GATE ECE (Electronics) Test: Double Sideband Suppressed Carrier (DSB-SC)

Concept:

The total transmitted power for an AM system is given by:

P_c = Carrier Power

μ = Modulation Index

Since the given modulation is DSB-SC, the carrier is suppressed. 
∴ The transmitted power will be:

Calculation: 

Given

μ = 0.4

Pc = 80 W

Concept:

Standard expression for AM expression for a message signal AmCosωmt is

S(t) = A_C(1 + μCosω_mt )cos ω_ct

Where;

ω_m → Modulating frequency (rad/sec) = 2πf_m

Bandwidth = 2f_m

Calculation:

Given;

Modulating frequency (f_m) = 5 KHz

Bandwidth = 2f_m = (2 × 5) = 10 KHz

Concept:

The expression of the full-carrier AM (DSB-full-carrier) = (E_c + m(t)) × sin(ωct) 

• ​E_c is peak amplitude of the carrier signal i.e  E_csin(ωct).
• ω_c is the frequency of the carrier signal in rad/sec.
• m(t) is modulating or the message signal.

Calculation:

Given, m(t) = E_msin(ωmt) , where E_m is the peak amplitude of the messege signal and ω_m is the frequency of the message signal in rad/sec.

.Equation for full-carrier AM signal:   

v(t) =  (E_c + m(t)) × sin(ωct) 

⇒ v(t) =  (E_c + E_msin (ω_mt) ) × sin(ωct) 

Hence, the equation of the full-carrier AM (DSB-full carrier) is v(t) = (E_c + E_m sin(ω_mt)) × sin(ω_ct).

Concept:

An amplitude modulation spectrum is represented as:

f_c = Carrier Frequency

f_m = Signal frequency

f_c + f_m = Upper Sideband

f_c - f_m = Lower Sideband

Analysis:

Given f_c = 40 kHz and f_m = 5 kHz

The upper and lower sidebands after Amplitude modulation will be:

Upper Sideband = fc + fm = 40 + 5 kHz = 45 kHz

Lower Sideband = fc - fm = 40 - 5 kHz = 35 kHz

In a balanced modulator, 2 AM-modulators are connected in a way that the resultant signal does not contain the carrier spectrum, i.e. to generate a Double Side-band suppressed carrier (DSB-SC).

Analysis:

The circuit diagram of a balanced modulator is as shown:

The resultant DSB signal is the difference of the two, i.e.

The frequency spectrum of the DSB-SC signal:

The standard DSB-SC equation is given as:

Assuming a single tone message signal as:

The DSB-SC signal can be written as:

∴ The output of balanced modulator contains the modulation frequency, LSB and USB.

The ring modulator and balanced modulator is generally used for the modulation of DSB SC signal.

SSB/SC is generated using a balanced modulator

DSB -SC:

1) It is a form of Amplitude modulation where only the sidebands are transmitted and the carrier is suppressed.

2) The advantage is 66% power saving compared to conventional AM.

3) The Bandwidth requirement is of 2f_m, with f_m being the maximum frequency component of the message signal.

4) It is used for transmitting stereo information.

Single Side Band (SSB SC):

1) SSB SC is a form of amplitude modulation where only a single sideband is transmitted and 1 sideband and carrier are suppressed.

2) The advantage of SSB is bandwidth and power-saving over both conventional AM and DSB SC, i.e. DSB SC has more power consumption than SSB SC.

3) The Bandwidth requirement is only fm, ∴ It needs half the bandwidth than the DSB-SC transmission.

Comparison:

Concept: Low pass filter allow the low frequency components ( up to Bandwidth of filter) to pass through it.

Calculation:

x(t) = m(t) cos (2πf_ct)

m(t) = 4 cos (1000 πt)

The output of the multiplier(s(t)) will be:

s(t) = x(t) cos (2π(f_c + 40)t)

s(t) = m(t) cos (2πf_ct) cos (2π(f_c + 40)t)

s(t) = 2 cos (100 πt)⋅cos (2π 40t) + 2 cos (1000 πt) cos (2π(2f_c + 40)t

Again using the same trigonometric identify, the above equation can be written as:

s(t) = cos (1000πt – 80πt) + cos (1000πt + 80πt) + cos (1000πt – 2π(2f_c + 40)t) + cos (1000πt + 2π (2f_c + 40)t)

s(t) = cos (920πt) + cos (1080πt) + cos (1000πt – 2π (2f_c - 40)t + cos(1000πt + 2π (2f_c + 40)t)

The frequency specimen of s(t) can be drawn as:

The output of the low pass filter will contain frequencies of 460 Hz only as shown:

∴ y(t) = cos (920 πt)

s₁(t) = A_c [1 + k_am(t) ] cosω_ct

s₂(t) = A_c [1 - k_am(t) ] cosω_ct

s(t) =  Ac [1 + kam(t) ] cosωct -  Ac [ 1 - kam(t) ] cosωct

= 2 A_c k_a m(t) cosωct

s(t) ∝ m(t) cosωct

Hence, Envelope of s(t) = |m(t)|

Concept:

A single sideband (SSB) AM Signal is represented as

m̂(t) → represent Hilbert transform of message signal

(-) → sign represent USB

(+) → sign represent LSB

Hilbert transform:

It is a specific Linear operator that takes a function u(t) of a real variable and produces another function of real variable H(u)(t)

→ It imparts a phase shift of ± 90° (π/2 radians) to every frequency component of a function.

Explanation:

From the given Block diagram.

I₁ = (Am sin ω_mt)(Ac cos ω_ct)

I₂ = (Am cos ω_mt)(Ac cos ω_ct)

V_out = I₁ + I₃

From I₁  

Option (A) correct choice.

Short trick:

Observe the V_out in block diagram i.e. V_out = I₁ + I₃

But in I₃ there is –π/2 phase change so the SSB becomes in the Form of

S(t) = m(t) cos ω_ct – m̂(t) sin ω_ct

And (-) sign represent USB only.

So the student can tick the correct answer.