The modulation index of an AM wave is given by :
\(m=\dfrac{V_{max}-V_{min}}{V_{max}+V_{min}}\)
The modulation index is the ratio of the message amplitude to the carrier amplitude, and both can be read off the envelope — giving option 3:
\(m=\dfrac{V_{max}-V_{min}}{V_{max}+V_{min}}\)
The derivation from the envelope. An AM wave's envelope swings between
\(V_{max}=V_{c}+V_{m},\qquad V_{min}=V_{c}-V_{m}\)
Adding and subtracting these two recovers the individual amplitudes:
\(V_{max}+V_{min}=2V_{c},\qquad V_{max}-V_{min}=2V_{m}\)
and dividing one by the other gives
\(\dfrac{V_{max}-V_{min}}{V_{max}+V_{min}}=\dfrac{2V_{m}}{2V_{c}}=\dfrac{V_{m}}{V_{c}}=m\)
Why the form must be a ratio of a difference to a sum. Two independent checks settle it without the algebra.
Dimensions. The index is a pure number, so both numerator and denominator must be voltages. Options 1 and 2 square one of them, giving units of volts and reciprocal volts respectively — neither is dimensionless, so both are impossible.
Range. The index must satisfy \(0\le m\le1\) for undistorted AM. In option 3, since \(V_{min}\ge0\), the numerator can never exceed the denominator, so m is automatically bounded by 1 — exactly right. Option 4 is its reciprocal and is therefore always at least 1, which no valid index can be.
Reading the two extremes off the formula.
| Case | Vmin | m |
|---|---|---|
| No modulation | Vmin = Vmax | 0 |
| Half modulation | Vmin = Vmax/3 | 0.5 |
| Full modulation | Vmin = 0 | 1 |
Why this particular form is the useful one. \(V_{max}\) and \(V_{min}\) are precisely what an oscilloscope displays when a trapezoidal or envelope pattern is examined, so the index can be measured directly from the screen without knowing the carrier or message amplitudes separately. Overmodulation shows itself immediately: the envelope would have to go negative for \(m\gt1\), so \(V_{min}\) flattens at zero and the envelope folds — visible distortion that a diode detector cannot undo.
Hence, m = (Vmax − Vmin)/(Vmax + Vmin).
Similar Questions
Assertion (A) : In amplitude modulation technique the modulation index should be close to 1.
Reason (R) : The power carried by message signal in the side bands increases with increase in modulation index.
Read the statements :
i. DSB has two side bands and SSB has one
ii. DSB has carrier and two side bands and SSB has a carrier and a side band
iii. DSB has carrier and two side bands and SSB without carrier and two different side bands.Which statements are correct ?
Assertion (A) : Amplitude modulation is wastage of power.
Reason (R) : Amplitude modulation is wastage of bandwidth.
The peak carrier voltage is 150 V. If the resistor is 200 Ω and the modulation index is 0.5, the total power in the AM signal is :
In amplitude modulation:
A. Amplitude of carrier is varied by modulating signal
B. Modulation index is between 0 and 1
C. Bandwidth is infinite
D. Bandwidth is twice of minimum modulating frequency.
Choose the correct answer from the options given below:
Match List I with List II
LIST I LIST II A. Power of AM Wave I. \(P_c\left(\frac{m^{2}}{4}\right)\) B. Power of VSB II. \(\frac{m^{2}}{4}P_c+F\left(\frac{m^{2}}{4}P_c\right)\) C. Power of SSB III. \(\frac{m^{2}}{4}\left(\frac{V_c^{2}}{2R}\right)+\frac{m^{2}}{4}\left(\frac{V_c^{2}}{2R}\right)\) D. Power of DSBSC IV. \(\frac{V_{carr}^{2}}{R}+\frac{V_{LSB}^{2}}{R}+\frac{V_{USB}^{2}}{R}\) Choose the correct answer from the options given below:
If 1 MHz carrier is amplitude modulated with a 5 kHz audio signal, the Upper Side Band (USB) and Lower Side Band (LSB) frequencies are ________ respectively.
Which of the following is associated with single balanced modulator circuit :
Consider an AM (amplitude modulated) signal
\(s(t) = 20\left[1 + 0.9\cos 2\pi \times 10^4 t\right]\cos 2\pi \times 10^6 t\)
The power efficiency (η) in the AM signal is
Important Questions from Amplitude Modulation
- An amplitude modulated voltage signal $x(t)$ drives a load of $1\ \Omega$. $$x(t) = K \cos(300\pi t) + L \cos(240\pi t) + L \cos(360\pi t).$$ If the efficiency is $60\%$, and the carrier power is $50\text{ W}$, the value of $L$ is ______ V.
A 2KW carrier is to be modulated to a 90% level. The total transmitted power would be__________.
Assertion (A) : In amplitude modulation technique the modulation index should be close to 1.
Reason (R) : The power carried by message signal in the side bands increases with increase in modulation index.
Read the statements :
i. DSB has two side bands and SSB has one
ii. DSB has carrier and two side bands and SSB has a carrier and a side band
iii. DSB has carrier and two side bands and SSB without carrier and two different side bands.Which statements are correct ?
Assertion (A) : Amplitude modulation is wastage of power.
Reason (R) : Amplitude modulation is wastage of bandwidth.
