A telecommunication system monitors 10 substations, each containing 5 signal channels. Each channel has a signal bandwidth of 100 Hz. The signals are quantized and encoded. The sampling rate is twice the Nyquist rate. The maximum quantization error is 0.25% of the peak amplitude m_p. The required overall bit rate of the system.

Step-by-step Solution:

Step 1: Calculate Thevenin's Equivalent Voltage (\(V_{th}\))

Terminals X-Y are open-circuited, so no current flows through the branch containing the \(10\,\Omega\) resistor and the \(X_C = 10\,\Omega\) capacitor connected to terminal X.

Therefore, the open-circuit voltage \(V_{th}\) across terminals X-Y is equal to the voltage drop across the parallel branch \((3 + j14)\,\Omega\).

Using the Voltage Divider Rule:

\(V_{th} = V_s \times \frac{3 + j14}{-j10 + (3 + j14)}\)

\(V_{th} = 100 \times \frac{3 + j14}{3 + j4}\)

Multiplying the numerator and denominator by the complex conjugate of the denominator \((3 - j4)\):

\(V_{th} = 100 \times \frac{(3 + j14)(3 - j4)}{3^2 + 4^2}\)

\(V_{th} = 100 \times \frac{9 - j12 + j42 + 56}{25}\)

\(V_{th} = 100 \times \frac{65 + j30}{25}\)

\(V_{th} = 4 \times (65 + j30) = 260 + j120\text{ V}\)

In polar form:

\(|V_{th}| = \sqrt{260^2 + 120^2} = 286.36\text{ V}\)

\(\theta = \tan^{-1}\left(\frac{120}{260}\right) = 24.78^\circ\)

\(V_{th} = 286.36 \angle 24.78^\circ\text{ V}\)

Step 2: Calculate Thevenin's Equivalent Impedance (\(Z_{th}\))

To find the equivalent impedance \(Z_{th}\), deactivate the independent voltage source by replacing it with a short circuit.

Looking back into terminals X-Y, the \(-j10\,\Omega\) capacitor is in parallel with the \((3 + j14)\,\Omega\) branch, and this combination is in series with the \((10 - j10)\,\Omega\) branch leading to terminal X.

\(Z_{th} = (10 - j10) + \left[ (-j10) \parallel (3 + j14) \right]\)

Calculating the parallel portion (\(Z_p\)):

\(Z_p = \frac{-j10 \times (3 + j14)}{-j10 + 3 + j14} = \frac{140 - j30}{3 + j4}\)

Multiplying by the conjugate \((3 - j4)\):

\(Z_p = \frac{(140 - j30)(3 - j4)}{3^2 + 4^2} = \frac{420 - j560 - j90 - 120}{25} = \frac{300 - j650}{25} = 12 - j26\,\Omega\)

Now, substitute \(Z_p\) back to calculate \(Z_{th}\):

\(Z_{th} = (10 - j10) + (12 - j26) = 22 - j36\,\Omega\)

In polar form:

\(|Z_{th}| = \sqrt{22^2 + (-36)^2} = 42.19\,\Omega\)

\(\theta = \tan^{-1}\left(\frac{-36}{22}\right) = -58.57^\circ\)

\(Z_{th} = 42.19 \angle -58.57^\circ\,\Omega\)

Final Answer:

The Thevenin's equivalent circuit across terminal X-Y consists of:

  
• Thevenin's Voltage (\(V_{th}\)): \(260 + j120\text{ V}\) or \(286.36 \angle 24.78^\circ\text{ V}\)
  
• Thevenin's Impedance (\(Z_{th}\)): \(22 - j36\,\Omega\) or \(42.19 \angle -58.57^\circ\,\Omega\)

Maximum Power Transfer Theorem :

In electrical engineering, the maximum power transfer theorem states that, to obtain maximum external power from a power source with internal resistance, the resistance of the load must equal the resistance of the source as viewed from its output terminals.

For t<0, the switch is closed.

Capacitor acts as open to DC.

Voltage across the capacitor,

     v(0^-)= {(12||4)×V}÷{(12||4)+6} = 8VFor t=0, the switch is opened,Voltage across the Capacitor cannot change instantaneously,So, v(0^-)= v(0) = 8VAt t>0, The capacitor is discharging.R_th = (12||4) = 3 ohmC= 1/6 FTime constant, Π = R_th×C = 0.5 So, v(t) = v(0)e^-t/Π  = 8e^-t/0.5 =8e^-2t V (Ans.)