Question

2. Using a well-known trigonometric identity involving the product of the sine of an angle and the cosine of another angle, demonstrate that the multiplication of the incoming FM wave s(t) by the locally generated FM wave r(t) produces two components (except for the scaling factor): 1. A high-frequency component, which is defined by the double-frequency term namely, where km is the multiplier gain. 2. A low-frequency component, which is defined by the diference-frequency term namely. kmAcA, sin[i(t) - 2(0)] Take s(t)-Ac sin[2ntat + φ1(t)] and r(t-Ac cos[2nat + φ2(t)].

Answer 1

The incoming wave is expressed as $s(t) = A_c \sin [2\pi f_ct + \phi_1(t)]$

The locally generated wave is expressed as $r(t) = A_v \cos [2\pi f_ct + \phi_2(t)]$

Note: In the problem, the amplitudes of waves are $A_c$ for incoming wave and $A_v$ for locally generated wave.

Let, the output of FM waves multiplication is expressed as

f (t)-t) * r(t)

The incoming wave is multiplied with the locally generated wave.

Replacing the waves variables with the actual expressions, we get

$f(t) = A_c \sin [2\pi f_ct + \phi_1(t)] * A_v \cos[2\pi f_ct + \phi_2(t)]$

$f(t) = A_c A_v \sin [2\pi f_ct + \phi_1(t)] \cos[2\pi f_ct + \phi_2(t)]$

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-------------- TRIGONOMETRY -----------------------------

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We know that the sine of sum of two angles is expressed as,

sin(A + B)- sin A cos B cos Asin B iln

Similarly the sine of difference of two angles is expressed as,

sin(A - B) sin A cos B -cos A sin B

Adding the above two sine expressions, we get

sin(A B) + sin(A- B)2 sin A cos B

or It can be expressed in other way as,

$\sin A \cos B = \frac{1}{2} \sin (A+B) + \frac{1}{2} \sin (A-B)$

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We have $A = 2\pi f_ct + \phi_1(t)$ and $B = 2\pi f_ct + \phi_2(t)$ in

f (t)

Now using the trigonometry,

$f(t) = A_c A_v [ \frac{1}{2} \sin (2\pi f_ct + \phi_1(t) + 2\pi f_ct + \phi_2(t)) + \frac{1}{2} \sin (2\pi f_ct + \phi_1(t) - 2\pi f_ct - \phi_2(t)) ]$

$f(t) = A_c A_v [ \frac{1}{2} \sin (4\pi f_ct + \phi_1(t) + \phi_2(t)) + \frac{1}{2} \sin (\phi_1(t) - \phi_2(t)) ]$

further, taking the amplitudes with the terms,

$f(t) = \frac{1}{2} A_c A_v \sin [4\pi f_ct + \phi_1(t) + \phi_2(t)] + \frac{1}{2} A_c A_v \sin [\phi_1(t) - \phi_2(t)]$

We have two components in the above expression with $k_m = \frac{1}{2}$ , the multiplier gain.

Thus, the multiplication of the two waves produces the two components. The first component is double frequency term and the second component is difference frequency term.