Communications Test 1

Analysis:

For a DSB modulator, the modulated signal is expressed as:

x₁(t) = c(t) m₁(t)

x₂(t) = c(t) m₂(t)

x₁(t) + x₂(t) = c(t) (m₁(t) + m₂(t))       ---(1)

When the combined signal m₁(t) and m₂(t) is passed, we get the DSB modulated signal as:

x_DSB(t) = c(t) (m₁(t) + m₂(t))        ---(2)

Since Equation (1) and (2) are giving the same signal, statement S₁ is correct.

When m₁(t) is FM modulated, the modulated wave will be:

\({x_1}\left( t \right) = {A_c}\cos \left[ {2\pi {f_c}t + 2\pi {k_t}\;\mathop \smallint \limits_{ - \infty }^t {m_1}\left( \tau \right)d\tau } \right]\) 

Similarly, for a signal m₂(t), FM output will be:

\({x_2}\left( t \right) = {A_c}\cos \left[ {2\pi {f_c}t + 2\pi {k_f}\;\mathop \smallint \limits_{ - \infty }^t {m_2}\left( \tau \right)d\tau } \right]\) 

When m₁(t) and m₂(t) are combined and are frequency modulated, we get:

\({x_{FM}}\left( t \right) = {A_c}\cos \left[ {2\pi {f_c}t + 2\pi {k_f}\mathop \smallint \limits_{ - \infty }^t \left( {{m_1}\left( t \right) + {m_2}\left( t \right)} \right)dt} \right]\;\) 

For FM, we observe that:

x₁(t) + x₂(t) ≠ x_FM(t)

∴ Statement 2 (S₂) is not correct.

Concept:

A general angle modulated wave is defined as:

S(t) = A_c cos (ω_c t + km(t))

The phase deviation is:

θ(t) = km(t)

and the frequency deviation is:

\({\rm{\Delta }}f = \frac{1}{{2\pi }}k\frac{d}{{dt}}\left( {m\left( t \right)} \right)\)

Application:

Given carrier frequency = 1 kHz, i.e.

ω_c = 2π × 1000 rad/sec

The output signal is given as:

x(t) = cos (2π 1100t)

This can be written as:  

x(t) = cos (2π 1000t + 2π 100t)

= cos (2πf_ct + 200πt)

So, we have the phase angle as:

θ(t) = 200πt

∴ The phase deviation as a function of time will be:

Δθ = θ(t) = 200 πt

And the frequency deviation will be:

\({\rm{\Delta }}f = \frac{1}{{2\pi }} \cdot \frac{{d\theta \left( t \right)}}{{dt}}\)

\( = \frac{1}{{2\pi }} \times 200\pi \)

Concept:

The general expression of an SSB modulated signal is:

S_SSB(t) = m(t) cosω_ct ± m̂(t) sinω_ct

A plus sign implies the Lower-sideband signal and minus sign implies an Upper sideband signal.

m(t) = Message signal

ω_c = Carrier signal frequency

m̂(t) = Hilbert transform of the message signal

Hilbert Transformer transforms the input signal as:

1) The magnitude of the frequency component present in x(t) remains unchanged when it is passed through the system.

2) The phase of the positive frequency components is shifted by -π/2 and that of negative frequency components is shifted by +π/2.

Application:

m(t) = cos 30t and ω_c = 850 rad/sec

The Hilbert transform of the above signal will be:

m̂(t) = cos (30t – 90°) = sin 30 t

For lower side band, the SSB expression will be:

s_L(t) = m(t) cos ω_c t + m̂(t) sin ω_ct

Putting on the respective function values, we get:

sL(t) = cos 30t cos 850t + sin 30t sin 850t

sL(t) = cos (850 - 30)t

The given modulating signal is:

x(t) = sin (200πt) + 2 sin (400πt)

Also, the modulation signal is given as:

g(t) = x(t) sin (400πt)

The above signal is then multiplied with sin 400πt to get:

g(t) sin 400 πt = x(t) sin (400 πt) sin (400πt)

= x(t) sin² 400 πt

Substituting x(t) in the above expression, we get:

x(t) sin² 400 πt = (sin (200πt) + 2 sin (400πt)) sin² 400 πt

\( = \left[ {\sin \left( {200\pi t} \right) + 2\sin \left( {400\pi t} \right)} \right]\left[ {\frac{{1 - \cos \left( {800\pi t} \right)}}{2}} \right]\) 

\( = \frac{1}{2}\left[ {\sin \left( {200\pi t} \right) + 2\sin \left( {400\pi t} \right) - \sin \left( {200\pi t} \right)\cos \left( {800\pi t} \right) - 2\sin \left( {400\;\pi t} \right)\cos \left( {800\pi t} \right)} \right]\) 

\( = \frac{1}{2}\sin \left( {200\pi t} \right) + \sin \left( {400\;\pi t} \right) - \frac{1}{4}[\sin \left( {1000\pi t} \right) - \sin \left( {600\pi t} \right)] - \frac{1}{2}\left[ {\sin \left( {1200\;\pi t - \sin \left( {400\pi t} \right)} \right)} \right]\) 

Now, the above signal is passed through a low pass filter with a passband gain of 2 and cut-off frequency ω₀ = 400π

∴ Only the frequency of 200π will be allowed to pass through the LPF.

So, the output signal will be:

\(y\left( t \right) = \left[ {\frac{1}{2}\sin \left( {200\pi t} \right)} \right] \times \left| {H\left( f \right)} \right|\) 

Where |H(f)| is the passband gain given as:

|H(f)| = 2

∴ The output will be:

\(y\left( t \right) = \left[ {\frac{1}{2}\sin \left( {200\pi t} \right)} \right] \times 2\) 

= sin (200πt)

Concept:

The general expression for an amplitude modulated signal, with single tone modulating signal, is defined as:

S_AM(t) = A_C (1 + cos ω_mt) cos ω_c t

= A_c cos ω_c + A_c cos ω_mt cos ω_ct

\( = {A_C}\cos {\omega _c}t + \frac{{{A_c}}}{2}\cos \left( {{\omega _c} - {\omega _m}} \right)t + \frac{{{A_c}}}{2}\cos \left( {{\omega _c} + {\omega _m}} \right)t\) 

Also, the modulation efficiency of an AM signal is defined as the ratio of the power with the sidebands to the total power i.e.

\(\eta = \frac{{{P_{sb}}}}{{{P_t}}} \times 100\) 

Analysis:

Given:

x_AM(t) = A cos (400πt) + B cos (380 πt) + B cos (420 πt) 

Comparing this with the general expression, the above can be written as:

x_AM(t) = A cos 400 πt + B cos (400 π – 20 π)t + B cos (400 π + 20 π)t

The carrier signal is, therefore:

c(t) = A cos 400πt

And the two sidebands are:

Sideband 1: B cos (380 πt)

Sideband 2: B cos (420 πt)

The carrier power will be:

\({P_c} = \frac{{{A^2}}}{2}\) 

Given P_c = 100, we can write:

\(100 = \frac{{{A^2}}}{2}\) 

A² = 200

A = 14.14

Now, the total sideband power will be:

\({P_{sb}} = \frac{{{B^2}}}{2} + \frac{{{B^2}}}{2}\) 

P_SB = B²

Given modulation efficiency = 40% = 0.4

We can write:

\(0.4 = \frac{{{P_{SB}}}}{{{P_t}}} = \frac{{{P_{SB}}}}{{{P_C} + {P_{SB}}}}\) 

\(0.4 = \frac{{{B^2}}}{{100 + {B^2}}}\) 

40 + 0.4 B² = B²

40 = 0.6 B²

\({B^2} = \frac{{400}}{6}\) 

B = 8.16

Alternate method:

The power efficiency of an AM signal with single tone modulation is given by:

\({\eta _{AM}} = \frac{{{\mu ^2}}}{{2 + {\mu ^2}}}\) 

μ = Modulation index

Given output of AM modulator is:

x(t) = A cos (400 πt) + B cos (380 πt) + B cos (240 πt)

Appying trigonometric property, the above equation becomes:

x(t) = A cos (400 πt) + 2 B cos (400 πt) cos (20 πt)

\( = A\left[ {1 + \frac{{2B}}{A}\cos \left( {20\pi t} \right)} \right]\cos \left( {400\;\pi t} \right)\) 

From the above, we get the carrier power as:

\({P_C} = \frac{{{A^2}}}{2}\) 

And the modulation index (μ) as:

\(\mu = \frac{{2B}}{A}\)         ---(1)

Now,

P_C = 100 W (Given)

\(\therefore 100 = \frac{{{A^2}}}{2}\) 

A = 14.14

η_AM = 40% = 0.4

\(0.4 = \frac{{{\mu ^2}}}{{2 + {\mu ^2}}}\) 

0.8 + 0.4 μ² = μ²

0.8 = 0.6 μ²

\({\mu ^2} = \;\frac{4}{3}\) 

\(\mu = \frac{2}{{\sqrt 3 }}\) 

Substituting this in equation (1), we get:

\(\frac{2}{{\sqrt 3 }} = \frac{{2B}}{A}\) 

\(B = \frac{A}{{\sqrt B }} = \frac{{14.14}}{{\sqrt 3 }}\) 

Concept:

The bandwidth of the multitone Angle modulated signal is given by:

B.W = 2(1 + β) f_max

Where, f_max = max (f_m1, f_{m2, …})

Calculation:

f_i = f_c­ + k_f m(t)

\(= \left( {4 \times {{10}^6}} \right) + 40 \times {10^3}\left( {\frac{{Hz}}{v}} \right)\left[ {2\cos \left( {2\pi \times {{10}^3}t} \right) + \cos \left( {3\pi \times {{10}^3}t} \right) + 5\cos \left( {8\pi \times {{10}^3}t} \right)} \right]\) 

\({f_i} = \left( {4 \times {{10}^6}} \right) + 80 \times {10^3}\cos \left( {2\pi \times {{10}^3}t} \right) + 40 \times {10^3}\left( {2\pi \times {{10}^3}t} \right) + 200 \times {10^3}\cos \left( {8\pi \times {{10}^3}t} \right)\) 

Δf = 80 × 10³ + 40 × 10³ + 200 × 10³

⇒ (320 × 10³) Hz

\(\beta = \frac{{{\rm{\Delta }}f}}{{{f_{max}}}} = \frac{{320 \times {{10}^3}}}{{4 \times {{10}^3}}} = 80\) 

Bandwidth = 2 (1 + 80) 4 × 10³