In a MOSFET-based amplifier circuit, the voltage gain of the amplifier depends on the type of configuration used. In this case, we are looking at common source and common gate amplifiers.
- Common Source Amplifier: This is an inverting amplifier configuration. The output voltage is inverted relative to the input voltage. The voltage gain for a common-source amplifier is typically negative, indicating inversion.
- Common Gate Amplifier: This configuration is also an inverting amplifier. The input signal is applied to the gate, and the output is taken from the drain. Similar to the common-source amplifier, the common-gate amplifier produces an inverted output.
Thus, the correct answer is (A) because both common source and common gate amplifiers are inverting amplifiers.
Selected data points of the step response of a stable first-order linear time-invariant (LTI) system are given below. The closest value of the time constant (in seconds) of the system is:
\[ \begin{array}{|c|c|} \hline \textbf{Time (sec)} & \textbf{Output} \\ \hline 0.6 & 0.78 \\ 1.6 & 2.8 \\ 2.6 & 2.98 \\ 10 & 3 \\ \infty & 3 \\ \hline \end{array} \]The op-amps in the following circuit are ideal. The voltage gain of the circuit is ……….. . (Round off to the nearest integer)
The maximum percentage error in the equivalent resistance of two parallel connected resistors of 100 ohm and 900 ohm, with each having a maximum 5% error, is:
A controller \( (1 + K_{DS}) \) is to be designed for the plant \[ G(s) = \frac{1000 \sqrt{2}}{s(s + 10)^2} \] The value of \( K_D \) that yields a phase margin of 45 degrees at the gain cross-over frequency of 10 rad/sec is ……… (round off to 1 decimal place).
Consider the state-space model:
\[ \dot{x}(t) = A x(t) + B u(t) \] \[ y(t) = C x(t) \] \[ A = \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix}, \quad B = \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \quad C = \begin{bmatrix} 1 & 0 \end{bmatrix} \]The sum of the magnitudes of the poles is: