(a) The capacitance of a piezoelectric transducer is 2000 pF and charge sensitivity is 30 × 10⁻³ C/m. Assume the capacitance of the connecting cable as 150 pF, when the oscilloscope used for readout has a readout input resistance of 1 MΩ with paralle

Question

(a) The capacitance of a piezoelectric transducer is 2000 pF and charge sensitivity is 30 × 10⁻³ C/m. Assume the capacitance of the connecting cable as 150 pF, when the oscilloscope used for readout has a readout input resistance of 1 MΩ with parallel capacitance of 100 pF.

Calculate the following : (20 marks)

(i) Sensitivity of transducer alone
(ii) High frequency sensitivity of the entire measuring system
(iii) Lowest frequency that can be measured with 5% amplitude error by the entire system
(iv) Value of the external shunt capacitance that can be connected in order to extend the range of 5% error down to 20 Hz

(b) Identify and explain briefly the addressing modes of 8085 microprocessor in the given instructions : (20 marks)

(i) ADD reg
(ii) MOV rd, M
(iii) CALL addr 16
(iv) LDA addr 16
(v) CMA

(c) The first order system and its response to unit step input are shown in Figure I and II respectively. Determine the system parameters 'a' and 'K'. (10 marks)

UPSC Answer Check · Accepted Answer

(a)
Let d = 30×10⁻³ C/m, C_t = 2000 pF = 2.000×10⁻⁹ F, C_c = 150 pF, C_o = 100 pF, R = 1 MΩ = 10⁶ Ω.
Total capacitance of the entire system is
C_total = C_t + C_c + C_o = 2000 + 150 + 100 = 2250 pF = 2.250×10⁻⁹ F.

(i) For a piezoelectric transducer, generated charge q = d x. The voltage sensitivity of the transducer alone is obtained by dividing charge sensitivity by its own capacitance.
S_t = d/C_t = (30×10⁻³)/(2000×10⁻¹²)
S_t = 15×10⁶ V/m = 15 MV/m.
Thus the charge sensitivity itself is 30×10⁻³ C/m, and the voltage sensitivity of the transducer alone is 15 MV/m.

(ii) At high frequency, the input resistance R is effectively open compared with the capacitive reactances, so all capacitances act in parallel. The high-frequency sensitivity of the entire measuring system is
S_HF = d/C_total = (30×10⁻³)/(2250×10⁻¹²)
S_HF = 13.333×10⁶ V/m = 13.33 MV/m.

(iii) For sinusoidal input, the low-frequency response relative to the high-frequency response is
S_low/S_HF = ωRC_total/√(1+(ωRC_total)²).
For 5% amplitude error, S_low = 0.95 S_HF. Hence
ωRC_total/√(1+(ωRC_total)²) = 0.95.
Let x = ωRC_total. Then
x²/(1+x²) = 0.95², so x² = 0.95²/(1−0.95²) = 361/39.
Therefore x = 19/√39 = 3.042.
Now RC_total = 10⁶ × 2.250×10⁻⁹ = 2.250×10⁻³ s.
f_low = x/(2πRC_total) = 3.042/(2π×2.250×10⁻³)
f_low ≈ 215.2 Hz.
So the lowest frequency measurable with 5% amplitude error is about 215 Hz.

(iv) To extend the 5% error range down to 20 Hz, the new total capacitance C_new must satisfy
x = 2πfRC_new at f = 20 Hz.
C_new = x/(2πfR) = 3.042/(2π×20×10⁶)
C_new = 24.21×10⁻⁹ F = 24.21 nF.
Therefore the required external shunt capacitance is
C_ext = C_new − C_total = 24.21 nF − 2.25 nF
C_ext = 21.96 nF ≈ 21960 pF.
It must be connected in parallel with the existing cable and oscilloscope capacitances.

(b)
(i) ADD reg: Register addressing mode. The operand is an 8-bit register specified in the instruction; the accumulator is the implied destination. Example: ADD B performs A ← A + B.

(ii) MOV rd, M: Register indirect addressing mode. Here M denotes the memory location whose address is stored in the HL register pair. The byte from memory address [HL] is copied to the destination register rd. Example: MOV B, M performs B ← [HL].

(iii) CALL addr16: Immediate addressing mode for the target address. The 16-bit subroutine address is embedded in the instruction itself. The microprocessor pushes the return address onto the stack and loads the program counter with addr16. Example: CALL 2050H.

(iv) LDA addr16: Direct addressing mode. The memory address is given directly in the instruction. The accumulator is loaded from that memory location. Example: LDA 2050H performs A ← [2050H].

(v) CMA: Implied or implicit addressing mode. No operand is specified in the instruction; the accumulator is implied. The operation complements the accumulator: A ← A̅.

(c)
The first-order system has transfer function
C(s)/U(s) = K/(s+a).
For a unit step input, U(s) = 1/s. Therefore
C(s) = K/[s(s+a)] = (K/a)[1/s − 1/(s+a)].
Taking inverse Laplace transform,
c(t) = (K/a)(1 − e^(−a t)).
The final value is c(∞) = K/a. From the graph, the response asymptotically approaches c(t) = 2.0, so
K/a = 2.0, hence K = 2a.

The initial slope of the response is
c'(0) = K.
The tangent drawn from the origin is therefore c = K t. It meets the final value c = 2.0 at t = 0.2 s. Thus
K(0.2) = 2.0, so K = 10 s⁻¹.
Then a = K/2 = 5 s⁻¹.

Final values:
a = 5 s⁻¹, K = 10 s⁻¹.
The system time constant is 1/a = 0.2 s, and the condition a > 0 ensures a stable first-order response.

(a) The capacitance of a piezoelectric transducer is 2000 pF and charge sensitivity is 30 × 10⁻³ C/m. Assume the capacitance of…

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