GATE ECE 2001 | Question: 17

Question

The periodic modulating signal $m(t)$ is shown in the figure. Using Carson's rule. estimate $B_{F M}$ (bandwidth of the FM signal) and $B_{P M}$ (bandwidth of the PM signal) for $k f=\pi \times 10^{4}$ and $k p=\frac{\pi}{4}$. Assume the essential bandwidth of $m(t)$ to consist only up to and including the third harmonic.

 

Answer 1

a. Bandwidth of FM Signal : 40 kHz
b. Bandwidth of PM Signal : 53.562 kHz

Answer 2

1. Analyze the Message Signal $m(t)$

• Waveform: Periodic triangular wave.

• Amplitude ($A_m$): The signal oscillates between $-1$ and $+1$, so the peak amplitude is $1 \text{ V}$.

• Period ($T$): The time between two consecutive peaks is given as $2 \times 10^{-4}$ seconds.

   

  $$T = 2 \times 10^{-4} \text{ s}$$

• Fundamental Frequency ($f_0$):

   

  $$f_0 = \frac{1}{T} = \frac{1}{2 \times 10^{-4}} = \frac{10^4}{2} = 5000 \text{ Hz} = 5 \text{ kHz}$$

• Message Bandwidth ($W$): The problem states the essential bandwidth includes up to the third harmonic.

   

  $$W = 3 \times f_0 = 3 \times 5 \text{ kHz} = 15 \text{ kHz}$$

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Part 1: Estimate FM Bandwidth ($B_{FM}$)

Given:

• Frequency Sensitivity: $k_f = \pi \times 10^4$ (Assuming units are rad/s/V based on the $\pi$ factor).

Step 1: Calculate Peak Frequency Deviation ($\Delta f$)

The frequency deviation in radians per second is $\Delta \omega = k_f \cdot \max|m(t)|$.

 

$$\Delta \omega = (\pi \times 10^4) \times 1 = \pi \times 10^4 \text{ rad/s}$$

Convert to Hertz ($\Delta f$):

 

$$\Delta f = \frac{\Delta \omega}{2\pi} = \frac{\pi \times 10^4}{2\pi} = 5000 \text{ Hz} = 5 \text{ kHz}$$

Step 2: Apply Carson's Rule

 

$$B_{FM} = 2(\Delta f + W)$$

$$B_{FM} = 2(5 \text{ kHz} + 15 \text{ kHz})$$

$$B_{FM} = 2(20 \text{ kHz})$$

$$B_{FM} = 40 \text{ kHz}$$

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Part 2: Estimate PM Bandwidth ($B_{PM}$)

Given:

• Phase Sensitivity: $k_p = \frac{\pi}{4}$ rad/V.

Step 1: Calculate Peak Frequency Deviation ($\Delta f$)

For Phase Modulation, the instantaneous frequency deviation is proportional to the slope (derivative) of the message signal: $\Delta \omega(t) = k_p \cdot \frac{dm(t)}{dt}$.

• Slope of $m(t)$: The signal goes from $-1$ to $+1$ (change of 2) in half a period ($T/2 = 10^{-4}$ s).

   

  $$\text{Slope} = \frac{\Delta \text{Amplitude}}{\Delta \text{Time}} = \frac{2}{10^{-4}} = 20,000 \text{ V/s}$$

• Calculate $\Delta \omega$:

   

  $$\Delta \omega = k_p \times \text{Slope} = \frac{\pi}{4} \times 20,000 = 5000\pi \text{ rad/s}$$

• Convert to Hertz ($\Delta f$):

   

  $$\Delta f = \frac{\Delta \omega}{2\pi} = \frac{5000\pi}{2\pi} = 2500 \text{ Hz} = 2.5 \text{ kHz}$$

Step 2: Apply Carson's Rule

 

$$B_{PM} = 2(\Delta f + W)$$

$$B_{PM} = 2(2.5 \text{ kHz} + 15 \text{ kHz})$$

$$B_{PM} = 2(17.5 \text{ kHz})$$

$$B_{PM} = 35 \text{ kHz}$$

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Final Answer

• $B_{FM} = 40 \text{ kHz}$

• $B_{PM} = 35 \text{ kHz}$