(a) The block diagram of a position control system is shown in the figure. Determine the sensitivity of the closed loop transfer…

(a) For the negative-feedback system,

T(s) = G(s)/[1 + G(s)H(s)].

The relative sensitivity functions are defined as

S_G = (∂T/T)/(∂G/G) = 1/[1 + G(s)H(s)]

and

S_H = (∂T/T)/(∂H/H) = −G(s)H(s)/[1 + G(s)H(s)].

Given:

G(s) = 10/[s(s + 1)], H(s) = 5.

At ω = 1 rad/s, s = j1. Therefore,

G(j1)H(j1) = [10/(j1(j1 + 1))] × 5 = 50/[j1(j1 + 1)].

Now j1(j1 + 1) = j1 + j² = −1 + j. Hence,

G(j1)H(j1) = 50/(−1 + j) = 50(−1 − j)/[(−1)² + 1²] = 50(−1 − j)/2 = −25 − j25.

Thus,

1 + G(j1)H(j1) = 1 − 25 − j25 = −24 − j25.

So,

S_G = 1/(−24 − j25) = (−24 + j25)/(24² + 25²) = (−24 + j25)/1201.

Similarly,

S_H = −(−25 − j25)/(−24 − j25) = (25 + j25)/(−24 − j25).

Multiplying numerator and denominator by the conjugate (−24 + j25):

S_H = [(25 + j25)(−24 + j25)]/1201 = (−1225 + j25)/1201.

Hence at 1 rad/s,

S_G = (−24 + j25)/1201 = −0.01998 + j0.02082

and

S_H = (−1225 + j25)/1201 = −1.01998 + j0.02082.

Their magnitudes are

|S_G| = 1/√1201 = 0.02886,

|S_H| = √(1225² + 25²)/1201 = 1.0202.

These are dimensionless sensitivity values, valid for small parameter variations in a linear system.

(b)(i) Energy consumed by the load is

Energy = V I cosφ × time.

For 110 V, 3 A, unity power factor, and 8 hours:

Energy = 110 × 3 × 1 × 8 = 2640 Wh = 2.64 kWh.

The disc makes 1320 revolutions for this energy. Therefore the meter constant is

K = revolutions/energy = 1320/2.64 = 500 rev/kWh.

Meter constant K = 500 rev/kWh.

(b)(ii) In the second test, the meter makes 750 revolutions. Using K = 500 rev/kWh,

Energy recorded = 750/500 = 1.5 kWh.

The apparent energy supplied is

V I t = 110 × 5 × 3 = 1650 Wh = 1.65 kWh.

Thus the load power factor is

cosφ = actual energy/apparent energy = 1.5/1.65 = 10/11 = 0.9091.

Load power factor = 10/11 = 0.9091 lagging (assuming the usual inductive load; the question gives only the magnitude).

(c) Taking bus 0 as the reference bus, the bus admittance matrix is formed for buses 1, 2, 3, 4. The current sources do not enter the Y-bus matrix; they contribute only to the current-injection vector.

The branch admittances are:

y₀₁ = y₀₂ = y₀₃ = 1/(−j0.3) = j10/3,

y₁₃ = 1/(−j8) = j1/8,

y₂₃ = 1/(−j4) = j1/4,

y₂₄ = 1/(−j6) = j1/6,

y₁₄ = 1/(−j6) = j1/6,

y₃₄ = 1/(−j12) = j1/12.

Using Y_ii = sum of admittances connected to bus i and Y_ij = −y_ij for i ≠ j:

Y₁₁ = j10/3 + j1/8 + j1/6 = j29/8,

Y₂₂ = j10/3 + j1/4 + j1/6 = j15/4,

Y₃₃ = j10/3 + j1/4 + j1/8 + j1/12 = j91/24,

Y₄₄ = j1/6 + j1/12 + j1/6 = j5/12.

Therefore, with bus order 1, 2, 3, 4,

Ybus = j ×

[ 29/8, 0, −1/8, −1/6 ; 0, 15/4, −1/4, −1/6 ; −1/8, −1/4, 91/24, −1/12 ; −1/6, −1/6, −1/12, 5/12 ]

This is the required bus admittance matrix in per unit. Bus 0 is the reference and hence does not appear as a row or column.

(d) For the ungraded single-core cable, the maximum electric stress is

E_max = V/[r ln(R/r)].

Here r = 1.12 cm, R = 2.75 cm, V = 14 kV.

R/r = 2.75/1.12 = 2.45536.

ln(R/r) = ln(2.45536) = 0.89827.

Thus,

E_max = 14/[1.12 × 0.89827] = 13.9156 kV/cm.

With one intersheath at radius r_m, let V₁ be the voltage from conductor to intersheath and V₂ be the voltage from intersheath to sheath. For equal maximum stresses in the two dielectric layers,

V₁ = E_max r ln(r_m/r),

V₂ = E_max r_m ln(R/r_m).

Total voltage is

V = E_max [ r ln(r_m/r) + r_m ln(R/r_m) ].

For the optimum or “suitable” intersheath radius, maximise the bracket with respect to r_m:

d/dr_m [ r ln(r_m/r) + r_m ln(R/r_m) ] = 0.

This gives

r/r_m + ln(R/r_m) − 1 = 0.

Solving numerically,

R/r_m = 1.4848,

so

r_m = R/1.4848 = 2.75/1.4848 = 1.852 cm.

The bracket then becomes

r ln(r_m/r) + r_m ln(R/r_m) = 1.29545 cm.

Therefore the maximum operating voltage is

V_max = E_max × 1.29545 = 13.9156 × 1.29545 = 18.03 kV.

Maximum operating voltage ≈ 18.03 kV (r.m.s.).

(e)(i) In the Internet model, information is passed from one layer to the next by encapsulation and decapsulation. At the sender, the application layer creates the message and passes it to the transport layer. The transport layer adds a TCP or UDP header, forming a segment or datagram, and passes it to the Internet layer. The Internet layer adds an IP header, forming a packet, and passes it to the network-access or link layer. The link layer adds its frame header and trailer, forming a frame, and passes it to the physical layer for transmission as bits.

At the receiver, the process is reversed. The physical layer receives bits and passes them to the link layer. The link layer removes the frame header and trailer and passes the IP packet to the Internet layer. The Internet layer removes the IP header and passes the segment or datagram to the transport layer. The transport layer removes the TCP or UDP header and delivers the data to the correct application process. Thus each layer provides a service to the layer above and uses the service of the layer below. Peer layers communicate logically through protocols, while actual data transfer between adjacent layers occurs through interfaces and service access points such as ports and protocol numbers.

(e)(ii) The TCP/IP Internet model is usually described with four layers, while the OSI model has seven layers. Their correlation is:

• Internet Application layer corresponds to OSI Application, Presentation and Session layers.
• Internet Transport layer corresponds to OSI Transport layer.
• Internet Internet layer corresponds to OSI Network layer.
• Internet Network Access or Link layer corresponds to OSI Data Link and Physical layers.

In the five-layer teaching version of the Internet model, the correspondence is:

• Application → OSI Application, Presentation, Session.
• Transport → OSI Transport.
• Network/Internet → OSI Network.
• Data Link → OSI Data Link.
• Physical → OSI Physical.

Thus the Internet model is more practical and protocol-oriented, while the OSI model is a more detailed reference model.