Question

The ring modulator circuit works by multiplying together the 125 kHz carrier signal, Vc, and the 10 kHz audio signal, Va, to produce the transmitted modulated signal output, Vo. The two signals and the output from the ring modulator are described as follows:

Vc= 4cos (2pi125*10^3t)
Va= 2cos (2pi10*10^3t)
Vo= [4 + 2cos (2pi10*10^3t)]cos (2pi125*10^3t)
Expanding the above equation for Vo, and using appropriate trigonometrical identities, show that the output from the modulator comprises frequency components at 115 kHz, 125 kHz, and 135 kHz. Subsequently, plot your alternative mathematical expansion to demonstrate that the two versions of the output signal, Vo, from Part 1 and Part 2, are identical.

Answer 1

The ring modulator circuit's output contains frequency components at 115 kHz, 125 kHz, and 135 kHz. The alternative mathematical expansion of the output equation confirms the equivalence of the two representations.

Step-by-step explanation:

To find the frequency components in the output signal, Vo, we need to expand the equation using trigonometric identities. Let's start by expanding the equation for Vo:

Vo = [4 + 2cos(2pi10*10^3t)]cos(2pi125*10^3t)

Using the identity cos(A)cos(B) = 0.5[cos(A+B) + cos(A-B)], we can rewrite the equation as:

Vo = [4*cos(2pi125*10^3t) + 2cos(2pi10*10^3t)*cos(2pi125*10^3t)]

Expanding the second term using the same identity, we get:

Vo = 4*cos(2pi125*10^3t) + cos(2pi(10*10^3 + 125*10^3)t) + cos(2pi(125*10^3 - 10*10^3)t)

From this expanded equation, we can see that the output signal, Vo, contains frequency components at 115 kHz, 125 kHz, and 135 kHz.

To show that the two versions of the output signal, Vo, from Part 1 and Part 2 are identical, we can plot the alternative mathematical expansion and compare it with the original equation. The plots will overlap, confirming the equivalence of the two representations.