, percent error = 20.42%

f=1kHz

Expected output voltage = 1257, measured out voltage = 1000, percent error = 22.43%

f=2kHz

Expected output voltage = 2513, measured out voltage = 2500, percent error = 0.5%

Summary:

Frequency (Hz)	Input Voltage (mv)	Expected Output Voltage (mv)	Measured Output Voltage (mv)	Percent Error%
500	200	500	628.3	20.42
1000	200	1257	1000.25	22.43
2000	200	2513	2500	0.52

Part 3: More Switching Functions/ 

The unit impulse function 

the unit ramp function r(t)

We did a practice to get familiar with the unit step function, the unit impulse function, and the unit ramp function. 

Step Response of an RC circuit

By using KCL, and the initial value of v, vc = v(t) as a function above. 

RC step response graphs, vc(t), ic(t) with V0=0V. 

v(t) = V _s (1 − e^{− t/τ})u(t)

i(t) = (V _s /R)*e^{− t/τ}u(t)

We continued to analyze the RC circuit step response, and let V0=0V. We deduce ic as above. 

We applied the theory of the RC circuit to solve for the RC circuit above, with the Vs = 5u(t).

Similarly, we studied about the step response of an RL circuit

The standard step response RL circuit  

From the KVL, and the intial value of i, we deduce iL=i as a function above

i(0-)=0, we deduce the iL=it as above

deducing from i(t), we get v(t) as function above 

The graphs of the step response RL circuit. 

We did a practice problem for the step response RL circuit.

Conclusion

Today, we discussed and studied some knowledge about the inverting differentiator, the step response RL, RC circuits,, the 3 new kinds functions: the unit step response, the unit impulse, the unit ramp. We did the inverting differentiator lab to verify the theory we had discussed. We also practiced some problem to grasp the theories of the step response RL, RC circuit. These kind of circuit can be analyzed through the KVL, KVC, the unit step response when t<0, and t>=0. The analysis of the behaviors of the circuit in each moment is very important. Remind some basic concepts: inductor resists the instantaneous change of current iL(0-)=iL(0+), the capacitor resists the instantaneous change of voltage vc(0-)=vc(0+). Time is infinite, the inductor acts as the short circuit, the capacitor acts as the open circuit. Some tranformation concept, and the Thervenin circuit are also important. For find Rthervenin, replace voltage source with a short circuit (imaging the standard circuit of Thevernin Vs and R in series), the current source with a open circuit. (imaging the standard circuit of Norton Is and R in parallel).