To find the upper and lower side frequencies when an information signal is modulated with a carrier wave, we use the following formulas:
where \( f_c \) is the carrier frequency and \( f_m \) is the frequency of the information signal.
Given:
We can calculate as follows:
Upper Side Frequency:
\( f_c + f_m = 3610 \text{ kHz} + 10 \text{ kHz} = 3620 \text{ kHz} \)
Lower Side Frequency:
\( f_c - f_m = 3610 \text{ kHz} - 10 \text{ kHz} = 3600 \text{ kHz} \)
Thus, the solution is that the upper side and lower side frequencies are 3620 kHz and 3600 kHz, respectively.
Given:
- Information signal frequency \( f_m = 10 \, {kHz} \)
- Carrier frequency \( f_c = 3.61 \, {MHz} = 3610 \, {kHz} \) When an information signal is modulated, the upper side frequency (USF) and lower side frequency (LSF) are given by the following formulas: \[ {USF} = f_c + f_m \quad {and} \quad {LSF} = f_c - f_m \] Substituting the given values: \[ {USF} = 3610 \, {kHz} + 10 \, {kHz} = 3620 \, {kHz} \] \[ {LSF} = 3610 \, {kHz} - 10 \, {kHz} = 3600 \, {kHz} \] Thus, the upper side frequency and lower side frequency are \( 3620 \, {kHz} \) and \( 3600 \, {kHz} \), respectively.
Which of the following are ambident nucleophiles?
[A.] CN$^{\,-}$
[B.] CH$_{3}$COO$^{\,-}$
[C.] NO$_{2}^{\,-}$
[D.] CH$_{3}$O$^{\,-}$
[E.] NH$_{3}$
Identify the anomers from the following.

The standard Gibbs free energy change \( \Delta G^\circ \) of a cell reaction is \(-301 { kJ/mol}\). What is \( E^\circ \) in volts?
(Given: \( F = 96500 { C/mol}\), \( n = 2 \))