Question

Q3) A pulse input was applied to a plant. The frequency response data is shown in the given table. Although the data seem noisy, find the transfer function of the plant. Then check the accuracy of your model by computing the amplitude ratio and phase at a 005 rad/sec, and 0.s rad/sec. dcg) Ratio 0,0010 42 0.0075 00200 -12 .27 0.0600 0.0800 0.1000 0.3000 0.5000 .52 32 05i 0.3 192

Answer 1

The data provided in the table is as follows,

f = [0.0010, 0.0050, 0.0075, 0.0200, 0.0400, 0.0600, 0.0800, 0.1000, 0.3000, 0.5000, 0.7000, 0.9000];
a = [4.4, 4.2, 4.4, 4.6, 4.3, 4.1, 3.6, 3.2, 1.1, 0.51, 0.3, 0.12];
p = [0, -4, -4, -12, -27, -43, -52, -68, -140, -162, -192, -206];

Calculating the amplitude value in decibels using the formula 20*log(a) for the actual amplitude we get

a in dB = [12.869, 12.465, 12.869, 13.255, 12.669, 12.256, 11.126, 10.103, 0.828, -5.849, -10.458, -18.416];

We observe that the magnitude is decreasing with the increase in frequency. Thus, the model is low pass filter.

Using the asymptotic properties of phase response and magnitude response, we can approximately identify the transfer function.

The asymptotic properties of magnitude response,

(1) Flat at low frequency

(2) From break point, the value starts decreasing

(3) High frequency -20db/decade for 1 pole and -40db/decade for 2 poles

The asymptotic properties of phase response,

(1) zero degrees at low frequency

(2) -90 degrees at break point for 2 poles, and -45 degrees for 1 pole

(3) -180 degrees at high frequency for 2 poles, and -90 degrees for 1 pole

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From the tabulated values for phase, we can conclude that the plant has 2 poles because at low frequency the value of phase is zero degrees and at high frequency the value of phase is -180 degrees.

To confirm that there are two poles, we find the difference in magnitude response. The difference between maximum and minimum value is 13.255 - (-18.416) = 31.671 dB, which is close to 40 dB. Hence, we can confirm that there are two poles.

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The transfer function for two pole plant model is

a2 H(s) = (s+ a)2

where is the break point.

We need to identify the break point from the phase and magnitude response table.

For two poles model, the break point is where phase is -90 degrees and the amplitude is -6db from maximum.

At f = 0.08, the phase is -52 degrees and the amplitude is -2.1 dB from maximum

At f = 0.1, the phase is -68 degrees and the amplitude is -3.12 dB from maximum

At f = 0.3, the phase is -140 degrees and the amplitude is -13 dB from maximum

Out of the three values, the most appropriate break point is f = 0.1 rad/sec, which fits the assumption.

Thus,

Now, The plant model is

0.12 H)0 s+0.1

In terms of 'w', the transfer function of plant model is

0.12 H(jw) w0.1)2

At w = 0,

H(jw) = 1

But the actual value of amplitude is around (4.4 + 4.2 + 4.4 + 4.6 + 4.3 + 4.1)/6 = 26/6 = 4.333.

We need to scale the value as 4.3333*0.7071 = 3.0641 for approximation

We need to multiply this scaling factor to the plant model. Thus, the transfer function of plant model is

3.0641 0.1 H(jw) (jw +0.1)2

Simplifying further

0.030641 Hjur) = (ju, +0.1)2 jw+0.1)2

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We need to calculate the amplitude and phase at w = 0.05 and w = 0.5

1) Substitute w = 0.05 in the above expression

0.030641 0.05 +0.1)2 HUjuv)lu'=0.05 =

0.030641 111, -0.05 =ー0.0025 + 0 H(ju)|w=

0.030641 0.00750.001 H(jw)|w=0.05=

The magnitude is

0.030641 H(ju)|w=0.05 = V 0.00752 ± 0.0012

0.030641 w-0.05- V0.000057273

$|H(jw) | _{w=0.05}= \frac{0.030641}{0.007566}$

(Ju)|w=0.05 4.0498

The phase value is

$\angle H(jw) = - \tan ^{-1} \left ( \frac{0.001}{0.0075} \right )$

The complex form is present in the denominator, so 'minus' sign is used in the expression

$\angle H(jw) = -7.59$ degrees

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2) Substitute w = 0.5 in the above expression

$H(jw) | _{w=0.5}= \frac{0.030641}{(j0.5+0.1)^2}$

$H(jw) | _{w=0.5}= \frac{0.030641}{-0.25+0.01+0.01j}$

$H(jw) | _{w=0.5}= \frac{0.030641}{-0.24+0.01j}$

The magnitude is

$|H(jw) | _{w=0.5}= \frac{0.030641}{\sqrt{-0.24^2+0.01^2}}$

$|H(jw) | _{w=0.5}= \frac{0.030641}{\sqrt{0.0577}}$

$|H(jw) | _{w=0.5}= \frac{0.030641}{0.24021}$

The phase value is

$\angle H(jw) = - \tan ^{-1} \left ( \frac{0.01}{-0.24} \right )$

The complex form is present in the denominator, so 'minus' sign is used in the expression

$\angle H(jw) = -177.61$ degrees

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The comparison of value from the table and the model is as follows

f	mag from table	phase from table	mag from model	phase from model
0.04	4.3	-27	-	-
0.06	4.1	-43	-	-
0.05	-	-	4.1	-7.59
0.5	0.51	-162	0.12756	-177.61

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From the comparison table, we can clearly observe that the transfer function of the plant closely resembles the values tabulated in the table.