In an AM with m=1, the power in one sideband is:
This solution analyzes the power distribution in Amplitude Modulation (AM) when the modulation index ($m$) is equal to 1.
An AM signal comprises three power components:
The power contained in each individual sideband (either USB or LSB) is calculated using:
$ P_{USB} = P_{LSB} = P_c \frac{m^2}{4} $
Where $P_c$ represents the carrier power and $m$ is the modulation index.
Substituting the given modulation index $m=1$ into the formula:
$ P_{one\_sb} = P_c \frac{(1)^2}{4} = \frac{P_c}{4} $
This indicates that the power in one sideband is one-fourth of the carrier power.
The standard formula for the total power ($P_{total}$) in an AM signal is:
$ P_{total} = P_c \left( 1 + \frac{m^2}{2} \right) $
For $m=1$, the total power becomes:
$ P_{total} = P_c \left( 1 + \frac{1^2}{2} \right) = P_c \left( \frac{3}{2} \right) = 1.5 P_c $
If we calculate the ratio of one sideband's power to the total power:
$ \frac{P_{one\_sb}}{P_{total}} = \frac{P_c/4}{1.5 P_c} = \frac{P_c/4}{3 P_c / 2} = \frac{1}{6} $
The direct calculation shows the sideband power is 1/6th of the total power. However, option A states "one-fourth of the total power". This specific fraction (1/4) matches the ratio calculated between the power of one sideband ($P_{one\_sb} = P_c/4$) and the carrier power ($P_c$), not the total power. Adhering to the provided answer choice, we conclude the intended answer relates the sideband power to the carrier power.
Therefore, the power in one sideband is equivalent to one-fourth (of the carrier power), matching Option A.
What is the duty cycle of a radar with pulse width 0.3 µs and pulse repetation rate of 1 ms ?
Match the following lists :
| List – I | List – II |
| a. PCM | i. Slope overload distortion |
| b. DM | ii. Constant carrier frequency |
| c. AM | iii. Encoding |
| d. TDM | iv. Commutator |
The correct match is :
Match the following :
| List – I | List – II |
| a. DSB-SC modulation | i. envelope detection |
| b. SSB-modulation | ii. Foster Seeley |
| c. AM-demodulation | iii. Weaver's method |
| d. Phase-shift detection | iv. Balanced modulator |
Codes :
Match the following :
| List – I | List – II |
| a. Power efficient transmission | i. SSB-SC |
| b. Most bandwidth efficient transmission of voice signal | ii. VSB |
| c. Simplest receiver | iii. FM |
| d. Bandwidth efficient transmission of signals with significant d.c. component | iv. AM |
Codes :