LAB #2 report. MAE 106. UCI. Winter 2005

Nasser Abbasi, LAB time: Thursday 1/20/2005 6 PM

Answer 1.

Explain time constant: This is the time the output will take to reach $63\%$ of its final value.

Answer 2.

Cut-off frequency is the frequency at which the output amplitude (Such as voltage) is $70.1\%$ of the input. This corresponds to a drop of from the input. It is also the frequency at which the output power is $50\%$ that of the input. This frequency is usually taken as the boundary frequency between the highpass region and the low pass region of the frequency response plot.

Answer 3

A heavy object has large inertia. Which means it will take time for it to accelerate when subjected to the same force as compared to an object of low mass. When the force fluctuates very quickly, the object will be slow to react due to its high mass, and by the time it starts to move in response to the force, the force will change its direction quickly, and the object will have to start to reverse its direction again in the direction of the force, and will again be slow in doing this new movement. So this mean the object will have a small motion amplitude of the same frequency as the input force. This is a low pass filter.

Answer 4.

I collected the data using LabJack. This below shows the first few lines of the text file collected:

Now I wrote a small script to plot the data and the theoretical response on the same plot. this is the result. I normalized both amplitudes to 1.

Answer 5.

For the circuit in figure 1, the ODE is . Take Laplace transform, we get

hence

hence,

and phase is

Hence for $C=10^{-6}F$ , and $R=10^{3}$ we get

$\$

and

The following is the data collected for P4, and the theoretical data based on above Laplace transform. This is a low pass filter.

Input Freq. (Hz) $Vpp\$ input output $\Delta t$ $\frac{180}{\pi }$ (degrees)

100 590 mV 662 mV 1.04 ms 10 ms -

200 1.7 V 1.156 V 740 $\mu s$ 4.9 ms -

500 1.59 V 562 mV 390 $\mu s$ 1.99 ms

1 K 1.562 V 312 mV 244 $\mu s$ 990 $\mu s$

2 K 1.56 V 171 mV 126 $\mu s$ 515 $\mu s$

4 K 2.8 V 500 mV 65 $\mu s$ 242 $\mu s$

Theortical:

Input Freq. (Hz) rees

100 $0.846\,73$ -

200 $0.622\,68$ - $\$

500 - -

1 K -

2 K

4 K

I now wrote a script to plot the needed plots as required by P4. This is the result. This shows that the amplitude plot involving the log is more clear and it shows the low pass filter, this is because when using the log scaling, the becomes straight lines.

The following is the data collected for P5.

Since $q=CV_{c}$ for the capacitor, and for the circuit in figure 3, we get . Take derivative, we get , but , hence ODE becomes , take Laplace transform we get

$\frac{sRC}{sRC+1}$

Let $s=j\omega$ , and

Hence for $C=10^{-6}F$ , and $R=10^{3}$ we get

$\$

and

This is a high pass filter

Input Freq. (Hz) $Vpp\$ output input $\Delta t$ (degrees)

10 800 mV 10 V 38 ms 97

50 3 V 9.8 V 15 ms 20 $\$

100 5 V 9.18 V 8.24 ms 10 $\ \$

200 6.6 V 8.25 V 4.34 ms 5.07 $\$ $\$

400 7.3 V 7.7 V 2.4 ms 2.52

1000 7.47 V 7.47 V 48 $\mu s$ 989 $\mu s$ $\$ $\$

2000 7.57 V 7.47 V 8 $\mu s$ 500 $\mu s$ $5.\,\allowbreak 76$ $\$

4000 7.5 V 7.3 V 4 $\mu s$ 247 $\mu s$ $\$

Theortical:

Input Freq. (Hz)

10 $\$ $\$

50 = $0.179\,83$

100 = $0.319\,21$

200

400

1000

2000

4000

Input Freq. (Hz)	$Vpp\$ input	output	$\Delta t$		$\frac{180}{\pi }$ (degrees)
100	590 mV	662 mV	1.04 ms	10 ms	-
200	1.7 V	1.156 V	740 $\mu s$	4.9 ms	-
500	1.59 V	562 mV	390 $\mu s$	1.99 ms
1 K	1.562 V	312 mV	244 $\mu s$	990 $\mu s$
2 K	1.56 V	171 mV	126 $\mu s$	515 $\mu s$
4 K	2.8 V	500 mV	65 $\mu s$	242 $\mu s$

Input Freq. (Hz)		rees
100	$0.846\,73$	-
200	$0.622\,68$	- $\$
500		- -
1 K		-
2 K
4 K

Input Freq. (Hz)	$Vpp\$ output	input	$\Delta t$		(degrees)
10	800 mV	10 V	38 ms	97
50	3 V	9.8 V	15 ms	20	$\$
100	5 V	9.18 V	8.24 ms	10	$\ \$
200	6.6 V	8.25 V	4.34 ms	5.07	$\$ $\$
400	7.3 V	7.7 V	2.4 ms	2.52
1000	7.47 V	7.47 V	48 $\mu s$	989 $\mu s$	$\$ $\$
2000	7.57 V	7.47 V	8 $\mu s$	500 $\mu s$	$5.\,\allowbreak 76$ $\$
4000	7.5 V	7.3 V	4 $\mu s$	247 $\mu s$	$\$

Input Freq. (Hz)
10		$\$ $\$
50	= $0.179\,83$
100	= $0.319\,21$
200
400
1000
2000
4000