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Front Panel of a Noise Reduction System

Amplitude modulation

In this section, we examine amplitude modulation and demodulation applications. For transmission purposes, signals are often modulated with a high-frequency carrier. A typical amplitude modulated signal can be described by

x ( t ) = x m ( t ) cos ( 2πf c t ) size 12{x \( t \) =x rSub { size 8{m} } \( t \) "cos" \( 2πf rSub { size 8{c} } t \) } {}

where x m ( t ) size 12{x rSub { size 8{m} } \( t \) } {} is called the message waveform, which contains the data of interest, and f c size 12{f rSub { size 8{c} } } {} is the carrier wave frequency. Using the fact that

cos ( 2πf c t ) = 1 2 ( e 2πf c t + e 2πf c t ) = 1 2 ( e ω c t + e ω c t ) size 12{"cos" \( 2πf rSub { size 8{c} } t \) = { {1} over {2} } \( e rSup { size 8{2πf rSub { size 6{c} } t} } +e rSup { - 2πf rSub { size 6{c} } t} size 12{ \) = { {1} over {2} } \( e rSup {ω rSub { size 6{c} } t} } size 12{+e rSup { - ω rSub { size 6{c} } t} } size 12{ \) }} {}

and the frequency shift property of CTFT, one can easily derive the CTFT ofto be

X ( ω ) = 1 2 X m ( ω ω c ) + X m ( ω + ω c ) size 12{X \( ω \) = { {1} over {2} } left (X rSub { size 8{m} } \( ω - ω rSub { size 8{c} } \) +X rSub { size 8{m} } \( ω+ω rSub { size 8{c} } \) right )} {}

At the receiver, some noisy version of this transmitted signal is received. The signal information resides in the envelope of the modulated signal, and thus an envelope detector can be used to recover the message signal.

[link] shows the completed block diagram of the amplitude modulation and demodulation system. In this example, use the combination of two sine waves to serve as a message signal. The signal is modulated with a high-frequency carrier, and some random noise is added. The frequency domain versions of the signals can also be observed using the function fft . As stated in Equation (3), the CTFT of the modulated signal is merely some frequency-shifted version of the original signal. In single sideband (SSB) modulation, only one side of the spectrum is transmitted due to symmetry. That is, just one side of the spectrum is taken and converted into a time signal using the function ifft .

[link] shows the completed front panel of this system. The Message signal, Modulated signal, Received signal (modulated signal with additional noise) and Demodulated signal are displayed in four waveform graphs in both the time and frequency domains.

Block Diagram of an Amplitude Modulation and Demodulation System

Front Panel of an Amplitude Modulation and Demodulation System

Lab exercises

Circuit Analysis

Find and plot the frequency response (both magnitude and phase spectrum) of each of the circuits shown in [link] . Set the values of R, L and C as controls.

Linear RLC Circuits

Insert Solution Text Here

Morse Coding

Consider a message containing some hidden information. Furthermore, to make it interesting, suppose the message contains a name. Assume that the message was coded using the amplitude modulation scheme as follows [link] :

x ( t ) = x m1 ( t ) cos ( 2πf 1 t ) + x m2 ( t ) cos ( 2πf 2 t ) + x m3 ( t ) cos ( 2πf 3 t ) size 12{x \( t \) =x rSub { size 8{m1} } \( t \) "cos" \( 2πf rSub { size 8{1} } t \) +x rSub { size 8{m2} } \( t \) "cos" \( 2πf rSub { size 8{2} } t \) +x rSub { size 8{m3} } \( t \) "cos" \( 2πf rSub { size 8{3} } t \) } {}

where x m1 ( t ) , x m2 ( t ) size 12{x rSub { size 8{m1} } \( t \) ,x rSub { size 8{m2} } \( t \) } {} and x m3 ( t ) size 12{x rSub { size 8{m3} } \( t \) } {} are the (message) signals containing the three letters of the name. More specifically, each of the signals, x m1 ( t ) , x m2 ( t ) size 12{x rSub { size 8{m1} } \( t \) ,x rSub { size 8{m2} } \( t \) } {} and x m3 ( t ) size 12{x rSub { size 8{m3} } \( t \) } {} , corresponds to a single letter of the alphabet. These letters are encoded using the International Morse Code as indicated below [7]:

A . H . . . . O V . . . B . . . I . . P . . W . C . . J . Q . X . . D . . K . R . . Y . E . L . . . S . . . Z . . F . . . M T G . N . U . . size 12{ matrix { A {} # "." - {} {} # {} # H {} # "." "." "." "." {} # {} # O {} # - - - {} {} # {} # V {} # "." "." "." - {} {} ##B {} # - "." "." "." {} # {} # I {} # "." "." {} # {} # P {} # "." - - "." {} # {} # W {} # "." - - {} {} ## C {} # - "." - "." {} # {} # J {} # "." - - - {} {} # {} # Q {} # - - "." - {} {} # {} # X {} # - "." "." - {} {} ##D {} # - "." "." {} # {} # K {} # - "." - {} {} # {} # R {} # "." - "." {} # {} # Y {} # - "." - - {} {} ## E {} # "." {} # {} # L {} # "." - "." "." {} # {} # S {} # "." "." "." {} # {} # Z {} # - - "." "." {} ##F {} # "." "." - "." {} # {} # M {} # - - {} {} # {} # T {} # - {} {} # {} # {} # {} ## G {} # - - "." {} # {} # N {} # - "." {} # {} # U {} # "." "." - {} {} # {} # {} # {}} } {}

Now to encode the letter A, one needs only a dot followed by a dash. That is, only two prototype signals are needed – one to represent the dash and one to represent the dot. Thus, for instance, to represent the letter A, set x m1 ( t ) = d ( t ) + dash ( t ) size 12{x rSub { size 8{m1} } \( t \) =d \( t \) + ital "dash" \( t \) } {} , where d ( t ) size 12{d \( t \) } {} represents the dot signal and dash ( t ) size 12{ ital "dash" \( t \) } {} the dash signal. Similarly, to represent the letter O, set x m1 ( t ) = 3 dash ( t ) size 12{x rSub { size 8{m1} } \( t \) =3 ital "dash" \( t \) } {} .

Find the prototype signals d ( t ) size 12{d \( t \) } {} and dash ( t ) size 12{ ital "dash" \( t \) } {} in the file morse.mat on the book website. After loading the file morse.mat

>>load morse

the signals d ( t ) size 12{d \( t \) } {} and dash ( t ) size 12{ ital "dash" \( t \) } {} can be located in the vectors dot and dash, respectively. The hidden signal, which is encoded, per Equation (4), containing the letters of the name, is in the vector xt size 12{ ital "xt"} {} Let the three modulation frequencies f 1 , f 2 size 12{f rSub { size 8{1} } ,f rSub { size 8{2} } } {} and f 3 size 12{f rSub { size 8{3} } } {} be 20, 40 and 80 Hz, respectively.

• Using the amplitude modulation property of the CTFT, determine the three possible letters and the hidden name. (Hint: Plot the CTFT of xt size 12{ ital "xt"} {} Use the values of T size 12{T} {} and τ au size 12{τ ital "au"} {} contained in the file.)

• Explain the strategy used to decode the message. Is the coding technique ambiguous? That is, is there a one-to-one mapping between the message waveforms ( x m1 ( t ) , x m2 ( t ) , x m3 ( t ) ) size 12{x rSub { size 8{m1} } \( t \) ,x rSub { size 8{m2} } \( t \) ,x rSub { size 8{m3} } \( t \) \) } {} ) and the alphabet letters? Or can you find multiple letters that correspond to the same message waveform?

Insert Solution Text Here

Doppler Effect

The Doppler effect phenomenon was covered in a previous chapter. In this exercise, let us examine the Doppler effect with a real sound wave rather than a periodic signal. The wave file firetrucksiren.wav on the book website contains a firetruck siren. Read the file using the LabVIEW MathScript function wavread and produce its upscale and downscale versions. Show the waves in the time and frequency domains (find the CTFT). Furthermore, play the sounds using the LabVIEW function Play Waveform . [link] shows a typical front panel for this system.

Front Panel of Doppler Effect System

Insert Solution Text Here

Diffraction of Light

The diffraction of light can be described as a Fourier transform [link] . Consider an opaque screen with a small slit being illuminated by a normally incident uniform light wave, as shown in [link] .

Diffraction of Light

Considering that d >> πl 1 2 / λ size 12{d">>"πl rSub { size 8{1} rSup { size 8{2} } } /λ} {} provides a good approximation for any l 1 size 12{l rSub { size 8{1} } } {} in the slit, the electric field strength of the light striking the viewing screen can be expressed as [link]

E 0 ( l 0 ) = K e j ( 2πd / λ ) jλd e j ( π / λd ) l 0 2 E 1 ( l 1 ) e j ( / λd ) l 0 l 1 dl 1 size 12{E rSub { size 8{0} } \( l rSub { size 8{0} } \) =K { {e rSup { size 8{j \( 2πd/λ \) } } } over {jλd} } e rSup { size 8{j \( π/λd \) l rSub { size 6{0} rSup { size 6{2} } } } } Int cSub { - infinity } cSup { infinity } {E rSub {1} size 12{ \( l rSub {1} } size 12{ \) e rSup { - j \( 2π/λd \) l rSub { size 6{0} } l rSub { size 6{1} } } } size 12{ ital "dl" rSub {1} }} } {}

where

E 1 size 12{E rSub { size 8{1} } } {} = field strength at diffraction screen

E 0 size 12{E rSub { size 8{0} } } {} = field strength at viewing screen

K size 12{K} {} = constant of proportionality

λ size 12{λ} {} = wavelength of light

The above integral is in fact Fourier transformation in a different notation. One can write the field strength at the viewing screen as [link]

( t ) E 1 f l 0 / λd E 0 ( l 0 ) = K e j ( 2πd / λ ) jλd e j ( π / λd ) l 0 2 CTFT size 12{E rSub { size 8{0} } \( l rSub { size 8{0} } \) =K { {e rSup { size 8{j \( 2πd/λ \) } } } over {jλd} } e rSup { size 8{j \( π/λd \) l rSub { size 6{0} rSup { size 6{2} } } } } ital "CTFT" lbrace E rSub {1} size 12{ \( t \) rbrace rSub {f rightarrow l rSub { size 6{0} } /λd} }} {}

The intensity I ( l 0 ) size 12{I \( l rSub { size 8{0} } \) } {} of the light at the viewing screen is the square of the magnitude of the field strength. That is,

I ( l 0 ) = E 0 ( l 0 ) 2 size 12{I \( l rSub { size 8{0} } \) = lline E rSub { size 8{0} } \( l rSub { size 8{0} } \) rline rSup { size 8{2} } } {}
  1. Plot the intensity of the light at the viewing screen. Set the slit width to this range (0.5 to 5 mm), the wavelength of light λ size 12{λ} {} to this range (300 to 800 nm), and the distance of the viewing screen d size 12{d} {} to this range (10 to 200 m) as controls. Assume the constant of proportionality is 10 3 size 12{"10" rSup { size 8{ - 3} } } {} , and the electric field strength at the diffraction screen is 1 V/m.
  2. Now replace the slit with two slits, each 0.1 mm in width, separated by 1 mm (center-to-center) and centered on the optical axis. Plot the intensity of light in the viewing screen by setting the parameters in part (1) as controls.

Insert Solution Text Here

Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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Differences Between Laspeyres and Paasche Indices
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J, combine like terms 7x-4y
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Asali
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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AMJAD
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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AMJAD
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AMJAD
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Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
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after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
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Prasenjit
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Damian
silver nanoparticles could handle the job?
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Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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Source:  OpenStax, An interactive approach to signals and systems laboratory. OpenStax CNX. Sep 06, 2012 Download for free at http://cnx.org/content/col10667/1.14
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