1. Home
  2. Introduction to analog and
  3. Introduction to analog and
  4. Analog communication

0.4 Analog communication  (Page 8/2)

<< Chapter < Page
  Introduction to analog and     Page 1 / 2
Chapter >> Page >
This module describes basic analog modulation techniques, including amplitude modulation (AM) with suppressed carrier, AM with a pilot tone or carrier tone, quadrature AM (QAM), vestigial sideband modulation (VSB), and frequency modulation (FM). Various demodulation techniques are also discussed, including envelope detection and the discriminator. Application examples include NTSC television and FM radio (both mono and stereo).
  1. Amplitude modulation (AM)
  2. Quadrature amplitude modulation (QAM)
  3. Vestigial sideband modulation (VSB)
  4. Frequency modulation (FM)

Am with “suppressed carrier”

AM of real-valued message m ( t ) (e.g., music) is

This figure contains one flowchart and two expressions. The flowchart begins with the variable m(t) with an arrow pointing to the right at a circle labeled x. Below the circle is a circle containing a tilde, labeled cos(2πf_ct). An arrow from the tilde circle points up at the x circle. To the right of the x circle is an arrow pointing to the right at the variable s(t). To the right of this flowchart are two expressions. The first reads s(t) = m(t)cost(2πf_ct). The second reads f_c = carrier freq. This figure contains one flowchart and two expressions. The flowchart begins with the variable m(t) with an arrow pointing to the right at a circle labeled x. Below the circle is a circle containing a tilde, labeled cos(2πf_ct). An arrow from the tilde circle points up at the x circle. To the right of the x circle is an arrow pointing to the right at the variable s(t). To the right of this flowchart are two expressions. The first reads s(t) = m(t)cost(2πf_ct). The second reads f_c = carrier freq.

Euler's cos ( 2 π f c t ) = 1 2 e j 2 π f c t + e - j 2 π f c t then implies

S ( f ) = ∫ - ∞ ∞ m ( t ) cos ( 2 π f c t ) e - j 2 π f t d t = 1 2 ∫ - ∞ ∞ m ( t ) e - j 2 π ( f - f c ) t d t + 1 2 ∫ - ∞ ∞ m ( t ) e - j 2 π ( f + f c ) t d t = 1 2 M ( f - f c ) + 1 2 M ( f + f c ) .
This figure contains two graphs. The first plots f on the horizontal axis, and |M(f)| on the vertical axis. There is a box of height 1 and width 2W on this graph. The base of the box sits on the horizontal axis, with the left side at value -W and the right side at W. To the right of this graph is an arrow pointing to the right at the second graph, which plots horizontal axis f against vertical axis |S(f)|. In the graph, there are two boxes, one centered on the horizontal axis at -f_c and the other centered at f_c. The width of these is measured as 2B. The height is measured as 1/2. This figure contains two graphs. The first plots f on the horizontal axis, and |M(f)| on the vertical axis. There is a box of height 1 and width 2W on this graph. The base of the box sits on the horizontal axis, with the left side at value -W and the right side at W. To the right of this graph is an arrow pointing to the right at the second graph, which plots horizontal axis f against vertical axis |S(f)|. In the graph, there are two boxes, one centered on the horizontal axis at -f_c and the other centered at f_c. The width of these is measured as 2B. The height is measured as 1/2.

Because m ( t ) ∈ R , know | M ( f ) | symmetric around f = 0 , implying the AM transmitted spectrum below f c is redundant! This motivates the QAM and VSB modulation schemes...

With f c known, AM demodulation can be accomplished by:

This figure contains one flowchart and two expressions. The flowchart begins with the variable r(t) with an arrow pointing to the right at a circle labeled x. Below the circle is a circle containing a tilde, labeled cos(2πf_ct). An arrow from the tilde circle points up at the x circle. To the right of the x circle is an arrow pointing to the right at a box labeled LPF. To the right is an arrow pointing to the right at the expression v(t). To the right of this flowchart is the expression v(t) = LPF{r(t)*2cos(2πf_ct)} This figure contains one flowchart and two expressions. The flowchart begins with the variable r(t) with an arrow pointing to the right at a circle labeled x. Below the circle is a circle containing a tilde, labeled cos(2πf_ct). An arrow from the tilde circle points up at the x circle. To the right of the x circle is an arrow pointing to the right at a box labeled LPF. To the right is an arrow pointing to the right at the expression v(t). To the right of this flowchart is the expression v(t) = LPF{r(t)*2cos(2πf_ct)}

For a trivial noiseless channel, we have r ( t ) = s ( t ) , so that

v ( t ) = LPF { s ( t ) · 2 cos ( 2 π f c t ) } = LPF { m ( t ) · 2 cos 2 ( 2 π f c t ) ︸ 1 + cos ( 2 π · 2 f c t ) } = LPF { m ( t ) + m ( t ) cos ( 2 π · 2 f c t ) } = m ( t ) ,

assuming a LPF with passband cutoff B p ≥ W Hz and stopband cutoff B s ≤ 2 f c - W Hz:

This figure is a wide graph with horizontal axis f of three shapes and two dashed lines. In the middle of the graph, centered at the vertical axis and the origin, is a box stretching from horizontal value -W to W. From the top of this box are two dashed lines that first start out horizontally and then decrease diagonally to the horizontal axis. The diagonal dashed lines land on horizontal values -B_s and B_s. Further outside are hash marks on either side of the dashed line, labeled -f_c and f_c. Beyond these are two more boxes, with bases on the horizontal axis. The inside corner of the boxes are labeled -2f_c + W and 2f_c - W, and the midpoints of the boxes are labeled -2f_c and 2f_c. Their height is not labeled but is approximately one-half the height of the box in the middle. This figure is a wide graph with horizontal axis f of three shapes and two dashed lines. In the middle of the graph, centered at the vertical axis and the origin, is a box stretching from horizontal value -W to W. From the top of this box are two dashed lines that first start out horizontally and then decrease diagonally to the horizontal axis. The diagonal dashed lines land on horizontal values -B_s and B_s. Further outside are hash marks on either side of the dashed line, labeled -f_c and f_c. Beyond these are two more boxes, with bases on the horizontal axis. The inside corner of the boxes are labeled -2f_c + W and 2f_c - W, and the midpoints of the boxes are labeled -2f_c and 2f_c. Their height is not labeled but is approximately one-half the height of the box in the middle.

Note that we've assumed perfectly synchronized oscillators!

When the receiver oscillator has {freq,phase} offset { γ , φ } :

v ( t ) = LPF m ( t ) cos ( 2 π f c t ) · 2 cos ( 2 π ( f c + γ ) t + φ ) ︸ cos ( 2 π γ t + φ ) + cos ( 2 π ( 2 f c + γ ) t + φ ) = m ( t ) cos ( 2 π γ t + φ ) ︸ time-varying attenuation! .
a freq offset of λ = ν f c c Hz can occur when there is relative velocity of ν m/s between transmitter and receiver.

Am with “pilot tone” or “carrier tone”

It's common to include a pilot/carrier tone with frequency| f c :

s ( t ) = m ( t ) cos ( 2 π f c t ) + A cos ( 2 π f c t ) ︸ pilot/carrier tone = [ m ( t ) + A ] cos ( 2 π f c t ) S ( f ) = 1 2 M ( f - f c ) + M ( f + f c ) + A δ ( f - f c ) + A δ ( f + f c )
This figure is comprised of two cartesian graphs. The graph on the left is rather small, plotting f on the horizontal axis and |M(f)| on the vertical axis. The graph contains one large rectangle, with base sitting on the horizontal axis, and with measured to be from -W to W. To the right of this graph is an arrow pointing to the right at a larger graph, also plotting f on the horizontal axis, but this time with a vertical axis |S(f)|. In this graph there are two rectangles, each approximately one-half the height of the rectangle in the first graph, but with a similar width. The bases are both on the horizontal axis, and the centers of the bases are labeled -f_c and f_c. The width of each of these is measured with an arrow above, labeled 2W. In the center of the bases of these rectangles are vertical arrows that begin below the horizontal axis and point directly up. This figure is comprised of two cartesian graphs. The graph on the left is rather small, plotting f on the horizontal axis and |M(f)| on the vertical axis. The graph contains one large rectangle, with base sitting on the horizontal axis, and with measured to be from -W to W. To the right of this graph is an arrow pointing to the right at a larger graph, also plotting f on the horizontal axis, but this time with a vertical axis |S(f)|. In this graph there are two rectangles, each approximately one-half the height of the rectangle in the first graph, but with a similar width. The bases are both on the horizontal axis, and the centers of the bases are labeled -f_c and f_c. The width of each of these is measured with an arrow above, labeled 2W. In the center of the bases of these rectangles are vertical arrows that begin below the horizontal axis and point directly up.

Advantage: aids receiver with carrier synchronization.

Disadvantage: consumes transmission power.

While modern systems choose A ≪ max | m ( t ) | , many older systems use A > max | m ( t ) | , known as “large carrier AM,” allowing reception based on envelope detection :

v ( t ) = π 2 LPF { | r ( t ) | } - A ≈ m ( t ) (with a trivial channel)

where | · | can be easily implemented using a diode.

The gain π 2 above makes up for the loss incurred when LPFing the rectified signal:

This figure is comprised of three graphs, one equation, and one statement. Beginning in the top-left is a graph plotting t on the horizontal axis. There are a series of peaks in the first quadrant of the graph, with a peak beginning on the vertical axis and continuing downward. The first peak is shaded tea. The point where the first peak reaches the horizontal axis is measured at a horizontal value 1/4f_c. Midway up the vertical axis is a horizontal dashed line labeled LPF output level. Along the peaks is another dashed horizontal line labeled desired output level. There are three complete peaks, one half-peak at the beginning, and one half-peak at the end. To the right of the graph is an equation. The equation begins with a large fraction on the top, the numerator being the integral of lower bound 0 and upper bound 1/4f_c of cosine (2 pi f_c t) dt, and the denominator 1/4f_c. This is equal to 2 / pi. Below this equation on the right side are two graphs, the first plotting time against amplitude and showing closely packed waves moving together with increasing, then decreasing, then increasing amplitudes. The graph is titled large-carrier AM. The second plots time against amplitude and contains two graphs that closely follow one another with a simple wave and is titled envelope-detected signal (and original message). To the left of these graphs is the statement MATLAB code here. This figure is comprised of three graphs, one equation, and one statement. Beginning in the top-left is a graph plotting t on the horizontal axis. There are a series of peaks in the first quadrant of the graph, with a peak beginning on the vertical axis and continuing downward. The first peak is shaded tea. The point where the first peak reaches the horizontal axis is measured at a horizontal value 1/4f_c. Midway up the vertical axis is a horizontal dashed line labeled LPF output level. Along the peaks is another dashed horizontal line labeled desired output level. There are three complete peaks, one half-peak at the beginning, and one half-peak at the end. To the right of the graph is an equation. The equation begins with a large fraction on the top, the numerator being the integral of lower bound 0 and upper bound 1/4f_c of cosine (2 pi f_c t) dt, and the denominator 1/4f_c. This is equal to 2 / pi. Below this equation on the right side are two graphs, the first plotting time against amplitude and showing closely packed waves moving together with increasing, then decreasing, then increasing amplitudes. The graph is titled large-carrier AM. The second plots time against amplitude and contains two graphs that closely follow one another with a simple wave and is titled envelope-detected signal (and original message). To the left of these graphs is the statement MATLAB code here.

Quadrature amplitude modulation (qam)

QAM is motivated by unwanted redundancy in the AM spectrum, which was symmetric around f c .

QAM sends two real-valued signals { m I ( t ) , m Q ( t ) } simultaneously, resulting in a non-symmetric spectrum.

This figure is a flowchart, beginning with an expression m_I(t), with an arrow pointing to the right at a circle containing an x. Below this is the expression m_Q(t), with an arrow pointing to the right at another circle containing an x. In between these two circles is a circle containing a tilde, with an arrow pointing up and another arrow pointing down at the two x-circles. These two arrows are labeled cos(2 pi f_c t) for the one pointing up, and sin (2 pi f_c t) for the one pointing down. To the right of both of the x-circles are arrows pointing at a single circle containing a plus sign, with the lower arrow labeled with a minus sign. The lower arrow is labeled quadrature, and the upper is labeled in phase. To the right of this plus-circle is an arrow pointing to the right at a large equation that reads s(t) = m_I(t)cos( 2 pi f_c t) - m_Q(t)sin (2 pi f_c t). This figure is a flowchart, beginning with an expression m_I(t), with an arrow pointing to the right at a circle containing an x. Below this is the expression m_Q(t), with an arrow pointing to the right at another circle containing an x. In between these two circles is a circle containing a tilde, with an arrow pointing up and another arrow pointing down at the two x-circles. These two arrows are labeled cos(2 pi f_c t) for the one pointing up, and sin (2 pi f_c t) for the one pointing down. To the right of both of the x-circles are arrows pointing at a single circle containing a plus sign, with the lower arrow labeled with a minus sign. The lower arrow is labeled quadrature, and the upper is labeled in phase. To the right of this plus-circle is an arrow pointing to the right at a large equation that reads s(t) = m_I(t)cos( 2 pi f_c t) - m_Q(t)sin (2 pi f_c t).

QAM demodulation is accomplished by:

This is a flowchart that starts with the expression r(t). From r(t) there are two arrows pointing at two separate circles containing an x, above and below one another. The arrow that points to the lower circle is labeled with a minus sign. In between the two circles is another circle with a tilde. Above and below this circle are arrows pointing up and down at the x-circles, and the arrows are labeled 2 cosine (2 pi f_c t) above and 2 sine (2 pi f_c t) below. To the right of the x-circles are two boxes labeled LPF. To the right of these boxes are two arrows pointing to the right at two final expressions, v_I(t) above and v_Q(t) below. This is a flowchart that starts with the expression r(t). From r(t) there are two arrows pointing at two separate circles containing an x, above and below one another. The arrow that points to the lower circle is labeled with a minus sign. In between the two circles is another circle with a tilde. Above and below this circle are arrows pointing up and down at the x-circles, and the arrows are labeled 2 cosine (2 pi f_c t) above and 2 sine (2 pi f_c t) below. To the right of the x-circles are two boxes labeled LPF. To the right of these boxes are two arrows pointing to the right at two final expressions, v_I(t) above and v_Q(t) below.

where the LPF specs are the same as in AM, i.e., passband edge B p ≥ W Hz and stopband edge B s ≤ 2 f c - W Hz.

For a trivial channel, we have r ( t ) = s ( t ) , so that

v I ( t ) = LPF { r ( t ) · 2 cos ( 2 π f c t ) } = LPF { m I ( t ) 2 cos 2 ( 2 π f c t ) ︸ 1 + cos ( 4 π f c t ) - m Q ( t ) 2 sin ( 2 π f c t ) cos ( 2 π f c t ) ︸ sin ( 4 π f c t ) } = m I ( t ) v Q ( t ) = LPF { - r ( t ) · 2 sin ( 2 π f c t ) } = LPF { - m I ( t ) 2 cos ( 2 π f c t ) sin ( 2 π f c t ) ︸ sin ( 4 π f c t ) + m Q ( t ) 2 sin 2 ( 2 π f c t ) ︸ 1 - cos ( 4 π f c t ) } = m Q ( t ) ,

assuming synchronized oscillators.

When the oscillators are not synchronized, one gets coupling between the I&Q components as well as attenuation of each. Writing the I&Q signals in the “complex-baseband” form

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
Aislinn Reply
cm
tijani
what is titration
John Reply
what is physics
Siyaka Reply
A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
Jude Reply
Can you compute that for me. Ty
Jude
what is the dimension formula of energy?
David Reply
what is viscosity?
David
what is inorganic
emma Reply
what is chemistry
Youesf Reply
what is inorganic
emma
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
Adjei
please, I'm a physics student and I need help in physics
Adjanou
chemistry could also be understood like the sexual attraction/repulsion of the male and female elements. the reaction varies depending on the energy differences of each given gender. + masculine -female.
Pedro
A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
Krampah Reply
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
Sahid Reply
you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
Ryan
what's motion
Maurice Reply
what are the types of wave
Maurice
answer
Magreth
progressive wave
Magreth
hello friend how are you
Muhammad Reply
fine, how about you?
Mohammed
hi
Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
yasuo Reply
Who can show me the full solution in this problem?
Reofrir Reply
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply
<< Chapter < Page Page > Chapter >>

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Introduction to analog and digital communications. OpenStax CNX. Sep 14, 2009 Download for free at http://cnx.org/content/col10968/1.2
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Introduction to analog and digital communications' conversation and receive update notifications?

Ask