The principal advantage of using Bode plots in control system analysis and design is their ability to convert multiplicative factors into additive factors. This transformation occurs because Bode plots represent gain in decibels (dB) and phase angles, which allows engineers to easily sketch and interpret complex frequency responses.
In a Bode plot, the logarithmic scale simplifies the multiplication of transfer function components, such as poles and zeros, by transforming them into a linear addition process. This aids in constructing a visual representation of system stability and performance by breaking the response into magnitude and phase plots.
Consider the unity-negative-feedback system shown in Figure (i) below, where gain \( K \geq 0 \). The root locus of this system is shown in Figure (ii) below.
For what value(s) of \( K \) will the system in Figure (i) have a pole at \( -1 + j1 \)?

Consider a message signal \( m(t) \) which is bandlimited to \( [-W, W] \), where \( W \) is in Hz. Consider the following two modulation schemes for the message signal:
• Double sideband-suppressed carrier (DSB-SC): \[ f_{DSB}(t) = A_c m(t) \cos(2\pi f_c t) \] • Amplitude modulation (AM): \[ f_{AM}(t) = A_c \left( 1 + \mu m(t) \right) \cos(2\pi f_c t) \] Here, \( A_c \) and \( f_c \) are the amplitude and frequency (in Hz) of the carrier, respectively. In the case of AM, \( \mu \) denotes the modulation index. Consider the following statements:
(i) An envelope detector can be used for demodulation in the DSB-SC scheme if \( m(t)>0 \) for all \( t \).
(ii) An envelope detector can be used for demodulation in the AM scheme only if \( m(t)>0 \) for all \( t \).
Which of the following options is/are correct?