The modulation index \( m \) is given by the ratio of the peak voltage of the modulating signal \( V_m \) to the peak voltage of the carrier wave \( V_c \): \[ m = \frac{V_m}{V_c} \] Given that \( m = 0.75 \) and \( V_c = 16 \, {V} \), we can solve for \( V_m \): \[ 0.75 = \frac{V_m}{16} \] \[ V_m = 0.75 \times 16 = 12 \, {V} \] Thus, the peak voltage of the modulating signal is 12 V.
Final Answer: 12 V.
What are X and Y respectively in the following reaction sequence?
Identify the amino acid which has:
The correct sequence of reactions involved in the following conversion is:
Given the function:
\[ f(x) = \begin{cases} \frac{(2x^2 - ax +1) - (ax^2 + 3bx + 2)}{x+1}, & \text{if } x \neq -1 \\ k, & \text{if } x = -1 \end{cases} \]
If \( a, b, k \in \mathbb{R} \) and \( f(x) \) is continuous for all \( x \), then the value of \( k \) is:
Given the function:
\[ f(x) = \begin{cases} \frac{2x e^{1/2x} - 3x e^{-1/2x}}{e^{1/2x} + 4e^{-1/2x}}, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases} \]
Determine the differentiability of \( f(x) \) at \( x = 0 \).