Quadrature amplitude modula.on

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zer Source Modulator analog sequence symbol sequence interface #8 EIE 67 Digital Communica.

on Interpolator message (output waveform) Table lookup Source receiver De Topics on (QAM; 6.

1 Quadrature amplitude modula.on message ( waveform) Today s topics concern with the and de transminer Watcharapan Suwansan.suk Sampler Quan.zer Source Modulator analog sequence symbol sequence interface #8 EIE 67 Digital Communica.ons King Mongkut s University of Technology Thonburi Chapter 6: s, modula.on, and demodula.on Interpolator message (output waveform) Table lookup Source receiver De Topics Quadrature amplitude modula.on (QAM; 6.) AXer this lecture, you will be able to Convert a sequence of bits into a waveform, and vice versa, using the QAM Describe a method to convert a waveform to a waveform, and vice versa Plot the and waveforms in.me- and frequency-domains for the noiseless channel

2 Recall that a turns a sequence of bits into a waveform (analog) value... Modulator value QAM: OVERVIEW time (sec).me (sec) 6 Also recall that a de turns a waveform into a sequence of bits Passband PAM has one problem: it wastes the transmission bandwidth ˆx(f) This picture is for real-valued ˆx(f) value... De value f c f c frequency f time (sec).me (sec) Wasteful, two iden.cal copies of the waveform appear at If the waveform components differ at +f c and f c, the transminer will send bits twice faster QAM speeds up the data rate by running two PAMs in parallel ±f c 7 8

3 In our design, the QAM and de are divided into blocks Modulator Intui.vely, QAM adds and subtracts the two waveform-components a a waveform from an output two sequences of real numbers de two waveforms to to waveform channel a b b b a + b waveform from a transmined waveform De Each pair of the opposing block performs a reversed role f c frequency f This picture is a mnemonic. See Example for precise drawing f c 9 Intui.vely, QAM de also add and subtract the waveform-components (con.nued) ( ) (a + b) + (a b) = a ( ) -- (a + b) (a b) = b f f f f f f One waveform is obtained by a sum The other waveform is obtained by a difference

4 QAM works by running PAMs in parallel QAM (bb) QAM output QAM (bb) QAM bb de QAM to to QAM to output de de to to QAM: SIGNAL ENCODER QAM bb de QAM to Input is a sequence; outputs are sequences of real numbers b,b,b,... a sequence of real numbers u,u,u,... another sequence of real numbers u,u,u,... Signal reads many bits at a.me Bits maps to point (, ) Bits maps to point (, ) Q-axis Bits maps to point (, ) I-axis Bits maps to point (, ) And map each group of bits into a point in dimensions Axes are usually called I (in-phase) and Q (quadrature) In the above examples, two bits are read at a.me 6

5 A constellaon is a picture that maps bits into points Q-axis I-axis -QAM constella.ons Q-axis I-axis (cont ) 6-QAM constella.on Q-axis... I-axis. Labels.. to.. can be arranged in any way, depending on the design of the 7 8 (cont ) 8-PSK constella.on Example : -QAM constella.on Q-axis Q-axis 8 I-axis I-axis PSK = phase shix keying Suppose the is Then the output at the is, and at the is, 9

6 In-class ac.vity Draw your own constella.on How many bits did the read at a.me? What is the output at the I- and es if the is below?... output QAM (bb) QAM bb de QAM to to QAM: SIGNAL DECODER QAM to The maps an point to the nearest point Q-axis output (.,.) I-axis Signal and use the same constella.on.. Example : -QAM constella.on Q-axis. (,.) output I-axis.. (.,.) The constella.on is the -QAM of example Suppose the to the I- and es are above Then, the output are

7 The may choose any nearest point to break the.e Q-axis either (,.) or output I-axis. output QAM (bb) QAM bb de QAM to to QAM to QAM: BASEBAND MODULATOR Suppose the is above Then, the output could be either or 6 Baseband consists of PAM s in parallel u,u,u,... u,u,u,... QAM (bb) T> is the symbol interval u I (t) = u Q (t) = X u k p(t p(t) is a shix-orthonormal pulse, with a limited bandwidth k= X k= u k p(t kt) kt) Recall the meanings of magnitude, phase, real part and imaginary part z = +i magnitude (length) = Real part of z imaginary axis Imaginary part of z phase (angle) =. radians = 7 real axis % Matlab! z = - + i*;! abs(z) % : magnitude! angle(z) %. : angle (radian)! real(z) % - : real part! imag(z) % : imaginary part! 7 8

8 Example : QAM (cont ) Time-domains and. u I (t) u Q (t) Baseband waveform at I and Q branches, u I (t) =p(t) p(t T ).. u I (t) u Q (t), u Q (t) = p(t) p(t T ). QAM (bb) Suppose the s at I- and es are above Then, the expressions of output are as shown Suppose t (sec) T = sec is the square root of raised cosine filter of p(t) =. 9 (con t) Frequency-domain waveforms (cont ) Magnitude and phase of. û I (f) The Fourier transform of u I (t) =p(t) p(t T ) is 6 Magnitude of ûi(f ) û I (f) =ˆp(f) ˆp(f)e ift =ˆp(f) e ift Magnitude The Fourier transform of u Q (t) = p(t) p(t T ) û Q (f) = ˆp(f) ˆp(f)e ift = ˆp(f) +e ift is Phase (rad).... Phase angle (radian) of ûi(f )....

9 (cont ) Real and imaginary parts of. û I (f) (cont ) Magnitude and phase of. û Q (f) Real part of ûi(f ) 6 Magnitude of ûq(f ) real part Imaginary part of ûi(f ) Phase angle (radian) of ûq(f ) imaginary part (cont ) Real and imaginary parts of. Real part of ûq(f ) û Q (f) QAM (bb) QAM.... output to to Imaginary part of ûq(f ) QAM bb de QAM to QAM: BASEBAND DEMODULATOR.... 6

.. de de v I (t) v Q (t) output QAM (bb) QAM bb de QAM to to QAM to Project onto each basis element: QAM (bb) de Z vk = v I (t)p(t Z vk = v Q (t)p(t kt)dt kt)dt QAM: BASEBAND TO PASSBAND 7 8 The

10 Baseband de consists of PAM des in parallel v,v,v,... v,v,v,... de de v I (t) v Q (t) output QAM (bb) QAM bb de QAM to to QAM to Project onto each basis element: QAM (bb) de Z vk = v I (t)p(t Z vk = v Q (t)p(t kt)dt kt)dt QAM: BASEBAND TO PASSBAND 7 8 The frequency is shixed up by mul.plica.ons of cosine and sine cos( f c t) This property of Fourier transform is central to modula.on using sine Let h(t) be an func.on, and ĥ(f) be its Fourier transform L u I (t) u Q (t) x I (t) x Q (t) x(t) ModulaNon property (for sine): if constant frequency, then f ĝ(f) = i ĥ(f + f ) g(t) =h(t)sin( f t) i ĥ(f f ), for any sin( f c t) is the carrier frequency: QAM f c f c Bandwidth W of pulse p(t) Interpretaon: if we mul.ple a func.on by a sine term of frequency, then f f its Fourier transform shix to the lex and to the right by, and the phases change 9

11 The waveforms in.me- and frequencydomains equal the following expressions cos( f c t) x I (t) =u I (t) cos( f c t) ˆx I (f) =û I (f + f c )+û I (f f c ) Ques.ons: True or false z imaginary axis z u I (t) x I (t) z u Q (t) x Q (t) x(t) =x I (t)+x Q (t) ˆx(f) =ˆx I (f)+ˆx Q (f) real axis sin( f c t) QAM x Q (t) = u Q (t)sin( f c t) ˆx Q (f) = iû Q (f + f c )+iû Q (f f c ) Let and denote any complex numbers Let z z z = z + z z denote the sum: (cont ) Ques.ons: True or false. The magnitude of z equals the magnitude of z plus the magnitude of z, i.e. z = z + z Answer: True False (cont ) Ques.ons: True or false. Real part of equals real part of plus real part of, i.e., Answer: True False z z z Re {z} =Re{z } +Re{z }. The phase of z equals the phase of z plus the phase of z, i.e., \z = \z + \z Answer: True False. Imaginary part of z equals imaginary part of z plus imaginary part of, i.e., z Im {z} =Im{z } +Im{z } Answer: True False

12 Example : cos( f c t) u I (t) =p(t) p(t T ) x I (t) x(t) u Q (t) = p(t) p(t T ) x Q (t) (con t) Time-domains and. x I (t) Passband waveform at I and Q branches x I (t) x Q (t) x Q (t) Suppose waveforms are the same as Example s Suppose f c = Hz Plot,, and x I (t) x Q (t) x(t) Plot,, and ˆx I (f) ˆx Q (f) ˆx(f) sin( f c t) t (sec) 6 (cont ) Magnitude and phase of. ˆx I (f) (cont ) Real and imaginary parts of. ˆx I (f) 6 Magnitude of ˆxI(f ) Real part of ˆxI(f ) Phase angle (radian) of ˆxI(f ) Imaginary part of ˆxI(f )

13 (cont ) Magnitude and phase of. ˆx Q (f) (cont ) Real and imaginary parts of. ˆx Q (f) 6 Magnitude of ˆxQ(f ) Real part of ˆxQ(f ) Phase angle (radian) of ˆxQ(f ) Imaginary part of ˆxQ(f ) (con t) Time-domain. x(t) =x I (t)+x Q (t) (cont ) Magnitude and phase of. ˆx(f) Passband waveform x(t) to the channel 8 Magnitude of ˆx(f ) x(t) 6 = (magnitude of ˆx I ) + (magnitude of ˆx Q ).... Phase angle (radian) of ˆx(f ) = (phase of ˆx I ) + (phase of ˆx Q ) t (sec)....

14 (cont ) Real and imaginary parts of. Real part of ˆx(f ) 8 6 ˆx(f) = (real part of ˆx I ) + (real part of ˆx Q ) QAM (bb) QAM.... output to to Imaginary part of ˆx(f ) = (imag. part of ˆx I ) + (image. part of ˆx Q ) QAM bb de QAM to QAM: PASSBAND TO BASEBAND.... The frequency is shixed down by mul.plica.ons of cos and sin and by filter v I (t) v Q (t) LPF = low-pass filter BW = bandwidth ideal LPF of BW. W ideal LPF of BW. W cos( f c t) g I (t) g Q (t) sin( f c t) y(t) QAM to If there is no noise, reliability is achieved: v I (t) =u I (t) and v Q (t) =u Q (t) u I (t) u Q (t) v I (t) v Q (t) cos( f c t) x I (t) sin( f c t) ideal LPF of BW. W ideal LPF of BW W. x Q (t) cos( f c t) g I (t) g Q (t) sin( f c t) QAM x(t) =y(t) QAM to 6

To verify that vi (t) = ui (t), apply trigonometry iden..es = x(t) (no noise) = ui (t) cos( fc t) (con.nued) Low-pass ﬁlter eliminates the components at high frequencies, ±fc.

15 To verify that vi (t) = ui (t), apply trigonometry iden..es = x(t) (no noise) = ui (t) cos( fc t) (con.nued) Low-pass ﬁlter eliminates the components at high frequencies, ±fc. uq (t) sin( fc t) uq (t) sin( fc t) cos( fc t) cos = + cos( ) sin cos = sin( ) gi (t) = y(t) cos( fc t) = ui (t) cos ( fc t) = ui (t) + ui (t) cos( fc t) g I (f ) = u I (f ) + h u I (f + fc ) + u I (f fc ) i ih u Q (f + fc ) u Q (f fc ) AXer the low-pass ﬁlter uq (t) sin( fc t) v I (f ) = u I (f ) Fourier transform g I (f ) = u I (f ) + i h u I (f + fc ) + u I (f fc ) ih u Q (f + fc ) u Q (f fc ) i 7 8 Example : to The other equality, vq (t) = uq (t), can be veriﬁed using a similar approach gq (t) = = sin = y(t) sin( fc t) cos( fc t) cos( ) vi (t) vq (t) ui (t) sin( fc t) cos( fc t) + uq (t) sin ( fc t) = uq (t) uq (t) cos( fc t) ui (t) sin( fc t) ideal LPF of BW W gi (t) ideal LPF of BW W gq (t) y(t) = x(t).. sin( fc t) Fourier transform g Q (f ) = u Q (f ) h u Q (f + fc ) + u Q (f QAM to fc ) i ih u I (f + fc ) u I (f fc ) AXer the low-pass ﬁlter v Q (f ) = u Q (f ) 9 i Suppose the channel is noiseless, and Suppose fc = (as in Example ) Plot gi (t) and Plot g I (f ) and y(t) equals x(t) in Ex gq (t) g Q (f ) 6 i

16 (con t) Time-domains and. g I (t) g Q (t) (cont ) Magnitude and phase of. ĝ I (f) Immediate waveform at I and Q branches 6 Magnitude of ĝi(f ) g I (t) g Q (t).... Phase angle (radian) of ĝi(f ) t (sec) (cont ) Real and imaginary parts of. ĝ I (f) (cont ) Magnitude and phase of. ĝ Q (f) Real part of ĝi(f ) 6 Magnitude of ĝq(f ) Imaginary part of ĝi(f ) Phase angle (radian) of ĝq(f )

17 (cont ) Real and imaginary parts of g Q (f ) Here is a summary of QAM. u, u, u,... Re al part of g Q(f ) b, b, b,... 6 u, u, u, ui (t) = uq (t) = X uk p(t kt ) cos( fc t) k= xi (t) x(t) xq (t) X uk p(t kt ) sin( fc t) k= I magi nary part of g Q(f ) cos( fc t) c, c, c, output v, v, v,... vi (t) ideal LPF de of BW W gi (t). v, v, v,... vq (t) ideal LPF gq (t) de of BW W. sin( fc t) 6 Here are acronym and symbols Meaning QAM quadrature amplitude modula.on PAM pulse amplitude modula.on bb LPF low-pass ﬁlter BW bandwidth Ideal LPF of BW W ideal low-pass ﬁlter of bandwidth W. The impulse response is hlp (t) = W sinc (W t) and x(t) = ag(t) + bh(t) $ x (f ) = ag (f ) + bh (f ) x(t) = g(t ) $ x (f ) = g (f )e Modula.on: g(t) = h(t) cos( f t) $ g (f ) = h (f + f ) + h (f f ) g(t) = h(t) sin( f t) $ g (f ) = i h (f + f ) f ) Time shix: h LP (f ) = T symbol interval, T > (sec) p(t) shix-orthogonal pulse W bandwidth of p(t), i.e., < W < (Hz) carrier frequency, fc Here are Fourier-transform proper.es Linearity: Acro/Symbol fc 66 (,, if i h (f f W otherwise W (Hz) 67 68

on (consists of PAM s) Baseband demodula.on (consists of PAM s) Baseband (mul.

18 Structure of the QAM / Summary / de -> / -> Methods of the QAM Signal (maps to points according to constella.on) Signal (decodes to a nearest point) Baseband modula.on (consists of PAM s) Baseband demodula.on (consists of PAM s) Baseband (mul.plies by cos and sin) Passband to (mul.plies by cos and sin; and filters) Examples (plots, Matlab) in.me- and frequency-domains 69

Pulse amplitude modula.on Baseband to passband and back

Pulse amplitude modula.on Baseband to passband and back Pulse ampliude modula.on Baseband o and back message inpu ) Today s opics concern wih he modulaor and demodulaor ransminer acharapan Suwansan.suk Sampler Quan.zer Source Channel Modulaor analog sequence