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

(a) Sensitivity of T(s) to G(s) and H(s) at ω = 1 rad/s

Definitions. For T = G/(1 + GH): S_G = (∂T/T)/(∂G/G) = (G/T)(∂T/∂G) = 1/(1 + GH) S_H = (∂T/T)/(∂H/H) = (H/T)(∂T/∂H) = −GH/(1 + GH) (Note that S_H = S_G − 1.)

Loop gain at s = j1. G(j1)H(j1) = 50/[j1(j1 + 1)] = 50/(−1 + j) = 50(−1 − j)/2 = −25 − j25 1 + G(j1)H(j1) = −24 − j25, with |1 + GH|² = 24² + 25² = 1201

Sensitivity to G. S_G = 1/(−24 − j25) = (−24 + j25)/1201 = −0.0200 + j0.0208

|S_G| = 1/√1201 = 0.0289

Sensitivity to H. S_H = (25 + j25)/(−24 − j25) = (25 + j25)(−24 + j25)/1201 = (−1225 + j25)/1201 = −1.020 + j0.021

|S_H| = 1.02

Interpretation. At 1 rad/s a 10% change in the forward gain G changes T by only about 0.3%, because the high loop gain (|GH| ≈ 35) suppresses it. A 10% change in the feedback element H changes T by about 10%. Feedback gives insensitivity to forward-path parameters but none to the feedback path, so the sensor H must be accurate and stable.

(b) Energy-meter constant and load power factor

Meter constant. Energy in the first test = VI cos φ × t = 110 × 3 × 1 × 8 = 2640 Wh = 2.64 kWh K = revolutions/energy = 1320/2.64 = 500 rev/kWh (that is, 2 Wh per revolution)

Load power factor. Energy recorded in the second test = 750/500 = 1.5 kWh (average power 500 W) Volt-ampere-hours = 110 × 5 × 3 = 1650 VAh = 1.65 kVAh cos φ = 1.5/1.65 = 0.909 (lagging, taking the usual inductive load; the meter reading gives only the magnitude)

(c) Bus admittance matrix

Reading the network. Every branch is drawn as an inductor, and its label is negative imaginary (−j0.3, −j4, …). An inductor's impedance is +jX, so the labels are the branch admittances in per unit, following the usual convention of Grainger and Stevenson. Bus 0 is the reference. The current sources between bus 0 and buses 1, 2 and 3 are injections: they form the current vector I = Y_bus V and do not enter Y_bus.

Branch admittances (per unit).

• To the reference: y₁₀ = y₂₀ = y₃₀ = −j0.3
• Between buses: y₂₃ = −j4, y₁₃ = −j8, y₂₄ = −j6, y₃₄ = −j12, y₁₄ = −j6; buses 1 and 2 are not connected, so y₁₂ = 0

Elements. Y_ii is the sum of the admittances connected to bus i, and Y_ij = −y_ij. Y₁₁ = −j(0.3 + 8 + 6) = −j14.3 Y₂₂ = −j(0.3 + 4 + 6) = −j10.3 Y₃₃ = −j(0.3 + 4 + 8 + 12) = −j24.3 Y₄₄ = −j(6 + 12 + 6) = −j24.0 Y₁₂ = 0, Y₁₃ = j8, Y₁₄ = j6, Y₂₃ = j4, Y₂₄ = j6, Y₃₄ = j12

Y_bus (bus order 1, 2, 3, 4; per unit) Row 1: [ −j14.3, 0, j8, j6 ] Row 2: [ 0, −j10.3, j4, j6 ] Row 3: [ j8, j4, −j24.3, j12 ] Row 4: [ j6, j6, j12, −j24.0 ]

Checks. The matrix is symmetric. Each row sum equals that bus's admittance to the reference: −j0.3 for buses 1, 2 and 3, and 0 for bus 4, which has no connection to bus 0. The zero Y₁₂ shows the sparsity that load-flow programs exploit. (If the labels were misread as impedances, the negative-j values would describe capacitors, contradicting the inductor symbols.)

(d) Maximum operating voltage with inter-sheath grading

Ungraded cable. r = 1.12 cm, R = 2.75 cm, V = 14 kV; ln(R/r) = ln 2.455 = 0.898 E_max = V/[r ln(R/r)] = 14/(1.12 × 0.898) = 13.92 kV/cm This is taken as the permissible working stress of the insulation.

One inter-sheath at radius r₁. Its potential is held so that both layers work at the same maximum stress E_max: V₁ = E_max r ln(r₁/r) (conductor to inter-sheath) V₂ = E_max r₁ ln(R/r₁) (inter-sheath to sheath) V = E_max[r ln(r₁/r) + r₁ ln(R/r₁)]

Best ("suitable") radius. Setting dV/dr₁ = 0: r/r₁ + ln(R/r₁) − 1 = 0 With x = R/r₁: 0.4073x + ln x = 1 ⇒ x = 1.485 ⇒ r₁ = 2.75/1.485 = 1.852 cm

Maximum voltage. r ln(r₁/r) = 1.12 × ln 1.654 = 0.563 cm ⇒ V₁ = 13.92 × 0.563 = 7.84 kV r₁ ln(R/r₁) = 1.852 × ln 1.485 = 0.732 cm ⇒ V₂ = 13.92 × 0.732 = 10.19 kV V_max = 7.84 + 10.19 ≈ 18.0 kV, with the inter-sheath held at about 10.2 kV above the sheath (7.84 kV below the conductor).

Grading lets the same cable work at about 29% higher voltage (18.0 kV against 14 kV), because the dielectric near the sheath is used at the full stress instead of being lightly stressed.

(e) Passing information between layers; Internet model versus OSI

Encapsulation at the sender. The application layer produces a message (for example an HTTP request). The transport layer adds a header with source and destination port numbers, sequence and checksum, forming a segment (TCP) or user datagram (UDP). The network (Internet) layer adds an IP header with source and destination IP addresses, forming a datagram or packet. The data link layer adds a header with MAC addresses and a trailer (CRC), forming a frame. The physical layer sends the frame as a stream of bits on the medium.

Decapsulation at the receiver. Each layer checks and removes its own header and hands the payload to the layer above. It chooses that layer from a field in the header: EtherType selects IP, the IP protocol field selects TCP or UDP, and the port number selects the application process.

Two kinds of communication. Peer layers on the two machines communicate logically through their headers, which is the protocol. Adjacent layers on one machine communicate through a service interface: each layer uses the services of the layer below and provides a service to the layer above.

Correspondence with OSI

• Internet application layer ↔ OSI application, presentation and session layers (HTTP, SMTP, DNS)
• Transport ↔ OSI transport (TCP, UDP)
• Network (Internet) ↔ OSI network (IP, ICMP)
• Data link ↔ OSI data link (Ethernet, PPP)
• Physical ↔ OSI physical In the four-layer description the last two are merged into a single network-access (link) layer.

OSI is a seven-layer reference model with strict layering. The Internet model is the protocol suite actually deployed, with a thinner structure and the session and presentation functions left to applications.